এ অধ্যায়ে আমরা যে বিষয়গুলি আলোচনা করব।
- নির্দিষ্ট যোগজীকরণ
- নির্দিষ্ট যোগজের বৈশিষ্ট্য
- নির্দিষ্ট যোগজীকরণের প্রয়োগ
- ক্ষেত্রফল থেকে যোগজের ধারণা
- উর্ধসীমা ও নিম্নসীমা
- নির্দিষ্ট যোগজীকরণে ধ্রুবক \( c\)-এর অপ্রয়োজনীয়তা
- \(\int_{a}^{b}{f(x)dx}\)-এর জ্যামিতিক ব্যাখ্যা
- \(\int_{a}^{b}{f(x)dx}\)-এর মান নির্ণয়ের ধাপসমুহ
- প্রতিস্থাপন পদ্ধতিতে নির্দিষ্ট যোগজ নির্ণয়ের ধাপসমুহ
- সমাধানকৃত উদাহরণমালা
- অতি সংক্ষিপ্ত প্রশ্ন-উত্তর
- সংক্ষিপ্ত প্রশ্ন-উত্তর
- বর্ণনামূলক প্রশ্ন-উত্তর
নির্দিষ্ট যোগজীকরণঃ
আমরা জেনেছি, যোগজীকরণ হলো অন্তরীকরণের বিপরীত প্রক্রিয়া। এখন, আমরা একে সমষ্টিকরণের পদ্ধতি হিসেবে সংজ্ঞায়িত করব। প্রকৃতপক্ষে যোগজীকরণের উৎপত্তিই হলো বক্ররেখা দ্বারা আবদ্ধ ক্ষেত্রের ক্ষেত্রফল নির্ণয়ের উদ্দেশ্য নিয়ে। এক্ষেত্রে ক্ষেত্রটিকে ক্ষুদ্র ক্ষুদ্র অংশে বিভক্ত করে, ক্ষুদ্র ক্ষুদ্র ক্ষেত্রের সমষ্টি থেকেই মূল ক্ষেত্রের ক্ষেত্রফল নির্ণয় করা হয়।
নির্দিষ্ট যোগজের বৈশিষ্ট্যঃ
\(1.\) \(\int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(t)dt}\)যেমনঃ \(\int_{0}^{1}{x^2dx}=\int_{0}^{1}{t^2dt}\)
\(2.\) \(\int_{a}^{b}{f(x)dx}=-\int_{b}^{a}{f(x)dx}\)
যেমনঃ \(\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}=-\int_{\frac{\pi}{2}}^{0}{\sin{x}dx}\)
যেহেতু,
\(\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{x}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{\frac{\pi}{2}}-\cos{0}\right]\)
\(=-\left[0-1\right]\) ➜ \(\because \cos{\frac{\pi}{2}}=0, \cos{0}=1\)
\(=-\left[-1\right]\)
\(=1\)
এবং
\(-\int_{\frac{\pi}{2}}^{0}{\sin{x}dx}\)
\(=-\left[-\cos{x}\right]_{\frac{\pi}{2}}^{0}\)
\(=\left[\cos{x}\right]_{\frac{\pi}{2}}^{0}\)
\(=\left[\cos{0}-\cos{\frac{\pi}{2}}\right]\)
\(=\left[1-0\right]\) ➜ \(\because \cos{0}=1, \cos{\frac{\pi}{2}}=0\)
\(=1\)
\(\therefore \int_{0}^{\frac{\pi}{2}}{\sin{x}dx}=-\int_{\frac{\pi}{2}}^{0}{\sin{x}dx}\)
\(3.\) \(\int_{0}^{a}{f(x)dx}=\int_{0}^{a}{f(a-x)dx}\)
যেমনঃ \(\int_{0}^{2}{x^3dx}=\int_{0}^{2}{(2-x)^3dx}\);
\(\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}=\int_{0}^{\frac{\pi}{2}}{\sin{\left(\frac{\pi}{2}-x\right)}dx}\)
ইত্যাদি।
নির্দিষ্ট যোগজ ও এর প্রয়োগঃ
নির্দিষ্ট যোগজের সাহায্যে আমরা কোনো সীমাবদ্ধ ক্ষেত্রের ক্ষেত্রফল নির্ণয় করতে পারি। বক্ররেখার দৈর্ঘ্য, বক্ররেখা দ্বারা পরিবেষ্টিত ক্ষেত্রের ক্ষেত্রফল, কোনো বস্তুর আয়তন এবং গতিবেগ ইত্যাদি নির্ণয়ে নির্দিষ্ট যোগজ বিশেষ ভূমিকা পালন করে।
নির্দিষ্ট যোগজের সাহায্যে আমরা কোনো সীমাবদ্ধ ক্ষেত্রের ক্ষেত্রফল নির্ণয় করতে পারি। বক্ররেখার দৈর্ঘ্য, বক্ররেখা দ্বারা পরিবেষ্টিত ক্ষেত্রের ক্ষেত্রফল, কোনো বস্তুর আয়তন এবং গতিবেগ ইত্যাদি নির্ণয়ে নির্দিষ্ট যোগজ বিশেষ ভূমিকা পালন করে।
ক্ষেত্রফল থেকে যোগজের ধারণাঃ
\(x=a, x=b, y=f(x)\) এবং \(y=0\) এ চারটি রেখা দ্বারা সীমাবদ্ধ ক্ষেত্রকে \(n\) সমানভাবে বিভক্ত করলে এবং প্রতিটি ভাগের দূরত্ব \(h\) হলে \(nh=b-a\) হবে। এখন \(nh=b-a\) হলে, \[\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\] কে নির্দিষ্ট যোগজ বলে। যাকে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা প্রকাশ করা হয়।\(\therefore \int_{a}^{b}{f(x)dx}\) \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\]
উর্ধসীমা ও নিম্নসীমাঃ
যদি \(f(x)\) ফাংশনের অনির্দিষ্ট যোগজ \(F(x)\) হয়, অর্থাৎ \(\int{f(x)dx}=F(x)\) হয়, তবে \([a, b]\) বদ্ধ ব্যবধিতে \(F(b)-F(a)\) কে \(f(x)\) ফাংশনের নির্দিষ্ট যোগজের মান বলা হয়, যাকে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা প্রকাশ করা হয় ।গাণিতিকভাবে, যদি \(f(x)\) এর যোগজ \(F(x)\) হয়
অর্থাৎ \(\int{f(x)dx}=F(x)\) হয় তবে \(\int_{a}^{b}{f(x)dx}=\left[F(x)\right]_{a}^{b}=F(b)-F(a)\).
এখানে \(a\) কে নিম্নপ্রান্ত (lower limit) , \(b\) কে উর্ধবপ্রান্ত (upper limit) এবং \([ a, b]\) ব্যবধিকে যোগজের রেঞ্জ বলে।
মন্তব্যঃ
স্বাধীন চলরাশি \(x\) এর মান পরিবর্তিত হলে, অনির্দিষ্ট যোগজটির মান যে নির্দিষ্ট পরিমাণ পার্থক্য সৃষ্টি হয়, তা হলো নির্দিষ্ট যোজিতফল বা নির্দিষ্ট যোগজ।
নির্দিষ্ট যোগজীকরণে ধ্রুবক \( c\)-এর অপ্রয়োজনীয়তাঃ
ব্যাখ্যাঃ \(\int{f(x)dx}=F(x)\) হয় তবে \(\int_{a}^{b}{f(x)dx}=\left[F(x)\right]_{a}^{b}=F(b)-F(a) ......(1)\)আবার,
\(\int{f(x)dx}=F(x)+c\) হয় তবে \(\int_{a}^{b}{f(x)dx}=\left[F(x)+c\right]_{a}^{b}=F(b)+c-F(a)-c=F(b)-F(a) ......(2)\)
অর্থাৎ নির্দিষ্ট যোগজে ধ্রুবক \(c\) সংযুক্ত করলে উর্ধবপ্রান্ত এবং নিম্নপ্রান্ত প্রয়োগের ফলে তা অপসারিত হয়।
\((1)\) ও \((2)\) থেকে ইহা স্পষ্ট যে, \(\int_{a}^{b}{f(x)dx}\) এর যোজিতফল \(F(x)\) এর সাথে \(c\) ব্যবহার না করলেও ফলাফল যা, ব্যবহার করেও ফলাফল একই থাকে। কাজেই \(\int_{a}^{b}{f(x)dx}\) এর মান ধ্রুবক \(c\) এর উপর নির্ভরশীল নয়। তাই নির্দিষ্ট যোগজে ধ্রুবক \(c\) এর ব্যবহার অপ্রয়োজনীয়।
\(\int_{a}^{b}{f(x)dx}\)-এর জ্যামিতিক ব্যাখ্যাঃ
\(y=f(x)\)বক্ররেখা এবং \(y=0, x=a\) এবং \(x=b\) রেখত্রয় দ্বারা সীমাবদ্ধ ক্ষেত্রের ক্ষেত্রফল \(ABCD\) কে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা প্রকাশ করা হয় ।অর্থাৎ \(ABCD\) এর ক্ষেত্রফল \(=\int_{a}^{b}{f(x)dx}\)
\(\int_{a}^{b}{f(x)dx}\)-এর মান নির্ণয়ের ধাপসমুহঃ
নির্দিষ্ট যোগজ নির্ণয়ের কতগুলি কৌশল অবলম্বন করা হয়। এই কৌশলের ধাপসমুহ নিম্নরূপঃ\((1)\) প্রদত্ত ফাংশনের অনির্দিষ্ট যোগজ নির্ণয় করতে হয়। অর্থাৎ \(\int{f(x)dx}\) এর অনির্দিষ্ট যোগজ \(F(x)\) হলে, \(F(x)\) বের করতে হয়।
\((2)\) \(F(x)\) এ \(x\) এর পরিবর্তে যথাক্রমে উর্ধবপ্রান্ত \(b\) এবং নিম্নপ্রান্ত \(a\) বসিয়ে \(F(b)\) থেকে \(F(a)\) বিয়োগ করতে হয়। এ বিয়োগফলই অর্থাৎ \(F(b)-F(a)\) হচ্ছে \(a\) এবং \(b\) সীমার মধ্যে \(f(x)\) এর যোজিতফল অর্থাৎ \(\int_{a}^{b}{f(x)dx}\) এর মান।
প্রতিস্থাপন পদ্ধতিতে নির্দিষ্ট যোগজ নির্ণয়ের ধাপসমুহঃ
নির্দিষ্ট যোগজ নির্ণয়ের ক্ষেত্রে প্রতিস্থাপন পদ্ধতি একটি অত্যন্ত প্রয়োজনীয় প্রক্রিয়া। এর ধাপসমুহ নিম্নরূপঃ\((1)\) যোজ্য ফাংশনকে নতুন চলরাশিতে প্রকাশ করা।
\((2)\) অন্তরজ \(dx\) কে নতুন চলরাশির অন্তরজে রূপান্তর করা।
\((2)\) নতুন চলরাশির সীমার মান নির্ণয় করা।
\(x=a, x=b, y=f(x)\) এবং \(y=0\) এ চারটি রেখা দ্বারা সীমাবদ্ধ ক্ষেত্রকে \(n\) সমানভাবে বিভক্ত করলে এবং প্রতিটি ভাগের দূরত্ব \(h\) হলে \(nh=b-a\) হবে। এখন \(nh=b-a\) হলে, \[\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\] কে নির্দিষ্ট যোগজ বলে। যাকে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা প্রকাশ করা হয়।
\(\therefore \int_{a}^{b}{f(x)dx}\) \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\]
\(\therefore \int_{a}^{b}{f(x)dx}\) \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\]
Proof:
ধরি,
\(x=a\) সরলরেখা \(y=f(x)\) বক্ররেখাকে \(B_{0}\) বিন্দুতে এবং \(x\)-অক্ষরেখাকে \(A_{0}\) বিন্দুতে ছেদ করে। এবং \(x=b\) সরলরেখা উক্ত বক্ররেখা এবং \(x\)-অক্ষরেখাকে যথাক্রমে \(B_{n}\) এবং \(A_{n}\) বিন্দুতে ছেদ করে। অতএব, \(y=f(x)\) বক্ররেখা \(x=a, x=b\) এবং \(y=0\) কতৃক গঠিত ক্ষেত্র \(A_{0}A_{n}B_{n}B_{0}\) এবং মনে করি, এর ক্ষেত্রফল \(S\). এখন \(A_{0}A_{n}\) রেখাংশকে \(A_{1} A_{2} A_{3} ......... A_{n}\) বিন্দু দ্বারা সমান \(n\) ভাগে ভাগ করি যেন \(A_{0}A_{n}=nh\) হয়, অর্থাৎ \(b-a=nh\) হয়। \(x=a\) এর সমান্তরাল করে \(A_{1} A_{2} A_{3} ......... A_{n}\) বিন্দুগুলির মধ্যদিয়ে অঙ্কিত রেখাগুলি \(y=f(x)\) বক্ররেখাকে যথাক্রমে \(B_{1} B_{2} B_{3} ......... B_{n}\) বিন্দুতে ছেদ করে।
এখানে,
\(B_{0}\) বিন্দুর ভুজ \(=a\) এবং কটি \(=A_{0}B_{0}=f(a)\).
\(B_{1}\) বিন্দুর ভুজ \(=a+h\) এবং কটি \(A_{1}B_{1}=f(a+h)\).
\(B_{2}\) বিন্দুর ভুজ \(=a+2h\) এবং কটি \(A_{2}B_{2}=f(a+2h)\) ইত্যাদি।
আবার ধরি,
\(S_{n}= A_{0}B_{0}.A_{0}A_{1}+A_{1}B_{1}.A_{1}A_{2}+A_{2}B_{2}.A_{2}A_{3}+ ....... +A_{n-1}B_{n-1}.A_{n-1}A_{n}\)
\(= f(a).h+f(a+h).h+f(a+2h).h+ ....... +f(a+\overline{n-1}.h).h\)
\(=h\{f(a)+f(a+h)+f(a+2h)+ ....... +f(a+\overline{n-1}.h)\}\)
\(h\) অতি ক্ষুদ্র হলে, \(B_{0}A_{0}A_{1}B_{1}, B_{1}A_{1}A_{2}B_{2} ..... \) ক্ষেত্রগুলি আয়তক্ষেত্রের আকার ধারণ করে।
অতএব, \(n\rightarrow\infty\) হলে \(n\rightarrow 0\) হবে এবং এক্ষেত্রে
\(A_{0}A_{n}B_{n}B_{0}\) এর ক্ষেত্রফল \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\] হবে।
এ সীমাস্থ মানকে নির্দিষ্ট যোগজ বলে, যাকে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা সূচিত করা হয়।
\(\therefore \int_{a}^{b}{f(x)dx}\) \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\]
(proved)
\(x=a\) সরলরেখা \(y=f(x)\) বক্ররেখাকে \(B_{0}\) বিন্দুতে এবং \(x\)-অক্ষরেখাকে \(A_{0}\) বিন্দুতে ছেদ করে। এবং \(x=b\) সরলরেখা উক্ত বক্ররেখা এবং \(x\)-অক্ষরেখাকে যথাক্রমে \(B_{n}\) এবং \(A_{n}\) বিন্দুতে ছেদ করে। অতএব, \(y=f(x)\) বক্ররেখা \(x=a, x=b\) এবং \(y=0\) কতৃক গঠিত ক্ষেত্র \(A_{0}A_{n}B_{n}B_{0}\) এবং মনে করি, এর ক্ষেত্রফল \(S\). এখন \(A_{0}A_{n}\) রেখাংশকে \(A_{1} A_{2} A_{3} ......... A_{n}\) বিন্দু দ্বারা সমান \(n\) ভাগে ভাগ করি যেন \(A_{0}A_{n}=nh\) হয়, অর্থাৎ \(b-a=nh\) হয়। \(x=a\) এর সমান্তরাল করে \(A_{1} A_{2} A_{3} ......... A_{n}\) বিন্দুগুলির মধ্যদিয়ে অঙ্কিত রেখাগুলি \(y=f(x)\) বক্ররেখাকে যথাক্রমে \(B_{1} B_{2} B_{3} ......... B_{n}\) বিন্দুতে ছেদ করে।
এখানে,
\(B_{0}\) বিন্দুর ভুজ \(=a\) এবং কটি \(=A_{0}B_{0}=f(a)\).
\(B_{1}\) বিন্দুর ভুজ \(=a+h\) এবং কটি \(A_{1}B_{1}=f(a+h)\).
\(B_{2}\) বিন্দুর ভুজ \(=a+2h\) এবং কটি \(A_{2}B_{2}=f(a+2h)\) ইত্যাদি।
আবার ধরি,
\(S_{n}= A_{0}B_{0}.A_{0}A_{1}+A_{1}B_{1}.A_{1}A_{2}+A_{2}B_{2}.A_{2}A_{3}+ ....... +A_{n-1}B_{n-1}.A_{n-1}A_{n}\)
\(= f(a).h+f(a+h).h+f(a+2h).h+ ....... +f(a+\overline{n-1}.h).h\)
\(=h\{f(a)+f(a+h)+f(a+2h)+ ....... +f(a+\overline{n-1}.h)\}\)
\(h\) অতি ক্ষুদ্র হলে, \(B_{0}A_{0}A_{1}B_{1}, B_{1}A_{1}A_{2}B_{2} ..... \) ক্ষেত্রগুলি আয়তক্ষেত্রের আকার ধারণ করে।
অতএব, \(n\rightarrow\infty\) হলে \(n\rightarrow 0\) হবে এবং এক্ষেত্রে
\(A_{0}A_{n}B_{n}B_{0}\) এর ক্ষেত্রফল \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\] হবে।
এ সীমাস্থ মানকে নির্দিষ্ট যোগজ বলে, যাকে \(\int_{a}^{b}{f(x)dx}\) প্রতীক দ্বারা সূচিত করা হয়।
\(\therefore \int_{a}^{b}{f(x)dx}\) \[=\lim_{a \rightarrow b}h\{f(a)+f(a+h)+f(a+2h)+f(a+3h)+ ..........+f(a+\overline{n-1}.h)\}\]
(proved)
অনুশীলনী \(10.G\) উদাহরণ সমুহ
নিচের যোগজগুলির মান নির্ণয় করঃ
\((1.)\) \(\int_{1}^{2}{\frac{(x^2-1)^2}{x^2}dx}\)
উত্তরঃ \(\frac{5}{6}\)
\((2.)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\((3.)\) \(\int_{1}^{3}{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(\ln{5}\)
\((4.)\) \(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\)
উত্তরঃ \(\sin{(\ln{3})}\)
\((5.)\) \(\int_{0}^{1}{\frac{1}{\sqrt{2x-x^2}}dx}\)
উত্তরঃ \(\frac{\pi}{2}\)
\((6.)\) \(\int_{0}^{1}{\ln{|1+x|}dx}\)
উত্তরঃ \(2\ln{2}-1\)
\((7.)\) \(\int_{0}^{\frac{\pi}{2}}{e^x(\sin{x}+\cos{x})dx}\)
উত্তরঃ \(e^{\frac{\pi}{2}}\)
উত্তরঃ \(\frac{5}{6}\)
\((2.)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\((3.)\) \(\int_{1}^{3}{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(\ln{5}\)
\((4.)\) \(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\)
উত্তরঃ \(\sin{(\ln{3})}\)
\((5.)\) \(\int_{0}^{1}{\frac{1}{\sqrt{2x-x^2}}dx}\)
উত্তরঃ \(\frac{\pi}{2}\)
\((6.)\) \(\int_{0}^{1}{\ln{|1+x|}dx}\)
উত্তরঃ \(2\ln{2}-1\)
\((7.)\) \(\int_{0}^{\frac{\pi}{2}}{e^x(\sin{x}+\cos{x})dx}\)
উত্তরঃ \(e^{\frac{\pi}{2}}\)
\((8.)\) \(\int_{0}^{4}{y\sqrt{4-y}dx}\)
উত্তরঃ \(\frac{128}{15}\)
\((9.)\) \(\int_{0}^{3}{(3-2x+x^2)dx}\)
উত্তরঃ \(9\)
\((10.)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\((11.)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{|\sin{x}|dx}\)
উত্তরঃ \(2\)
\((12.)\) \(\int_{2}^{8}{|x-5|dx}\)
উত্তরঃ \(9\)
\((13.)\) \(\int_{0}^{3}{\sqrt{9-x^2}dx}\)
উত্তরঃ \(\frac{9\pi}{4}\)
উত্তরঃ \(\frac{128}{15}\)
\((9.)\) \(\int_{0}^{3}{(3-2x+x^2)dx}\)
উত্তরঃ \(9\)
\((10.)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\((11.)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{|\sin{x}|dx}\)
উত্তরঃ \(2\)
\((12.)\) \(\int_{2}^{8}{|x-5|dx}\)
উত্তরঃ \(9\)
\((13.)\) \(\int_{0}^{3}{\sqrt{9-x^2}dx}\)
উত্তরঃ \(\frac{9\pi}{4}\)
\((1.)\) \(\int_{1}^{2}{\frac{(x^2-1)^2}{x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{5}{6}\)
উত্তরঃ \(\frac{5}{6}\)
সমাধানঃ
\(\int_{1}^{2}{\frac{(x^2-1)^2}{x^2}dx}\)
\(=\int_{1}^{2}{\frac{x^4-2x^2+1}{x^2}dx}\) ➜ \(\because (a+b)^2=a^2-2ab+b^2\)
\(=\int_{1}^{2}{\left(\frac{x^4}{x^2}-2\frac{x^2}{x^2}+\frac{1}{x^2}\right)dx}\)
\(=\int_{1}^{2}{\left(x^2-2+x^{-2}\right)dx}\)
\(=\int_{1}^{2}{x^2dx}-2\int_{1}^{2}{1dx}+\int_{1}^{2}{x^{-2}dx}\)
\(=\left[\frac{x^{2+1}}{2+1}\right]_{1}^{2}-2\left[x\right]_{1}^{2}+\left[\frac{x^{-2+1}}{-2+1}\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{1dx}=x\)
\(=\left[\frac{x^{3}}{3}\right]_{1}^{2}-2\left[2-1\right]+\left[\frac{x^{-1}}{-1}\right]_{1}^{2}\)
\(=\frac{1}{3}\left[x^{3}\right]_{1}^{2}-2\left[2-1\right]-\left[\frac{1}{x}\right]_{1}^{2}\)
\(=\frac{1}{3}\left[2^{3}-1^3\right]-2.1-\left[\frac{1}{2}-\frac{1}{1}\right]\)
\(=\frac{1}{3}\left[8-1\right]-2-\left[\frac{1}{2}-1\right]\)
\(=\frac{1}{3}\times{7}-2-\frac{1-2}{2}\)
\(=\frac{7}{3}-2-\frac{-1}{2}\)
\(=\frac{7}{3}-2+\frac{1}{2}\)
\(=\frac{14-12+3}{6}\)
\(=\frac{17-12}{6}\)
\(=\frac{5}{6}\)
\(=\int_{1}^{2}{\frac{x^4-2x^2+1}{x^2}dx}\) ➜ \(\because (a+b)^2=a^2-2ab+b^2\)
\(=\int_{1}^{2}{\left(\frac{x^4}{x^2}-2\frac{x^2}{x^2}+\frac{1}{x^2}\right)dx}\)
\(=\int_{1}^{2}{\left(x^2-2+x^{-2}\right)dx}\)
\(=\int_{1}^{2}{x^2dx}-2\int_{1}^{2}{1dx}+\int_{1}^{2}{x^{-2}dx}\)
\(=\left[\frac{x^{2+1}}{2+1}\right]_{1}^{2}-2\left[x\right]_{1}^{2}+\left[\frac{x^{-2+1}}{-2+1}\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{1dx}=x\)
\(=\left[\frac{x^{3}}{3}\right]_{1}^{2}-2\left[2-1\right]+\left[\frac{x^{-1}}{-1}\right]_{1}^{2}\)
\(=\frac{1}{3}\left[x^{3}\right]_{1}^{2}-2\left[2-1\right]-\left[\frac{1}{x}\right]_{1}^{2}\)
\(=\frac{1}{3}\left[2^{3}-1^3\right]-2.1-\left[\frac{1}{2}-\frac{1}{1}\right]\)
\(=\frac{1}{3}\left[8-1\right]-2-\left[\frac{1}{2}-1\right]\)
\(=\frac{1}{3}\times{7}-2-\frac{1-2}{2}\)
\(=\frac{7}{3}-2-\frac{-1}{2}\)
\(=\frac{7}{3}-2+\frac{1}{2}\)
\(=\frac{14-12+3}{6}\)
\(=\frac{17-12}{6}\)
\(=\frac{5}{6}\)
\((2.)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}\right]+\frac{1}{4}\left[\sin{\pi}-\sin{0}\right]\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0, \sin{0}=0 \)
\(=\frac{\pi}{4}+\frac{1}{4}.0\)
\(=\frac{\pi}{4}+0\)
\(=\frac{\pi}{4}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}\right]+\frac{1}{4}\left[\sin{\pi}-\sin{0}\right]\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0, \sin{0}=0 \)
\(=\frac{\pi}{4}+\frac{1}{4}.0\)
\(=\frac{\pi}{4}+0\)
\(=\frac{\pi}{4}\)
\((3.)\) \(\int_{1}^{3}{\frac{2x}{1+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{5}\)
উত্তরঃ \(\ln{5}\)
সমাধানঃ
\(\int_{1}^{3}{\frac{2x}{1+x^2}dx}\)
\(=\int_{1}^{3}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\ln{(1+x^2)}\right]_{1}^{3}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\left[\ln{(1+3^2)}-\ln{(1+1^2)}\right]\)
\(=\ln{(1+9)}-\ln{(1+1)}\)
\(=\ln{(10)}-\ln{(2)}\)
\(=\ln{\left(\frac{10}{2}\right)}\) ➜ \(\because \ln{(M)}-ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(=\ln{(5)}\)
\(=\int_{1}^{3}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\ln{(1+x^2)}\right]_{1}^{3}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\left[\ln{(1+3^2)}-\ln{(1+1^2)}\right]\)
\(=\ln{(1+9)}-\ln{(1+1)}\)
\(=\ln{(10)}-\ln{(2)}\)
\(=\ln{\left(\frac{10}{2}\right)}\) ➜ \(\because \ln{(M)}-ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(=\ln{(5)}\)
\((4.)\) \(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\sin{(\ln{3})}\)
উত্তরঃ \(\sin{(\ln{3})}\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=\ln{|x|}\)
\(\Rightarrow t=\ln{(1)}\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\therefore t=\ln{(3)}\), যখন \(x=3\)
\(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(3\) |
| \(t\) | \(0\) | \(\ln{3}\) |
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\therefore t=\ln{(3)}\), যখন \(x=3\)
\(=\int_{1}^{3}{\cos{(\ln{|x|})}.\frac{1}{x}dx}\)
\(=\int_{0}^{\ln{(3)}}{\cos{t}.dt}\)
\(=\left[\sin{t}\right]_{0}^{\ln{(3)}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=\left[\sin{(\ln{3})}-\sin{0}\right]\)
\(=\sin{(\ln{3})}-0\) ➜ \(\because \sin{0}=0\)
\(=\sin{(\ln{3})}\)
\((5.)\) \(\int_{0}^{1}{\frac{1}{\sqrt{2x-x^2}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}\)
উত্তরঃ \(\frac{\pi}{2}\)
সমাধানঃ
\(\int_{0}^{1}{\frac{1}{\sqrt{2x-x^2}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-1+2x-x^2}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(1-2x+x^2)}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(x^2-2x+1)}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(x-1)^2}}dx}\)
\(=\left[\sin^{-1}(x-1)\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\)
\(=\sin^{-1}(1-1)-\sin^{-1}(0-1)\)
\(=\sin^{-1}(0)-\sin^{-1}(-1)\)
\(=0+\sin^{-1}(1)\)
\(=0+\sin^{-1}\left\{\sin{\left(\frac{\pi}{2}\right)}\right\}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=0+\frac{\pi}{2}\)
\(=\frac{\pi}{2}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-1+2x-x^2}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(1-2x+x^2)}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(x^2-2x+1)}}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-(x-1)^2}}dx}\)
\(=\left[\sin^{-1}(x-1)\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\)
\(=\sin^{-1}(1-1)-\sin^{-1}(0-1)\)
\(=\sin^{-1}(0)-\sin^{-1}(-1)\)
\(=0+\sin^{-1}(1)\)
\(=0+\sin^{-1}\left\{\sin{\left(\frac{\pi}{2}\right)}\right\}\) ➜ \(\because 1=\sin{\frac{\pi}{2}}\)
\(=0+\frac{\pi}{2}\)
\(=\frac{\pi}{2}\)
\((6.)\) \(\int_{0}^{1}{\ln{|1+x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\ln{2}-1\)
উত্তরঃ \(2\ln{2}-1\)
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|1+x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|1+x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{\ln{|1+x|}dx}\)\(A=1, \ \ L=\ln{|1+x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|1+x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{1}{\ln{|1+x|}.1dx}\)
\(=\left[\ln{|1+x|}.x\right]_{0}^{1}-\int_{0}^{1}{\left\{\frac{d}{dx}(\ln{|1+x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[x\ln{|1+x|}\right]_{0}^{1}-\int_{0}^{1}{\frac{1}{1+x}.xdx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{x})=\frac{1}{x}\)
\(=\left[1\ln{|1+1|}-0\ln{|1+0|}\right]-\int_{0}^{1}{\frac{x}{1+x}dx}\)
\(=\left[\ln{2}-0\right]-\int_{0}^{1}{\frac{1+x-1}{1+x}dx}\)
\(=\ln{2}-\int_{0}^{1}{\left(\frac{1+x}{1+x}-\frac{1}{1+x}\right)dx}\)
\(=\ln{2}-\int_{0}^{1}{\left(1-\frac{1}{1+x}\right)dx}\)
\(=\ln{2}-\int_{0}^{1}{1dx}+\int_{0}^{1}{\frac{1}{1+x}dx}\)
\(=\ln{2}-\left[x\right]_{0}^{1}+\left[\ln{|1+x|}\right]_{0}^{1}\) ➜ \(\because \int{1dx}=x, \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\ln{2}-\left[1-0\right]+\left[\ln{|1+1|}-\ln{|1+0|}\right]\)
\(=\ln{2}-1+\left[\ln{2}-\ln{1}\right]\)
\(=\ln{2}-1+\left[\ln{2}-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\ln{2}-1+\ln{2}\)
\(=2\ln{2}-1\)
\((7.)\) \(\int_{0}^{\frac{\pi}{2}}{e^x(\sin{x}+\cos{x})dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(e^{\frac{\pi}{2}}\)
উত্তরঃ \(e^{\frac{\pi}{2}}\)
সমাধানঃ
এখানে, LIATE শব্দের
\(T=\sin{x}, \ \ E=e^x\)
যেহেতু T, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\sin{x}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\frac{\pi}{2}}{e^x(\sin{x}+\cos{x})dx}\)\(T=\sin{x}, \ \ E=e^x\)
যেহেতু T, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\sin{x}\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{\frac{\pi}{2}}{(e^x\sin{x}+e^x\cos{x})dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{e^x\sin{x}dx}+\int_{0}^{\frac{\pi}{2}}{e^x\cos{x}dx}\)
\(=\left[\sin{x}.e^x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\left\{\frac{d}{dx}(\sin{x})\int{e^xdx}\right\}dx}+\int_{0}^{\frac{\pi}{2}}{e^x\cos{x}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\left[e^x\sin{x}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\cos{x}.e^xdx}+\int_{0}^{\frac{\pi}{2}}{e^x\cos{x}dx}\) ➜ \(\because \int{e^xdx}=e^x, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(=\left[e^{\frac{\pi}{2}}\sin{\frac{\pi}{2}}-e^{0}\sin{0}\right]-\int_{0}^{\frac{\pi}{2}}{e^x\cos{x}dx}+\int_{0}^{\frac{\pi}{2}}{e^x\cos{x}dx}\)
\(=e^{\frac{\pi}{2}}.1-1.0\) ➜ \(\because \sin{\frac{\pi}{2}}=1, e^{0}=1, \sin{0}=0\)
\(=e^{\frac{\pi}{2}}-0\)
\(=e^{\frac{\pi}{2}}\)
\((8.)\) \(\int_{0}^{4}{y\sqrt{4-y}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{128}{15}\)
উত্তরঃ \(\frac{128}{15}\)
সমাধানঃ
ধরি,
\(\sqrt{4-y}=t\)
\(\Rightarrow 4-y=t^2\)
\(\Rightarrow 4-t^2=y\)
\(\Rightarrow y=4-t^2\)
\(\Rightarrow \frac{d}{dy}(y)=\frac{d}{dy}(4-t^2)\)
\(\Rightarrow 1=0-2t\frac{dt}{dy}\)
\(\therefore dy=-2tdt\)
➜ \(\because t=\sqrt{4-y}\)
\(\Rightarrow t=\sqrt{4-0}\), যখন \(y=0\)
\(\Rightarrow t=\sqrt{4}\)
\(\therefore t=2\)
আবার,
\(t=\sqrt{4-y}\)
\(\Rightarrow t=\sqrt{4-4}\), যখন \(y=4\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(\int_{0}^{4}{y\sqrt{4-y}dx}\)\(\sqrt{4-y}=t\)
\(\Rightarrow 4-y=t^2\)
\(\Rightarrow 4-t^2=y\)
\(\Rightarrow y=4-t^2\)
\(\Rightarrow \frac{d}{dy}(y)=\frac{d}{dy}(4-t^2)\)
\(\Rightarrow 1=0-2t\frac{dt}{dy}\)
\(\therefore dy=-2tdt\)
| \(y\) | \(0\) | \(4\) |
| \(t\) | \(2\) | \(0\) |
\(\Rightarrow t=\sqrt{4}\)
\(\therefore t=2\)
আবার,
\(t=\sqrt{4-y}\)
\(\Rightarrow t=\sqrt{4-4}\), যখন \(y=4\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(=\int_{2}^{0}{(4-t^2).t\times{-2tdt}}\)
\(=\int_{2}^{0}{(4-t^2)\times{-2t^2dt}}\)
\(=\int_{2}^{0}{(-8t^2+2t^4)dt}\)
\(=-8\int_{2}^{0}{t^2dt}+2\int_{2}^{0}{t^4dt}\)
\(=-8\left[\frac{t^3}{3}\right]_{2}^{0}+2\left[\frac{t^5}{5}\right]_{2}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{8}{3}\left[t^3\right]_{2}^{0}+\frac{2}{5}\left[t^5\right]_{2}^{0}\)
\(=-\frac{8}{3}\left[0^3-2^3\right]+\frac{2}{5}\left[0^5-2^5\right]\)
\(=-\frac{8}{3}\left[0-8\right]+\frac{2}{5}\left[0-32\right]\)
\(=\frac{64}{3}-\frac{64}{5}\)
\(=\frac{320-192}{15}\)
\(=\frac{128}{15}\)
\((9.)\) \(\int_{0}^{3}{(3-2x+x^2)dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(9\)
উত্তরঃ \(9\)
সমাধানঃ
\(\int_{0}^{3}{(3-2x+x^2)dx}\)
\(=\left[3x-2.\frac{x^2}{2}+\frac{x^3}{3}\right]_{0}^{3}\) ➜ \(\because \int{1dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[3x-x^2+\frac{x^3}{3}\right]_{0}^{3}\)
\(=3.3-3^2+\frac{3^3}{3}-3.0+0^2-\frac{0^3}{3}\)
\(=9-9+\frac{27}{3}-0+0-0\)
\(=9\)
\(=\left[3x-2.\frac{x^2}{2}+\frac{x^3}{3}\right]_{0}^{3}\) ➜ \(\because \int{1dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[3x-x^2+\frac{x^3}{3}\right]_{0}^{3}\)
\(=3.3-3^2+\frac{3^3}{3}-3.0+0^2-\frac{0^3}{3}\)
\(=9-9+\frac{27}{3}-0+0-0\)
\(=9\)
\((10.)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{2\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{(\sin{x}+\cos{x})+(\sin{x}-\cos{x})}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left\{\frac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}}+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left\{1+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}-\sin{x}}{\sin{x}+\cos{x}}dx}\) ➜ \(\because \int{1dx}=x\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{d(\sin{x}+\cos{x})}{\sin{x}+\cos{x}}}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{|\sin{x}+\cos{x}|}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{|\sin{\left(\frac{\pi}{2}\right)}+\cos{\left(\frac{\pi}{2}\right)}|}-\ln{|\sin{0}+\cos{0}|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{\left(1+0\right)}-\ln{|0+1|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(1)}-\ln{(1)}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}.0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{2\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{(\sin{x}+\cos{x})+(\sin{x}-\cos{x})}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left\{\frac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}}+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left\{1+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}-\sin{x}}{\sin{x}+\cos{x}}dx}\) ➜ \(\because \int{1dx}=x\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{d(\sin{x}+\cos{x})}{\sin{x}+\cos{x}}}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{|\sin{x}+\cos{x}|}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{|\sin{\left(\frac{\pi}{2}\right)}+\cos{\left(\frac{\pi}{2}\right)}|}-\ln{|\sin{0}+\cos{0}|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{\left(1+0\right)}-\ln{|0+1|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(1)}-\ln{(1)}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}.0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\((11.)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{|\sin{x}|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
উত্তরঃ \(2\)
সমাধানঃ
আমরা জানি,
\(1>\sin{x}>-1\)
\(\Rightarrow \sin{\left(\frac{\pi}{2}\right)}>\sin{x}>\sin{\left(-\frac{\pi}{2}\right)}\)
\(\therefore \frac{\pi}{2}>x>-\frac{\pi}{2}\)
আবার,
\(\sin{x}=0\)
\(\Rightarrow \sin{x}=\sin{0}\)
\(\therefore x=0\)
এখন,
\(0>\sin{x}\)
\(\Rightarrow 0>x\)
\(\therefore 0>x>-\frac{\pi}{2}\) ব্যাবধিতে \(|\sin{x}|=-\sin{x}\)
আবার,
\(\sin{x}>0\)
\(\Rightarrow x>0\)
\(\therefore \frac{\pi}{2}>x>0\) ব্যাবধিতে \(|\sin{x}|=\sin{x}\)
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{|\sin{x}|dx}\)\(1>\sin{x}>-1\)
\(\Rightarrow \sin{\left(\frac{\pi}{2}\right)}>\sin{x}>\sin{\left(-\frac{\pi}{2}\right)}\)
\(\therefore \frac{\pi}{2}>x>-\frac{\pi}{2}\)
আবার,
\(\sin{x}=0\)
\(\Rightarrow \sin{x}=\sin{0}\)
\(\therefore x=0\)
এখন,
\(0>\sin{x}\)
\(\Rightarrow 0>x\)
\(\therefore 0>x>-\frac{\pi}{2}\) ব্যাবধিতে \(|\sin{x}|=-\sin{x}\)
আবার,
\(\sin{x}>0\)
\(\Rightarrow x>0\)
\(\therefore \frac{\pi}{2}>x>0\) ব্যাবধিতে \(|\sin{x}|=\sin{x}\)
\(=\int_{-\frac{\pi}{2}}^{0}{(-\sin{x})dx}+\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=-\int_{-\frac{\pi}{2}}^{0}{\sin{x}dx}+\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=-\left[-\cos{x}\right]_{-\frac{\pi}{2}}^{0}+\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\)
\(=\left[\cos{x}\right]_{-\frac{\pi}{2}}^{0}-\left[\cos{x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\left[\cos{0}-\cos{\left(-\frac{\pi}{2}\right)}\right]-\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\)
\(=\left[\cos{0}-\cos{\left(\frac{\pi}{2}\right)}\right]-\left[0-1\right]\)
\(=\left[1-0\right]+1\)
\(=1+1\)
\(=2\)
\((12.)\) \(\int_{2}^{8}{|x-5|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(9\)
উত্তরঃ \(9\)
সমাধানঃ
ধরি,
\(0>x-5\)
\(\Rightarrow 5>x\)
\(\Rightarrow |x-5|=-(x-5)\)
\(\therefore 5>x>2\) ব্যাবধিতে \(|x-5|=-(x-5)\)
আবার,
\(x-5>0\)
\(\Rightarrow x>5\)
\(\therefore 8>x>5\) ব্যাবধিতে \(|x-5|=x-5\)
\(\int_{2}^{8}{|x-5|dx}\)\(0>x-5\)
\(\Rightarrow 5>x\)
\(\Rightarrow |x-5|=-(x-5)\)
\(\therefore 5>x>2\) ব্যাবধিতে \(|x-5|=-(x-5)\)
আবার,
\(x-5>0\)
\(\Rightarrow x>5\)
\(\therefore 8>x>5\) ব্যাবধিতে \(|x-5|=x-5\)
\(=\int_{2}^{5}{(5-x)dx}+\int_{5}^{8}{(x-5)dx}\)
\(=\left[5x-\frac{x^2}{2}\right]_{2}^{5}+\left[\frac{x^2}{2}-5x\right]_{5}^{8}\) ➜ \(\because \int{5dx}=5x, \int{xdx}=\frac{x^2}{2}\)
\(=5.5-\frac{5^2}{2}-5.2+\frac{2^2}{2}+\frac{8^2}{2}-5.8-\frac{5^2}{2}+5.5\)
\(=25-\frac{25}{2}-10+\frac{4}{2}+\frac{64}{2}-40-\frac{25}{2}+25\)
\(=50-50-2\frac{25}{2}+2+32\)
\(=-25+34\)
\(=9\)
\((13.)\) \(\int_{0}^{3}{\sqrt{9-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{9\pi}{4}\)
উত্তরঃ \(\frac{9\pi}{4}\)
সমাধানঃ
\(\int_{0}^{3}{\sqrt{9-x^2}dx}\)
\(=\int_{0}^{3}{\sqrt{3^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{3^2-x^2}}{2}+\frac{3^2}{2}\sin^{-1}{\left(\frac{x}{3}\right)}\right]_{0}^{3}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}{\left(\frac{x}{3}\right)}\right]_{0}^{3}\)
\(=\frac{3\sqrt{9-3^2}}{2}+\frac{9}{2}\sin^{-1}{\left(\frac{3}{3}\right)}-\frac{0.\sqrt{9-0^2}}{2}-\frac{9}{2}\sin^{-1}{\left(\frac{0}{3}\right)}\)
\(=\frac{3\sqrt{0}}{2}+\frac{9}{2}\sin^{-1}{1}-\frac{0}{2}-\frac{9}{2}\sin^{-1}{0}\)
\(=\frac{3.0}{2}+\frac{9}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{9}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{9}{2}\frac{\pi}{2}-\frac{9}{2}.0\)
\(=0+\frac{9\pi}{4}-0\)
\(=\frac{9\pi}{4}\)
\(=\int_{0}^{3}{\sqrt{3^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{3^2-x^2}}{2}+\frac{3^2}{2}\sin^{-1}{\left(\frac{x}{3}\right)}\right]_{0}^{3}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}{\left(\frac{x}{3}\right)}\right]_{0}^{3}\)
\(=\frac{3\sqrt{9-3^2}}{2}+\frac{9}{2}\sin^{-1}{\left(\frac{3}{3}\right)}-\frac{0.\sqrt{9-0^2}}{2}-\frac{9}{2}\sin^{-1}{\left(\frac{0}{3}\right)}\)
\(=\frac{3\sqrt{0}}{2}+\frac{9}{2}\sin^{-1}{1}-\frac{0}{2}-\frac{9}{2}\sin^{-1}{0}\)
\(=\frac{3.0}{2}+\frac{9}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{9}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{9}{2}\frac{\pi}{2}-\frac{9}{2}.0\)
\(=0+\frac{9\pi}{4}-0\)
\(=\frac{9\pi}{4}\)
অনুশীলনী \(10.G / Q.1\)-এর অতি সংক্ষিপ্ত প্রশ্নসমুহ
নিচের যোগজগুলির মান নির্ণয় করঃ
\(Q.1.(i)\) \(\int_{0}^{3}{(3-2x+x^2)dx}\)
উত্তরঃ \(9\)
[ কুঃ ২০০৭,২০০৬; বঃ২০০৪ ]
\(Q.1.(ii)\) \(\int_{0}^{1}{x(1-\sqrt{x})^2dx}\)
উত্তরঃ \(\frac{1}{30}\)
\(Q.1.(iii)\) \(\int_{-1}^{1}{|x|dx}\)
উত্তরঃ \(1\)
\(Q.1.(iv)\) \(\int_{0}^{4}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx}\)
উত্তরঃ \(\frac{28}{3}\)
\(Q.1.(v)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
উত্তরঃ \(2\)
[ চঃ২০০৪]
\(Q.1.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}d\theta}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
[ চঃ২০১৪; রাঃ২০০৩ ]
\(Q.1.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\cos{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৩ ]
\(Q.1.(viii)\) \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{1-\cos{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sqrt{3}-1)\)
[ দিঃ,চঃ২০১২ ]
\(Q.1.(ix)\) \(\int_{0}^{\frac{\pi}{3}}{\frac{dx}{1-\sin{x}}}\)
উত্তরঃ \(\sqrt{3}+1\)
[ ঢাঃ২০১৩,২০০৯,২০০৮; রাঃ২০১৩; সিঃ২০১০; কুঃ,যঃ২০০৯ ]
\(Q.1.(x)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{dx}{1+\sin{x}}}\)
উত্তরঃ \(2-\sqrt{2}\)
[ বঃ২০১৪,২০১২,২০০৯,২০০৬,২০০৩; দিঃ২০১৪,২০১০; সিঃ২০১৪,২০০৮,২০০৪; কুঃ২০১৪,২০০৫; ঢাঃ২০১২,২০১০; রাঃ২০১০,২০০৮; চঃ২০০৮,২০০৬; যঃ২০০৮,২০০৩ ]
\(Q.1.(xi)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\)
উত্তরঃ \(\pi+2\)
[ যঃ২০১৩,২০০৬,২০০৩ ]
\(Q.1.(xii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2x}}{\cos^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
\(Q.1.(xiii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}dx}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
[ বঃ২০১১; কুঃ২০০৯ ]
\(Q.1.(xiv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x+1}+\sqrt{x}}}\)
উত্তরঃ \(\frac{4}{3}(\sqrt{2}-1)\)
\(Q.1.(xv)\) \(\int_{0}^{4}{\frac{dx}{\sqrt{2x+1}}}\)
উত্তরঃ \(2\)
\(Q.1.(xvi)\) \(\int_{1}^{4}{\frac{dx}{(2+3x)^2}}\)
উত্তরঃ \(\frac{3}{70}\)
[ যঃ২০০৭ ]
\(Q.1.(xvii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\)
উত্তরঃ \(0\)
[ রাঃ২০০৪; কুঃ২০০৬ ]
\(Q.1.(xviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cos{x}}}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১১; বঃ২০০৮ ]
\(Q.1.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\cos{x}}dx}\)
উত্তরঃ \(2\)
\(Q.1.(xx)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}dx}\)
উত্তরঃ \(6\sqrt{2}\)
[ কুঃ২০০৮ ]
\(Q.1.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{x}}dx}\)
উত্তরঃ \(2\)
[ বঃ২০১১ ]
\(Q.1.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{\theta}}d\theta}\)
উত্তরঃ \(2\)
[ বঃ২০১৭ ]
\(Q.1.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ সিঃ২০০৫ ]
উত্তরঃ \(9\)
[ কুঃ ২০০৭,২০০৬; বঃ২০০৪ ]
\(Q.1.(ii)\) \(\int_{0}^{1}{x(1-\sqrt{x})^2dx}\)
উত্তরঃ \(\frac{1}{30}\)
\(Q.1.(iii)\) \(\int_{-1}^{1}{|x|dx}\)
উত্তরঃ \(1\)
\(Q.1.(iv)\) \(\int_{0}^{4}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx}\)
উত্তরঃ \(\frac{28}{3}\)
\(Q.1.(v)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
উত্তরঃ \(2\)
[ চঃ২০০৪]
\(Q.1.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}d\theta}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
[ চঃ২০১৪; রাঃ২০০৩ ]
\(Q.1.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\cos{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৩ ]
\(Q.1.(viii)\) \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{1-\cos{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sqrt{3}-1)\)
[ দিঃ,চঃ২০১২ ]
\(Q.1.(ix)\) \(\int_{0}^{\frac{\pi}{3}}{\frac{dx}{1-\sin{x}}}\)
উত্তরঃ \(\sqrt{3}+1\)
[ ঢাঃ২০১৩,২০০৯,২০০৮; রাঃ২০১৩; সিঃ২০১০; কুঃ,যঃ২০০৯ ]
\(Q.1.(x)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{dx}{1+\sin{x}}}\)
উত্তরঃ \(2-\sqrt{2}\)
[ বঃ২০১৪,২০১২,২০০৯,২০০৬,২০০৩; দিঃ২০১৪,২০১০; সিঃ২০১৪,২০০৮,২০০৪; কুঃ২০১৪,২০০৫; ঢাঃ২০১২,২০১০; রাঃ২০১০,২০০৮; চঃ২০০৮,২০০৬; যঃ২০০৮,২০০৩ ]
\(Q.1.(xi)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\)
উত্তরঃ \(\pi+2\)
[ যঃ২০১৩,২০০৬,২০০৩ ]
\(Q.1.(xii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2x}}{\cos^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
\(Q.1.(xiii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}dx}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
[ বঃ২০১১; কুঃ২০০৯ ]
\(Q.1.(xiv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x+1}+\sqrt{x}}}\)
উত্তরঃ \(\frac{4}{3}(\sqrt{2}-1)\)
\(Q.1.(xv)\) \(\int_{0}^{4}{\frac{dx}{\sqrt{2x+1}}}\)
উত্তরঃ \(2\)
\(Q.1.(xvi)\) \(\int_{1}^{4}{\frac{dx}{(2+3x)^2}}\)
উত্তরঃ \(\frac{3}{70}\)
[ যঃ২০০৭ ]
\(Q.1.(xvii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\)
উত্তরঃ \(0\)
[ রাঃ২০০৪; কুঃ২০০৬ ]
\(Q.1.(xviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cos{x}}}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১১; বঃ২০০৮ ]
\(Q.1.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\cos{x}}dx}\)
উত্তরঃ \(2\)
\(Q.1.(xx)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}dx}\)
উত্তরঃ \(6\sqrt{2}\)
[ কুঃ২০০৮ ]
\(Q.1.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{x}}dx}\)
উত্তরঃ \(2\)
[ বঃ২০১১ ]
\(Q.1.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{\theta}}d\theta}\)
উত্তরঃ \(2\)
[ বঃ২০১৭ ]
\(Q.1.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ সিঃ২০০৫ ]
\(Q.1.(xxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ২০০৯,২০০৫; সিঃ২০১১; চঃ২০০৪ ]
\(Q.1.(xxv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{2\theta}d\theta}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ মাঃ২০০৯ ]
\(Q.1.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০১৩,২০০৯,২০০৭; দিঃ২০১৩; বঃ২০০৮; সিঃ২০১২,২০০৬,২০০৫ ]
\(Q.1.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{\theta}d\theta}\)
উত্তরঃ \(\frac{2}{3}\)
\(Q.1.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{x}+b\sin^2{x})dx}\)
উত্তরঃ \(\frac{1}{4}(a+b)\pi\)
[ চঃ২০০৩ ]
\(Q.1.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^4{x}dx}\)
উত্তরঃ \(\frac{3\pi}{16}\)
[ যঃ২০০৪ ]
\(Q.1.(xxx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^4{x}dx}\)
উত্তরঃ \(\frac{3\pi-8}{32}\)
[ চঃ২০০৩ ]
\(Q.1.(xxxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{x}\sin{2x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ রাঃ২০০৮; দিঃ২০১৩; যঃ২০০৮; চঃ২০০৬; বঃ২০০৬,২০০৪ ]
\(Q.1.(xxxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{2x}\cos{x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০০৫ ]
\(Q.1.(xxxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{3\theta}\cos{2\theta}d\theta}\)
উত্তরঃ \(\frac{3}{5}\)
[ ঢাঃ২০১৪; চঃ২০০৩ ]
\(Q.1.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}\sin{3x}dx}\)
উত্তরঃ \(-\frac{2}{15}\)
[ সিঃ২০০৩; বঃ২০০৫; যঃ২০১৪; রাঃ২০০৮; মাঃ২০০৩,২০০৫ ]
\(Q.1.(xxxv)\) \(\int_{1}^{4}{\frac{dx}{(2x+3)^2}}\)
উত্তরঃ \(\frac{3}{55}\)
\(Q.1.(xxxvi)\) \(\int_{3}^{1}{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(-\ln{(5)}\)
[ কুঃ২০০৩; সিঃ২০০৬ ]
\(Q.1.(xxxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{6}\)
[ বুয়েটঃ২০০৭-২০০৮ ]
\(Q.1.(xxxviii)\) \(\int_{2}^{5}{\frac{7x}{\sqrt{x^2+3}}dx}\)
উত্তরঃ \(7\sqrt{7}\)
\(Q.1.(xxxix)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{\cos{x}}{\sqrt{12+\sin{x}}}dx}\)
উত্তরঃ \(5\sqrt{2}-4\sqrt{3}\)
[ সাস্টঃ২০০৭-২০০৮ ]
\(Q.1.(xL)\) \(\int_{-\frac{\pi}{4}}^{0}{\tan{\left(\frac{\pi}{4}+x\right)}dx}\)
উত্তরঃ \(\frac{1}{2}\ln{(2)}\)
\(Q.1.(xLi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2\theta}}{\cos^2{\theta}}d\theta}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
\(Q.1.(xLii)\) \(\int_{\frac{1}{4}}^{1}{|2x-1|dx}\)
উত্তরঃ \(\frac{5}{16}\)
\(Q.1.(xLiii)\) \(\int_{0}^{\frac{\pi}{2}}{|\cos{2x}|dx}\)
উত্তরঃ \(1\)
\(Q.1.(xLiv)\) \(\int_{-\pi}^{\pi}{|\cos{x}|dx}\)
উত্তরঃ \(4\)
\(Q.1.(xLv)\) \(\int_{1}^{4}{\frac{(2-x)^2}{\sqrt{x}}dx}\)
উত্তরঃ \(1\frac{11}{15}\)
\(Q.1.(xLvi)\) \(\int_{0}^{4}{\frac{dx}{(2x+3)^2}}\)
উত্তরঃ \(\frac{4}{33}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ২০০৯,২০০৫; সিঃ২০১১; চঃ২০০৪ ]
\(Q.1.(xxv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{2\theta}d\theta}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ মাঃ২০০৯ ]
\(Q.1.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০১৩,২০০৯,২০০৭; দিঃ২০১৩; বঃ২০০৮; সিঃ২০১২,২০০৬,২০০৫ ]
\(Q.1.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{\theta}d\theta}\)
উত্তরঃ \(\frac{2}{3}\)
\(Q.1.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{x}+b\sin^2{x})dx}\)
উত্তরঃ \(\frac{1}{4}(a+b)\pi\)
[ চঃ২০০৩ ]
\(Q.1.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^4{x}dx}\)
উত্তরঃ \(\frac{3\pi}{16}\)
[ যঃ২০০৪ ]
\(Q.1.(xxx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^4{x}dx}\)
উত্তরঃ \(\frac{3\pi-8}{32}\)
[ চঃ২০০৩ ]
\(Q.1.(xxxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{x}\sin{2x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ রাঃ২০০৮; দিঃ২০১৩; যঃ২০০৮; চঃ২০০৬; বঃ২০০৬,২০০৪ ]
\(Q.1.(xxxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{2x}\cos{x}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০০৫ ]
\(Q.1.(xxxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{3\theta}\cos{2\theta}d\theta}\)
উত্তরঃ \(\frac{3}{5}\)
[ ঢাঃ২০১৪; চঃ২০০৩ ]
\(Q.1.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}\sin{3x}dx}\)
উত্তরঃ \(-\frac{2}{15}\)
[ সিঃ২০০৩; বঃ২০০৫; যঃ২০১৪; রাঃ২০০৮; মাঃ২০০৩,২০০৫ ]
\(Q.1.(xxxv)\) \(\int_{1}^{4}{\frac{dx}{(2x+3)^2}}\)
উত্তরঃ \(\frac{3}{55}\)
\(Q.1.(xxxvi)\) \(\int_{3}^{1}{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(-\ln{(5)}\)
[ কুঃ২০০৩; সিঃ২০০৬ ]
\(Q.1.(xxxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{6}\)
[ বুয়েটঃ২০০৭-২০০৮ ]
\(Q.1.(xxxviii)\) \(\int_{2}^{5}{\frac{7x}{\sqrt{x^2+3}}dx}\)
উত্তরঃ \(7\sqrt{7}\)
\(Q.1.(xxxix)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{\cos{x}}{\sqrt{12+\sin{x}}}dx}\)
উত্তরঃ \(5\sqrt{2}-4\sqrt{3}\)
[ সাস্টঃ২০০৭-২০০৮ ]
\(Q.1.(xL)\) \(\int_{-\frac{\pi}{4}}^{0}{\tan{\left(\frac{\pi}{4}+x\right)}dx}\)
উত্তরঃ \(\frac{1}{2}\ln{(2)}\)
\(Q.1.(xLi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2\theta}}{\cos^2{\theta}}d\theta}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
\(Q.1.(xLii)\) \(\int_{\frac{1}{4}}^{1}{|2x-1|dx}\)
উত্তরঃ \(\frac{5}{16}\)
\(Q.1.(xLiii)\) \(\int_{0}^{\frac{\pi}{2}}{|\cos{2x}|dx}\)
উত্তরঃ \(1\)
\(Q.1.(xLiv)\) \(\int_{-\pi}^{\pi}{|\cos{x}|dx}\)
উত্তরঃ \(4\)
\(Q.1.(xLv)\) \(\int_{1}^{4}{\frac{(2-x)^2}{\sqrt{x}}dx}\)
উত্তরঃ \(1\frac{11}{15}\)
\(Q.1.(xLvi)\) \(\int_{0}^{4}{\frac{dx}{(2x+3)^2}}\)
উত্তরঃ \(\frac{4}{33}\)
\(Q.1.(i)\) \(\int_{0}^{3}{(3-2x+x^2)dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(9\)
[ কুঃ ২০০৭,২০০৬; বঃ২০০৪ ]
উত্তরঃ \(9\)
[ কুঃ ২০০৭,২০০৬; বঃ২০০৪ ]
সমাধানঃ
\(\int_{0}^{3}{(3-2x+x^2)dx}\)
\(=3\int_{0}^{3}{1dx}-2\int_{0}^{3}{xdx}+\int_{0}^{3}{x^2dx}\)
\(=3\left[x\right]_{0}^{3}-2\left[\frac{x^2}{2}\right]_{0}^{3}+\left[\frac{x^3}{3}\right]_{0}^{3}\) ➜ \(\because \int{1dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=3\left[x\right]_{0}^{3}-\left[x^2\right]_{0}^{3}+\frac{1}{3}\left[x^3\right]_{0}^{3}\)
\(=3\left[3-0\right]-\left[3^2-0^2\right]+\frac{1}{3}\left[3^3-0^3\right]\)
\(=9-\left[9-0\right]+\frac{1}{3}\left[27-0\right]\)
\(=9-9+9\)
\(=9\)
\(=3\int_{0}^{3}{1dx}-2\int_{0}^{3}{xdx}+\int_{0}^{3}{x^2dx}\)
\(=3\left[x\right]_{0}^{3}-2\left[\frac{x^2}{2}\right]_{0}^{3}+\left[\frac{x^3}{3}\right]_{0}^{3}\) ➜ \(\because \int{1dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=3\left[x\right]_{0}^{3}-\left[x^2\right]_{0}^{3}+\frac{1}{3}\left[x^3\right]_{0}^{3}\)
\(=3\left[3-0\right]-\left[3^2-0^2\right]+\frac{1}{3}\left[3^3-0^3\right]\)
\(=9-\left[9-0\right]+\frac{1}{3}\left[27-0\right]\)
\(=9-9+9\)
\(=9\)
\(Q.1.(ii)\) \(\int_{0}^{1}{x(1-\sqrt{x})^2dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{30}\)
উত্তরঃ \(\frac{1}{30}\)
সমাধানঃ
\(\int_{0}^{1}{x(1-\sqrt{x})^2dx}\)
\(=\int_{0}^{1}{x\{1-2\sqrt{x}+(\sqrt{x})^2\}dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int_{0}^{1}{x\{1-2\sqrt{x}+x\}dx}\)
\(=\int_{0}^{1}{\{x-2x.x^{\frac{1}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{1}{2}+1}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{1+2}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{3}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{xdx}-2\int_{0}^{1}{x^{\frac{3}{2}}dx}+\int_{0}^{1}{x^2dx}\)
\(=\left[\frac{x^2}{2}\right]_{0}^{1}-2\left[\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]_{0}^{1}+\left[\frac{x^3}{3}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[x^2\right]_{0}^{1}-2\left[\frac{x^{\frac{3+2}{2}}}{\frac{3+2}{2}}\right]_{0}^{1}+\frac{1}{3}\left[x^3\right]_{0}^{1}\)
\(=\frac{1}{2}\left[1^2-0^2\right]-2\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{1}+\frac{1}{3}\left[1^3-0^3\right]\)
\(=\frac{1}{2}\left[1-0\right]-2\times{\frac{2}{5}}\left[x^{\frac{5}{2}}\right]_{0}^{1}+\frac{1}{3}\left[1-0\right]\)
\(=\frac{1}{2}-\frac{4}{5}\left[1^{\frac{5}{2}}-0^{\frac{5}{2}}\right]+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{4}{5}\left[1-0\right]+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{4}{5}+\frac{1}{3}\)
\(=\frac{15-24+10}{30}\)
\(=\frac{25-24}{30}\)
\(=\frac{1}{30}\)
\(=\int_{0}^{1}{x\{1-2\sqrt{x}+(\sqrt{x})^2\}dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int_{0}^{1}{x\{1-2\sqrt{x}+x\}dx}\)
\(=\int_{0}^{1}{\{x-2x.x^{\frac{1}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{1}{2}+1}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{1+2}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{\{x-2x^{\frac{3}{2}}+x^2\}dx}\)
\(=\int_{0}^{1}{xdx}-2\int_{0}^{1}{x^{\frac{3}{2}}dx}+\int_{0}^{1}{x^2dx}\)
\(=\left[\frac{x^2}{2}\right]_{0}^{1}-2\left[\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]_{0}^{1}+\left[\frac{x^3}{3}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[x^2\right]_{0}^{1}-2\left[\frac{x^{\frac{3+2}{2}}}{\frac{3+2}{2}}\right]_{0}^{1}+\frac{1}{3}\left[x^3\right]_{0}^{1}\)
\(=\frac{1}{2}\left[1^2-0^2\right]-2\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{1}+\frac{1}{3}\left[1^3-0^3\right]\)
\(=\frac{1}{2}\left[1-0\right]-2\times{\frac{2}{5}}\left[x^{\frac{5}{2}}\right]_{0}^{1}+\frac{1}{3}\left[1-0\right]\)
\(=\frac{1}{2}-\frac{4}{5}\left[1^{\frac{5}{2}}-0^{\frac{5}{2}}\right]+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{4}{5}\left[1-0\right]+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{4}{5}+\frac{1}{3}\)
\(=\frac{15-24+10}{30}\)
\(=\frac{25-24}{30}\)
\(=\frac{1}{30}\)
\(Q.1.(iii)\) \(\int_{-1}^{1}{|x|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
উত্তরঃ \(1\)
সমাধানঃ
ধরি,
\(0>x\)
\(\Rightarrow |x|=-x\)
\(\therefore 0>x>-1\) ব্যাবধিতে \(|x|=-x\)
আবার,
\(x>0\)
\(\Rightarrow |x|=x\)
\(\therefore 1>x>0\) ব্যাবধিতে \(|x|=-x\)
\(\int_{-1}^{1}{|x|dx}\)\(0>x\)
\(\Rightarrow |x|=-x\)
\(\therefore 0>x>-1\) ব্যাবধিতে \(|x|=-x\)
আবার,
\(x>0\)
\(\Rightarrow |x|=x\)
\(\therefore 1>x>0\) ব্যাবধিতে \(|x|=-x\)
\(=\int_{-1}^{0}{|x|dx}+\int_{0}^{1}{|x|dx}\) ➜ \(\because \int_{a}^{b}{f(x)dx}=\int_{a}^{0}{f(x)dx}+\int_{0}^{b}{f(x)dx}\)
\(=\int_{-1}^{0}{(-x)dx}+\int_{0}^{1}{xdx}\)
\(=-\int_{-1}^{0}{xdx}+\int_{0}^{1}{xdx}\)
\(=-\left[\frac{x^2}{2}\right]_{-1}^{0}+\left[\frac{x^2}{2}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{2}\left[x^2\right]_{-1}^{0}+\frac{1}{2}\left[x^2\right]_{0}^{1}\)
\(=-\frac{1}{2}\left[0^2-(-1)^2\right]+\frac{1}{2}\left[1^2-0^2\right]\)
\(=-\frac{1}{2}\left[0-1\right]+\frac{1}{2}\left[1-0\right]\)
\(=\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1}{2}\)
\(=\frac{2}{2}\)
\(=1\)
\(Q.1.(iv)\) \(\int_{0}^{4}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{28}{3}\)
উত্তরঃ \(\frac{28}{3}\)
সমাধানঃ
\(\int_{0}^{4}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx}\)
\(=\int_{0}^{4}{\left(x^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}}\right)dx}\)
\(=\int_{0}^{4}{x^{\frac{1}{2}}dx}+\int_{0}^{4}{x^{-\frac{1}{2}}dx}\)
\(=\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{4}+\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{0}^{4}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{4}+\left[\frac{x^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}\right]_{0}^{4}\)
\(=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}+\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{0}^{4}\)
\(=\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{4}+2\left[x^{\frac{1}{2}}\right]_{0}^{4}\)
\(=\frac{2}{3}\left[4^{\frac{3}{2}}-0^{\frac{3}{2}}\right]+2\left[4^{\frac{1}{2}}-0^{\frac{1}{2}}\right]\)
\(=\frac{2}{3}\left[(\sqrt{4})^{3}-0\right]+2\left[\sqrt{4}-0\right]\)
\(=\frac{2}{3}\left[(2)^{3}-0\right]+2\left[2-0\right]\)
\(=\frac{2}{3}.(2)^{3}+2.2\)
\(=\frac{2}{3}.8+4\)
\(=\frac{16}{3}+4\)
\(=\frac{16+12}{3}\)
\(=\frac{28}{3}\)
\(=\int_{0}^{4}{\left(x^{\frac{1}{2}}+\frac{1}{x^{\frac{1}{2}}}\right)dx}\)
\(=\int_{0}^{4}{x^{\frac{1}{2}}dx}+\int_{0}^{4}{x^{-\frac{1}{2}}dx}\)
\(=\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{4}+\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{0}^{4}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{4}+\left[\frac{x^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}\right]_{0}^{4}\)
\(=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}+\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{0}^{4}\)
\(=\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{4}+2\left[x^{\frac{1}{2}}\right]_{0}^{4}\)
\(=\frac{2}{3}\left[4^{\frac{3}{2}}-0^{\frac{3}{2}}\right]+2\left[4^{\frac{1}{2}}-0^{\frac{1}{2}}\right]\)
\(=\frac{2}{3}\left[(\sqrt{4})^{3}-0\right]+2\left[\sqrt{4}-0\right]\)
\(=\frac{2}{3}\left[(2)^{3}-0\right]+2\left[2-0\right]\)
\(=\frac{2}{3}.(2)^{3}+2.2\)
\(=\frac{2}{3}.8+4\)
\(=\frac{16}{3}+4\)
\(=\frac{16+12}{3}\)
\(=\frac{28}{3}\)
\(Q.1.(v)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
[ চঃ২০০৪]
উত্তরঃ \(2\)
[ চঃ২০০৪]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}\)
\(=\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\)
\(=-\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]\)
\(=-\left[0-1\right]+\left[1-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\), \(\sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=1+1\)
\(=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}\)
\(=\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\)
\(=-\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]\)
\(=-\left[0-1\right]+\left[1-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\), \(\sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=1+1\)
\(=2\)
\(Q.1.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1-\frac{\pi}{4}\)
[ চঃ২০১৪; রাঃ২০০৩ ]
উত্তরঃ \(1-\frac{\pi}{4}\)
[ চঃ২০১৪; রাঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}d\theta}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}, 1+\cos{2A}=2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{\sin^2{\theta}}{\cos^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\tan^2{\theta}d\theta}\) ➜ \(\because \frac{\sin^2{A}}{\cos^2{A}}=\tan^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{(\sec^2{\theta}-1)d\theta}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{\theta}d\theta}-\int_{0}^{\frac{\pi}{4}}{1d\theta}\)
\(=\left[\tan{\theta}\right]_{0}^{\frac{\pi}{4}}-\left[\theta\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{1dx}=x\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\frac{\pi}{4}-0\right]\)
\(=\left[1-0\right]-\frac{\pi}{4}\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=1-\frac{\pi}{4}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}d\theta}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}, 1+\cos{2A}=2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{\sin^2{\theta}}{\cos^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\tan^2{\theta}d\theta}\) ➜ \(\because \frac{\sin^2{A}}{\cos^2{A}}=\tan^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{(\sec^2{\theta}-1)d\theta}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{\theta}d\theta}-\int_{0}^{\frac{\pi}{4}}{1d\theta}\)
\(=\left[\tan{\theta}\right]_{0}^{\frac{\pi}{4}}-\left[\theta\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{1dx}=x\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\frac{\pi}{4}-0\right]\)
\(=\left[1-0\right]-\frac{\pi}{4}\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=1-\frac{\pi}{4}\)
\(Q.1.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\cos{2x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৩ ]
উত্তরঃ \(\frac{1}{2}\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\cos{2x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{2\cos^2{x}}dx}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{2}.\frac{1}{\cos^2{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\frac{1}{2}\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\)
\(=\frac{1}{2}\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]\)
\(=\frac{1}{2}\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=\frac{1}{2}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{2\cos^2{x}}dx}\) ➜ \(\because 1+\cos{2A}=2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{2}.\frac{1}{\cos^2{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\frac{1}{2}\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\)
\(=\frac{1}{2}\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]\)
\(=\frac{1}{2}\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=\frac{1}{2}\)
\(Q.1.(viii)\) \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{1-\cos{2x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\sqrt{3}-1)\)
[ দিঃ,চঃ২০১২ ]
উত্তরঃ \(\frac{1}{2}(\sqrt{3}-1)\)
[ দিঃ,চঃ২০১২ ]
সমাধানঃ
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{1-\cos{2x}}dx}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{2\sin^2{x}}dx}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{2}.\frac{1}{\sin^2{x}}dx}\)
\(=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{cosec^2{x}dx}\) ➜ \(\because \frac{1}{\sin^2{A}}=cosec^2{A}\)
\(=\frac{1}{2}\left[-\cot{x}\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}\)
\(=-\frac{1}{2}\left[\cot{\left(\frac{\pi}{4}\right)}-\cot{\frac{\pi}{6}}\right]\)
\(=-\frac{1}{2}\left[1-\sqrt{3}\right]\) ➜ \(\because \cot{\frac{\pi}{4}}=1, \cot{\frac{\pi}{6}}=\sqrt{3}\)
\(=\frac{1}{2}(\sqrt{3}-1)\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{2\sin^2{x}}dx}\) ➜ \(\because 1-\cos{2A}=2\sin^2{A}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{\frac{1}{2}.\frac{1}{\sin^2{x}}dx}\)
\(=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}{cosec^2{x}dx}\) ➜ \(\because \frac{1}{\sin^2{A}}=cosec^2{A}\)
\(=\frac{1}{2}\left[-\cot{x}\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}\)
\(=-\frac{1}{2}\left[\cot{\left(\frac{\pi}{4}\right)}-\cot{\frac{\pi}{6}}\right]\)
\(=-\frac{1}{2}\left[1-\sqrt{3}\right]\) ➜ \(\because \cot{\frac{\pi}{4}}=1, \cot{\frac{\pi}{6}}=\sqrt{3}\)
\(=\frac{1}{2}(\sqrt{3}-1)\)
\(Q.1.(ix)\) \(\int_{0}^{\frac{\pi}{3}}{\frac{dx}{1-\sin{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\sqrt{3}+1\)
[ ঢাঃ২০১৩,২০০৯,২০০৮; রাঃ২০১৩; সিঃ২০১০; কুঃ,যঃ২০০৯ ]
উত্তরঃ \(\sqrt{3}+1\)
[ ঢাঃ২০১৩,২০০৯,২০০৮; রাঃ২০১৩; সিঃ২০১০; কুঃ,যঃ২০০৯ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{3}}{\frac{dx}{1-\sin{x}}}\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1}{1-\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{(1-\sin{x})(1+\sin{x})}dx}\) ➜ লব ও হরের সহিত \((1+\sin{x})\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a-b)(a+b)=a^2-b^2\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\frac{1}{\cos^2{x}}+\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\sec^2{x}+\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\sec^2{x}+\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos{A}}=\sec{A}, \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\sec^2{x}dx}+\int_{0}^{\frac{\pi}{3}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{3}}+\left[\sec{x}\right]_{0}^{\frac{\pi}{3}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\tan{\left(\frac{\pi}{3}\right)}-\tan{0}\right]+\left[\sec{\left(\frac{\pi}{3}\right)}-\sec{0}\right]\)
\(=\left[\sqrt{3}-0\right]+\left[2-1\right]\) ➜ \(\because \tan{\frac{\pi}{3}}=\sqrt{3}, \tan{0}=0\), \(\sec{\frac{\pi}{3}}=2, \sec{0}=1\)
\(=\sqrt{3}+1\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1}{1-\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{(1-\sin{x})(1+\sin{x})}dx}\) ➜ লব ও হরের সহিত \((1+\sin{x})\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a-b)(a+b)=a^2-b^2\)
\(=\int_{0}^{\frac{\pi}{3}}{\frac{1+\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\frac{1}{\cos^2{x}}+\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\sec^2{x}+\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\left(\sec^2{x}+\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos{A}}=\sec{A}, \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{3}}{\sec^2{x}dx}+\int_{0}^{\frac{\pi}{3}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{3}}+\left[\sec{x}\right]_{0}^{\frac{\pi}{3}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\tan{\left(\frac{\pi}{3}\right)}-\tan{0}\right]+\left[\sec{\left(\frac{\pi}{3}\right)}-\sec{0}\right]\)
\(=\left[\sqrt{3}-0\right]+\left[2-1\right]\) ➜ \(\because \tan{\frac{\pi}{3}}=\sqrt{3}, \tan{0}=0\), \(\sec{\frac{\pi}{3}}=2, \sec{0}=1\)
\(=\sqrt{3}+1\)
\(Q.1.(x)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{dx}{1+\sin{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2-\sqrt{2}\)
[ বঃ২০১৪,২০১২,২০০৯,২০০৬,২০০৩; দিঃ২০১৪,২০১০; সিঃ২০১৪,২০০৮,২০০৪; কুঃ২০১৪,২০০৫; ঢাঃ২০১২,২০১০; রাঃ২০১০,২০০৮; চঃ২০০৮,২০০৬; যঃ২০০৮,২০০৩ ]
উত্তরঃ \(2-\sqrt{2}\)
[ বঃ২০১৪,২০১২,২০০৯,২০০৬,২০০৩; দিঃ২০১৪,২০১০; সিঃ২০১৪,২০০৮,২০০৪; কুঃ২০১৪,২০০৫; ঢাঃ২০১২,২০১০; রাঃ২০১০,২০০৮; চঃ২০০৮,২০০৬; যঃ২০০৮,২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\frac{dx}{1+\sin{x}}}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{(1-\sin{x})(1+\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a-b)(a+b)=a^2-b^2\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos{A}}=\sec{A}, \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[\sec{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\sec{\left(\frac{\pi}{4}\right)}-\sec{0}\right]\)
\(=\left[1-0\right]-\left[\sqrt{2}-1\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\), \(\sec{\frac{\pi}{4}}=\sqrt{2}, \sec{0}=1\)
\(=1-\sqrt{2}+1\)
\(=2-\sqrt{2}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{(1-\sin{x})(1+\sin{x})}dx}\) ➜ লব ও হরের সহিত \(1-\sin{x}\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\) ➜ \(\because (a-b)(a+b)=a^2-b^2\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜ \(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos{A}}=\sec{A}, \frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[\sec{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\sec{\left(\frac{\pi}{4}\right)}-\sec{0}\right]\)
\(=\left[1-0\right]-\left[\sqrt{2}-1\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\), \(\sec{\frac{\pi}{4}}=\sqrt{2}, \sec{0}=1\)
\(=1-\sqrt{2}+1\)
\(=2-\sqrt{2}\)
\(Q.1.(xi)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi+2\)
[ যঃ২০১৩,২০০৬,২০০৩ ]
উত্তরঃ \(\pi+2\)
[ যঃ২০১৩,২০০৬,২০০৩ ]
সমাধানঃ
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(\frac{\sec{x}}{\sec{x}}+\frac{1}{\sec{x}}\right)dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(1+\cos{x}\right)dx}\) ➜ \(\because \frac{1}{\sec{A}}=\cos{A}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{1dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{x}dx}\)
\(=\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[\sin{x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{x}dx}=\sin{x}\)
\(=\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{\left(-\frac{\pi}{2}\right)}\right]\)
\(=\left[\frac{\pi}{2}+\frac{\pi}{2}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}+\sin{\left(\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi+\pi}{2}+\left[1+1\right]\) ➜ \(\because \sin{\frac{\pi}{2}}=1\)
\(=\frac{2\pi}{2}+2\)
\(=\pi+2\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\sec{x}+1}{\sec{x}}dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(\frac{\sec{x}}{\sec{x}}+\frac{1}{\sec{x}}\right)dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(1+\cos{x}\right)dx}\) ➜ \(\because \frac{1}{\sec{A}}=\cos{A}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{1dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{x}dx}\)
\(=\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[\sin{x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{x}dx}=\sin{x}\)
\(=\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{\left(-\frac{\pi}{2}\right)}\right]\)
\(=\left[\frac{\pi}{2}+\frac{\pi}{2}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}+\sin{\left(\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi+\pi}{2}+\left[1+1\right]\) ➜ \(\because \sin{\frac{\pi}{2}}=1\)
\(=\frac{2\pi}{2}+2\)
\(=\pi+2\)
\(Q.1.(xii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2x}}{\cos^2{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2x}}{\cos^2{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\cos^2{x}-1}{\cos^2{x}}dx}\) ➜ \(\because \cos{2A}=1+2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2\frac{\cos^2{x}}{\cos^2{x}}-\frac{1}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2-\sec^2{x}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=2\int_{0}^{\frac{\pi}{4}}{1dx}-\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}\)
\(=2\left[x\right]_{0}^{\frac{\pi}{4}}-\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=tan{x}, \int{1dx}=x\)
\(=2\left[\frac{\pi}{4}-0\right]-\left[\tan{\left(\frac{\pi}{4}\right)-\tan{0}}\right]\)
\(=2.\frac{\pi}{4}-\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\)
\(=\frac{\pi}{2}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\cos^2{x}-1}{\cos^2{x}}dx}\) ➜ \(\because \cos{2A}=1+2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2\frac{\cos^2{x}}{\cos^2{x}}-\frac{1}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2-\sec^2{x}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=2\int_{0}^{\frac{\pi}{4}}{1dx}-\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}\)
\(=2\left[x\right]_{0}^{\frac{\pi}{4}}-\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=tan{x}, \int{1dx}=x\)
\(=2\left[\frac{\pi}{4}-0\right]-\left[\tan{\left(\frac{\pi}{4}\right)-\tan{0}}\right]\)
\(=2.\frac{\pi}{4}-\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\)
\(=\frac{\pi}{2}-1\)
\(Q.1.(xiii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1-\frac{\pi}{4}\)
[ বঃ২০১১; কুঃ২০০৯ ]
উত্তরঃ \(1-\frac{\pi}{4}\)
[ বঃ২০১১; কুঃ২০০৯ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{(\sec^2{x}-1)dx}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{1dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[x\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{1dx}=x\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\frac{\pi}{4}-0\right]\)
\(=\left[1-0\right]-\frac{\pi}{4}\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=1-\frac{\pi}{4}\)
\(=\int_{0}^{\frac{\pi}{4}}{(\sec^2{x}-1)dx}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{1dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[x\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{1dx}=x\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-\tan{0}\right]-\left[\frac{\pi}{4}-0\right]\)
\(=\left[1-0\right]-\frac{\pi}{4}\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=1-\frac{\pi}{4}\)
\(Q.1.(xiv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x+1}+\sqrt{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{4}{3}(\sqrt{2}-1)\)
উত্তরঃ \(\frac{4}{3}(\sqrt{2}-1)\)
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{\sqrt{x+1}+\sqrt{x}}}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{x+1}+\sqrt{x}}dx}\)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x})\) গুণ করে।
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1})^2-(\sqrt{x})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2 \)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}dx}\)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{1}dx}\)
\(=\int_{0}^{1}{\left(\sqrt{x+1}-\sqrt{x}\right)dx}\)
\(=\int_{0}^{1}{\sqrt{x+1}dx}-\int_{0}^{1}{\sqrt{x}dx}\)
\(=\int_{0}^{1}{(x+1)^{\frac{1}{2}}dx}-\int_{0}^{1}{(x)^{\frac{1}{2}}dx}\)
\(=\left[\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}-\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{(x+1)^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{1}-\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{1}\)
\(=\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}-\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3}\left[(x+1)^{\frac{3}{2}}\right]_{0}^{1}-\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3}\left[(1+1)^{\frac{3}{2}}-(0+1)^{\frac{3}{2}}\right]-\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]\)
\(=\frac{2}{3}\left[(2)^{\frac{3}{2}}-1\right]-\frac{2}{3}\left[1-0\right]\)
\(=\frac{2}{3}\left[(\sqrt{2})^{3}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}\left[(\sqrt{2})^{2}\sqrt{2}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}\left[2\sqrt{2}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}(2\sqrt{2}-1-1)\)
\(=\frac{2}{3}(2\sqrt{2}-2)\)
\(=\frac{2}{3}.2(\sqrt{2}-1)\)
\(=\frac{4}{3}(\sqrt{2}-1)\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{x+1}+\sqrt{x}}dx}\)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x})(\sqrt{x+1}-\sqrt{x})}dx}\) ➜ লব ও হরের সহিত \((\sqrt{x+1}-\sqrt{x})\) গুণ করে।
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1})^2-(\sqrt{x})^2}dx}\) ➜ \(\because (a+b)(a-b)=a^2-b^2 \)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{x+1-x}dx}\)
\(=\int_{0}^{1}{\frac{\sqrt{x+1}-\sqrt{x}}{1}dx}\)
\(=\int_{0}^{1}{\left(\sqrt{x+1}-\sqrt{x}\right)dx}\)
\(=\int_{0}^{1}{\sqrt{x+1}dx}-\int_{0}^{1}{\sqrt{x}dx}\)
\(=\int_{0}^{1}{(x+1)^{\frac{1}{2}}dx}-\int_{0}^{1}{(x)^{\frac{1}{2}}dx}\)
\(=\left[\frac{(x+1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}-\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{(x+1)^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{1}-\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{0}^{1}\)
\(=\left[\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}-\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3}\left[(x+1)^{\frac{3}{2}}\right]_{0}^{1}-\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3}\left[(1+1)^{\frac{3}{2}}-(0+1)^{\frac{3}{2}}\right]-\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]\)
\(=\frac{2}{3}\left[(2)^{\frac{3}{2}}-1\right]-\frac{2}{3}\left[1-0\right]\)
\(=\frac{2}{3}\left[(\sqrt{2})^{3}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}\left[(\sqrt{2})^{2}\sqrt{2}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}\left[2\sqrt{2}-1\right]-\frac{2}{3}\)
\(=\frac{2}{3}(2\sqrt{2}-1-1)\)
\(=\frac{2}{3}(2\sqrt{2}-2)\)
\(=\frac{2}{3}.2(\sqrt{2}-1)\)
\(=\frac{4}{3}(\sqrt{2}-1)\)
\(Q.1.(xv)\) \(\int_{0}^{4}{\frac{dx}{\sqrt{2x+1}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
উত্তরঃ \(2\)
সমাধানঃ
\(\int_{0}^{4}{\frac{dx}{\sqrt{2x+1}}}\)
\(=\int_{0}^{4}{\frac{1}{(2x+1)^{\frac{1}{2}}}dx}\)
\(=\int_{0}^{4}{(2x+1)^{-\frac{1}{2}}dx}\)
\(=\left[\frac{1}{2}.\frac{(2x+1)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\right]_{0}^{4}\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}\)
\(=\left[\frac{(2x+1)^{\frac{-1+2}{2}}}{2\left(\frac{-1+2}{2}\right)}\right]_{0}^{4}\)
\(=\left[\frac{(2x+1)^{\frac{1}{2}}}{2\left(\frac{1}{2}\right)}\right]_{0}^{4}\)
\(=\left[\frac{(2x+1)^{\frac{1}{2}}}{1}\right]_{0}^{4}\)
\(=\left[(2x+1)^{\frac{1}{2}}\right]_{0}^{4}\)
\(=(2.4+1)^{\frac{1}{2}}-(2.0+1)^{\frac{1}{2}}\)
\(=(8+1)^{\frac{1}{2}}-(0+1)^{\frac{1}{2}}\)
\(=(9)^{\frac{1}{2}}-1^{\frac{1}{2}}\)
\(=\sqrt{9}-\sqrt{1}\)
\(=3-1\)
\(=2\)
\(=\int_{0}^{4}{\frac{1}{(2x+1)^{\frac{1}{2}}}dx}\)
\(=\int_{0}^{4}{(2x+1)^{-\frac{1}{2}}dx}\)
\(=\left[\frac{1}{2}.\frac{(2x+1)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\right]_{0}^{4}\) ➜ \(\because \int{(ax+b)^ndx}=\frac{1}{a}.\frac{(ax+b)^{n+1}}{n+1}\)
\(=\left[\frac{(2x+1)^{\frac{-1+2}{2}}}{2\left(\frac{-1+2}{2}\right)}\right]_{0}^{4}\)
\(=\left[\frac{(2x+1)^{\frac{1}{2}}}{2\left(\frac{1}{2}\right)}\right]_{0}^{4}\)
\(=\left[\frac{(2x+1)^{\frac{1}{2}}}{1}\right]_{0}^{4}\)
\(=\left[(2x+1)^{\frac{1}{2}}\right]_{0}^{4}\)
\(=(2.4+1)^{\frac{1}{2}}-(2.0+1)^{\frac{1}{2}}\)
\(=(8+1)^{\frac{1}{2}}-(0+1)^{\frac{1}{2}}\)
\(=(9)^{\frac{1}{2}}-1^{\frac{1}{2}}\)
\(=\sqrt{9}-\sqrt{1}\)
\(=3-1\)
\(=2\)
\(Q.1.(xvi)\) \(\int_{1}^{4}{\frac{dx}{(2+3x)^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{70}\)
[ যঃ২০০৭ ]
উত্তরঃ \(\frac{3}{70}\)
[ যঃ২০০৭ ]
সমাধানঃ
\(\int_{1}^{4}{\frac{dx}{(2+3x)^2}}\)
\(=\left[-\frac{1}{3}.\frac{1}{2+3x}\right]_{1}^{4}\) ➜ \(\because \int{\frac{dx}{(ax+b)^2}}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{3}\left[\frac{1}{2+3x}\right]_{1}^{4}\)
\(=-\frac{1}{3}\left[\frac{1}{2+3.4}-\frac{1}{2+3.1}\right]\)
\(=-\frac{1}{3}\left[\frac{1}{2+12}-\frac{1}{2+3}\right]\)
\(=-\frac{1}{3}\left[\frac{1}{14}-\frac{1}{5}\right]\)
\(=-\frac{1}{3}\times{\frac{5-14}{70}}\)
\(=-\frac{1}{3}\times{\frac{-9}{70}}\)
\(=\frac{3}{70}\)
\(=\left[-\frac{1}{3}.\frac{1}{2+3x}\right]_{1}^{4}\) ➜ \(\because \int{\frac{dx}{(ax+b)^2}}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{3}\left[\frac{1}{2+3x}\right]_{1}^{4}\)
\(=-\frac{1}{3}\left[\frac{1}{2+3.4}-\frac{1}{2+3.1}\right]\)
\(=-\frac{1}{3}\left[\frac{1}{2+12}-\frac{1}{2+3}\right]\)
\(=-\frac{1}{3}\left[\frac{1}{14}-\frac{1}{5}\right]\)
\(=-\frac{1}{3}\times{\frac{5-14}{70}}\)
\(=-\frac{1}{3}\times{\frac{-9}{70}}\)
\(=\frac{3}{70}\)
\(Q.1.(xvii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(0\)
[ রাঃ২০০৪; কুঃ২০০৬ ]
উত্তরঃ \(0\)
[ রাঃ২০০৪; কুঃ২০০৬ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\)
\(=\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{4}\left[\sin{4x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}\left[\sin{\left(4\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=\frac{1}{4}\left[\sin{\left(2\pi\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{(2\pi)}=0\)
\(=\frac{1}{4}.0\)
\(=0\)
\(=\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{4}\left[\sin{4x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}\left[\sin{\left(4\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=\frac{1}{4}\left[\sin{\left(2\pi\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{(2\pi)}=0\)
\(=\frac{1}{4}.0\)
\(=0\)
\(Q.1.(xviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cos{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
[ ঢাঃ২০১১; বঃ২০০৮ ]
উত্তরঃ \(1\)
[ ঢাঃ২০১১; বঃ২০০৮ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cos{x}}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\cos{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{2\cos^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\left(\frac{A}{2}\right)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{\cos^2{\left(\frac{x}{2}\right)}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sec^2{\left(\frac{x}{2}\right)}dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\frac{1}{2}\left[\frac{1}{\frac{1}{2}}\tan{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}\)
\(=\left[\tan{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=\left[\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\tan{0}\right]\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \tan{0}=0\)
\(=\left[1-0\right]\) ➜ \(\because \tan{\left(\frac{\pi}{4}\right)}=1\)
\(=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\cos{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{2\cos^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\left(\frac{A}{2}\right)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{\cos^2{\left(\frac{x}{2}\right)}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sec^2{\left(\frac{x}{2}\right)}dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=\frac{1}{2}\left[\frac{1}{\frac{1}{2}}\tan{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sec^2{ax}dx}=\frac{1}{a}\tan{ax}\)
\(=\left[\tan{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=\left[\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\tan{0}\right]\)
\(=\left[\tan{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \tan{0}=0\)
\(=\left[1-0\right]\) ➜ \(\because \tan{\left(\frac{\pi}{4}\right)}=1\)
\(=1\)
\(Q.1.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\cos{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
উত্তরঃ \(2\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\cos{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{2\cos^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\left(\frac{A}{2}\right)}\)
\(=\sqrt{2}\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos^2{\left(\frac{x}{2}\right)}}dx}\)
\(=\sqrt{2}\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{x}{2}\right)}dx}\)
\(=\sqrt{2}\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=2\sqrt{2}\left[\sin{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=2\sqrt{2}\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=2\sqrt{2}\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=2\sqrt{2}\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=2\sqrt{2}\times{\frac{1}{\sqrt{2}}}\)
\(=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{2\cos^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\cos{A}=2\cos^2{\left(\frac{A}{2}\right)}\)
\(=\sqrt{2}\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos^2{\left(\frac{x}{2}\right)}}dx}\)
\(=\sqrt{2}\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{x}{2}\right)}dx}\)
\(=\sqrt{2}\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=2\sqrt{2}\left[\sin{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=2\sqrt{2}\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=2\sqrt{2}\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=2\sqrt{2}\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=2\sqrt{2}\times{\frac{1}{\sqrt{2}}}\)
\(=2\)
\(Q.1.(xx)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(6\sqrt{2}\)
[ কুঃ২০০৮ ]
উত্তরঃ \(6\sqrt{2}\)
[ কুঃ২০০৮ ]
সমাধানঃ
\(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}dx}\)
\(=\int_{0}^{\pi}{3\sqrt{2\sin^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\left(\frac{A}{2}\right)}\)
\(=3\sqrt{2}\int_{0}^{\pi}{\sqrt{\sin^2{\left(\frac{x}{2}\right)}}dx}\)
\(=3\sqrt{2}\int_{0}^{\pi}{\sin{\left(\frac{x}{2}\right)}dx}\)
\(=3\sqrt{2}\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\pi}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-6\sqrt{2}\left[\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\pi}\)
\(=-6\sqrt{2}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\)
\(=-6\sqrt{2}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=6\sqrt{2}\)
\(=\int_{0}^{\pi}{3\sqrt{2\sin^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1-\cos{A}=2\sin^2{\left(\frac{A}{2}\right)}\)
\(=3\sqrt{2}\int_{0}^{\pi}{\sqrt{\sin^2{\left(\frac{x}{2}\right)}}dx}\)
\(=3\sqrt{2}\int_{0}^{\pi}{\sin{\left(\frac{x}{2}\right)}dx}\)
\(=3\sqrt{2}\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\pi}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-6\sqrt{2}\left[\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\pi}\)
\(=-6\sqrt{2}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\)
\(=-6\sqrt{2}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=6\sqrt{2}\)
\(Q.1.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
[ বঃ২০১১ ]
উত্তরঃ \(2\)
[ বঃ২০১১ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\sin^2{\left(\frac{x}{2}\right)}+\cos^2{\left(\frac{x}{2}\right)}+2\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1=\sin^2{\left(\frac{A}{2}\right)}+\cos^2{\left(\frac{A}{2}\right)}, \sin{A}=2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\left\{\sin{\left(\frac{x}{2}\right)}+\cos{\left(\frac{x}{2}\right)}\right\}^2}dx}\) ➜ \(\because a^2+2ab+b^2=(a+b)^2\)
\(=\int_{0}^{\frac{\pi}{2}}{\left\{\sin{\left(\frac{x}{2}\right)}+\cos{\left(\frac{x}{2}\right)}\right\}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\left(\frac{x}{2}\right)}dx}+\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{x}{2}\right)}dx}\)
\(=\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{(ax)}dx}=-\frac{1}{a}\cos{(ax)}, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=-2\left[\cos{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=-2\left[\cos{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\cos{0}\right]+2\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=-2\left[\cos{\left(\frac{\pi}{4}\right)}-1\right]+2\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \cos{0}=1, \sin{0}=0\)
\(=-2\left[\frac{1}{\sqrt{2}}-1\right]+2\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{4}\right)}=\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\frac{2}{\sqrt{2}}+2+\frac{2}{\sqrt{2}}\)
\(=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\sin^2{\left(\frac{x}{2}\right)}+\cos^2{\left(\frac{x}{2}\right)}+2\sin{\left(\frac{x}{2}\right)}\cos{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1=\sin^2{\left(\frac{A}{2}\right)}+\cos^2{\left(\frac{A}{2}\right)}, \sin{A}=2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\left\{\sin{\left(\frac{x}{2}\right)}+\cos{\left(\frac{x}{2}\right)}\right\}^2}dx}\) ➜ \(\because a^2+2ab+b^2=(a+b)^2\)
\(=\int_{0}^{\frac{\pi}{2}}{\left\{\sin{\left(\frac{x}{2}\right)}+\cos{\left(\frac{x}{2}\right)}\right\}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\left(\frac{x}{2}\right)}dx}+\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{x}{2}\right)}dx}\)
\(=\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{x}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{(ax)}dx}=-\frac{1}{a}\cos{(ax)}, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=-2\left[\cos{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin{\left(\frac{1}{2}x\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=-2\left[\cos{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\cos{0}\right]+2\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=-2\left[\cos{\left(\frac{\pi}{4}\right)}-1\right]+2\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \cos{0}=1, \sin{0}=0\)
\(=-2\left[\frac{1}{\sqrt{2}}-1\right]+2\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{4}\right)}=\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\frac{2}{\sqrt{2}}+2+\frac{2}{\sqrt{2}}\)
\(=2\)
\(Q.1.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{\theta}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
[ বঃ২০১৭ ]
উত্তরঃ \(2\)
[ বঃ২০১৭ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sqrt{1+\sin{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\sin^2{\left(\frac{\theta}{2}\right)}+\cos^2{\left(\frac{\theta}{2}\right)}+2\sin{\left(\frac{\theta}{2}\right)}\cos{\left(\frac{\theta}{2}\right)}}d\theta}\) ➜ \(\because 1=\sin^2{\left(\frac{A}{2}\right)}+\cos^2{\left(\frac{A}{2}\right)}, \sin{A}=2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\left\{\sin{\left(\frac{\theta}{2}\right)}+\cos{\left(\frac{\theta}{2}\right)}\right\}^2}d\theta}\) ➜ \(\because a^2+2ab+b^2=(a+b)^2\)
\(=\int_{0}^{\frac{\pi}{2}}{\left\{\sin{\left(\frac{\theta}{2}\right)}+\cos{\left(\frac{\theta}{2}\right)}\right\}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\left(\frac{\theta}{2}\right)}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{\theta}{2}\right)}d\theta}\)
\(=\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{\theta}{2}\right)}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{\theta}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{(ax)}dx}=-\frac{1}{a}\cos{(ax)}, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=-2\left[\cos{\left(\frac{1}{2}\theta\right)}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin{\left(\frac{1}{2}\theta\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=-2\left[\cos{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\cos{0}\right]+2\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=-2\left[\cos{\left(\frac{\pi}{4}\right)}-1\right]+2\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \cos{0}=1, \sin{0}=0\)
\(=-2\left[\frac{1}{\sqrt{2}}-1\right]+2\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{4}\right)}=\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\frac{2}{\sqrt{2}}+2+\frac{2}{\sqrt{2}}\)
\(=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\sin^2{\left(\frac{\theta}{2}\right)}+\cos^2{\left(\frac{\theta}{2}\right)}+2\sin{\left(\frac{\theta}{2}\right)}\cos{\left(\frac{\theta}{2}\right)}}d\theta}\) ➜ \(\because 1=\sin^2{\left(\frac{A}{2}\right)}+\cos^2{\left(\frac{A}{2}\right)}, \sin{A}=2\sin{\left(\frac{A}{2}\right)}\cos{\left(\frac{A}{2}\right)}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sqrt{\left\{\sin{\left(\frac{\theta}{2}\right)}+\cos{\left(\frac{\theta}{2}\right)}\right\}^2}d\theta}\) ➜ \(\because a^2+2ab+b^2=(a+b)^2\)
\(=\int_{0}^{\frac{\pi}{2}}{\left\{\sin{\left(\frac{\theta}{2}\right)}+\cos{\left(\frac{\theta}{2}\right)}\right\}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\left(\frac{\theta}{2}\right)}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\left(\frac{\theta}{2}\right)}d\theta}\)
\(=\left[-\frac{1}{\frac{1}{2}}\cos{\left(\frac{\theta}{2}\right)}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{1}{\frac{1}{2}}\sin{\left(\frac{\theta}{2}\right)}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{(ax)}dx}=-\frac{1}{a}\cos{(ax)}, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=-2\left[\cos{\left(\frac{1}{2}\theta\right)}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin{\left(\frac{1}{2}\theta\right)}\right]_{0}^{\frac{\pi}{2}}\)
\(=-2\left[\cos{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\cos{0}\right]+2\left[\sin{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=-2\left[\cos{\left(\frac{\pi}{4}\right)}-1\right]+2\left[\sin{\left(\frac{\pi}{4}\right)}-0\right]\) ➜ \(\because \cos{0}=1, \sin{0}=0\)
\(=-2\left[\frac{1}{\sqrt{2}}-1\right]+2\left[\frac{1}{\sqrt{2}}-0\right]\) ➜ \(\because \cos{\left(\frac{\pi}{4}\right)}=\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\)
\(=-\frac{2}{\sqrt{2}}+2+\frac{2}{\sqrt{2}}\)
\(=2\)
\(Q.1.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
[ সিঃ২০০৫ ]
উত্তরঃ \(\frac{\pi}{4}\)
[ সিঃ২০০৫ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[\sin{2\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[\sin{\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[\sin{2\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[\sin{\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}-\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(Q.1.(xxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ২০০৯,২০০৫; সিঃ২০১১; চঃ২০০৪ ]
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ২০০৯,২০০৫; সিঃ২০১১; চঃ২০০৪ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[\sin{2\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[\sin{\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0\)
\(=\frac{\pi}{4}+0\)
\(=\frac{\pi}{4}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[\sin{2\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[\sin{\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}+\frac{1}{4}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0\)
\(=\frac{\pi}{4}+0\)
\(=\frac{\pi}{4}\)
\(Q.1.(xxv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{2\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
[ মাঃ২০০৯ ]
উত্তরঃ \(\frac{\pi}{4}\)
[ মাঃ২০০৯ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin^2{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2.2\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1d\theta}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{4\theta}d\theta}\)
\(=\frac{1}{2}\left[\theta\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{4}\sin{4\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{8}\left[\sin{4\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[\sin{4\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[\sin{2\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[0-0\right]\) ➜ \(\because \sin{2\pi}=0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2.2\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1d\theta}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{4\theta}d\theta}\)
\(=\frac{1}{2}\left[\theta\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{4}\sin{4\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{8}\left[\sin{4\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[\sin{4\times{\frac{\pi}{2}}}-\sin{0}\right]\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[\sin{2\pi}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{\pi}{4}-\frac{1}{8}\left[0-0\right]\) ➜ \(\because \sin{2\pi}=0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(Q.1.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০১৩,২০০৯,২০০৭; দিঃ২০১৩; বঃ২০০৮; সিঃ২০১২,২০০৬,২০০৫ ]
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০১৩,২০০৯,২০০৭; দিঃ২০১৩; বঃ২০০৮; সিঃ২০১২,২০০৬,২০০৫ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{4\cos^3{x}dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(\cos{3x}+3\cos{x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\cos{3x}dx}+\frac{3}{4}\int_{0}^{\frac{\pi}{2}}{\cos{x}dx}\)
\(=\frac{1}{4}\left[\frac{1}{3}\sin{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}\left[\sin{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}, \int{\cos{x}dx}=\sin{x}\)
\(=\frac{1}{12}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{12}\left[\sin{\left(3\times{\frac{\pi}{2}}\right)}-\sin{(3.0)}\right]+\frac{3}{4}\left[1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{12}\left[\sin{\left(\frac{3\pi}{2}\right)}-\sin{0}\right]+\frac{3}{4}\)
\(=\frac{1}{12}\left[\sin{\left(\pi+\frac{\pi}{2}\right)}-0\right]+\frac{3}{4}\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{12}\left[-\sin{\left(\frac{\pi}{2}\right)}-0\right]+\frac{3}{4}\) ➜ \(\because \sin{\left(\frac{\pi}{2}+A\right)}=-\sin{A}\)
\(=\frac{1}{12}\left[-1\right]+\frac{3}{4}\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=-\frac{1}{12}+\frac{3}{4}\)
\(=\frac{-1+9}{12}\)
\(=\frac{8}{12}\)
\(=\frac{2}{3}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{4\cos^3{x}dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(\cos{3x}+3\cos{x})dx}\) ➜ \(\because 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\cos{3x}dx}+\frac{3}{4}\int_{0}^{\frac{\pi}{2}}{\cos{x}dx}\)
\(=\frac{1}{4}\left[\frac{1}{3}\sin{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}\left[\sin{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}, \int{\cos{x}dx}=\sin{x}\)
\(=\frac{1}{12}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{12}\left[\sin{\left(3\times{\frac{\pi}{2}}\right)}-\sin{(3.0)}\right]+\frac{3}{4}\left[1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{12}\left[\sin{\left(\frac{3\pi}{2}\right)}-\sin{0}\right]+\frac{3}{4}\)
\(=\frac{1}{12}\left[\sin{\left(\pi+\frac{\pi}{2}\right)}-0\right]+\frac{3}{4}\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{12}\left[-\sin{\left(\frac{\pi}{2}\right)}-0\right]+\frac{3}{4}\) ➜ \(\because \sin{\left(\frac{\pi}{2}+A\right)}=-\sin{A}\)
\(=\frac{1}{12}\left[-1\right]+\frac{3}{4}\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=-\frac{1}{12}+\frac{3}{4}\)
\(=\frac{-1+9}{12}\)
\(=\frac{8}{12}\)
\(=\frac{2}{3}\)
\(Q.1.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\)
উত্তরঃ \(\frac{2}{3}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{4\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(3\sin{\theta}-\sin{3\theta})d\theta}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{3\theta}d\theta}\)
\(=\frac{3}{4}\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[-\frac{1}{3}\cos{3\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-\frac{3}{4}\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{12}\left[\cos{3\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\frac{3}{4}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]+\frac{1}{12}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{0}\right]\)
\(=-\frac{3}{4}\left[0-1\right]+\frac{1}{12}\left[\cos{\left(\frac{3\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{3}{4}+\frac{1}{12}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]\)
\(=\frac{3}{4}+\frac{1}{12}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{3}{4}+\frac{1}{12}\left[-0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{3}{4}-\frac{1}{12}\)
\(=\frac{9-1}{12}\)
\(=\frac{8}{12}\)
\(=\frac{2}{3}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{4\sin^3{\theta}d\theta}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(3\sin{\theta}-\sin{3\theta})d\theta}\) ➜ \(\because 4\sin^3{A}=3\sin{A}-\sin{3A}\)
\(=\frac{3}{4}\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{3\theta}d\theta}\)
\(=\frac{3}{4}\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[-\frac{1}{3}\cos{3\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-\frac{3}{4}\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{12}\left[\cos{3\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\frac{3}{4}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]+\frac{1}{12}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{0}\right]\)
\(=-\frac{3}{4}\left[0-1\right]+\frac{1}{12}\left[\cos{\left(\frac{3\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{3}{4}+\frac{1}{12}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]\)
\(=\frac{3}{4}+\frac{1}{12}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{3}{4}+\frac{1}{12}\left[-0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{3}{4}-\frac{1}{12}\)
\(=\frac{9-1}{12}\)
\(=\frac{8}{12}\)
\(=\frac{2}{3}\)
\(Q.1.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{x}+b\sin^2{x})dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}(a+b)\pi\)
[ চঃ২০০৩ ]
উত্তরঃ \(\frac{1}{4}(a+b)\pi\)
[ চঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{x}+b\sin^2{x})dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a.2\cos^2{x}+b.2\sin^2{x})dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a(1+\cos{2x})+b(1-\cos{2x})\}dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a+a\cos{2x}+b-b\cos{2x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{(a+b)+(a-b)\cos{2x}\}dx}\)
\(=\frac{1}{2}(a+b)\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}(a-b)\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}(a+b)\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}(a-b)\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}(a+b)\left[\frac{\pi}{2}-0\right]+\frac{1}{4}(a-b)\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(2\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(\pi\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[0-0\right]\) ➜ \(\because \sin{\left(\pi\right)}=0\)
\(=\frac{1}{4}(a+b)\pi+0\)
\(=\frac{1}{4}(a+b)\pi\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a.2\cos^2{x}+b.2\sin^2{x})dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a(1+\cos{2x})+b(1-\cos{2x})\}dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a+a\cos{2x}+b-b\cos{2x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{(a+b)+(a-b)\cos{2x}\}dx}\)
\(=\frac{1}{2}(a+b)\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}(a-b)\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}\)
\(=\frac{1}{2}(a+b)\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}(a-b)\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}(a+b)\left[\frac{\pi}{2}-0\right]+\frac{1}{4}(a-b)\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(2\times{\frac{\pi}{2}}\right)}-\sin{0}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(\pi\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[0-0\right]\) ➜ \(\because \sin{\left(\pi\right)}=0\)
\(=\frac{1}{4}(a+b)\pi+0\)
\(=\frac{1}{4}(a+b)\pi\)
\(Q.1.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^4{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3\pi}{16}\)
[ যঃ২০০৪ ]
উত্তরঃ \(\frac{3\pi}{16}\)
[ যঃ২০০৪ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos^4{x}dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(2\cos^2{x})^2dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[\frac{\pi}{2}-0\right]+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{\pi}{8}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\)
\(=\frac{\pi}{8}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{\pi}{8}+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{\pi}{2}-0\right]+\frac{1}{32}\left[\sin{4x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{8}+\frac{1}{4}\left[\sin{\left(2\times{\frac{\pi}{2}}\right)}-\sin{2.0}\right]+\frac{\pi}{16}+\frac{1}{32}\left[\sin{\left(4\times{\frac{\pi}{2}}\right)}-\sin{4.0}\right]\)
\(=\frac{\pi}{8}+\frac{\pi}{16}+\frac{1}{4}\left[\sin{(\pi)}-\sin{0}\right]+\frac{1}{32}\left[\sin{(2\pi)}-\sin{0}\right]\)
\(=\frac{2\pi+\pi}{16}+\frac{1}{4}\left[0-0\right]+\frac{1}{32}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0' \sin{2\pi}=0, \sin{0}=0\)
\(=\frac{3\pi}{16}+0+0\)
\(=\frac{3\pi}{16}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(2\cos^2{x})^2dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[\frac{\pi}{2}-0\right]+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{\pi}{8}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{2}}{\cos{4x}dx}\)
\(=\frac{\pi}{8}+\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{\pi}{8}+\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{\pi}{2}-0\right]+\frac{1}{32}\left[\sin{4x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{8}+\frac{1}{4}\left[\sin{\left(2\times{\frac{\pi}{2}}\right)}-\sin{2.0}\right]+\frac{\pi}{16}+\frac{1}{32}\left[\sin{\left(4\times{\frac{\pi}{2}}\right)}-\sin{4.0}\right]\)
\(=\frac{\pi}{8}+\frac{\pi}{16}+\frac{1}{4}\left[\sin{(\pi)}-\sin{0}\right]+\frac{1}{32}\left[\sin{(2\pi)}-\sin{0}\right]\)
\(=\frac{2\pi+\pi}{16}+\frac{1}{4}\left[0-0\right]+\frac{1}{32}\left[0-0\right]\) ➜ \(\because \sin{\pi}=0' \sin{2\pi}=0, \sin{0}=0\)
\(=\frac{3\pi}{16}+0+0\)
\(=\frac{3\pi}{16}\)
\(Q.1.(xxx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^4{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3\pi-8}{32}\)
[ চঃ২০০৩ ]
উত্তরঃ \(\frac{3\pi-8}{32}\)
[ চঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\sin^4{x}dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(2\sin^2{x})^2dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(1-\cos{2x})^2dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(1-2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{1dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[x\right]_{0}^{\frac{\pi}{4}}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[\frac{\pi}{4}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{\pi}{16}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{1dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{\cos{4x}dx}\)
\(=\frac{\pi}{16}-\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[x\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{\pi}{16}-\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[\frac{\pi}{4}-0\right]+\frac{1}{32}\left[\sin{4x}\right]_{0}^{\frac{\pi}{4}}\)
\(=\frac{\pi}{16}-\frac{1}{4}\left[\sin{\left(2\times{\frac{\pi}{4}}\right)}-\sin{2.0}\right]+\frac{\pi}{32}+\frac{1}{32}\left[\sin{\left(4\times{\frac{\pi}{4}}\right)}-\sin{4.0}\right]\)
\(=\frac{\pi}{16}+\frac{\pi}{32}-\frac{1}{4}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]+\frac{1}{32}\left[\sin{(\pi)}-\sin{0}\right]\)
\(=\frac{2\pi+\pi}{32}-\frac{1}{4}\left[1-0\right]+\frac{1}{32}\left[0-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1' \sin{\pi}=0, \sin{0}=0\)
\(=\frac{3\pi}{32}-\frac{1}{4}+0\)
\(=\frac{3\pi-8}{32}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(2\sin^2{x})^2dx}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(1-\cos{2x})^2dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{(1-2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{1dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{4}\int_{0}^{\frac{\pi}{4}}{\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[x\right]_{0}^{\frac{\pi}{4}}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\left[\frac{\pi}{4}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{\pi}{16}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{1dx}+\frac{1}{8}\int_{0}^{\frac{\pi}{4}}{\cos{4x}dx}\)
\(=\frac{\pi}{16}-\frac{1}{2}\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[x\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[\frac{1}{4}\sin{4x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{\pi}{16}-\frac{1}{4}\left[\sin{2x}\right]_{0}^{\frac{\pi}{4}}+\frac{1}{8}\left[\frac{\pi}{4}-0\right]+\frac{1}{32}\left[\sin{4x}\right]_{0}^{\frac{\pi}{4}}\)
\(=\frac{\pi}{16}-\frac{1}{4}\left[\sin{\left(2\times{\frac{\pi}{4}}\right)}-\sin{2.0}\right]+\frac{\pi}{32}+\frac{1}{32}\left[\sin{\left(4\times{\frac{\pi}{4}}\right)}-\sin{4.0}\right]\)
\(=\frac{\pi}{16}+\frac{\pi}{32}-\frac{1}{4}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]+\frac{1}{32}\left[\sin{(\pi)}-\sin{0}\right]\)
\(=\frac{2\pi+\pi}{32}-\frac{1}{4}\left[1-0\right]+\frac{1}{32}\left[0-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1' \sin{\pi}=0, \sin{0}=0\)
\(=\frac{3\pi}{32}-\frac{1}{4}+0\)
\(=\frac{3\pi-8}{32}\)
\(Q.1.(xxxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{x}\sin{2x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\)
[ রাঃ২০০৮; দিঃ২০১৩; যঃ২০০৮; চঃ২০০৬; বঃ২০০৬,২০০৪ ]
উত্তরঃ \(\frac{2}{3}\)
[ রাঃ২০০৮; দিঃ২০১৩; যঃ২০০৮; চঃ২০০৬; বঃ২০০৬,২০০৪ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin{x}\sin{2x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin{2x}\sin{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{(2x-x)}-\cos{(2x+x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{x}-\cos{3x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{x}dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{3x}dx}\)
\(=\frac{1}{2}\left[\sin{x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{3}\sin{3x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]-\frac{1}{6}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[1-0\right]-\frac{1}{6}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(3\times{\frac{\pi}{2}}\right)}-\sin{3.0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(\frac{3\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(\pi+\frac{\pi}{2}\right)}-0\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[-\sin{\left(\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{\left(\pi+\frac{\pi}{2}\right)}=-\sin{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}-\frac{1}{6}\left[-1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=\frac{1}{2}+\frac{1}{6}\)
\(=\frac{3+1}{6}\)
\(=\frac{4}{6}\)
\(=\frac{2}{3}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin{2x}\sin{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{(2x-x)}-\cos{(2x+x)}\}dx}\) ➜ \(\because 2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{x}-\cos{3x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{x}dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{3x}dx}\)
\(=\frac{1}{2}\left[\sin{x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{1}{3}\sin{3x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]-\frac{1}{6}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[1-0\right]-\frac{1}{6}\left[\sin{3x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(3\times{\frac{\pi}{2}}\right)}-\sin{3.0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(\frac{3\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\sin{\left(\pi+\frac{\pi}{2}\right)}-0\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[-\sin{\left(\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{\left(\pi+\frac{\pi}{2}\right)}=-\sin{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}-\frac{1}{6}\left[-1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=\frac{1}{2}+\frac{1}{6}\)
\(=\frac{3+1}{6}\)
\(=\frac{4}{6}\)
\(=\frac{2}{3}\)
\(Q.1.(xxxii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin{2x}\cos{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০০৫ ]
উত্তরঃ \(\frac{2}{3}\)
[ যঃ২০০৫ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin{2x}\cos{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin{2x}\cos{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\sin{(2x+x)}+\sin{(2x-x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\sin{3x}+\sin{x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=\frac{1}{2}\left[-\frac{1}{3}\cos{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}, \int{\sin{x}dx}=-\cos{x}\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=-\frac{1}{6}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{3.0}\right]+\frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{6}\left[\cos{\left(\frac{3\pi}{2}\right)}-\cos{0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}-\frac{1}{6}\left[-0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{2}+\frac{1}{6}\)
\(=\frac{3+1}{6}\)
\(=\frac{4}{6}\)
\(=\frac{2}{3}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin{2x}\cos{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\sin{(2x+x)}+\sin{(2x-x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\sin{3x}+\sin{x}\}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=\frac{1}{2}\left[-\frac{1}{3}\cos{3x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}, \int{\sin{x}dx}=-\cos{x}\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=-\frac{1}{6}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{3.0}\right]+\frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{6}\left[\cos{\left(\frac{3\pi}{2}\right)}-\cos{0}\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]\)
\(=\frac{1}{2}-\frac{1}{6}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}-\frac{1}{6}\left[-0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{2}+\frac{1}{6}\)
\(=\frac{3+1}{6}\)
\(=\frac{4}{6}\)
\(=\frac{2}{3}\)
\(Q.1.(xxxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos{3\theta}\cos{2\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{5}\)
[ ঢাঃ২০১৪; চঃ২০০৩ ]
উত্তরঃ \(\frac{3}{5}\)
[ ঢাঃ২০১৪; চঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\cos{3\theta}\cos{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos{3\theta}\cos{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{(3\theta-2\theta)}+\cos{(3\theta+2\theta)}\}d\theta}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{\theta}+\cos{5\theta}\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{5\theta}d\theta}\)
\(=\frac{1}{2}\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{5}\sin{5\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]+\frac{1}{10}\left[\cos{5\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[1-0\right]+\frac{1}{10}\left[\sin{\left(5\times{\frac{\pi}{2}}\right)}-\sin{5.0}\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(\frac{5\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(2\pi+\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{\left(2\pi+\frac{\pi}{2}\right)}=\sin{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}+\frac{1}{10}\left[1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=\frac{5+1}{10}\)
\(=\frac{6}{10}\)
\(=\frac{3}{5}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\cos{3\theta}\cos{2\theta}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{(3\theta-2\theta)}+\cos{(3\theta+2\theta)}\}d\theta}\) ➜ \(\because 2\cos{A}\cos{B}=\cos{(A-B)}+\cos{(A+B)}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{\cos{\theta}+\cos{5\theta}\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\cos{5\theta}d\theta}\)
\(=\frac{1}{2}\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{1}{5}\sin{5\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]+\frac{1}{10}\left[\cos{5\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[1-0\right]+\frac{1}{10}\left[\sin{\left(5\times{\frac{\pi}{2}}\right)}-\sin{5.0}\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(\frac{5\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(2\pi+\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{0}=0\)
\(=\frac{1}{2}+\frac{1}{10}\left[\sin{\left(\frac{\pi}{2}\right)}-0\right]\) ➜ \(\because \sin{\left(2\pi+\frac{\pi}{2}\right)}=\sin{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{2}+\frac{1}{10}\left[1-0\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1\)
\(=\frac{5+1}{10}\)
\(=\frac{6}{10}\)
\(=\frac{3}{5}\)
\(Q.1.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}\sin{3x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(-\frac{2}{15}\)
[ সিঃ২০০৩; বঃ২০০৫; যঃ২০১৪; রাঃ২০০৮; মাঃ২০০৩,২০০৫ ]
উত্তরঃ \(-\frac{2}{15}\)
[ সিঃ২০০৩; বঃ২০০৫; যঃ২০১৪; রাঃ২০০৮; মাঃ২০০৩,২০০৫ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\sin^2{x}\sin{3x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{x}\sin{3x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2x})\sin{3x}dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(\sin{3x}-\sin{3x}\cos{2x})dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}\cos{2x})dx}\)
\(=\frac{1}{2}\left[-\frac{1}{3}\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{2\sin{3x}\cos{2x})dx}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\{\sin{(3x+2x)}+\sin{(3x-2x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=-\frac{1}{6}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{(3.0)}\right]-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\{\sin{5x}+\sin{x}\}dx}\)
\(=-\frac{1}{6}\left[\cos{\left(\frac{3\pi}{2}\right)}-\cos{0}\right]-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{5x}dx}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=-\frac{1}{6}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\left[-\frac{1}{5}\cos{5x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\), \( \int{\sin{x}dx}=-\cos{x}, \cos{0}=1\)
\(=-\frac{1}{6}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]+\frac{1}{20}\left[\cos{5x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{4}\left[\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=-\frac{1}{6}\left[-0-1\right]+\frac{1}{20}\left[\cos{\left(5\times{\frac{\pi}{2}}\right)}-\cos{5.0}\right]+\frac{1}{4}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(\frac{5\pi}{2}\right)}-\cos{0}\right]+\frac{1}{4}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(2\pi+\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(2\pi+\frac{\pi}{2}\right)}=\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{6}+\frac{1}{20}\left[0-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{6}-\frac{1}{20}-\frac{1}{4}\)
\(=\frac{10-3-15}{60}\)
\(=\frac{10-18}{60}\)
\(=\frac{-8}{60}\)
\(=-\frac{2}{15}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{2\sin^2{x}\sin{3x}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(1-\cos{2x})\sin{3x}dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(\sin{3x}-\sin{3x}\cos{2x})dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}dx}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\sin{3x}\cos{2x})dx}\)
\(=\frac{1}{2}\left[-\frac{1}{3}\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{2\sin{3x}\cos{2x})dx}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=-\frac{1}{6}\left[\cos{3x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\{\sin{(3x+2x)}+\sin{(3x-2x)}\}dx}\) ➜ \(\because 2\sin{A}\cos{B}=\sin{(A+B)}+\sin{(A-B)}\)
\(=-\frac{1}{6}\left[\cos{\left(3\times{\frac{\pi}{2}}\right)}-\cos{(3.0)}\right]-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\{\sin{5x}+\sin{x}\}dx}\)
\(=-\frac{1}{6}\left[\cos{\left(\frac{3\pi}{2}\right)}-\cos{0}\right]-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{5x}dx}-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}{\sin{x}dx}\)
\(=-\frac{1}{6}\left[\cos{\left(\pi+\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\left[-\frac{1}{5}\cos{5x}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[-\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\), \( \int{\sin{x}dx}=-\cos{x}, \cos{0}=1\)
\(=-\frac{1}{6}\left[-\cos{\left(\frac{\pi}{2}\right)}-1\right]+\frac{1}{20}\left[\cos{5x}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{4}\left[\cos{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \cos{\left(\pi+\frac{\pi}{2}\right)}=-\cos{\left(\frac{\pi}{2}\right)}\)
\(=-\frac{1}{6}\left[-0-1\right]+\frac{1}{20}\left[\cos{\left(5\times{\frac{\pi}{2}}\right)}-\cos{5.0}\right]+\frac{1}{4}\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{0}\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(\frac{5\pi}{2}\right)}-\cos{0}\right]+\frac{1}{4}\left[0-1\right]\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(2\pi+\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0, \cos{0}=1\)
\(=\frac{1}{6}+\frac{1}{20}\left[\cos{\left(\frac{\pi}{2}\right)}-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(2\pi+\frac{\pi}{2}\right)}=\cos{\left(\frac{\pi}{2}\right)}\)
\(=\frac{1}{6}+\frac{1}{20}\left[0-1\right]-\frac{1}{4}\) ➜ \(\because \cos{\left(\frac{\pi}{2}\right)}=0\)
\(=\frac{1}{6}-\frac{1}{20}-\frac{1}{4}\)
\(=\frac{10-3-15}{60}\)
\(=\frac{10-18}{60}\)
\(=\frac{-8}{60}\)
\(=-\frac{2}{15}\)
\(Q.1.(xxxv)\) \(\int_{1}^{4}{\frac{dx}{(2x+3)^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{55}\)
উত্তরঃ \(\frac{3}{55}\)
সমাধানঃ
\(\int_{1}^{4}{\frac{dx}{(2x+3)^2}}\)
\(=\left[-\frac{1}{2}.\frac{1}{2x+3}\right]_{1}^{4}\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{2}\left[\frac{1}{2x+3}\right]_{1}^{4}\)
\(=-\frac{1}{2}\left[\frac{1}{2.4+3}-\frac{1}{2.1+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{8+3}-\frac{1}{2+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{11}-\frac{1}{5}\right]\)
\(=-\frac{1}{2}\times{\frac{5-11}{55}}\)
\(=-\frac{1}{2}\times{\frac{-6}{55}}\)
\(=\frac{3}{55}\)
\(=\left[-\frac{1}{2}.\frac{1}{2x+3}\right]_{1}^{4}\) ➜ \(\because \int{\frac{1}{(ax+b)^2}dx}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{2}\left[\frac{1}{2x+3}\right]_{1}^{4}\)
\(=-\frac{1}{2}\left[\frac{1}{2.4+3}-\frac{1}{2.1+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{8+3}-\frac{1}{2+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{11}-\frac{1}{5}\right]\)
\(=-\frac{1}{2}\times{\frac{5-11}{55}}\)
\(=-\frac{1}{2}\times{\frac{-6}{55}}\)
\(=\frac{3}{55}\)
\(Q.1.(xxxvi)\) \(\int_{3}^{1}{\frac{2x}{1+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(-\ln{(5)}\)
[ কুঃ২০০৩; সিঃ২০০৬ ]
উত্তরঃ \(-\ln{(5)}\)
[ কুঃ২০০৩; সিঃ২০০৬ ]
সমাধানঃ
\(\int_{3}^{1}{\frac{2x}{1+x^2}dx}\)
\(=\int_{3}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\int_{3}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\ln{|1+x^2|}\right]_{3}^{1}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\ln{(1+1^2)}-\ln{(1+3^2)}\)
\(=\ln{(1+1)}-\ln{(1+9)}\)
\(=\ln{(2)}-\ln{(10)}\)
\(=\ln{\left(\frac{2}{10}\right)}\)
\(=\ln{\left(\frac{1}{5}\right)}\)
\(=\ln{(5)^{-1}}\)
\(=-\ln{(5)}\)
\(=\int_{3}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\int_{3}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\ln{|1+x^2|}\right]_{3}^{1}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\ln{(1+1^2)}-\ln{(1+3^2)}\)
\(=\ln{(1+1)}-\ln{(1+9)}\)
\(=\ln{(2)}-\ln{(10)}\)
\(=\ln{\left(\frac{2}{10}\right)}\)
\(=\ln{\left(\frac{1}{5}\right)}\)
\(=\ln{(5)^{-1}}\)
\(=-\ln{(5)}\)
\(Q.1.(xxxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos{\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}\)
[ বুয়েটঃ২০০৭-২০০৮ ]
উত্তরঃ \(\frac{1}{6}\)
[ বুয়েটঃ২০০৭-২০০৮ ]
সমাধানঃ
ধরি,
\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
➜ \(\because t=\sin{\theta}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos{\theta}d\theta}\)\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{1}{t^5dt}\) \(=\left[\frac{t^6}{6}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{6}\left[t^6\right]_{0}^{1}\)
\(=\frac{1}{6}\left[1^6-0^6\right]\)
\(=\frac{1}{6}\left[1-0\right]\)
\(=\frac{1}{6}\)
\(Q.1.(xxxviii)\) \(\int_{2}^{5}{\frac{7x}{\sqrt{x^2+3}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(7\sqrt{7}\)
উত্তরঃ \(7\sqrt{7}\)
সমাধানঃ
ধরি,
\(\sqrt{x^2+3}=t\)
\(\Rightarrow x^2+3=t^2\)
\(\Rightarrow \frac{d}{dx}(x^2+3)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 2x+0=2t\frac{dt}{dx}\)
\(\Rightarrow 2x=2t\frac{dt}{dx}\)
\(\Rightarrow xdx=t\frac{dt}{dx}\)
\(\therefore xdx=tdt\)
➜ \(\because t=\sqrt{x^2+3}\)
\(\Rightarrow t=\sqrt{2^2+3}\), যখন \(x=2\)
\(\Rightarrow t=\sqrt{4+3}\)
\(\therefore t=\sqrt{7}\)
আবার,
\(t=\sqrt{x^2+3}\)
\(\Rightarrow t=\sqrt{5^2+3}\), যখন \(x=5\)
\(\Rightarrow t=\sqrt{25+3}\)
\(\Rightarrow t=\sqrt{28}\)
\(\Rightarrow t=\sqrt{4\times{7}}\)
\(\therefore t=2\sqrt{7}\)
\(\int_{2}^{5}{\frac{7x}{\sqrt{x^2+3}}dx}\)\(\sqrt{x^2+3}=t\)
\(\Rightarrow x^2+3=t^2\)
\(\Rightarrow \frac{d}{dx}(x^2+3)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 2x+0=2t\frac{dt}{dx}\)
\(\Rightarrow 2x=2t\frac{dt}{dx}\)
\(\Rightarrow xdx=t\frac{dt}{dx}\)
\(\therefore xdx=tdt\)
| \(x\) | \(2\) | \(5\) |
| \(t\) | \(\sqrt{7}\) | \(2\sqrt{7}\) |
\(\Rightarrow t=\sqrt{4+3}\)
\(\therefore t=\sqrt{7}\)
আবার,
\(t=\sqrt{x^2+3}\)
\(\Rightarrow t=\sqrt{5^2+3}\), যখন \(x=5\)
\(\Rightarrow t=\sqrt{25+3}\)
\(\Rightarrow t=\sqrt{28}\)
\(\Rightarrow t=\sqrt{4\times{7}}\)
\(\therefore t=2\sqrt{7}\)
\(=7\int_{2}^{5}{\frac{x}{\sqrt{x^2+3}}dx}\)
\(=7\int_{2}^{5}{\frac{1}{\sqrt{x^2+3}}.xdx}\)
\(=7\int_{\sqrt{7}}^{2\sqrt{7}}{\frac{1}{t}.tdt}\)
\(=7\int_{\sqrt{7}}^{2\sqrt{7}}{1dt}\)
\(=7\left[t\right]_{\sqrt{7}}^{2\sqrt{7}}\) ➜ \(\because \int{1dx}=x\)
\(=7\left[2\sqrt{7}-\sqrt{7}\right]\)
\(=7\times{\sqrt{7}}\)
\(=7\sqrt{7}\)
\(Q.1.(xxxix)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{\cos{x}}{\sqrt{12+\sin{x}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(5\sqrt{2}-4\sqrt{3}\)
[ সাস্টঃ২০০৭-২০০৮ ]
উত্তরঃ \(5\sqrt{2}-4\sqrt{3}\)
[ সাস্টঃ২০০৭-২০০৮ ]
সমাধানঃ
ধরি,
\(\sqrt{12+\sin{x}}=t\)
\(\Rightarrow 12+\sin{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(12+\sin{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+\cos{x}=2t\frac{dt}{dx}\)
\(\Rightarrow \cos{x}=2t\frac{dt}{dx}\)
\(\therefore \cos{x}dx=2tdt\)
➜ \(\because t=\sqrt{12+\sin{x}}\)
\(\Rightarrow t=\sqrt{12+\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{12+0}\)
\(\Rightarrow t=\sqrt{12}\)
\(\Rightarrow t=\sqrt{4\times{3}}\)
\(\therefore t=2\sqrt{3}\)
আবার,
\(t=\sqrt{12+\sin{x}}\)
\(\Rightarrow t=\sqrt{12+\sin{\left(\frac{\pi}{6}\right)}}\), যখন \(x=\frac{\pi}{6}\)
\(\Rightarrow t=\sqrt{12+\frac{1}{2}}\)
\(\Rightarrow t=\sqrt{\frac{24+1}{2}}\)
\(\Rightarrow t=\sqrt{\frac{25}{2}}\)
\(\therefore t=\frac{5}{\sqrt{2}}\)
\(\int_{0}^{\frac{\pi}{6}}{\frac{\cos{x}}{\sqrt{12+\sin{x}}}dx}\)\(\sqrt{12+\sin{x}}=t\)
\(\Rightarrow 12+\sin{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(12+\sin{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+\cos{x}=2t\frac{dt}{dx}\)
\(\Rightarrow \cos{x}=2t\frac{dt}{dx}\)
\(\therefore \cos{x}dx=2tdt\)
| \(x\) | \(0\) | \(\frac{\pi}{6}\) |
| \(t\) | \(2\sqrt{3}\) | \(\frac{5}{\sqrt{2}}\) |
\(\Rightarrow t=\sqrt{12+0}\)
\(\Rightarrow t=\sqrt{12}\)
\(\Rightarrow t=\sqrt{4\times{3}}\)
\(\therefore t=2\sqrt{3}\)
আবার,
\(t=\sqrt{12+\sin{x}}\)
\(\Rightarrow t=\sqrt{12+\sin{\left(\frac{\pi}{6}\right)}}\), যখন \(x=\frac{\pi}{6}\)
\(\Rightarrow t=\sqrt{12+\frac{1}{2}}\)
\(\Rightarrow t=\sqrt{\frac{24+1}{2}}\)
\(\Rightarrow t=\sqrt{\frac{25}{2}}\)
\(\therefore t=\frac{5}{\sqrt{2}}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{\sqrt{12+\sin{x}}}.\cos{x}dx}\)
\(=\int_{2\sqrt{3}}^{\frac{5}{\sqrt{2}}}{\frac{1}{t}.2tdt}\)
\(=2\int_{2\sqrt{3}}^{\frac{5}{\sqrt{2}}}{1dt}\)
\(=2\left[t\right]_{2\sqrt{3}}^{\frac{5}{\sqrt{2}}}\) ➜ \(\because \int{1dx}=x\)
\(=2\left[\frac{5}{\sqrt{2}}-2\sqrt{3}\right]\)
\(=2\frac{5}{\sqrt{2}}-4\sqrt{3}\)
\(=\sqrt{2}.\sqrt{2}\frac{5}{\sqrt{2}}-4\sqrt{3}\)
\(=5\sqrt{2}-4\sqrt{3}\)
\(Q.1.(xL)\) \(\int_{-\frac{\pi}{4}}^{0}{\tan{\left(\frac{\pi}{4}+x\right)}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\ln{(2)}\)
উত্তরঃ \(\frac{1}{2}\ln{(2)}\)
সমাধানঃ
\(\int_{-\frac{\pi}{4}}^{0}{\tan{\left(\frac{\pi}{4}+x\right)}dx}\)
\(=\left[\frac{1}{1}\ln{|\sec{\left(\frac{\pi}{4}+x\right)}|}\right]_{-\frac{\pi}{4}}^{0}\) ➜ \(\because \int{\tan{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}|}\)
\(=\left[\ln{|\sec{\left(\frac{\pi}{4}+x\right)}|}\right]_{-\frac{\pi}{4}}^{0}\)
\(=\left[\ln{|\sec{\left(\frac{\pi}{4}+0\right)}|}-\ln{|\sec{\left(\frac{\pi}{4}-\frac{\pi}{4}\right)}|}\right]\)
\(=\ln{|\sec{\left(\frac{\pi}{4}\right)}|}-\ln{|\sec{0}|}\)
\(=\ln{(\sqrt{2})}-\ln{(1)}\) ➜ \(\because \sec{\left(\frac{\pi}{4}\right)}=1\)
\(=\ln{(\sqrt{2})}-0\) ➜ \(\because \ln{(1)}=0\)
\(=\ln{(2)^{\frac{1}{2}}}\)
\(=\frac{1}{2}\ln{(2)}\)
\(=\left[\frac{1}{1}\ln{|\sec{\left(\frac{\pi}{4}+x\right)}|}\right]_{-\frac{\pi}{4}}^{0}\) ➜ \(\because \int{\tan{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}|}\)
\(=\left[\ln{|\sec{\left(\frac{\pi}{4}+x\right)}|}\right]_{-\frac{\pi}{4}}^{0}\)
\(=\left[\ln{|\sec{\left(\frac{\pi}{4}+0\right)}|}-\ln{|\sec{\left(\frac{\pi}{4}-\frac{\pi}{4}\right)}|}\right]\)
\(=\ln{|\sec{\left(\frac{\pi}{4}\right)}|}-\ln{|\sec{0}|}\)
\(=\ln{(\sqrt{2})}-\ln{(1)}\) ➜ \(\because \sec{\left(\frac{\pi}{4}\right)}=1\)
\(=\ln{(\sqrt{2})}-0\) ➜ \(\because \ln{(1)}=0\)
\(=\ln{(2)^{\frac{1}{2}}}\)
\(=\frac{1}{2}\ln{(2)}\)
\(Q.1.(xLi)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2\theta}}{\cos^2{\theta}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
উত্তরঃ \(\frac{\pi}{2}-1\)
[ বঃ২০১১; চঃ২০০৪; রাঃ২০০৯,২০০৫; সিঃ২০১১ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\frac{\cos{2\theta}}{\cos^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\cos^2{\theta}-1}{\cos^2{\theta}}d\theta}\) ➜ \(\because \cos{2A}=1+2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2\frac{\cos^2{\theta}}{\cos^2{\theta}}-\frac{1}{\cos^2{\theta}}\right)d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2-\sec^2{\theta}\right)d\theta}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=2\int_{0}^{\frac{\pi}{4}}{1d\theta}-\int_{0}^{\frac{\pi}{4}}{\sec^2{\theta}d\theta}\)
\(=2\left[\theta\right]_{0}^{\frac{\pi}{4}}-\left[\tan{\theta}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=tan{x}, \int{1dx}=x\)
\(=2\left[\frac{\pi}{4}-0\right]-\left[\tan{\left(\frac{\pi}{4}\right)-\tan{0}}\right]\)
\(=2.\frac{\pi}{4}-\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\)
\(=\frac{\pi}{2}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\cos^2{\theta}-1}{\cos^2{\theta}}d\theta}\) ➜ \(\because \cos{2A}=1+2\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2\frac{\cos^2{\theta}}{\cos^2{\theta}}-\frac{1}{\cos^2{\theta}}\right)d\theta}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(2-\sec^2{\theta}\right)d\theta}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A}\)
\(=2\int_{0}^{\frac{\pi}{4}}{1d\theta}-\int_{0}^{\frac{\pi}{4}}{\sec^2{\theta}d\theta}\)
\(=2\left[\theta\right]_{0}^{\frac{\pi}{4}}-\left[\tan{\theta}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=tan{x}, \int{1dx}=x\)
\(=2\left[\frac{\pi}{4}-0\right]-\left[\tan{\left(\frac{\pi}{4}\right)-\tan{0}}\right]\)
\(=2.\frac{\pi}{4}-\left[1-0\right]\) ➜ \(\because \tan{\frac{\pi}{4}}=1, \tan{0}=0\)
\(=\frac{\pi}{2}-1\)
\(Q.1.(xLii)\) \(\int_{\frac{1}{4}}^{1}{|2x-1|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{5}{16}\)
উত্তরঃ \(\frac{5}{16}\)
সমাধানঃ
ধরি,
\(0>2x-1\)
\(\Rightarrow 1>2x\)
\(\Rightarrow \frac{1}{2}>x\)
\(\Rightarrow |2x-1|=-(2x-1)\)
\(\therefore |2x-1|=1-2x\)
আবার,
\(2x-1>0\)
\(\Rightarrow 2x>1\)
\(\Rightarrow x>\frac{1}{2}\)
\(\therefore |2x-1|=2x-1\)
\(\int_{\frac{1}{4}}^{1}{|2x-1|dx}\)\(0>2x-1\)
\(\Rightarrow 1>2x\)
\(\Rightarrow \frac{1}{2}>x\)
\(\Rightarrow |2x-1|=-(2x-1)\)
\(\therefore |2x-1|=1-2x\)
আবার,
\(2x-1>0\)
\(\Rightarrow 2x>1\)
\(\Rightarrow x>\frac{1}{2}\)
\(\therefore |2x-1|=2x-1\)
\(=\int_{\frac{1}{4}}^{\frac{1}{2}}{(1-2x)dx}+\int_{\frac{1}{2}}^{1}{(2x-1)dx}\)
\(=\left[x-2.\frac{x^2}{2}\right]_{\frac{1}{4}}^{\frac{1}{2}}+\left[2.\frac{x^2}{2}-x\right]_{\frac{1}{2}}^{1}\) ➜ \(\because \int{1dx}=x, \int{xdx}=\frac{x^2}{2}\)
\(=\left[x-x^2\right]_{\frac{1}{4}}^{\frac{1}{2}}+\left[x^2-x\right]_{\frac{1}{2}}^{1}\)
\(=\frac{1}{2}-\left(\frac{1}{2}\right)^2-\frac{1}{4}+\left(\frac{1}{4}\right)^2+1^2-1-\left(\frac{1}{2}\right)^2+\frac{1}{2}\)
\(=2.\frac{1}{2}-3\frac{1}{4}+\frac{1}{16}\)
\(=1-\frac{3}{4}+\frac{1}{16}\)
\(=\frac{16-12+1}{16}\)
\(=\frac{17-12}{16}\)
\(=\frac{5}{16}\)
\(Q.1.(xLiii)\) \(\int_{0}^{\frac{\pi}{2}}{|\cos{2x}|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
উত্তরঃ \(1\)
সমাধানঃ
ধরি,
\(\cos{2x}=0\)
\(\Rightarrow \cos{2x}=\cos{\left(\frac{\pi}{2}\right)}\)
\(\Rightarrow 2x=\frac{\pi}{2}\)
\(\therefore x=\frac{\pi}{4}\)
এখন,
\(0>\cos{2x}\)
\(\Rightarrow \frac{\pi}{2}\ge{x}>\frac{\pi}{4}\) ব্যাবধিতে \( |\cos{2x}|=-\cos{2x}\)
আবার,
\(\cos{2x}>0\)
\(\Rightarrow \frac{\pi}{4}>x\ge{0}\) ব্যাবধিতে \(|\cos{2x}|=\cos{2x}\)
\(\int_{0}^{\frac{\pi}{2}}{|\cos{2x}|dx}\)\(\cos{2x}=0\)
\(\Rightarrow \cos{2x}=\cos{\left(\frac{\pi}{2}\right)}\)
\(\Rightarrow 2x=\frac{\pi}{2}\)
\(\therefore x=\frac{\pi}{4}\)
এখন,
\(0>\cos{2x}\)
\(\Rightarrow \frac{\pi}{2}\ge{x}>\frac{\pi}{4}\) ব্যাবধিতে \( |\cos{2x}|=-\cos{2x}\)
আবার,
\(\cos{2x}>0\)
\(\Rightarrow \frac{\pi}{4}>x\ge{0}\) ব্যাবধিতে \(|\cos{2x}|=\cos{2x}\)
\(=\int_{0}^{\frac{\pi}{4}}{\cos{2x}dx}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{(-\cos{2x})dx}\)
\(=\left[\frac{1}{2}\sin{2x}\right]_{0}^{\frac{\pi}{4}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{(\cos{2x})dx}\)
\(=\frac{1}{2}\left[\sin{2x}\right]_{0}^{\frac{\pi}{4}}-\left[\frac{1}{2}\sin{2x}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{(ax)}dx}=\frac{1}{a}\sin{(ax)}\)
\(=\frac{1}{2}\left[\sin{2\left(\frac{\pi}{4}\right)}-\sin{0}\right]-\frac{1}{2}\left[\sin{2\left(\frac{\pi}{2}\right)}-\sin{2\left(\frac{\pi}{4}\right)}\right]\)
\(=\frac{1}{2}\left[\sin{\left(\frac{\pi}{2}\right)}-0\right]-\frac{1}{2}\left[\sin{(\pi)}-\sin{\left(\frac{\pi}{2}\right)}\right]\)
\(=\frac{1}{2}\left[1-0\right]-\frac{1}{2}\left[0-1\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{0}=0, \sin{(\pi)}=0\)
\(=\frac{1}{2}+\frac{1}{2}\)
\(=\frac{2}{2}\)
\(=1\)
\(Q.1.(xLiv)\) \(\int_{-\pi}^{\pi}{|\cos{x}|dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\)
উত্তরঃ \(4\)
সমাধানঃ
ধরি,
\(\cos{x}=0\)
\(\Rightarrow \cos{x}=\cos{\left(\frac{\pi}{2}\right)}\)
\(\therefore x=\frac{\pi}{2}\)
এখন,
\(0>\cos{x}\)
\(\Rightarrow -\pi\ge{x}>-\frac{\pi}{2}\) ব্যাবধিতে \(|\cos{x}|=-\cos{x}\)
আবার,
\(\cos{x}>0\)
\(\Rightarrow -\frac{\pi}{2}>x\ge{\frac{\pi}{2}}\) ব্যাবধিতে \(|\cos{x}|=\cos{x}\)
আবার,
\(0>\cos{x}\)
\(\Rightarrow \frac{\pi}{2}\ge{x}>\pi\) ব্যাবধিতে \(|\cos{x}|=-\cos{x}\)
\(\int_{-\pi}^{\pi}{|\cos{x}|dx}\)\(\cos{x}=0\)
\(\Rightarrow \cos{x}=\cos{\left(\frac{\pi}{2}\right)}\)
\(\therefore x=\frac{\pi}{2}\)
এখন,
\(0>\cos{x}\)
\(\Rightarrow -\pi\ge{x}>-\frac{\pi}{2}\) ব্যাবধিতে \(|\cos{x}|=-\cos{x}\)
আবার,
\(\cos{x}>0\)
\(\Rightarrow -\frac{\pi}{2}>x\ge{\frac{\pi}{2}}\) ব্যাবধিতে \(|\cos{x}|=\cos{x}\)
আবার,
\(0>\cos{x}\)
\(\Rightarrow \frac{\pi}{2}\ge{x}>\pi\) ব্যাবধিতে \(|\cos{x}|=-\cos{x}\)
\(=\int_{-\pi}^{-\frac{\pi}{2}}{(-\cos{x})dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{x}dx}+\int_{\frac{\pi}{2}}^{\pi}{(-\cos{x})dx}\)
\(=-\int_{-\pi}^{-\frac{\pi}{2}}{\cos{x}dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos{x}dx}-\int_{\frac{\pi}{2}}^{\pi}{\cos{x}dx}\)
\(=-\left[\sin{x}\right]_{-\pi}^{-\frac{\pi}{2}}+\left[\sin{x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}-\left[\sin{x}\right]_{\frac{\pi}{2}}^{\pi}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=-\left[\sin{\left(-\frac{\pi}{2}\right)}-\sin{(-\pi)}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{\left(-\frac{\pi}{2}\right)}\right]-\left[\sin{(\pi)}-\sin{\left(\frac{\pi}{2}\right)}\right]\)
\(=-\left[-\sin{\left(\frac{\pi}{2}\right)}+\sin{(\pi)}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}+\sin{\left(\frac{\pi}{2}\right)}\right]-\left[\sin{(\pi)}-\sin{\left(\frac{\pi}{2}\right)}\right]\)
\(=-\left[-1+0\right]+\left[1+1\right]-\left[0-1\right]\) ➜ \(\because \sin{\left(\frac{\pi}{2}\right)}=1, \sin{(\pi)}=0\)
\(=1+2+1\)
\(=4\)
\(Q.1.(xLv)\) \(\int_{1}^{4}{\frac{(2-x)^2}{\sqrt{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\frac{11}{15}\)
উত্তরঃ \(1\frac{11}{15}\)
সমাধানঃ
\(\int_{1}^{4}{\frac{(2-x)^2}{\sqrt{x}}dx}\)
\(=\int_{1}^{4}{\frac{2^2-2.2.x+x^2}{\sqrt{x}}dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int_{1}^{4}{\frac{4-4x+x^2}{x^{\frac{1}{2}}}dx}\)
\(=\int_{1}^{4}{\left(\frac{4}{x^{\frac{1}{2}}}-\frac{4x}{x^{\frac{1}{2}}}+\frac{x^2}{x^{\frac{1}{2}}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{1-\frac{1}{2}}+x^{2-\frac{1}{2}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{\frac{2-1}{2}}+x^{\frac{4-1}{2}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{\frac{1}{2}}+x^{\frac{3}{2}}\right)dx}\)
\(=4\int_{1}^{4}{x^{-\frac{1}{2}}dx}-4\int_{1}^{4}{x^{\frac{1}{2}}dx}+\int_{1}^{4}{x^{\frac{3}{2}}dx}\)
\(=4\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{1}^{4}-4\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{4}+\left[\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]_{1}^{4}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=4\left[\frac{x^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}\right]_{1}^{4}-4\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{1}^{4}+\left[\frac{x^{\frac{3+2}{2}}}{\frac{3+2}{2}}\right]_{1}^{4}\)
\(=4\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{1}^{4}-4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}+\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{1}^{4}\)
\(=4\left[2x^{\frac{1}{2}}\right]_{1}^{4}-4\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{1}^{4}+\left[\frac{2}{5}x^{\frac{5}{2}}\right]_{1}^{4}\)
\(=8\left[x^{\frac{1}{2}}\right]_{1}^{4}-\frac{8}{3}\left[x^{\frac{3}{2}}\right]_{1}^{4}+\frac{2}{5}\left[x^{\frac{5}{2}}\right]_{1}^{4}\)
\(=8\left[4^{\frac{1}{2}}-1^{\frac{1}{2}}\right]-\frac{8}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{5}\left[4^{\frac{5}{2}}-1^{\frac{5}{2}}\right]\)
\(=8\left[2-1\right]-\frac{8}{3}\left[2^3-1\right]+\frac{2}{5}\left[2^5-1\right]\)
\(=8\times{1}-\frac{8}{3}\left[8-1\right]+\frac{2}{5}\left[32-1\right]\)
\(=8-\frac{8}{3}\times{7}+\frac{2}{5}\times{31}\)
\(=8-\frac{56}{3}+\frac{62}{5}\)
\(=\frac{120-280+186}{15}\)
\(=\frac{306-280}{15}\)
\(=\frac{26}{15}\)
\(=1\frac{11}{15}\)
\(=\int_{1}^{4}{\frac{2^2-2.2.x+x^2}{\sqrt{x}}dx}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int_{1}^{4}{\frac{4-4x+x^2}{x^{\frac{1}{2}}}dx}\)
\(=\int_{1}^{4}{\left(\frac{4}{x^{\frac{1}{2}}}-\frac{4x}{x^{\frac{1}{2}}}+\frac{x^2}{x^{\frac{1}{2}}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{1-\frac{1}{2}}+x^{2-\frac{1}{2}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{\frac{2-1}{2}}+x^{\frac{4-1}{2}}\right)dx}\)
\(=\int_{1}^{4}{\left(4x^{-\frac{1}{2}}-4x^{\frac{1}{2}}+x^{\frac{3}{2}}\right)dx}\)
\(=4\int_{1}^{4}{x^{-\frac{1}{2}}dx}-4\int_{1}^{4}{x^{\frac{1}{2}}dx}+\int_{1}^{4}{x^{\frac{3}{2}}dx}\)
\(=4\left[\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{1}^{4}-4\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{4}+\left[\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]_{1}^{4}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=4\left[\frac{x^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}\right]_{1}^{4}-4\left[\frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]_{1}^{4}+\left[\frac{x^{\frac{3+2}{2}}}{\frac{3+2}{2}}\right]_{1}^{4}\)
\(=4\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{1}^{4}-4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}+\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_{1}^{4}\)
\(=4\left[2x^{\frac{1}{2}}\right]_{1}^{4}-4\left[\frac{2}{3}x^{\frac{3}{2}}\right]_{1}^{4}+\left[\frac{2}{5}x^{\frac{5}{2}}\right]_{1}^{4}\)
\(=8\left[x^{\frac{1}{2}}\right]_{1}^{4}-\frac{8}{3}\left[x^{\frac{3}{2}}\right]_{1}^{4}+\frac{2}{5}\left[x^{\frac{5}{2}}\right]_{1}^{4}\)
\(=8\left[4^{\frac{1}{2}}-1^{\frac{1}{2}}\right]-\frac{8}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{5}\left[4^{\frac{5}{2}}-1^{\frac{5}{2}}\right]\)
\(=8\left[2-1\right]-\frac{8}{3}\left[2^3-1\right]+\frac{2}{5}\left[2^5-1\right]\)
\(=8\times{1}-\frac{8}{3}\left[8-1\right]+\frac{2}{5}\left[32-1\right]\)
\(=8-\frac{8}{3}\times{7}+\frac{2}{5}\times{31}\)
\(=8-\frac{56}{3}+\frac{62}{5}\)
\(=\frac{120-280+186}{15}\)
\(=\frac{306-280}{15}\)
\(=\frac{26}{15}\)
\(=1\frac{11}{15}\)
\(Q.1.(xLvi)\) \(\int_{0}^{4}{\frac{dx}{(2x+3)^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{4}{33}\)
উত্তরঃ \(\frac{4}{33}\)
সমাধানঃ
\(\int_{0}^{4}{\frac{dx}{(2x+3)^2}}\)
\(=\int_{0}^{4}{\frac{1}{(2x+3)^2}dx}\)
\(=\left[-\frac{1}{2}.\frac{1}{2x+3}\right]_{0}^{4}\) ➜ \(\because \int{\frac{1}{ax+b}dx}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{2}\left[\frac{1}{2x+3}\right]_{0}^{4}\)
\(=-\frac{1}{2}\left[\frac{1}{2.4+3}-\frac{1}{2.0+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{8+3}-\frac{1}{0+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{11}-\frac{1}{3}\right]\)
\(=-\frac{1}{2}\times{\frac{3-11}{33}}\)
\(=-\frac{1}{2}\times{\frac{-8}{33}}\)
\(=\frac{4}{33}\)
\(=\int_{0}^{4}{\frac{1}{(2x+3)^2}dx}\)
\(=\left[-\frac{1}{2}.\frac{1}{2x+3}\right]_{0}^{4}\) ➜ \(\because \int{\frac{1}{ax+b}dx}=-\frac{1}{a}.\frac{1}{ax+b}\)
\(=-\frac{1}{2}\left[\frac{1}{2x+3}\right]_{0}^{4}\)
\(=-\frac{1}{2}\left[\frac{1}{2.4+3}-\frac{1}{2.0+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{8+3}-\frac{1}{0+3}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{11}-\frac{1}{3}\right]\)
\(=-\frac{1}{2}\times{\frac{3-11}{33}}\)
\(=-\frac{1}{2}\times{\frac{-8}{33}}\)
\(=\frac{4}{33}\)
অনুশীলনী \(10.G / Q.2\)-এর সংক্ষিপ্ত প্রশ্নসমুহ
নিচের যোগজগুলির মান নির্ণয় করঃ
\(\int{f\{g(x)\}g^\prime(x)dx}\) আকার।
\(Q.2.(i)\) \(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\)
উত্তরঃ \(\sin{(\ln{3})}\)
[ ঢাঃ২০১২,২০০৮; দিঃ২০১১; বুয়েটঃ২০০৯; চঃ২০১৩; কুঃ২০১৪,২০০৮; বঃ২০১২ ]
\(Q.2.(ii)\) \(\int_{1}^{2}{\frac{1}{z}\cos{(\ln{|z|})}dz}\)
উত্তরঃ \(\sin{(\ln{2})}\)
[ দিঃ২০১৭ ]
\(\int{\{f(x)\}^nf^\prime(x)dx}\) আকার।
\(Q.2.(iii)\) \(\int_{0}^{1}{x^3\sqrt{1+3x^4}dx}\)
উত্তরঃ \(\frac{7}{18}\)
[ রাঃ২০৯,২০০৭,২০০৫; কুঃ২০১০,২০০৭; চঃ২০০৮,২০০৫; সিঃ২০০৮,২০১২; যঃ২০১২; বঃ২০০৯,২০০৪; মাঃ ২০১১ ]
\(Q.2.(iv)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}\sin{x}dx}\)
উত্তরঃ \(4\sqrt{2}\)
[ কুঃ২০৪]
\(Q.2.(v)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{(1+\sin{\theta})^3}d\theta}\)
উত্তরঃ \(\frac{3}{8}\)
\(Q.2.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{3}\)
[ ঢাঃ২০১৩,২০০৫,২০০৩; চঃ২০১১,২০০৪; রাঃ২০০৫; কুঃ২০০৬,২০০৪ ]
\(Q.2.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{4}\)
[ সিঃ২০১৩ ]
\(Q.2.(viii)\) \(\int_{0}^{\frac{\pi}{4}}{4\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১১; কুঃ২০০৯; দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
\(Q.2.(ix)\) \(\int_{0}^{\frac{\pi}{4}}{(\tan^3{x}+\tan{x})dx}\)
উত্তরঃ \(\frac{1}{2}\)
[ কুঃ২০০৮; যঃ২০০৫ ]
\(Q.2.(x)\) \(\int_{0}^{\frac{\pi}{4}}{(1+\cos{x})^2\sin{x}dx}\)
উত্তরঃ \(\frac{7}{3}\)
[ চঃ২০১১; সিঃ২০০৫; বুয়েটঃ ২০০৮-২০০৯ ]
\(Q.2.(xi)\) \(\int_{0}^{1}{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^2}{8}\)
[ দিঃ২০০৯; সিঃ২০০৭; যঃ ২০০৪; বঃ২০০৮; মাঃ২০১২ ]
\(Q.2.(xii)\) \(\int_{0}^{1}{\frac{(\cos^{-1}{x})^3}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^4}{64}\)
[ রাঃ,যঃ২০০৩ ]
\(Q.2.(xiii)\) \(\int_{0}^{1}{\frac{(\sin^{-1}{x})^2}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^3}{24}\)
[ রাঃ২০১১; কুঃ২০০৬,২০০৩; ঢাঃ২০০৪ ]
\(Q.2.(xiv)\) \(\int_{0}^{1}{\frac{(\tan^{-1}{x})^2}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi^3}{192}\)
[ ঢাঃ২০১১,২০০৫; কুঃ২০১১,২০০৮; রাঃ২০০৭; সিঃ২০১০,২০০৬; চঃ,যঃ২০১৩,২০১০; বঃ২০১২,২০০৬,২০০৩ ]
\(Q.2.(xv)\) \(\int_{0}^{1}{\frac{\tan^{-1}{x}}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi^2}{32}\)
[ মাঃ২০১২; দিঃ২০৯; বঃ২০০৮; সিঃ২০০৭; যঃ২০০৪ ]
\(Q.2.(xvi)\) \(\int_{0}^{1}{\frac{2x(\tan^{-1}{x^2})^3}{1+x^4}dx}\)
উত্তরঃ \(\frac{\pi^4}{1024}\)
\(Q.2.(xvii)\) \(\int_{1}^{e^{2}}{\frac{\ln{|x|}}{x}dx}+\int_{1}^{2}{e^xdx}\)
উত্তরঃ \(2+e^2-e\)
[ দিঃ২০১৭ ]
\(Q.2.(xviii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})^2}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ ঢাঃ২০১৪,২০০৮,২০০৬; রাঃ২০১৩,২০০৯; কুঃ২০০৯; যঃ২০১২,২০১০,২০০৬; চঃ২০১৩,২০০৭,২০০৫; দিঃ২০১৪; সিঃ২০১০,২০০৪; মাঃ২০১৪,২০১০ ]
\(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকার।
\(Q.2.(xix)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos^5{x}}{\sin^7{x}}dx}\)
উত্তরঃ \(\frac{1}{162}\)
[ ঢাঃ২০১২; রাঃ২০০৭; যঃ২০০৫; চঃ২০০৮; দিঃ২০১১ ]
\(Q.2.(xx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^3{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{16}\)
[ ঢাঃ২০১১; কুঃ,দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
\(Q.2.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^6{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{7}\)
[ সিঃ২০১৭ ]
\(Q.2.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^5{x}\sin{x}dx}\)
উত্তরঃ \(\frac{1}{6}\)
[ দিঃ২০১০; যঃ২০১১ ]
\(Q.2.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{\theta}\sin{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{3}\)
[ দিঃ২০১০; যঃ২০১১; ঢাঃ২০০৩ ]
\(Q.2.(xxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos^4{\theta}d\theta}\)
উত্তরঃ \(\frac{8}{315}\)
\(\int{e^{f(x)}f^{\prime}{x}dx}\) আকার।
\(Q.2.(xxv)\) \(\int_{0}^{1}{xe^{x^2}dx}\)
উত্তরঃ \(\frac{1}{2}(e-1)\)
[ ঢাঃ২০১৩,২০০৯,২০০৫; কুঃ২০১৩,২০১২; যঃ২০১৩,২০০৮,২০০৬; চঃ২০১২,২০০৬,২০০৪,২০০৩; সিঃ২০১০,২০০৭,২০০৩; বঃ২০০৫; দিঃ২০১২ ]
\(Q.2.(xxvi)\) \(\int_{1}^{2}{x^2e^{x^3}dx}\)
উত্তরঃ \(\frac{1}{3}e(e^7-1)\)
[ ঢাঃ২০০৯ বঃ২০১০; রাঃ২০০৬,২০০৪ ]
\(Q.2.(xxvii)\) \(\int_{0}^{1}{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2(e-1)\)
\(\int{\frac{f^{\prime}{x}}{\sqrt{f(x)}}dx}\) আকার।
\(Q.2.(xxviii)\) \(\int_{4}^{8}{\frac{xdx}{\sqrt{x^2-15}}}\)
উত্তরঃ \(6\)
উত্তরঃ \(\sin{(\ln{3})}\)
[ ঢাঃ২০১২,২০০৮; দিঃ২০১১; বুয়েটঃ২০০৯; চঃ২০১৩; কুঃ২০১৪,২০০৮; বঃ২০১২ ]
\(Q.2.(ii)\) \(\int_{1}^{2}{\frac{1}{z}\cos{(\ln{|z|})}dz}\)
উত্তরঃ \(\sin{(\ln{2})}\)
[ দিঃ২০১৭ ]
\(\int{\{f(x)\}^nf^\prime(x)dx}\) আকার।
\(Q.2.(iii)\) \(\int_{0}^{1}{x^3\sqrt{1+3x^4}dx}\)
উত্তরঃ \(\frac{7}{18}\)
[ রাঃ২০৯,২০০৭,২০০৫; কুঃ২০১০,২০০৭; চঃ২০০৮,২০০৫; সিঃ২০০৮,২০১২; যঃ২০১২; বঃ২০০৯,২০০৪; মাঃ ২০১১ ]
\(Q.2.(iv)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}\sin{x}dx}\)
উত্তরঃ \(4\sqrt{2}\)
[ কুঃ২০৪]
\(Q.2.(v)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{(1+\sin{\theta})^3}d\theta}\)
উত্তরঃ \(\frac{3}{8}\)
\(Q.2.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{3}\)
[ ঢাঃ২০১৩,২০০৫,২০০৩; চঃ২০১১,২০০৪; রাঃ২০০৫; কুঃ২০০৬,২০০৪ ]
\(Q.2.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{4}\)
[ সিঃ২০১৩ ]
\(Q.2.(viii)\) \(\int_{0}^{\frac{\pi}{4}}{4\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১১; কুঃ২০০৯; দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
\(Q.2.(ix)\) \(\int_{0}^{\frac{\pi}{4}}{(\tan^3{x}+\tan{x})dx}\)
উত্তরঃ \(\frac{1}{2}\)
[ কুঃ২০০৮; যঃ২০০৫ ]
\(Q.2.(x)\) \(\int_{0}^{\frac{\pi}{4}}{(1+\cos{x})^2\sin{x}dx}\)
উত্তরঃ \(\frac{7}{3}\)
[ চঃ২০১১; সিঃ২০০৫; বুয়েটঃ ২০০৮-২০০৯ ]
\(Q.2.(xi)\) \(\int_{0}^{1}{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^2}{8}\)
[ দিঃ২০০৯; সিঃ২০০৭; যঃ ২০০৪; বঃ২০০৮; মাঃ২০১২ ]
\(Q.2.(xii)\) \(\int_{0}^{1}{\frac{(\cos^{-1}{x})^3}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^4}{64}\)
[ রাঃ,যঃ২০০৩ ]
\(Q.2.(xiii)\) \(\int_{0}^{1}{\frac{(\sin^{-1}{x})^2}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{\pi^3}{24}\)
[ রাঃ২০১১; কুঃ২০০৬,২০০৩; ঢাঃ২০০৪ ]
\(Q.2.(xiv)\) \(\int_{0}^{1}{\frac{(\tan^{-1}{x})^2}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi^3}{192}\)
[ ঢাঃ২০১১,২০০৫; কুঃ২০১১,২০০৮; রাঃ২০০৭; সিঃ২০১০,২০০৬; চঃ,যঃ২০১৩,২০১০; বঃ২০১২,২০০৬,২০০৩ ]
\(Q.2.(xv)\) \(\int_{0}^{1}{\frac{\tan^{-1}{x}}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi^2}{32}\)
[ মাঃ২০১২; দিঃ২০৯; বঃ২০০৮; সিঃ২০০৭; যঃ২০০৪ ]
\(Q.2.(xvi)\) \(\int_{0}^{1}{\frac{2x(\tan^{-1}{x^2})^3}{1+x^4}dx}\)
উত্তরঃ \(\frac{\pi^4}{1024}\)
\(Q.2.(xvii)\) \(\int_{1}^{e^{2}}{\frac{\ln{|x|}}{x}dx}+\int_{1}^{2}{e^xdx}\)
উত্তরঃ \(2+e^2-e\)
[ দিঃ২০১৭ ]
\(Q.2.(xviii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})^2}dx}\)
উত্তরঃ \(\frac{2}{3}\)
[ ঢাঃ২০১৪,২০০৮,২০০৬; রাঃ২০১৩,২০০৯; কুঃ২০০৯; যঃ২০১২,২০১০,২০০৬; চঃ২০১৩,২০০৭,২০০৫; দিঃ২০১৪; সিঃ২০১০,২০০৪; মাঃ২০১৪,২০১০ ]
\(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকার।
\(Q.2.(xix)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos^5{x}}{\sin^7{x}}dx}\)
উত্তরঃ \(\frac{1}{162}\)
[ ঢাঃ২০১২; রাঃ২০০৭; যঃ২০০৫; চঃ২০০৮; দিঃ২০১১ ]
\(Q.2.(xx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^3{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{16}\)
[ ঢাঃ২০১১; কুঃ,দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
\(Q.2.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^6{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{7}\)
[ সিঃ২০১৭ ]
\(Q.2.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^5{x}\sin{x}dx}\)
উত্তরঃ \(\frac{1}{6}\)
[ দিঃ২০১০; যঃ২০১১ ]
\(Q.2.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{\theta}\sin{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{3}\)
[ দিঃ২০১০; যঃ২০১১; ঢাঃ২০০৩ ]
\(Q.2.(xxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos^4{\theta}d\theta}\)
উত্তরঃ \(\frac{8}{315}\)
\(\int{e^{f(x)}f^{\prime}{x}dx}\) আকার।
\(Q.2.(xxv)\) \(\int_{0}^{1}{xe^{x^2}dx}\)
উত্তরঃ \(\frac{1}{2}(e-1)\)
[ ঢাঃ২০১৩,২০০৯,২০০৫; কুঃ২০১৩,২০১২; যঃ২০১৩,২০০৮,২০০৬; চঃ২০১২,২০০৬,২০০৪,২০০৩; সিঃ২০১০,২০০৭,২০০৩; বঃ২০০৫; দিঃ২০১২ ]
\(Q.2.(xxvi)\) \(\int_{1}^{2}{x^2e^{x^3}dx}\)
উত্তরঃ \(\frac{1}{3}e(e^7-1)\)
[ ঢাঃ২০০৯ বঃ২০১০; রাঃ২০০৬,২০০৪ ]
\(Q.2.(xxvii)\) \(\int_{0}^{1}{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2(e-1)\)
\(\int{\frac{f^{\prime}{x}}{\sqrt{f(x)}}dx}\) আকার।
\(Q.2.(xxviii)\) \(\int_{4}^{8}{\frac{xdx}{\sqrt{x^2-15}}}\)
উত্তরঃ \(6\)
\(Q.2.(xxix)\) \(\int_{0}^{1}{\frac{xdx}{\sqrt{1-x^2}}}\)
উত্তরঃ \(1\)
[ ঢাঃ২০০৭; রাঃ২০১২; যঃ২০০৭ ]
\(Q.2.(xxx)\) \(\int_{0}^{2}{\frac{xdx}{\sqrt{9-2x^2}}}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১৫; রাঃ২০১২; সিঃ,চঃ২০১৪; দিঃ২০১৩ ]
\(Q.2.(xxxi)\) \(\int_{0}^{1}{\frac{xdx}{\sqrt{4-x^2}}}\)
উত্তরঃ \(2-\sqrt{3}\)
[ ঢাঃ, রাঃ২০১০; কুঃ২০১০,২০০৫; দিঃ২০১৩ ]
\(\int{\frac{f^{\prime}{x}}{f(x)}dx}\) আকার।
\(Q.2.(xxxii)\) \(\int_{0}^{1}{\frac{e^x}{1+e^x}dx}\)
উত্তরঃ \(\ln{(\frac{3}{2})}\)
[ ঢাঃ২০১০,২০০৭; রাঃ২০১০,২০০৬,২০০৪; কুঃ২০১৭,২০১৩,২০১০,২০০৭,২০০৫; যঃ২০১৪,২০১১,২০০৯,২০০৭; দিঃ২০০৯; চঃ২০১১,২০০৯,২০০৫; সিঃ২০১২,২০০৮; বঃ২০১৪,২০১১,২০০৬; মাঃ২০১৪,২০১১ ]
\(Q.2.(xxxiii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})}}\)
উত্তরঃ \(\ln{3}\)
[ দিঃ২০১১; বঃ২০৭,২০০৪ ]
\(Q.2.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cot{x}}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.2.(xxxv)\) \(\int_{0}^{1}{\frac{dx}{1+x^2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.2.(xxxvi)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{1-x^2}}}\)
উত্তরঃ \(\frac{\pi}{2}\)
\(Q.2.(xxxvii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{4-3x^2}}}\)
উত্তরঃ \(\frac{\pi}{3\sqrt{3}}\)
[ ঢাঃ২০০৩ ]
\(Q.2.(xxxviii)\) \(\int_{3}^{4}{\frac{dx}{25-x^2}}\)
উত্তরঃ \(\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
[ বঃ২০১৩ ]
\(Q.2.(xxxix)\) \(\int_{0}^{3}{\frac{dx}{\sqrt{3x-x^2}}}\)
উত্তরঃ \(\pi\)
[ চঃ২০১০ ]
\(Q.2.(xL)\) \(\int_{0}^{1}{\frac{1+x}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
[ সিঃ২০১২,২০০৫; কুঃ২০১২; রাঃ২০০৯,২০০৬,২০০৪; ঢাঃ২০০৯; বঃ২০০৭; দিঃ,চঃ,যঃ২০১১ ]
\(Q.2.(xLi)\) \(\int_{0}^{1}{\frac{xdx}{1+x^4}}\)
উত্তরঃ \(\frac{\pi}{8}\)
[ রাঃ২০১১ ]
\(Q.2.(xLii)\) \(\int_{0}^{1}{\frac{dx}{e^x+e^{-x}}}\)
উত্তরঃ \(\tan^{-1}{e}-\frac{\pi}{4}\)
[ রাঃ২০১২,২০০৩; কুঃ২০০৮; বঃ২০১৩; সিঃ২০০৭,২০০৪; মাঃ২০১০ ]
\(Q.2.(xLiii)\) \(\int_{0}^{\pi}{\frac{\sin{x}}{1+\cos^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃ২০০৭ ]
\(Q.2.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{1+\sin^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ,সিঃ২০১৩ ]
\(Q.2.(xLv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{\sqrt{4-\sin^2{\theta}}}d\theta}\)
উত্তরঃ \(\frac{\pi}{6}\)
[ রাঃ,সিঃ২০১৩ ]
\(Q.2.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{9-\sin^2{x}}dx}\)
উত্তরঃ \(\frac{1}{6}\ln{2}\)
[ ঢাঃ২০০৫; কুঃ২০১০; দিঃ২০১৩; চঃ২০০৯; সিঃ২০১৩,২০০৯; বঃ২০১০ ]
\(Q.2.(xLvii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\sin{2x}}{\cos^4{x}+\sin^4{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ বুয়েটঃ২০০৮-২০০৯ ]
\(Q.2.(xLviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{a^2\cos^2{x}+b^2\sin^2{x}}}\)
উত্তরঃ \(\frac{\pi}{2ab}\)
[ রুয়েটঃ২০১১-২০১২ ]
\(Q.2.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a\sin^2{\theta}+b\cos^2{\theta}}}\)
উত্তরঃ \(\frac{\pi}{2\sqrt{ab}}\)
[ রাঃ২০১১ ]
\(Q.2.(L)\) \(\int_{0}^{1}{\frac{\ln{|x+1|}}{x+1}dx}\)
উত্তরঃ \(\frac{1}{2}(\ln{2})^2\)
[ ঢাঃবিঃ২০১৪-২০১৫ ]
\(Q.2.(Li)\) \(\int_{1}^{e^2}{\frac{1+\ln{|x|}}{x}dx}\)
উত্তরঃ \(4\)
\(Q.2.(Lii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x-x^2}}}\)
উত্তরঃ \(\pi\)
[ বুয়েটঃ২০১৩-২০১৪ ]
\(Q.2.(Liii)\) \(\int_{2}^{5}{\frac{dx}{x^2-4x+13}}\)
উত্তরঃ \(\frac{\pi}{12}\)
\(Q.2.(Liv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{2x-x^2}}}\)
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃবিঃ২০০৮-২০০৯ ]
\(Q.2.(Lv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\sqrt{4-\sin^2{x}}}dx}\)
উত্তরঃ \(\frac{\pi}{6}\)
\(Q.2.(Lvi)\) \(\int_{2}^{3}{\frac{dx}{9x^2-16}}\)
উত্তরঃ \(\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
উত্তরঃ \(1\)
[ ঢাঃ২০০৭; রাঃ২০১২; যঃ২০০৭ ]
\(Q.2.(xxx)\) \(\int_{0}^{2}{\frac{xdx}{\sqrt{9-2x^2}}}\)
উত্তরঃ \(1\)
[ ঢাঃ২০১৫; রাঃ২০১২; সিঃ,চঃ২০১৪; দিঃ২০১৩ ]
\(Q.2.(xxxi)\) \(\int_{0}^{1}{\frac{xdx}{\sqrt{4-x^2}}}\)
উত্তরঃ \(2-\sqrt{3}\)
[ ঢাঃ, রাঃ২০১০; কুঃ২০১০,২০০৫; দিঃ২০১৩ ]
\(\int{\frac{f^{\prime}{x}}{f(x)}dx}\) আকার।
\(Q.2.(xxxii)\) \(\int_{0}^{1}{\frac{e^x}{1+e^x}dx}\)
উত্তরঃ \(\ln{(\frac{3}{2})}\)
[ ঢাঃ২০১০,২০০৭; রাঃ২০১০,২০০৬,২০০৪; কুঃ২০১৭,২০১৩,২০১০,২০০৭,২০০৫; যঃ২০১৪,২০১১,২০০৯,২০০৭; দিঃ২০০৯; চঃ২০১১,২০০৯,২০০৫; সিঃ২০১২,২০০৮; বঃ২০১৪,২০১১,২০০৬; মাঃ২০১৪,২০১১ ]
\(Q.2.(xxxiii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})}}\)
উত্তরঃ \(\ln{3}\)
[ দিঃ২০১১; বঃ২০৭,২০০৪ ]
\(Q.2.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cot{x}}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.2.(xxxv)\) \(\int_{0}^{1}{\frac{dx}{1+x^2}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.2.(xxxvi)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{1-x^2}}}\)
উত্তরঃ \(\frac{\pi}{2}\)
\(Q.2.(xxxvii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{4-3x^2}}}\)
উত্তরঃ \(\frac{\pi}{3\sqrt{3}}\)
[ ঢাঃ২০০৩ ]
\(Q.2.(xxxviii)\) \(\int_{3}^{4}{\frac{dx}{25-x^2}}\)
উত্তরঃ \(\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
[ বঃ২০১৩ ]
\(Q.2.(xxxix)\) \(\int_{0}^{3}{\frac{dx}{\sqrt{3x-x^2}}}\)
উত্তরঃ \(\pi\)
[ চঃ২০১০ ]
\(Q.2.(xL)\) \(\int_{0}^{1}{\frac{1+x}{1+x^2}dx}\)
উত্তরঃ \(\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
[ সিঃ২০১২,২০০৫; কুঃ২০১২; রাঃ২০০৯,২০০৬,২০০৪; ঢাঃ২০০৯; বঃ২০০৭; দিঃ,চঃ,যঃ২০১১ ]
\(Q.2.(xLi)\) \(\int_{0}^{1}{\frac{xdx}{1+x^4}}\)
উত্তরঃ \(\frac{\pi}{8}\)
[ রাঃ২০১১ ]
\(Q.2.(xLii)\) \(\int_{0}^{1}{\frac{dx}{e^x+e^{-x}}}\)
উত্তরঃ \(\tan^{-1}{e}-\frac{\pi}{4}\)
[ রাঃ২০১২,২০০৩; কুঃ২০০৮; বঃ২০১৩; সিঃ২০০৭,২০০৪; মাঃ২০১০ ]
\(Q.2.(xLiii)\) \(\int_{0}^{\pi}{\frac{\sin{x}}{1+\cos^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃ২০০৭ ]
\(Q.2.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{1+\sin^2{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ,সিঃ২০১৩ ]
\(Q.2.(xLv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{\sqrt{4-\sin^2{\theta}}}d\theta}\)
উত্তরঃ \(\frac{\pi}{6}\)
[ রাঃ,সিঃ২০১৩ ]
\(Q.2.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{9-\sin^2{x}}dx}\)
উত্তরঃ \(\frac{1}{6}\ln{2}\)
[ ঢাঃ২০০৫; কুঃ২০১০; দিঃ২০১৩; চঃ২০০৯; সিঃ২০১৩,২০০৯; বঃ২০১০ ]
\(Q.2.(xLvii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\sin{2x}}{\cos^4{x}+\sin^4{x}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
[ বুয়েটঃ২০০৮-২০০৯ ]
\(Q.2.(xLviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{a^2\cos^2{x}+b^2\sin^2{x}}}\)
উত্তরঃ \(\frac{\pi}{2ab}\)
[ রুয়েটঃ২০১১-২০১২ ]
\(Q.2.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a\sin^2{\theta}+b\cos^2{\theta}}}\)
উত্তরঃ \(\frac{\pi}{2\sqrt{ab}}\)
[ রাঃ২০১১ ]
\(Q.2.(L)\) \(\int_{0}^{1}{\frac{\ln{|x+1|}}{x+1}dx}\)
উত্তরঃ \(\frac{1}{2}(\ln{2})^2\)
[ ঢাঃবিঃ২০১৪-২০১৫ ]
\(Q.2.(Li)\) \(\int_{1}^{e^2}{\frac{1+\ln{|x|}}{x}dx}\)
উত্তরঃ \(4\)
\(Q.2.(Lii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x-x^2}}}\)
উত্তরঃ \(\pi\)
[ বুয়েটঃ২০১৩-২০১৪ ]
\(Q.2.(Liii)\) \(\int_{2}^{5}{\frac{dx}{x^2-4x+13}}\)
উত্তরঃ \(\frac{\pi}{12}\)
\(Q.2.(Liv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{2x-x^2}}}\)
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃবিঃ২০০৮-২০০৯ ]
\(Q.2.(Lv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\sqrt{4-\sin^2{x}}}dx}\)
উত্তরঃ \(\frac{\pi}{6}\)
\(Q.2.(Lvi)\) \(\int_{2}^{3}{\frac{dx}{9x^2-16}}\)
উত্তরঃ \(\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
\(Q.2.(i)\) \(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\sin{(\ln{3})}\)
[ ঢাঃ২০১২,২০০৮; দিঃ২০১১; বুয়েটঃ২০০৯; চঃ২০১৩; কুঃ২০১৪,২০০৮; বঃ২০১২ ]
উত্তরঃ \(\sin{(\ln{3})}\)
[ ঢাঃ২০১২,২০০৮; দিঃ২০১১; বুয়েটঃ২০০৯; চঃ২০১৩; কুঃ২০১৪,২০০৮; বঃ২০১২ ]
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=\ln{|x|}\)
\(\Rightarrow t=\ln{|1|}\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{|3|}\), যখন \(x=3\)
\(\therefore t=\ln{(3)}\)
\(\int_{1}^{3}{\frac{1}{x}\cos{(\ln{|x|})}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(3\) |
| \(t\) | \(0\) | \(\ln{3}\) |
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{|3|}\), যখন \(x=3\)
\(\therefore t=\ln{(3)}\)
\(=\int_{1}^{3}{\cos{(\ln{|x|})}.\frac{1}{x}dx}\)
\(=\int_{0}^{\ln{(3)}}{\cos{t}.dt}\)
\(=\left[\sin{t}\right]_{0}^{\ln{(3)}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=\left[\sin{(\ln{3})}-\sin{0}\right]\)
\(=\sin{(\ln{3})}-0\) ➜ \(\because \sin{0}=0\)
\(=\sin{(\ln{3})}\)
\(Q.2.(ii)\) \(\int_{1}^{2}{\frac{1}{z}\cos{(\ln{|z|})}dz}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\sin{(\ln{2})}\)
[ দিঃ২০১৭ ]
উত্তরঃ \(\sin{(\ln{2})}\)
[ দিঃ২০১৭ ]
সমাধানঃ
ধরি,
\(\ln{|z|}=t\)
\(\Rightarrow \frac{d}{dz}(\ln{|z|})=\frac{d}{dz}(t)\)
\(\Rightarrow \frac{1}{z}=\frac{dt}{dz}\)
\(\therefore \frac{1}{z}dx=dt\)
➜ \(\because t=\ln{|x|}\)
\(\Rightarrow t=\ln{|1|}\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{|2|}\), যখন \(x=2\)
\(\therefore t=\ln{(2)}\)
\(\int_{1}^{2}{\frac{1}{z}\cos{(\ln{|z|})}dz}\)\(\ln{|z|}=t\)
\(\Rightarrow \frac{d}{dz}(\ln{|z|})=\frac{d}{dz}(t)\)
\(\Rightarrow \frac{1}{z}=\frac{dt}{dz}\)
\(\therefore \frac{1}{z}dx=dt\)
| \(z\) | \(1\) | \(2\) |
| \(t\) | \(0\) | \(\ln{2}\) |
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{|2|}\), যখন \(x=2\)
\(\therefore t=\ln{(2)}\)
\(=\int_{1}^{2}{\cos{(\ln{|z|})}.\frac{1}{z}dz}\)
\(=\int_{0}^{\ln{(2)}}{\cos{t}.dt}\)
\(=\left[\sin{t}\right]_{0}^{\ln{(2)}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=\left[\sin{(\ln{2})}-\sin{0}\right]\)
\(=\sin{(\ln{2})}-0\) ➜ \(\because \sin{0}=0\)
\(=\sin{(\ln{2})}\)
\(Q.2.(iii)\) \(\int_{0}^{1}{x^3\sqrt{1+3x^4}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{7}{18}\)
[ রাঃ২০৯,২০০৭,২০০৫; কুঃ২০১০,২০০৭; চঃ২০০৮,২০০৫; সিঃ২০০৮,২০১২; যঃ২০১২; বঃ২০০৯,২০০৪; মাঃ ২০১১ ]
উত্তরঃ \(\frac{7}{18}\)
[ রাঃ২০৯,২০০৭,২০০৫; কুঃ২০১০,২০০৭; চঃ২০০৮,২০০৫; সিঃ২০০৮,২০১২; যঃ২০১২; বঃ২০০৯,২০০৪; মাঃ ২০১১ ]
সমাধানঃ
ধরি,
\(\sqrt{1+3x^4}=t\)
\(\Rightarrow 1+3x^4=t^2\)
\(\Rightarrow \frac{d}{dx}(1+3x^4)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+12x^3=2t\frac{dt}{dx}\)
\(\Rightarrow 12x^3=2t\frac{dt}{dx}\)
\(\Rightarrow 12x^3dx=2tdt\)
\(\Rightarrow x^3dx=\frac{2}{12}tdt\)
\(\therefore x^3dx=\frac{1}{6}tdt\)
➜ \(\because t=\sqrt{1+3x^4}\)
\(\Rightarrow t=\sqrt{1+3.0^4}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{1+3x^4}\)
\(\Rightarrow t=\sqrt{1+3.1^4}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{4}\)
\(\therefore t=2\)
\(\int_{0}^{1}{x^3\sqrt{1+3x^4}dx}\)\(\sqrt{1+3x^4}=t\)
\(\Rightarrow 1+3x^4=t^2\)
\(\Rightarrow \frac{d}{dx}(1+3x^4)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+12x^3=2t\frac{dt}{dx}\)
\(\Rightarrow 12x^3=2t\frac{dt}{dx}\)
\(\Rightarrow 12x^3dx=2tdt\)
\(\Rightarrow x^3dx=\frac{2}{12}tdt\)
\(\therefore x^3dx=\frac{1}{6}tdt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(1\) | \(2\) |
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{1+3x^4}\)
\(\Rightarrow t=\sqrt{1+3.1^4}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{4}\)
\(\therefore t=2\)
\(=\int_{0}^{1}{\sqrt{1+3x^4}.x^3dx}\)
\(=\int_{1}^{2}{t.\frac{1}{6}tdt}\)
\(=\frac{1}{6}\int_{1}^{2}{t^2dt}\)
\(=\frac{1}{6}\left[\frac{t^3}{3}\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{18}\left[t^3\right]_{1}^{2}\)
\(=\frac{1}{18}\left[2^3-1^3\right]\)
\(=\frac{1}{18}\left[8-1\right]\)
\(=\frac{1}{18}\left[7\right]\)
\(=\frac{7}{18}\)
\(Q.2.(iv)\) \(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}\sin{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\sqrt{2}\)
[ কুঃ২০৪]
উত্তরঃ \(4\sqrt{2}\)
[ কুঃ২০৪]
সমাধানঃ
ধরি,
\(\sqrt{1-\cos{x}}=t\)
\(\Rightarrow 1-\cos{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(1-\cos{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+\sin{x}=2t\frac{dt}{dx}\)
\(\therefore \sin{x}dx=2tdt\)
➜ \(\because t=\sqrt{1-\cos{x}}\)
\(\Rightarrow t=\sqrt{1-\cos{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{1-\cos{x}}\)
\(\Rightarrow t=\sqrt{1-\cos{\pi}}\), যখন \(x=\pi\)
\(\Rightarrow t=\sqrt{1+1}\)
\(\therefore t=\sqrt{2}\)
\(\int_{0}^{\pi}{3\sqrt{1-\cos{x}}\sin{x}dx}\)\(\sqrt{1-\cos{x}}=t\)
\(\Rightarrow 1-\cos{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(1-\cos{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0+\sin{x}=2t\frac{dt}{dx}\)
\(\therefore \sin{x}dx=2tdt\)
| \(x\) | \(0\) | \(\pi\) |
| \(t\) | \(0\) | \(\sqrt{2}\) |
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{1-\cos{x}}\)
\(\Rightarrow t=\sqrt{1-\cos{\pi}}\), যখন \(x=\pi\)
\(\Rightarrow t=\sqrt{1+1}\)
\(\therefore t=\sqrt{2}\)
\(=3\int_{0}^{\pi}{\sqrt{1-\cos{x}}\sin{x}dx}\)
\(=3\int_{0}^{\sqrt{2}}{t.2tdt}\)
\(=6\int_{0}^{\sqrt{2}}{t^2dt}\)
\(=6\left[\frac{t^3}{3}\right]_{0}^{\sqrt{2}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=6.\frac{1}{3}\left[t^3\right]_{0}^{\sqrt{2}}\)
\(=2\left[(\sqrt{2})^3-0^3\right]\)
\(=2\left[\sqrt{2}(\sqrt{2})^2-0\right]\)
\(=2\left[\sqrt{2}.2\right]\)
\(=4\sqrt{2}\)
\(Q.2.(v)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{(1+\sin{\theta})^3}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{8}\)
উত্তরঃ \(\frac{3}{8}\)
সমাধানঃ
ধরি,
\(1+\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(1+\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow 0+\cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
➜ \(\because t=1+\sin{\theta}\)
\(\Rightarrow t=1+\sin{0}\), যখন \(x=0\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\sin{\theta}\)
\(\Rightarrow t=1+\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{(1+\sin{\theta})^3}d\theta}\)\(1+\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(1+\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow 0+\cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(2\) |
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\sin{\theta}\)
\(\Rightarrow t=1+\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{(1+\sin{\theta})^3}.\cos{\theta}d\theta}\)
\(=\int_{1}^{2}{\frac{1}{t^3}dt}\)
\(=\int_{1}^{2}{t^{-3}dt}\)
\(=\left[\frac{t^{-3+1}}{-3+1}\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{t^{-2}}{-2}\right]_{1}^{2}\)
\(=\frac{1}{-2}\left[t^{-2}\right]_{1}^{2}\)
\(=-\frac{1}{2}\left[\frac{1}{t^2}\right]_{1}^{2}\)
\(=-\frac{1}{2}\left[\frac{1}{2^2}-\frac{1}{1^2}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{4}-\frac{1}{1}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{4}-1\right]\)
\(=-\frac{1}{2}\times{\frac{1-4}{4}}\)
\(=-\frac{1}{2}\times{\frac{-3}{4}}\)
\(=\frac{3}{8}\)
\(Q.2.(vi)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}\sec^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\)
[ ঢাঃ২০১৩,২০০৫,২০০৩; চঃ২০১১,২০০৪; রাঃ২০০৫; কুঃ২০০৬,২০০৪ ]
উত্তরঃ \(\frac{1}{3}\)
[ ঢাঃ২০১৩,২০০৫,২০০৩; চঃ২০১১,২০০৪; রাঃ২০০৫; কুঃ২০০৬,২০০৪ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
➜ \(\because t=\tan{x}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{\tan^2{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(=\int_{0}^{1}{t^2dt}\)
\(=\left[\frac{t^3}{3}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{3}\left[t^3\right]_{0}^{1}\)
\(=\frac{1}{3}\left[1^3-0^3\right]\)
\(=\frac{1}{3}\left[1-0\right]\)
\(=\frac{1}{3}\)
\(Q.2.(vii)\) \(\int_{0}^{\frac{\pi}{4}}{\tan^3{x}\sec^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\)
[ সিঃ২০১৩ ]
উত্তরঃ \(\frac{1}{4}\)
[ সিঃ২০১৩ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
➜ \(\because t=\tan{x}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{\tan^3{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(=\int_{0}^{1}{t^3dt}\)
\(=\left[\frac{t^4}{4}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{4}\left[t^4\right]_{0}^{1}\)
\(=\frac{1}{4}\left[1^4-0^4\right]\)
\(=\frac{1}{4}\left[1-0\right]\)
\(=\frac{1}{4}\)
\(Q.2.(viii)\) \(\int_{0}^{\frac{\pi}{4}}{4\tan^3{x}\sec^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
[ ঢাঃ২০১১; কুঃ২০০৯; দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
উত্তরঃ \(1\)
[ ঢাঃ২০১১; কুঃ২০০৯; দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
➜ \(\because t=\tan{x}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{4\tan^3{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(=4\int_{0}^{\frac{\pi}{4}}{\tan^3{x}\sec^2{x}dx}\)
\(=4\int_{0}^{1}{t^3dt}\)
\(=4\left[\frac{t^4}{4}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=4.\frac{1}{4}\left[t^4\right]_{0}^{1}\)
\(=\left[1^4-0^4\right]\)
\(=\left[1-0\right]\)
\(=1\)
\(Q.2.(ix)\) \(\int_{0}^{\frac{\pi}{4}}{(\tan^3{x}+\tan{x})dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\)
[ কুঃ২০০৮; যঃ২০০৫ ]
উত্তরঃ \(\frac{1}{2}\)
[ কুঃ২০০৮; যঃ২০০৫ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
➜ \(\because t=\tan{x}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{(\tan^3{x}+\tan{x})dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\frac{\pi}{4}}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{4}}{\tan{x}(\tan^1{x}+1)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\tan{x}sec^2{x}dx}\)
\(=\int_{0}^{1}{tdt}\)
\(=\left[\frac{t^2}{2}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[t^2\right]_{0}^{1}\)
\(=\frac{1}{2}\left[1^2-0^2\right]\)
\(=\frac{1}{2}\left[1-0\right]\)
\(=\frac{1}{2}\)
\(Q.2.(x)\) \(\int_{0}^{\frac{\pi}{2}}{(1+\cos{x})^2\sin{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{7}{3}\)
[ চঃ২০১১; সিঃ২০০৫; বুয়েটঃ ২০০৮-২০০৯ ]
উত্তরঃ \(\frac{7}{3}\)
[ চঃ২০১১; সিঃ২০০৫; বুয়েটঃ ২০০৮-২০০৯ ]
সমাধানঃ
ধরি,
\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
➜ \(\because t=1+\cos{x}\)
\(\Rightarrow t=1+\cos{0}\), যখন \(x=0\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
আবার,
\(t=1+\cos{x}\)
\(\Rightarrow t=1+\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{(1+\cos{x})^2\sin{x}dx}\)\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(2\) | \(1\) |
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
আবার,
\(t=1+\cos{x}\)
\(\Rightarrow t=1+\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
\(=\int_{2}^{1}{t^2\times{-dt}}\)
\(=-\int_{2}^{1}{t^2dt}\)
\(=-\left[\frac{t^3}{3}\right]_{2}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{3}\left[t^3\right]_{2}^{1}\)
\(=-\frac{1}{3}\left[1^3-2^3\right]\)
\(=-\frac{1}{3}\left[1-8\right]\)
\(=-\frac{1}{3}\left[-7\right]\)
\(=\frac{7}{3}\)
\(Q.2.(xi)\) \(\int_{0}^{1}{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^2}{8}\)
[ দিঃ২০০৯; সিঃ২০০৭; যঃ ২০০৪; বঃ২০০৮; মাঃ২০১২ ]
উত্তরঃ \(\frac{\pi^2}{8}\)
[ দিঃ২০০৯; সিঃ২০০৭; যঃ ২০০৪; বঃ২০০৮; মাঃ২০১২ ]
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
➜ \(\because t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{1}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(\int_{0}^{1}{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{2}\) |
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{1}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(=\int_{0}^{1}{\sin^{-1}{x}.\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{tdt}\)
\(=\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[t^2\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^2-0^2\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{4}-0\right]\)
\(=\frac{\pi^2}{8}\)
\(Q.2.(xii)\) \(\int_{0}^{1}{\frac{(\cos^{-1}{x})^3}{\sqrt{1-x^2}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^4}{64}\)
[ রাঃ,যঃ২০০৩ ]
উত্তরঃ \(\frac{\pi^4}{64}\)
[ রাঃ,যঃ২০০৩ ]
সমাধানঃ
ধরি,
\(\cos^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=-\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=-dt\)
➜ \(\because t=\cos^{-1}{x}\)
\(\Rightarrow t=\cos^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\cos^{-1}{\cos{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
আবার,
\(t=\cos^{-1}{x}\)
\(\Rightarrow t=\cos^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\cos^{-1}{\cos{0}}\)
\(\therefore t=0\)
\(\int_{0}^{1}{\frac{(\cos^{-1}{x})^3}{\sqrt{1-x^2}}dx}\)\(\cos^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=-\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=-dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(\frac{\pi}{2}\) | \(0\) |
\(\Rightarrow t=\cos^{-1}{\cos{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
আবার,
\(t=\cos^{-1}{x}\)
\(\Rightarrow t=\cos^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\cos^{-1}{\cos{0}}\)
\(\therefore t=0\)
\(=\int_{0}^{1}{(\cos^{-1}{x})^3.\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int_{\frac{\pi}{2}}^{0}{t^3\times{-dt}}\)
\(=-\int_{\frac{\pi}{2}}^{0}{t^3dt}\)
\(=-\left[\frac{t^4}{4}\right]_{\frac{\pi}{2}}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{4}\left[t^4\right]_{\frac{\pi}{2}}^{0}\)
\(=-\frac{1}{4}\left[0^4-\left(\frac{\pi}{2}\right)^4\right]\)
\(=-\frac{1}{4}\left[0-\frac{\pi^4}{16}\right]\)
\(=\frac{\pi^4}{64}\)
\(Q.2.(xiii)\) \(\int_{0}^{1}{\frac{(\sin^{-1}{x})^2}{\sqrt{1-x^2}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^3}{24}\)
[ রাঃ২০১১; কুঃ২০০৬,২০০৩; ঢাঃ২০০৪ ]
উত্তরঃ \(\frac{\pi^3}{24}\)
[ রাঃ২০১১; কুঃ২০০৬,২০০৩; ঢাঃ২০০৪ ]
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
➜ \(\because t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(\int_{0}^{1}{\frac{(\sin^{-1}{x})^2}{\sqrt{1-x^2}}dx}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{2}\) |
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(=\int_{0}^{1}{(\sin^{-1}{x})^2.\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{t^2dt}\)
\(=\left[\frac{t^3}{3}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{3}\left[t^3\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{3}\left[\left(\frac{\pi}{2}\right)^3-0^3\right]\)
\(=\frac{1}{3}\left[\frac{\pi^3}{8}-0\right]\)
\(=\frac{\pi^3}{24}\)
\(Q.2.(xiv)\) \(\int_{0}^{1}{\frac{(\tan^{-1}{x})^2}{1+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^3}{192}\)
[ ঢাঃ২০১১,২০০৫; কুঃ২০১১,২০০৮; রাঃ২০০৭; সিঃ২০১০,২০০৬; চঃ,যঃ২০১৩,২০১০; বঃ২০১২,২০০৬,২০০৩ ]
উত্তরঃ \(\frac{\pi^3}{192}\)
[ ঢাঃ২০১১,২০০৫; কুঃ২০১১,২০০৮; রাঃ২০০৭; সিঃ২০১০,২০০৬; চঃ,যঃ২০১৩,২০১০; বঃ২০১২,২০০৬,২০০৩ ]
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
➜ \(\because t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(\int_{0}^{1}{\frac{(\tan^{-1}{x})^2}{1+x^2}dx}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{4}\) |
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(=\int_{0}^{1}{(\tan^{-1}{x})^2.\frac{1}{1+x^2}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{t^2dt}\)
\(=\left[\frac{t^3}{3}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{3}\left[t^3\right]_{0}^{\frac{\pi}{4}}\)
\(=\frac{1}{3}\left[\left(\frac{\pi}{4}\right)^3-0^3\right]\)
\(=\frac{1}{3}\left[\frac{\pi^3}{64}-0\right]\)
\(=\frac{\pi^3}{192}\)
\(Q.2.(xv)\) \(\int_{0}^{1}{\frac{\tan^{-1}{x}}{1+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^2}{32}\)
[ মাঃ২০১২; দিঃ২০৯; বঃ২০০৮; সিঃ২০০৭; যঃ২০০৪ ]
উত্তরঃ \(\frac{\pi^2}{32}\)
[ মাঃ২০১২; দিঃ২০৯; বঃ২০০৮; সিঃ২০০৭; যঃ২০০৪ ]
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
➜ \(\because t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(\int_{0}^{1}{\frac{\tan^{-1}{x}}{1+x^2}dx}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{4}\) |
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x}\)
\(\Rightarrow t=\tan^{-1}{1}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(=\int_{0}^{1}{\tan^{-1}{x}.\frac{1}{1+x^2}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{tdt}\)
\(=\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[t^2\right]_{0}^{\frac{\pi}{4}}\)
\(=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^2-0^2\right]\)
\(=\frac{1}{2}\left[\frac{\pi^2}{16}-0\right]\)
\(=\frac{\pi^2}{32}\)
\(Q.2.(xvi)\) \(\int_{0}^{1}{\frac{2x(\tan^{-1}{x^2})^3}{1+x^4}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^4}{1024}\)
উত্তরঃ \(\frac{\pi^4}{1024}\)
সমাধানঃ
ধরি,
\(\tan^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^2)^2}\frac{d}{dx}(x^2)=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x^4}2x=\frac{dt}{dx}\)
\(\therefore \frac{2x}{1+x^4}dx=dt\)
➜ \(\because t=\tan^{-1}{x^2}\)
\(\Rightarrow t=\tan^{-1}{0^2}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{0}\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x^2}\)
\(\Rightarrow t=\tan^{-1}{1^2}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{1}\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(\int_{0}^{1}{\frac{2x(\tan^{-1}{x^2})^3}{1+x^4}dx}\)\(\tan^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^2)^2}\frac{d}{dx}(x^2)=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x^4}2x=\frac{dt}{dx}\)
\(\therefore \frac{2x}{1+x^4}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{4}\) |
\(\Rightarrow t=\tan^{-1}{0}\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{x^2}\)
\(\Rightarrow t=\tan^{-1}{1^2}\), যখন \(x=1\)
\(\Rightarrow t=\tan^{-1}{1}\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(=\int_{0}^{1}{(\tan^{-1}{x^2})^3.\frac{2x}{1+x^4}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{t^3dt}\)
\(=\left[\frac{t^4}{4}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{4}\left[t^4\right]_{0}^{\frac{\pi}{4}}\)
\(=\frac{1}{4}\left[\left(\frac{\pi}{4}\right)^4-0^4\right]\)
\(=\frac{1}{4}\left[\frac{\pi^4}{256}-0\right]\)
\(=\frac{\pi^4}{1024}\)
\(Q.2.(xvii)\) \(\int_{1}^{e^{2}}{\frac{\ln{|x|}}{x}dx}+\int_{1}^{2}{e^xdx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2+e^2-e\)
[ দিঃ২০১৭ ]
উত্তরঃ \(2+e^2-e\)
[ দিঃ২০১৭ ]
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=\ln{|x|}\)
\(\Rightarrow t=\ln{|1|}\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{e^2}\), যখন \(x=e^2\)
\(\Rightarrow t=2\ln{e}\)
\(\therefore t=2.1=2\)
\(\int_{1}^{e^{2}}{\frac{\ln{|x|}}{x}dx}+\int_{1}^{2}{e^xdx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(e^{2}\) |
| \(t\) | \(0\) | \(2\) |
\(\therefore t=0\)
আবার,
\(t=\ln{|x|}\)
\(\Rightarrow t=\ln{e^2}\), যখন \(x=e^2\)
\(\Rightarrow t=2\ln{e}\)
\(\therefore t=2.1=2\)
\(=\int_{1}^{e^{2}}{\ln{|x|}.\frac{1}{x}dx}+\int_{1}^{2}{e^xdx}\)
\(=\int_{0}^{2}{tdt}+\int_{1}^{2}{e^xdx}\)
\(=\left[\frac{t^2}{2}\right]_{0}^{2}+\left[e^x\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{e^xdx}=e^x\)
\(=\frac{1}{2}\left[t^2\right]_{0}^{2}+\left[e^x\right]_{1}^{2}\)
\(=\frac{1}{2}\left[2^2-0^2\right]+\left[e^2-e^1\right]\)
\(=\frac{1}{2}\left[4-0\right]+\left[e^2-e\right]\)
\(=2+e^2-e\)
\(Q.2.(xviii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\)
[ ঢাঃ২০১৪,২০০৮,২০০৬; রাঃ২০১৩,২০০৯; কুঃ২০০৯; যঃ২০১২,২০১০,২০০৬; চঃ২০১৩,২০০৭,২০০৫; দিঃ২০১৪; সিঃ২০১০,২০০৪; মাঃ২০১৪,২০১০ ]
উত্তরঃ \(\frac{2}{3}\)
[ ঢাঃ২০১৪,২০০৮,২০০৬; রাঃ২০১৩,২০০৯; কুঃ২০০৯; যঃ২০১২,২০১০,২০০৬; চঃ২০১৩,২০০৭,২০০৫; দিঃ২০১৪; সিঃ২০১০,২০০৪; মাঃ২০১৪,২০১০ ]
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{|1|}\), যখন \(x=1\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{e^2}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{e}\)
\(\Rightarrow t=1+2.1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})^2}}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(e^{2}\) |
| \(t\) | \(1\) | \(3\) |
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{e^2}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{e}\)
\(\Rightarrow t=1+2.1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(=\int_{1}^{e^{2}}{\frac{1}{(1+\ln{|x|})^2}.\frac{dx}{x}dx}\)
\(=\int_{1}^{3}{\frac{1}{t^2}dt}\)
\(=\left[-\frac{1}{t}\right]_{1}^{3}\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\)
\(=-\left[\frac{1}{t}\right]_{1}^{3}\)
\(=-\left[\frac{1}{3}-\frac{1}{1}\right]\)
\(=-\left[\frac{1}{3}-1\right]\)
\(=-\frac{1-3}{3}\)
\(=-\frac{-2}{3}\)
\(=\frac{2}{3}\)
\(Q.2.(xix)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos^5{x}}{\sin^7{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{162}\)
[ ঢাঃ২০১২; রাঃ২০০৭; যঃ২০০৫; চঃ২০০৮; দিঃ২০১১ ]
উত্তরঃ \(\frac{1}{162}\)
[ ঢাঃ২০১২; রাঃ২০০৭; যঃ২০০৫; চঃ২০০৮; দিঃ২০১১ ]
সমাধানঃ
ধরি,
\(\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}=\frac{dt}{dx}\)
\(\therefore -cosec \ {x}dx=dt\)
➜ \(\because t=\cot{x}\)
\(\Rightarrow t=\cot{\frac{\pi}{3}}\), যখন \(x=\frac{\pi}{3}\)
\(\therefore t=\frac{1}{\sqrt{3}}\)
আবার,
\(t=\cot{x}\)
\(\Rightarrow t=\cot{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos^5{x}}{\sin^7{x}}dx}\)\(\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}=\frac{dt}{dx}\)
\(\therefore -cosec \ {x}dx=dt\)
| \(x\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) |
| \(t\) | \(\frac{1}{\sqrt{3}}\) | \(0\) |
\(\therefore t=\frac{1}{\sqrt{3}}\)
আবার,
\(t=\cot{x}\)
\(\Rightarrow t=\cot{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\cos^5{x}}{\sin^5{x}}.\frac{1}{\sin^2{x}}dx}\)
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\cot^5{x} \ cosec^2{x}dx}\) ➜ \(\because \frac{\cos^5{A}}{\sin^5{A}}=\cot^5{A}, \frac{1}{\sin^2{A}}= \ cosec^2 \ {A}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{0}{t^5\times{-dt}}\)
\(=-\int_{\frac{1}{\sqrt{3}}}^{0}{t^5dt}\)
\(=-\left[\frac{t^6}{6}\right]_{\frac{1}{\sqrt{3}}}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{6}\left[t^6\right]_{\frac{1}{\sqrt{3}}}^{0}\)
\(=-\frac{1}{6}\left[0^6-\left(\frac{1}{\sqrt{3}}\right)^6\right]\)
\(=-\frac{1}{6}\left[0-\frac{1}{(\sqrt{3})^6}\right]\)
\(=\frac{1}{6}\times{\frac{1}{\{(\sqrt{3})^2\}^3}}\)
\(=\frac{1}{6}\times{\frac{1}{\{3\}^3}}\)
\(=\frac{1}{6}\times{\frac{1}{27}}\)
\(=\frac{1}{162}\)
\(Q.2.(xx)\) \(\int_{0}^{\frac{\pi}{4}}{\sin^3{\theta}\cos{\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}\)
[ ঢাঃ২০১১; কুঃ,দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
উত্তরঃ \(\frac{1}{16}\)
[ ঢাঃ২০১১; কুঃ,দিঃ২০০৯; যঃ২০০৬; সিঃ২০১৩,২০০৯; বঃ২০১১ ]
সমাধানঃ
ধরি,
\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
➜ \(\because t=\sin{\theta}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{4}\right)}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=\frac{1}{\sqrt{2}}\)
\(\int_{0}^{\frac{\pi}{4}}{\sin^3{\theta}\cos{\theta}d\theta}\)\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(\frac{1}{\sqrt{2}}\) |
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{4}\right)}\), যখন \(x=\frac{\pi}{4}\)
\(\therefore t=\frac{1}{\sqrt{2}}\)
\(=\int_{0}^{\frac{1}{\sqrt{2}}}{t^3dt}\)
\(=\left[\frac{t^4}{4}\right]_{0}^{\frac{1}{\sqrt{2}}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{4}\left[t^4\right]_{0}^{\frac{1}{\sqrt{2}}}\)
\(=\frac{1}{4}\left[\left(\frac{1}{\sqrt{2}}\right)^4-0^4\right]\)
\(=\frac{1}{4}\left[\frac{1}{(\sqrt{2})^4}-0\right]\)
\(=\frac{1}{4}\times{\frac{1}{\{(\sqrt{2})^2\}^2}}\)
\(=\frac{1}{4}\times{\frac{1}{\{2\}^2}}\)
\(=\frac{1}{4}\times{\frac{1}{4}}\)
\(=\frac{1}{16}\)
\(Q.2.(xxi)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^6{x}\cos{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{7}\)
[ সিঃ২০১৭ ]
উত্তরঃ \(\frac{1}{7}\)
[ সিঃ২০১৭ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=\sin{x}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\sin^6{x}\cos{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{1}{t^6dt}\)
\(=\left[\frac{t^7}{7}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{7}\left[t^7\right]_{0}^{\frac{1}{\sqrt{2}}}\)
\(=\frac{1}{7}\left[1^7-0^7\right]\)
\(=\frac{1}{7}\left[1-0\right]\)
\(=\frac{1}{7}\)
\(Q.2.(xxii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^5{x}\sin{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}\)
[ দিঃ২০১০; যঃ২০১১ ]
উত্তরঃ \(\frac{1}{6}\)
[ দিঃ২০১০; যঃ২০১১ ]
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
➜ \(\because t=\cos{x}\)
\(\Rightarrow t=\cos{0}\), যখন \(x=0\)
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(\int_{0}^{\frac{\pi}{4}}{\cos^5{x}\sin{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(0\) |
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(=\int_{1}^{0}{t^5\times{-dt}}\)
\(=-\int_{1}^{0}{t^5dt}\)
\(=-\left[\frac{t^6}{6}\right]_{1}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{6}\left[t^6\right]_{1}^{0}\)
\(=-\frac{1}{6}\left[0^6-1^6\right]\)
\(=-\frac{1}{6}\left[0-1\right]\)
\(=\frac{1}{6}\)
\(Q.2.(xxiii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^2{\theta}\sin{\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\)
[ দিঃ২০১০; যঃ২০১১; ঢাঃ২০০৩ ]
উত্তরঃ \(\frac{1}{3}\)
[ দিঃ২০১০; যঃ২০১১; ঢাঃ২০০৩ ]
সমাধানঃ
ধরি,
\(\cos{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\cos{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow -\sin{\theta}=\frac{dt}{d\theta}\)
\(\Rightarrow \sin{\theta}=-\frac{dt}{d\theta}\)
\(\therefore \sin{\theta}d\theta=-dt\)
➜ \(\because t=\cos{x}\)
\(\Rightarrow t=\cos{0}\), যখন \(x=0\)
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(\int_{0}^{\frac{\pi}{2}}{\cos^2{\theta}\sin{\theta}d\theta}\)\(\cos{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\cos{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow -\sin{\theta}=\frac{dt}{d\theta}\)
\(\Rightarrow \sin{\theta}=-\frac{dt}{d\theta}\)
\(\therefore \sin{\theta}d\theta=-dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(0\) |
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(=\int_{1}^{0}{t^2\times{-dt}}\)
\(=-\int_{1}^{0}{t^2dt}\)
\(=-\left[\frac{t^3}{3}\right]_{1}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\frac{1}{3}\left[t^3\right]_{1}^{0}\)
\(=-\frac{1}{3}\left[0^3-1^3\right]\)
\(=-\frac{1}{3}\left[0-1\right]\)
\(=\frac{1}{3}\)
\(Q.2.(xxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos^4{\theta}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{8}{315}\)
উত্তরঃ \(\frac{8}{315}\)
সমাধানঃ
ধরি,
\(\cos{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\cos{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow -\sin{\theta}=\frac{dt}{d\theta}\)
\(\Rightarrow \sin{\theta}=-\frac{dt}{d\theta}\)
\(\therefore \sin{\theta}d\theta=-dt\)
➜ \(\because t=\cos{x}\)
\(\Rightarrow t=\cos{0}\), যখন \(x=0\)
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(\int_{0}^{\frac{\pi}{2}}{\sin^5{\theta}\cos^4{\theta}d\theta}\)\(\cos{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\cos{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow -\sin{\theta}=\frac{dt}{d\theta}\)
\(\Rightarrow \sin{\theta}=-\frac{dt}{d\theta}\)
\(\therefore \sin{\theta}d\theta=-dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(0\) |
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=0\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin^4{\theta}\cos^4{\theta}.\sin{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{(\sin^2{\theta})^2\cos^4{\theta}.\sin{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{(1-\cos^2{\theta})^2\cos^4{\theta}.\sin{\theta}d\theta}\)
\(=\int_{1}^{0}{(1-t^2)^2t^4\times{-dt}}\)
\(=-\int_{1}^{0}{(1-2t^2+t^4)t^4dt}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=-\int_{1}^{0}{(t^4-2t^6+t^8)dt}\)
\(=-\left[\frac{t^5}{5}-2\frac{t^7}{7}+\frac{t^9}{9}\right]_{1}^{0}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=-\left[\left(\frac{0^5}{5}-2\frac{0^7}{7}+\frac{0^9}{9}\right)-\left(\frac{1^5}{5}-2\frac{1^7}{7}+\frac{1^9}{9}\right)\right]\)
\(=-\left[\left(0-0+0\right)-\left(\frac{1}{5}-\frac{2}{7}+\frac{1}{9}\right)\right]\)
\(=-\left[0-\frac{63-90+35}{315}\right]\)
\(=\frac{98-90}{315}\)
\(=\frac{8}{315}\)
\(Q.2.(xxv)\) \(\int_{0}^{1}{xe^{x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(e-1)\)
[ ঢাঃ২০১৩,২০০৯,২০০৫; কুঃ২০১৩,২০১২; যঃ২০১৩,২০০৮,২০০৬; চঃ২০১২,২০০৬,২০০৪,২০০৩; সিঃ২০১০,২০০৭,২০০৩; বঃ২০০৫; দিঃ২০১২ ]
উত্তরঃ \(\frac{1}{2}(e-1)\)
[ ঢাঃ২০১৩,২০০৯,২০০৫; কুঃ২০১৩,২০১২; যঃ২০১৩,২০০৮,২০০৬; চঃ২০১২,২০০৬,২০০৪,২০০৩; সিঃ২০১০,২০০৭,২০০৩; বঃ২০০৫; দিঃ২০১২ ]
সমাধানঃ
ধরি,
\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
➜ \(\because t=x^2\)
\(\Rightarrow t=0^2\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=x^2\)
\(\Rightarrow t=1^2\), যখন \(x=1\)
\(\therefore t=1\)
\(\int_{0}^{1}{xe^{x^2}dx}\)\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=x^2\)
\(\Rightarrow t=1^2\), যখন \(x=1\)
\(\therefore t=1\)
\(=\int_{0}^{1}{e^{x^2}.xdx}\)
\(=\int_{0}^{1}{e^{t}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int_{0}^{1}{e^{t}dt}\)
\(=\frac{1}{2}\left[e^{t}\right]_{0}^{1}\) ➜ \(\because \int{e^xdx}=e^x\)
\(=\frac{1}{2}\left[e^{1}-e^{0}\right]\)
\(=\frac{1}{2}\left[e-1\right]\) ➜ \(\because e^{0}=1\)
\(=\frac{1}{2}(e-1)\)
\(Q.2.(xxvi)\) \(\int_{1}^{2}{x^2e^{x^3}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}e(e^7-1)\)
[ ঢাঃ২০০৯ বঃ২০১০; রাঃ২০০৬,২০০৪ ]
উত্তরঃ \(\frac{1}{3}e(e^7-1)\)
[ ঢাঃ২০০৯ বঃ২০১০; রাঃ২০০৬,২০০৪ ]
সমাধানঃ
ধরি,
\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
➜ \(\because t=x^3\)
\(\Rightarrow t=1^3\), যখন \(x=1\)
\(\therefore t=1\)
আবার,
\(t=x^3\)
\(\Rightarrow t=2^3\), যখন \(x=2\)
\(\therefore t=8\)
\(\int_{1}^{2}{x^2e^{x^3}dx}\)\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
| \(x\) | \(1\) | \(2\) |
| \(t\) | \(1\) | \(8\) |
\(\therefore t=1\)
আবার,
\(t=x^3\)
\(\Rightarrow t=2^3\), যখন \(x=2\)
\(\therefore t=8\)
\(=\int_{1}^{2}{e^{x^3}.x^2dx}\)
\(=\int_{1}^{8}{e^{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int_{1}^{8}{e^{t}dt}\)
\(=\frac{1}{3}\left[e^{t}\right]_{1}^{8}\) ➜ \(\because \int{e^xdx}=e^x\)
\(=\frac{1}{3}\left[e^{8}-e^{1}\right]\)
\(=\frac{1}{3}(e^8-e)\)
\(=\frac{1}{3}e(e^7-1)\)
\(Q.2.(xxvii)\) \(\int_{0}^{1}{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2(e-1)\)
উত্তরঃ \(2(e-1)\)
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{x}}=2\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
➜ \(\because t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{1}\), যখন \(x=1\)
\(\therefore t=1\)
\(\int_{0}^{1}{\frac{e^{\sqrt{x}}}{\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{x}}=2\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{1}\), যখন \(x=1\)
\(\therefore t=1\)
\(=\int_{0}^{1}{e^{\sqrt{x}}.\frac{1}{\sqrt{x}}dx}\)
\(=\int_{0}^{1}{e^t\times{2dt}}\)
\(=2\int_{0}^{1}{e^tdt}\)
\(=2\left[e^t\right]_{0}^{1}\) ➜ \(\because \int{e^xdx}=e^x\)
\(=2\left[e^1-e^0\right]\)
\(=2\left[e-1\right]\) ➜ \(\because e^0=1\)
\(=2(e-1)\)
\(Q.2.(xxviii)\) \(\int_{4}^{8}{\frac{xdx}{\sqrt{x^2-15}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(6\)
উত্তরঃ \(6\)
সমাধানঃ
ধরি,
\(\sqrt{x^2-15}=t\)
\(\Rightarrow x^2-15=t^2\)
\(\Rightarrow \frac{d}{dx}(x^2-15)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 2x-0=2t\frac{dt}{dx}\)
\(\Rightarrow 2xdx=2tdt\)
\(\therefore xdx=tdt\)
➜ \(\because t=\sqrt{x^2-15}\)
\(\Rightarrow t=\sqrt{4^2-15}\), যখন \(x=4\)
\(\Rightarrow t=\sqrt{16-15}\)
\(\therefore t=\sqrt{1}=1\)
আবার,
\(t=\sqrt{x^2-15}\)
\(\Rightarrow t=\sqrt{8^2-15}\), যখন \(x=8\)
\(\Rightarrow t=\sqrt{64-15}\)
\(\therefore t=\sqrt{49}=7\)
\(\int_{4}^{8}{\frac{xdx}{\sqrt{x^2-15}}}\)\(\sqrt{x^2-15}=t\)
\(\Rightarrow x^2-15=t^2\)
\(\Rightarrow \frac{d}{dx}(x^2-15)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 2x-0=2t\frac{dt}{dx}\)
\(\Rightarrow 2xdx=2tdt\)
\(\therefore xdx=tdt\)
| \(x\) | \(4\) | \(8\) |
| \(t\) | \(1\) | \(7\) |
\(\Rightarrow t=\sqrt{16-15}\)
\(\therefore t=\sqrt{1}=1\)
আবার,
\(t=\sqrt{x^2-15}\)
\(\Rightarrow t=\sqrt{8^2-15}\), যখন \(x=8\)
\(\Rightarrow t=\sqrt{64-15}\)
\(\therefore t=\sqrt{49}=7\)
\(=\int_{4}^{8}{\frac{1}{\sqrt{x^2-15}}.xdx}\)
\(=\int_{1}^{7}{\frac{1}{t}.tdt}\)
\(=\int_{1}^{7}{1dt}\)
\(=\left[t\right]_{1}^{7}\) ➜ \(\because \int{1dx}=x\)
\(=\left[7-1\right]\)
\(=6\)
\(Q.2.(xxix)\) \(\int_{0}^{1}{\frac{xdx}{\sqrt{1-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
[ ঢাঃ২০০৭; রাঃ২০১২; যঃ২০০৭ ]
উত্তরঃ \(1\)
[ ঢাঃ২০০৭; রাঃ২০১২; যঃ২০০৭ ]
সমাধানঃ
ধরি,
\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\Rightarrow -xdx=tdt\)
\(\therefore xdx=-tdt\)
➜ \(\because t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-0^2}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1-0}\)
\(\therefore t=\sqrt{1}=1\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\therefore t=\sqrt{0}=0\)
\(\int_{0}^{1}{\frac{xdx}{\sqrt{1-x^2}}}\)\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\Rightarrow -xdx=tdt\)
\(\therefore xdx=-tdt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(1\) | \(0\) |
\(\Rightarrow t=\sqrt{1-0}\)
\(\therefore t=\sqrt{1}=1\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\therefore t=\sqrt{0}=0\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{1-x^2}}.xdx}\)
\(=\int_{1}^{0}{\frac{1}{t}\times{-tdt}}\)
\(=-\int_{1}^{0}{1dt}\)
\(=-\left[t\right]_{1}^{0}\) ➜ \(\because \int{1dx}=x\)
\(=-\left[0-1\right]\)
\(=1\)
\(Q.2.(xxx)\) \(\int_{0}^{2}{\frac{xdx}{\sqrt{9-2x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
[ ঢাঃ২০১৫; রাঃ২০১২; সিঃ,চঃ২০১৪; দিঃ২০১৩ ]
উত্তরঃ \(1\)
[ ঢাঃ২০১৫; রাঃ২০১২; সিঃ,চঃ২০১৪; দিঃ২০১৩ ]
সমাধানঃ
ধরি,
\(\sqrt{9-2x^2}=t\)
\(\Rightarrow 9-2x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(9-2x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-4x=2t\frac{dt}{dx}\)
\(\Rightarrow -4xdx=2tdt\)
\(\Rightarrow xdx=-\frac{2}{4}tdt\)
\(\therefore xdx=-\frac{1}{2}tdt\)
➜ \(\because t=\sqrt{9-2x^2}\)
\(\Rightarrow t=\sqrt{9-2.0^2}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{9-0}\)
\(\therefore t=\sqrt{9}=3\)
আবার,
\(t=\sqrt{9-2x^2}\)
\(\Rightarrow t=\sqrt{9-2.2^2}\), যখন \(x=2\)
\(\Rightarrow t=\sqrt{9-2.4}\)
\(\Rightarrow t=\sqrt{9-8}=3\)
\(\therefore t=\sqrt{1}=1\)
\(\int_{0}^{2}{\frac{xdx}{\sqrt{9-2x^2}}}\)\(\sqrt{9-2x^2}=t\)
\(\Rightarrow 9-2x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(9-2x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-4x=2t\frac{dt}{dx}\)
\(\Rightarrow -4xdx=2tdt\)
\(\Rightarrow xdx=-\frac{2}{4}tdt\)
\(\therefore xdx=-\frac{1}{2}tdt\)
| \(x\) | \(0\) | \(2\) |
| \(t\) | \(3\) | \(1\) |
\(\Rightarrow t=\sqrt{9-0}\)
\(\therefore t=\sqrt{9}=3\)
আবার,
\(t=\sqrt{9-2x^2}\)
\(\Rightarrow t=\sqrt{9-2.2^2}\), যখন \(x=2\)
\(\Rightarrow t=\sqrt{9-2.4}\)
\(\Rightarrow t=\sqrt{9-8}=3\)
\(\therefore t=\sqrt{1}=1\)
\(=\int_{0}^{2}{\frac{1}{\sqrt{9-2x^2}}.xdx}\)
\(=\int_{3}^{1}{\frac{1}{t}\times{-\frac{1}{2}tdt}}\)
\(=-\frac{1}{2}\int_{3}^{1}{1dt}\)
\(=-\frac{1}{2}\left[t\right]_{3}^{1}\) ➜ \(\because \int{1dx}=x\)
\(=-\frac{1}{2}\left[1-3\right]\)
\(=-\frac{1}{2}\times{-2}\)
\(=1\)
\(Q.2.(xxxi)\) \(\int_{0}^{1}{\frac{xdx}{\sqrt{4-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2-\sqrt{3}\)
[ ঢাঃ, রাঃ২০১০; কুঃ২০১০,২০০৫; দিঃ২০১৩ ]
উত্তরঃ \(2-\sqrt{3}\)
[ ঢাঃ, রাঃ২০১০; কুঃ২০১০,২০০৫; দিঃ২০১৩ ]
সমাধানঃ
ধরি,
\(\sqrt{4-x^2}=t\)
\(\Rightarrow 4-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(4-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\Rightarrow -xdx=tdt\)
\(\therefore xdx=-tdt\)
➜ \(\because t=\sqrt{4-x^2}\)
\(\Rightarrow t=\sqrt{4-0^2}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{4-0}\)
\(\therefore t=\sqrt{4}=2\)
আবার,
\(t=\sqrt{4-x^2}\)
\(\Rightarrow t=\sqrt{4-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{4-1}\)
\(\therefore t=\sqrt{3}\)
\(\int_{0}^{1}{\frac{xdx}{\sqrt{4-x^2}}}\)\(\sqrt{4-x^2}=t\)
\(\Rightarrow 4-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(4-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\Rightarrow -xdx=tdt\)
\(\therefore xdx=-tdt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(2\) | \(\sqrt{3}\) |
\(\Rightarrow t=\sqrt{4-0}\)
\(\therefore t=\sqrt{4}=2\)
আবার,
\(t=\sqrt{4-x^2}\)
\(\Rightarrow t=\sqrt{4-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{4-1}\)
\(\therefore t=\sqrt{3}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{4-x^2}}.xdx}\)
\(=\int_{2}^{\sqrt{3}}{\frac{1}{t}\times{-tdt}}\)
\(=-\int_{2}^{\sqrt{3}}{1dt}\)
\(=-\left[t\right]_{2}^{\sqrt{3}}\) ➜ \(\because \int{1dx}=x\)
\(=-\left[\sqrt{3}-2\right]\)
\(=2-\sqrt{3}\)
\(Q.2.(xxxii)\) \(\int_{0}^{\ln{2}}{\frac{e^x}{1+e^x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{\left(\frac{3}{2}\right)}\)
[ ঢাঃ২০১০,২০০৭; রাঃ২০১০,২০০৬,২০০৪; কুঃ২০১৭,২০১৩,২০১০,২০০৭,২০০৫; যঃ২০১৪,২০১১,২০০৯,২০০৭; দিঃ২০০৯; চঃ২০১১,২০০৯,২০০৫; সিঃ২০১২,২০০৮; বঃ২০১৪,২০১১,২০০৬; মাঃ২০১৪,২০১১ ]
উত্তরঃ \(\ln{\left(\frac{3}{2}\right)}\)
[ ঢাঃ২০১০,২০০৭; রাঃ২০১০,২০০৬,২০০৪; কুঃ২০১৭,২০১৩,২০১০,২০০৭,২০০৫; যঃ২০১৪,২০১১,২০০৯,২০০৭; দিঃ২০০৯; চঃ২০১১,২০০৯,২০০৫; সিঃ২০১২,২০০৮; বঃ২০১৪,২০১১,২০০৬; মাঃ২০১৪,২০১১ ]
সমাধানঃ
ধরি,
\(1+e^x=t\)
\(\Rightarrow \frac{d}{dx}(1+e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0+e^x=\frac{dt}{dx}\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
➜ \(\because t=1+e^x\)
\(\Rightarrow t=1+e^0\), যখন \(x=0\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
আবার,
\(t=1+e^x\)
\(\Rightarrow t=1+e^{\ln{2}}\), যখন \(x=\ln{2}\)
\(\Rightarrow t=1+2\), \(\because e^{\ln{x}}=x\)
\(\therefore t=3\)
\(\int_{0}^{1}{\frac{e^x}{1+e^x}dx}\)\(1+e^x=t\)
\(\Rightarrow \frac{d}{dx}(1+e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0+e^x=\frac{dt}{dx}\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
| \(x\) | \(0\) | \(\ln{2}\) |
| \(t\) | \(2\) | \(3\) |
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
আবার,
\(t=1+e^x\)
\(\Rightarrow t=1+e^{\ln{2}}\), যখন \(x=\ln{2}\)
\(\Rightarrow t=1+2\), \(\because e^{\ln{x}}=x\)
\(\therefore t=3\)
\(=\int_{0}^{1}{\frac{1}{1+e^x}e^xdx}\)
\(=\int_{2}^{3}{\frac{1}{t}dt}\)
\(=\left[\ln{|t|}\right]_{2}^{3}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\left[\ln{|3|}-\ln{|2|}\right]\)
\(=\ln{3}-\ln{2}\)
\(=\ln{\left(\frac{3}{2}\right)}\)
\(Q.2.(xxxiii)\) \(\int_{1}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{3}\)
[ দিঃ২০১১; বঃ২০৭,২০০৪ ]
উত্তরঃ \(\ln{3}\)
[ দিঃ২০১১; বঃ২০৭,২০০৪ ]
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{|1|}\), যখন \(x=1\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{e^{2}}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{e}\)
\(\Rightarrow t=1+2.1\), \(\because \ln{e}=1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(\int_{0}^{e^{2}}{\frac{dx}{x(1+\ln{|x|})}}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(e^{2}\) |
| \(t\) | \(1\) | \(3\) |
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{e^{2}}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{e}\)
\(\Rightarrow t=1+2.1\), \(\because \ln{e}=1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(=\int_{0}^{e^{2}}{\frac{1}{1+\ln{|x|}}.\frac{1}{x}dx}\)
\(=\int_{1}^{3}{\frac{1}{t}dt}\)
\(=\left[\ln{|t|}\right]_{1}^{3}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\left[\ln{|3|}-\ln{|1|}\right]\)
\(=\ln{3}-0\) ➜ \(\because \ln{1}=0\)
\(=\ln{3}\)
\(Q.2.(xxxiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cot{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\cot{x}}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\frac{\cos{x}}{\sin{x}}}dx}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\) ➜ লব ও হরের সহিত \(\sin{x}\) গুণ করে।
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{2\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{(\sin{x}+\cos{x})+(\sin{x}-\cos{x})}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left(\frac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}}+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left(1+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}-\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{d(\sin{x}+\cos{x})}{\sin{x}+\cos{x}}}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|\sin{x}+\cos{x}|}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|\sin{\frac{\pi}{2}}+\cos{\frac{\pi}{2}}|}-\ln{(|\sin{0}+\cos{0}|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|1+0|}-\ln{(|0+1|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{1}-\ln{1}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[0-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\frac{\pi}{4}-\frac{1}{2}.0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\frac{\cos{x}}{\sin{x}}}dx}\) ➜ \(\because \cot{A}=\frac{\cos{A}}{\sin{A}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\) ➜ লব ও হরের সহিত \(\sin{x}\) গুণ করে।
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{2\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{(\sin{x}+\cos{x})+(\sin{x}-\cos{x})}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left(\frac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}}+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\left(1+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{1dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}-\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-0\right]-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{d(\sin{x}+\cos{x})}{\sin{x}+\cos{x}}}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|\sin{x}+\cos{x}|}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|\sin{\frac{\pi}{2}}+\cos{\frac{\pi}{2}}|}-\ln{(|\sin{0}+\cos{0}|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{(|1+0|}-\ln{(|0+1|}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[\ln{1}-\ln{1}\right]\)
\(=\frac{\pi}{4}-\frac{1}{2}\left[0-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\frac{\pi}{4}-\frac{1}{2}.0\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(Q.2.(xxxv)\) \(\int_{0}^{1}{\frac{dx}{1+x^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{1+x^2}}\)
\(=\left[\tan^{-1}{x}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}\)
\(=\left[\tan^{-1}{1}-\tan^{-1}{0}\right]\)
\(=\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\left[\frac{\pi}{4}-0\right]\)
\(=\frac{\pi}{4}\)
\(=\left[\tan^{-1}{x}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}\)
\(=\left[\tan^{-1}{1}-\tan^{-1}{0}\right]\)
\(=\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\left[\frac{\pi}{4}-0\right]\)
\(=\frac{\pi}{4}\)
\(Q.2.(xxxvi)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{1-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}\)
উত্তরঃ \(\frac{\pi}{2}\)
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{\sqrt{1-x^2}}}\)
\(=\left[\sin^{-1}{x}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{\sqrt{1-x^2}}}=\sin^{-1}{x}\)
\(=\left[\sin^{-1}{1}-\sin^{-1}{0}\right]\)
\(=\left[\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-\sin^{-1}{\sin{0}}\right]\)
\(=\left[\frac{\pi}{2}-0\right]\)
\(=\frac{\pi}{2}\)
\(=\left[\sin^{-1}{x}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{\sqrt{1-x^2}}}=\sin^{-1}{x}\)
\(=\left[\sin^{-1}{1}-\sin^{-1}{0}\right]\)
\(=\left[\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-\sin^{-1}{\sin{0}}\right]\)
\(=\left[\frac{\pi}{2}-0\right]\)
\(=\frac{\pi}{2}\)
\(Q.2.(xxxvii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{4-3x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{3\sqrt{3}}\)
[ ঢাঃ২০০৩ ]
উত্তরঃ \(\frac{\pi}{3\sqrt{3}}\)
[ ঢাঃ২০০৩ ]
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{\sqrt{4-3x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{3}\sqrt{\frac{4}{3}-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\int_{0}^{1}{\frac{dx}{\sqrt{\frac{4}{3}-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\int_{0}^{1}{\frac{dx}{\sqrt{\left(\frac{2}{\sqrt{3}}\right)^2-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{x}{\frac{2}{\sqrt{3}}}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{\sqrt{a^2-x^2}}}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}x}{2}\right)}\right]_{0}^{1}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}.1}{2}\right)}-\sin^{-1}{\left(\frac{\sqrt{3}.0}{2}\right)}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}}{2}\right)}-\sin^{-1}{0}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\sin{\left(\frac{\pi}{3}\right)}}-\sin^{-1}{\sin{0}}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\frac{\pi}{3}-0\right]\)
\(=\frac{\pi}{3\sqrt{3}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{3}\sqrt{\frac{4}{3}-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\int_{0}^{1}{\frac{dx}{\sqrt{\frac{4}{3}-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\int_{0}^{1}{\frac{dx}{\sqrt{\left(\frac{2}{\sqrt{3}}\right)^2-x^2}}}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{x}{\frac{2}{\sqrt{3}}}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{\sqrt{a^2-x^2}}}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}x}{2}\right)}\right]_{0}^{1}\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}.1}{2}\right)}-\sin^{-1}{\left(\frac{\sqrt{3}.0}{2}\right)}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\left(\frac{\sqrt{3}}{2}\right)}-\sin^{-1}{0}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\sin^{-1}{\sin{\left(\frac{\pi}{3}\right)}}-\sin^{-1}{\sin{0}}\right]\)
\(=\frac{1}{\sqrt{3}}\left[\frac{\pi}{3}-0\right]\)
\(=\frac{\pi}{3\sqrt{3}}\)
\(Q.2.(xxxviii)\) \(\int_{3}^{4}{\frac{dx}{25-x^2}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
[ বঃ২০১৩ ]
উত্তরঃ \(\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
[ বঃ২০১৩ ]
সমাধানঃ
\(\int_{3}^{4}{\frac{dx}{25-x^2}}\)
\(=\int_{3}^{4}{\frac{dx}{5^2-x^2}}\)
\(=\left[\frac{1}{2.5}\ln{\left|\frac{5+x}{5-x}\right|}\right]_{3}^{4}\) ➜ \(\because \int{\frac{dx}{a^2-x^2}}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\left[\frac{1}{10}\ln{\left|\frac{5+x}{5-x}\right|)}\right]_{3}^{4}\)
\(=\frac{1}{10}\left[\ln{\left|\frac{5+x}{5-x}\right|}\right]_{3}^{4}\)
\(=\frac{1}{10}\left[\ln{\left|\frac{5+4}{5-4}\right|}-\ln{\left|\frac{5+3}{5-3}\right|}\right]\)
\(=\frac{1}{10}\left[\ln{\frac{9}{1}}-\ln{\frac{8}{2}}\right]\)
\(=\frac{1}{10}\left[\ln{9}-\ln{4}\right]\)
\(=\frac{1}{10}\times{\ln{\left(\frac{9}{4}\right)}}\)
\(=\frac{1}{10}\times{\ln{\left(\frac{3}{2}\right)^2}}\)
\(=\frac{1}{10}\times{2\ln{\left(\frac{3}{2}\right)}}\)
\(=\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
\(=\int_{3}^{4}{\frac{dx}{5^2-x^2}}\)
\(=\left[\frac{1}{2.5}\ln{\left|\frac{5+x}{5-x}\right|}\right]_{3}^{4}\) ➜ \(\because \int{\frac{dx}{a^2-x^2}}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\left[\frac{1}{10}\ln{\left|\frac{5+x}{5-x}\right|)}\right]_{3}^{4}\)
\(=\frac{1}{10}\left[\ln{\left|\frac{5+x}{5-x}\right|}\right]_{3}^{4}\)
\(=\frac{1}{10}\left[\ln{\left|\frac{5+4}{5-4}\right|}-\ln{\left|\frac{5+3}{5-3}\right|}\right]\)
\(=\frac{1}{10}\left[\ln{\frac{9}{1}}-\ln{\frac{8}{2}}\right]\)
\(=\frac{1}{10}\left[\ln{9}-\ln{4}\right]\)
\(=\frac{1}{10}\times{\ln{\left(\frac{9}{4}\right)}}\)
\(=\frac{1}{10}\times{\ln{\left(\frac{3}{2}\right)^2}}\)
\(=\frac{1}{10}\times{2\ln{\left(\frac{3}{2}\right)}}\)
\(=\frac{1}{5}\ln{\left(\frac{3}{2}\right)}\)
\(Q.2.(xxxix)\) \(\int_{0}^{3}{\frac{dx}{\sqrt{3x-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi\)
[ চঃ২০১০ ]
উত্তরঃ \(\pi\)
[ চঃ২০১০ ]
সমাধানঃ
\(\int_{0}^{3}{\frac{dx}{\sqrt{3x-x^2}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left(\frac{9}{4}-3x+x^2\right)}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left\{x^2-3x+\frac{9}{4}\right\}}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left\{x^2-2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2\right\}}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-\frac{3}{2}}{\frac{3}{2}}\right)}\right]_{0}^{3}\) ➜ \(\because \int{\frac{dx}{\sqrt{a^2-x^2}}}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{\left(\frac{2x-3}{3}\right)}\right]_{0}^{3}\) ➜ লব ও হরের সহিত \(2\) গুণ করে।
\(=\sin^{-1}{\left(\frac{2.3-3}{3}\right)}-\sin^{-1}{\left(\frac{2.0-3}{3}\right)}\)
\(=\sin^{-1}{\left(\frac{6-3}{3}\right)}-\sin^{-1}{\left(\frac{0-3}{3}\right)}\)
\(=\sin^{-1}{\left(\frac{3}{3}\right)}-\sin^{-1}{\left(\frac{-3}{3}\right)}\)
\(=\sin^{-1}{\left(1\right)}+\sin^{-1}{\left(1\right)}\)
\(=2\sin^{-1}{\left(1\right)}\)
\(=2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=2\frac{\pi}{2}\)
\(=\pi\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left(\frac{9}{4}-3x+x^2\right)}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left\{x^2-3x+\frac{9}{4}\right\}}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\frac{9}{4}-\left\{x^2-2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2\right\}}}}\)
\(=\int_{0}^{3}{\frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-\frac{3}{2}}{\frac{3}{2}}\right)}\right]_{0}^{3}\) ➜ \(\because \int{\frac{dx}{\sqrt{a^2-x^2}}}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{\left(\frac{2x-3}{3}\right)}\right]_{0}^{3}\) ➜ লব ও হরের সহিত \(2\) গুণ করে।
\(=\sin^{-1}{\left(\frac{2.3-3}{3}\right)}-\sin^{-1}{\left(\frac{2.0-3}{3}\right)}\)
\(=\sin^{-1}{\left(\frac{6-3}{3}\right)}-\sin^{-1}{\left(\frac{0-3}{3}\right)}\)
\(=\sin^{-1}{\left(\frac{3}{3}\right)}-\sin^{-1}{\left(\frac{-3}{3}\right)}\)
\(=\sin^{-1}{\left(1\right)}+\sin^{-1}{\left(1\right)}\)
\(=2\sin^{-1}{\left(1\right)}\)
\(=2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=2\frac{\pi}{2}\)
\(=\pi\)
\(Q.2.(xL)\) \(\int_{0}^{1}{\frac{1+x}{1+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
[ সিঃ২০১২,২০০৫; কুঃ২০১২; রাঃ২০০৯,২০০৬,২০০৪; ঢাঃ২০০৯; বঃ২০০৭; দিঃ,চঃ,যঃ২০১১ ]
উত্তরঃ \(\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
[ সিঃ২০১২,২০০৫; কুঃ২০১২; রাঃ২০০৯,২০০৬,২০০৪; ঢাঃ২০০৯; বঃ২০০৭; দিঃ,চঃ,যঃ২০১১ ]
সমাধানঃ
\(\int_{0}^{1}{\frac{1+x}{1+x^2}dx}\)
\(=\int_{0}^{1}{\left(\frac{1}{1+x^2}+\frac{x}{1+x^2}\right)dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\int_{0}^{1}{\frac{x}{1+x^2}dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\frac{1}{2}\int_{0}^{1}{\frac{2x}{1+x^2}dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\frac{1}{2}\int_{0}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\tan^{-1}{x}\right]_{0}^{1}+\frac{1}{2}\left[\ln{(1+x^2)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}, \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\tan^{-1}{1}-\tan^{-1}{0}+\frac{1}{2}\left[\ln{(1+1^2)}-\ln{(1+0^2)}\right]\)
\(=\tan^{-1}{\tan{\frac{\pi}{4}}}-\tan^{-1}{\tan{0}}+\frac{1}{2}\left[\ln{(1+1)}-\ln{(1+0)}\right]\)
\(=\frac{\pi}{4}-0+\frac{1}{2}\left[\ln{2}-\ln{1}\right]\)
\(=\frac{\pi}{4}+\frac{1}{2}\left[\ln{2}-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
\(=\int_{0}^{1}{\left(\frac{1}{1+x^2}+\frac{x}{1+x^2}\right)dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\int_{0}^{1}{\frac{x}{1+x^2}dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\frac{1}{2}\int_{0}^{1}{\frac{2x}{1+x^2}dx}\)
\(=\int_{0}^{1}{\frac{dx}{1+x^2}}+\frac{1}{2}\int_{0}^{1}{\frac{d(1+x^2)}{1+x^2}}\)
\(=\left[\tan^{-1}{x}\right]_{0}^{1}+\frac{1}{2}\left[\ln{(1+x^2)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}, \int{\frac{dx}{x}}=\ln{|x|}\)
\(=\tan^{-1}{1}-\tan^{-1}{0}+\frac{1}{2}\left[\ln{(1+1^2)}-\ln{(1+0^2)}\right]\)
\(=\tan^{-1}{\tan{\frac{\pi}{4}}}-\tan^{-1}{\tan{0}}+\frac{1}{2}\left[\ln{(1+1)}-\ln{(1+0)}\right]\)
\(=\frac{\pi}{4}-0+\frac{1}{2}\left[\ln{2}-\ln{1}\right]\)
\(=\frac{\pi}{4}+\frac{1}{2}\left[\ln{2}-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\frac{\pi}{4}+\frac{1}{2}\ln{2}\)
\(Q.2.(xLi)\) \(\int_{0}^{1}{\frac{xdx}{1+x^4}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{8}\)
[ রাঃ২০১১ ]
উত্তরঃ \(\frac{\pi}{8}\)
[ রাঃ২০১১ ]
সমাধানঃ
ধরি,
\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
➜ \(\because t=x^2\)
\(\Rightarrow t=0^2\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=x^2\)
\(\Rightarrow t=1^2\), যখন \(x=1\)
\(\therefore t=1\)
\(\int_{0}^{1}{\frac{xdx}{1+x^4}}\)\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=x^2\)
\(\Rightarrow t=1^2\), যখন \(x=1\)
\(\therefore t=1\)
\(=\int_{0}^{1}{\frac{1}{1+(x^2)^2}.xdx}\)
\(=\int_{0}^{1}{\frac{1}{1+t^2}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int_{0}^{1}{\frac{dt}{1+t^2}}\)
\(=\frac{1}{2}\left[\tan^{-1}{t}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}\)
\(=\frac{1}{2}\left[\tan^{-1}{1}-\tan^{-1}{0}\right]\)
\(=\frac{1}{2}\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{2}\left[\frac{\pi}{4}-0\right]\)
\(=\frac{\pi}{8}\)
\(Q.2.(xLii)\) \(\int_{0}^{1}{\frac{dx}{e^x+e^{-x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\tan^{-1}{e}-\frac{\pi}{4}\)
[ রাঃ২০১২,২০০৩; কুঃ২০০৮; বঃ২০১৩; সিঃ২০০৭,২০০৪; মাঃ২০১০ ]
উত্তরঃ \(\tan^{-1}{e}-\frac{\pi}{4}\)
[ রাঃ২০১২,২০০৩; কুঃ২০০৮; বঃ২০১৩; সিঃ২০০৭,২০০৪; মাঃ২০১০ ]
সমাধানঃ
ধরি,
\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
➜ \(\because t=e^x\)
\(\Rightarrow t=e^0\), যখন \(x=0\)
\(\therefore t=1\)
আবার,
\(t=e^x\)
\(\Rightarrow t=e^1\), যখন \(x=1\)
\(\therefore t=e\)
\(\int_{0}^{1}{\frac{dx}{e^x+e^{-x}}}\)\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(1\) | \(e\) |
\(\therefore t=1\)
আবার,
\(t=e^x\)
\(\Rightarrow t=e^1\), যখন \(x=1\)
\(\therefore t=e\)
\(=\int_{0}^{1}{\frac{1}{e^x+e^{-x}}dx}\)
\(=\int_{0}^{1}{\frac{e^x}{(e^x)^2+e^{-x}e^x}dx}\) ➜ লব ও হরের সহিত \(e^x\) গুণ করে।
\(=\int_{0}^{1}{\frac{e^x}{(e^x)^2+e^{-x+x}}dx}\)
\(=\int_{0}^{1}{\frac{e^x}{(e^x)^2+e^{0}}dx}\)
\(=\int_{0}^{1}{\frac{e^x}{(e^x)^2+1}dx}\)
\(=\int_{0}^{1}{\frac{1}{1+(e^x)^2}.e^xdx}\)
\(=\int_{1}^{e}{\frac{1}{1+t^2}dt}\)
\(=\left[\tan^{-1}{t}\right]_{1}^{e}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\tan^{-1}{e}-\tan^{-1}{1}\)
\(=\tan^{-1}{e}-\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(=\tan^{-1}{e}-\frac{\pi}{4}\)
\(Q.2.(xLiii)\) \(\int_{0}^{\pi}{\frac{\sin{x}}{1+\cos^2{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃ২০০৭ ]
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃ২০০৭ ]
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
➜ \(\because t=\cos{x}\)
\(\Rightarrow t=\cos{0}\), যখন \(x=0\)
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\pi}\), যখন \(x=\pi\)
\(\therefore t=-1\)
\(\int_{0}^{\pi}{\frac{\sin{x}}{1+\cos^2{x}}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
| \(x\) | \(0\) | \(\pi\) |
| \(t\) | \(1\) | \(-1\) |
\(\therefore t=1\)
আবার,
\(t=\cos{x}\)
\(\Rightarrow t=\cos{\pi}\), যখন \(x=\pi\)
\(\therefore t=-1\)
\(=\int_{0}^{\pi}{\frac{1}{1+\cos^2{x}}.\sin{x}dx}\)
\(=\int_{1}^{-1}{\frac{1}{1+t^2}\times{-dt}}\)
\(=-\int_{1}^{-1}{\frac{dt}{1+t^2}}\)
\(=-\left[\tan^{-1}{t}\right]_{1}^{-1}\) ➜ \(\because \int{\frac{dx}{1+x^2}}=\tan^{-1}{x}\)
\(=-\left[\tan^{-1}{(-1)}-\tan^{-1}{1}\right]\)
\(=-\left[-\tan^{-1}{1}-\tan^{-1}{1}\right]\)
\(=-\left[-2\tan^{-1}{1}\right]\)
\(=2\tan^{-1}{1}\)
\(=2\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(=2\times{\frac{\pi}{4}}\)
\(=\frac{\pi}{2}\)
\(Q.2.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{1+\sin^2{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ,সিঃ২০১৩ ]
উত্তরঃ \(\frac{\pi}{4}\)
[ রাঃ,সিঃ২০১৩ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=\sin{x}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{1+\sin^2{x}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\sin^2{x}}.\cos{x}}\)
\(=\int_{0}^{1}{\frac{1}{1+t^2}dt}\)
\(=\left[\tan^{-1}{t}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\tan^{-1}{1}-\tan^{-1}{0}\)
\(=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(Q.2.(xLv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{\sqrt{4-\sin^2{\theta}}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{6}\)
[ রাঃ,সিঃ২০১৩ ]
উত্তরঃ \(\frac{\pi}{6}\)
[ রাঃ,সিঃ২০১৩ ]
সমাধানঃ
ধরি,
\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
➜ \(\because t=\sin{\theta}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\theta}}{\sqrt{4-\sin^2{\theta}}}d\theta}\)\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{\theta}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{\sqrt{4-\sin^2{\theta}}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{4-t^2}}dt}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{2^2-t^2}}dt}\)
\(=\left[\sin^{-1}{\left(\frac{t}{2}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\sin^{-1}{\left(\frac{1}{2}\right)}-\sin^{-1}{\left(\frac{0}{2}\right)}\)
\(=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}-\sin^{-1}{\sin{0}}\)
\(=\frac{\pi}{6}-0\)
\(=\frac{\pi}{6}\)
\(Q.2.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{9-\sin^2{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}\ln{2}\)
[ ঢাঃ২০০৫; কুঃ২০১০; দিঃ২০১৩; চঃ২০০৯; সিঃ২০১৩,২০০৯; বঃ২০১০ ]
উত্তরঃ \(\frac{1}{6}\ln{2}\)
[ ঢাঃ২০০৫; কুঃ২০১০; দিঃ২০১৩; চঃ২০০৯; সিঃ২০১৩,২০০৯; বঃ২০১০ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=\sin{x}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{9-\sin^2{x}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{9-\sin^2{x}}.\cos{x}dx}\)
\(=\int_{0}^{1}{\frac{1}{9-t^2}dt}\)
\(=\int_{0}^{1}{\frac{dt}{3^2-t^2}}\)
\(=\left[\frac{1}{2.3}\ln{\left|\frac{3+t}{3-t}\right|}\right]_{0}^{1}\) ➜ \(\because \int{\frac{dx}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\left[\frac{1}{6}\ln{\left|\frac{3+t}{3-t}\right|}\right]_{0}^{1}\)
\(=\frac{1}{6}\left[\ln{\left|\frac{3+t}{3-t}\right|}\right]_{0}^{1}\)
\(=\frac{1}{6}\left[\ln{\frac{3+1}{3-1}}-\ln{\frac{3+0}{3-0}}\right]\)
\(=\frac{1}{6}\left[\ln{\frac{4}{2}}-\ln{\frac{3}{3}}\right]\)
\(=\frac{1}{6}\left[\ln{2}-\ln{1}\right]\)
\(=\frac{1}{6}\left[\ln{2}-0\right]\) ➜ \(\because \ln{1}=0\)
\(=\frac{1}{6}\ln{2}\)
\(Q.2.(xLvii)\) \(\int_{0}^{\frac{\pi}{4}}{\frac{\sin{2x}}{\cos^4{x}+\sin^4{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
[ বুয়েটঃ২০০৮-২০০৯ ]
উত্তরঃ \(\frac{\pi}{4}\)
[ বুয়েটঃ২০০৮-২০০৯ ]
সমাধানঃ
ধরি,
\(\tan^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 2\tan{x}\frac{d}{dx}(\tan{x})=\frac{dt}{dx}\)
\(\Rightarrow 2\tan{x}\sec^2{x}=\frac{dt}{dx}\)
\(\therefore 2\tan{x}\sec^2{x}dx=dt\)
➜ \(\because t=\tan^2{x}\)
\(\Rightarrow t=\tan^2{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan^2{x}\)
\(\Rightarrow t=\tan^2{\left(\frac{\pi}{4}\right)}\), যখন \(x=\frac{\pi}{4}\)
\(\Rightarrow t=1^2\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{4}}{\frac{\sin{2x}}{\cos^4{x}+\sin^4{x}}dx}\)\(\tan^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 2\tan{x}\frac{d}{dx}(\tan{x})=\frac{dt}{dx}\)
\(\Rightarrow 2\tan{x}\sec^2{x}=\frac{dt}{dx}\)
\(\therefore 2\tan{x}\sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{4}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\tan^2{x}\)
\(\Rightarrow t=\tan^2{\left(\frac{\pi}{4}\right)}\), যখন \(x=\frac{\pi}{4}\)
\(\Rightarrow t=1^2\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{2\sin{x}\cos{x}}{\cos^4{x}+\sin^4{x}}dx}\) ➜ \(\because \sin{2A}=2\sin{A}\cos{A}\)
\(=2\int_{0}^{\frac{\pi}{4}}{\frac{\sin{x}\cos{x}}{\cos^4{x}+\sin^4{x}}dx}\)
\(=2\int_{0}^{\frac{\pi}{4}}{\frac{\frac{\sin{x}\cos{x}}{\cos^4{x}}}{1+\frac{\sin^4{x}}{\cos^4{x}}}dx}\) ➜ লব ও হরের সহিত \(\cos^4{x}\) ভাগ করে।
\(=2\int_{0}^{\frac{\pi}{4}}{\frac{\frac{\sin{x}}{\cos^3{x}}}{1+\tan^4{x}}dx}\)
\(=2\int_{0}^{\frac{\pi}{4}}{\frac{\frac{\sin{x}}{\cos{x}}.\frac{1}{\cos^2{x}}}{1+\tan^4{x}}dx}\)
\(=2\int_{0}^{\frac{\pi}{4}}{\frac{\tan{x}\sec^2{x}}{1+\tan^4{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+(\tan^2{x})^2}.2\tan{x}\sec^2{x}dx}\)
\(=\int_{0}^{1}{\frac{1}{1+t^2}dt}\)
\(=\left[\tan^{-1}{t}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\)
\(=\frac{\pi}{4}-0\)
\(=\frac{\pi}{4}\)
\(Q.2.(xLviii)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{a^2\cos^2{x}+b^2\sin^2{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2ab}\)
[ রুয়েটঃ২০১১-২০১২ ]
উত্তরঃ \(\frac{\pi}{2ab}\)
[ রুয়েটঃ২০১১-২০১২ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
➜ \(\because t=\tan{x}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{a^2\cos^2{x}+b^2\sin^2{x}}}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(\infty\) |
\(\therefore t=0\)
আবার,
\(t=\tan{x}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{a^2\cos^2{x}+b^2\sin^2{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\frac{1}{\cos^2{x}}}{a^2+b^2\frac{\sin^2{x}}{\cos^2{x}}}dx}\) ➜ লব ও হরের সহিত \(\cos^2{x}\) ভাগ করে।
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{x}}{a^2+b^2\tan^2{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{x}}{b^2\left(\frac{a^2}{b^2}+\tan^2{x}\right)}dx}\)
\(=\frac{1}{b^2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{\frac{a^2}{b^2}+\tan^2{x}}\sec^2{x}dx}\)
\(=\frac{1}{b^2}\int_{0}^{\infty}{\frac{1}{\left(\frac{a}{b}\right)^2+t^2}dt}\)
\(=\frac{1}{b^2}\left[\frac{1}{\frac{a}{b}}\tan^{-1}{\left(\frac{t}{\frac{a}{b}}\right)}\right]_{0}^{\infty}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{b^2}.\frac{b}{a}\left[\tan^{-1}{\left(\frac{bt}{a}\right)}\right]_{0}^{\infty}\)
\(=\frac{1}{ab}\left[\tan^{-1}{\left(\frac{b\times{\infty}}{a}\right)}-\tan^{-1}{\left(\frac{b.0}{a}\right)}\right]\)
\(=\frac{1}{ab}\left[\tan^{-1}{\left(\infty\right)}-\tan^{-1}{0}\right]\)
\(=\frac{1}{ab}\left[\tan^{-1}{\tan{\left(\frac{\pi}{2}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{ab}\left[\frac{\pi}{2}-0\right]\)
\(=\frac{\pi}{2ab}\)
\(Q.2.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a\sin^2{\theta}+b\cos^2{\theta}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2\sqrt{ab}}\)
[ রাঃ২০১১ ]
উত্তরঃ \(\frac{\pi}{2\sqrt{ab}}\)
[ রাঃ২০১১ ]
সমাধানঃ
ধরি,
\(\tan{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\tan{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \sec^2{\theta}=\frac{dt}{d\theta}\)
\(\therefore \sec^2{\theta}d\theta=dt\)
➜ \(\because t=\tan{\theta}\)
\(\Rightarrow t=\tan{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{\theta}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a\sin^2{\theta}+b\cos^2{\theta}}}\)\(\tan{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\tan{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \sec^2{\theta}=\frac{dt}{d\theta}\)
\(\therefore \sec^2{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(\infty\) |
\(\therefore t=0\)
আবার,
\(t=\tan{\theta}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{a\cos^2{\theta}+b\sin^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\frac{1}{\cos^2{\theta}}}{a+b\frac{\sin^2{\theta}}{\cos^2{\theta}}}d\theta}\) ➜ লব ও হরের সহিত \(\cos^2{\theta}\) ভাগ করে।
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{\theta}}{a+b\tan^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{\theta}}{b\left(\frac{a}{b}+\tan^2{\theta}\right)}d\theta}\)
\(=\frac{1}{b}\int_{0}^{\frac{\pi}{2}}{\frac{1}{\frac{a}{b}+\tan^2{\theta}}\sec^2{\theta}d\theta}\)
\(=\frac{1}{b}\int_{0}^{\infty}{\frac{1}{\left(\sqrt{\frac{a}{b}}\right)^2+t^2}dt}\)
\(=\frac{1}{b}\left[\frac{1}{\sqrt{\frac{a}{b}}}\tan^{-1}{\left(\frac{t}{\sqrt{\frac{a}{b}}}\right)}\right]_{0}^{\infty}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{b}.\sqrt{\frac{b}{a}}\left[\tan^{-1}{\left(\sqrt{\frac{b}{a}}t\right)}\right]_{0}^{\infty}\)
\(=\frac{1}{\sqrt{b}\sqrt{b}}.\frac{\sqrt{b}}{\sqrt{a}}\left[\tan^{-1}{\left(\sqrt{\frac{b}{a}}t\right)}\right]_{0}^{\infty}\)
\(=\frac{1}{\sqrt{ab}}\left[\tan^{-1}{\left(\sqrt{\frac{b}{a}}\times{\infty}\right)}-\tan^{-1}{\left(\sqrt{\frac{b}{a}}.0\right)}\right]\)
\(=\frac{1}{\sqrt{ab}}\left[\tan^{-1}{\left(\infty\right)}-\tan^{-1}{0}\right]\)
\(=\frac{1}{\sqrt{ab}}\left[\tan^{-1}{\tan{\left(\frac{\pi}{2}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{\sqrt{ab}}\left[\frac{\pi}{2}-0\right]\)
\(=\frac{\pi}{2\sqrt{ab}}\)
\(Q.2.(L)\) \(\int_{0}^{1}{\frac{\ln{|x+1|}}{x+1}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\ln{2})^2\)
[ ঢাঃবিঃ২০১৪-২০১৫ ]
উত্তরঃ \(\frac{1}{2}(\ln{2})^2\)
[ ঢাঃবিঃ২০১৪-২০১৫ ]
সমাধানঃ
ধরি,
\(\ln{|x+1|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x+1|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x+1}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x+1}dx=dt\)
➜ \(\because t=\ln{|x+1|}\)
\(\Rightarrow t=\ln{(1+0)}\), যখন \(x=0\)
\(\Rightarrow t=\ln{(1)}\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x+1|}\)
\(\Rightarrow t=\ln{(1+1)}\), যখন \(x=1\)
\(\therefore t=\ln{(2)}\)
\(\int_{0}^{1}{\frac{\ln{|x+1|}}{x+1}dx}\)\(\ln{|x+1|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x+1|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x+1}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x+1}dx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\ln{(2)}\) |
\(\Rightarrow t=\ln{(1)}\)
\(\therefore t=0\)
আবার,
\(t=\ln{|x+1|}\)
\(\Rightarrow t=\ln{(1+1)}\), যখন \(x=1\)
\(\therefore t=\ln{(2)}\)
\(=\int_{0}^{1}{\ln{|x+1|}.\frac{1}{x+1}dx}\)
\(=\int_{0}^{\ln{(2)}}{tdt}\)
\(=\left[\frac{t^2}{2}\right]_{0}^{\ln{(2)}}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[t^2\right]_{0}^{\ln{(2)}}\)
\(=\frac{1}{2}\left[(\ln{2})^2-0^2\right]\)
\(=\frac{1}{2}\left[(\ln{2})^2-0\right]\)
\(=\frac{1}{2}(\ln{2})^2\)
\(Q.2.(Li)\) \(\int_{1}^{e^2}{\frac{1+\ln{|x|}}{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\)
উত্তরঃ \(4\)
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
➜ \(\because t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{|1|}\), যখন \(x=1\)
\(\Rightarrow t=1+\ln{(1)}\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{(e)^2}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{(e)}\)
\(\Rightarrow t=1+2.1\), \(\because \ln{(e)}=1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(\int_{1}^{e^2}{\frac{1+\ln{|x|}}{x}dx}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
| \(x\) | \(1\) | \(e^2\) |
| \(t\) | \(1\) | \(3\) |
\(\Rightarrow t=1+\ln{(1)}\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\ln{|x|}\)
\(\Rightarrow t=1+\ln{(e)^2}\), যখন \(x=e^2\)
\(\Rightarrow t=1+2\ln{(e)}\)
\(\Rightarrow t=1+2.1\), \(\because \ln{(e)}=1\)
\(\Rightarrow t=1+2\)
\(\therefore t=3\)
\(=\int_{1}^{e^2}{1+\ln{|x|}.\frac{1}{x}dx}\)
\(=\int_{1}^{3}{tdt}\)
\(=\left[\frac{t^2}{2}\right]_{1}^{3}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\frac{1}{2}\left[t^2\right]_{1}^{3}\)
\(=\frac{1}{2}\left[3^2-1^2\right]\)
\(=\frac{1}{2}\left[9-1\right]\)
\(=\frac{1}{2}\times{8}\)
\(=4\)
\(Q.2.(Lii)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{x-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi\)
উত্তরঃ \(\pi\)
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{\sqrt{x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\left\{x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right\}}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{\left(\frac{2x-1}{1}\right)}\right]_{0}^{1}\) ➜ লব ও হরের সহিত \(2\) গুণ করে।
\(=\left[\sin^{-1}{(2x-1)}\right]_{0}^{1}\)
\(=\sin^{-1}{(2.1-1)}-\sin^{-1}{(2.0-1)}\)
\(=\sin^{-1}{(2-1)}-\sin^{-1}{(0-1)}\)
\(=\sin^{-1}{(1)}-\sin^{-1}{(-1)}\)
\(=\sin^{-1}{(1)}+\sin^{-1}{(1)}\)
\(=2\sin^{-1}{(1)}\)
\(=2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=2\times{\frac{\pi}{2}}\)
\(=\pi\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\frac{1}{4}-\left\{x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right\}}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{\left(\frac{2x-1}{1}\right)}\right]_{0}^{1}\) ➜ লব ও হরের সহিত \(2\) গুণ করে।
\(=\left[\sin^{-1}{(2x-1)}\right]_{0}^{1}\)
\(=\sin^{-1}{(2.1-1)}-\sin^{-1}{(2.0-1)}\)
\(=\sin^{-1}{(2-1)}-\sin^{-1}{(0-1)}\)
\(=\sin^{-1}{(1)}-\sin^{-1}{(-1)}\)
\(=\sin^{-1}{(1)}+\sin^{-1}{(1)}\)
\(=2\sin^{-1}{(1)}\)
\(=2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=2\times{\frac{\pi}{2}}\)
\(=\pi\)
\(Q.2.(Liii)\) \(\int_{2}^{5}{\frac{dx}{x^2-4x+13}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{12}\)
উত্তরঃ \(\frac{\pi}{12}\)
সমাধানঃ
\(\int_{2}^{5}{\frac{dx}{x^2-4x+13}}\)
\(=\int_{2}^{5}{\frac{dx}{x^2-4x+4+9}}\)
\(=\int_{2}^{5}{\frac{dx}{x^2-2.x.2+2^2+9}}\)
\(=\int_{2}^{5}{\frac{dx}{(x-2)^2+3^2}}\)
\(=\int_{2}^{5}{\frac{dx}{3^2+(x-2)^2}}\)
\(=\left[\frac{1}{3}\tan^{-1}{\left(\frac{x-2}{3}\right)}\right]_{2}^{5}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{x-2}{3}\right)}\right]_{2}^{5}\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{5-2}{3}\right)}-\tan^{-1}{\left(\frac{2-2}{3}\right)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{3}{3}\right)}-\tan^{-1}{\left(\frac{0}{3}\right)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{(1)}-\tan^{-1}{(0)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{3}\left[\frac{\pi}{4}-0\right]\)
\(=\frac{\pi}{12}\)
\(=\int_{2}^{5}{\frac{dx}{x^2-4x+4+9}}\)
\(=\int_{2}^{5}{\frac{dx}{x^2-2.x.2+2^2+9}}\)
\(=\int_{2}^{5}{\frac{dx}{(x-2)^2+3^2}}\)
\(=\int_{2}^{5}{\frac{dx}{3^2+(x-2)^2}}\)
\(=\left[\frac{1}{3}\tan^{-1}{\left(\frac{x-2}{3}\right)}\right]_{2}^{5}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{x-2}{3}\right)}\right]_{2}^{5}\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{5-2}{3}\right)}-\tan^{-1}{\left(\frac{2-2}{3}\right)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{\left(\frac{3}{3}\right)}-\tan^{-1}{\left(\frac{0}{3}\right)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{(1)}-\tan^{-1}{(0)}\right]\)
\(=\frac{1}{3}\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{3}\left[\frac{\pi}{4}-0\right]\)
\(=\frac{\pi}{12}\)
\(Q.2.(Liv)\) \(\int_{0}^{1}{\frac{dx}{\sqrt{2x-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃবিঃ২০০৮-২০০৯ ]
উত্তরঃ \(\frac{\pi}{2}\)
[ ঢাঃবিঃ২০০৮-২০০৯ ]
সমাধানঃ
\(\int_{0}^{1}{\frac{dx}{\sqrt{2x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-1+2x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-(x^2-2x+1)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-(x^2-2.x.1+1^2)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1^2-(x-1)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-1}{1}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{(x-1)}\right]_{0}^{1}\)
\(=\sin^{-1}{(1-1)}-\sin^{-1}{(0-1)}\)
\(=\sin^{-1}{(0)}-\sin^{-1}{(-1)}\)
\(=\sin^{-1}{\sin{0}}+\sin^{-1}{(1)}\)
\(=0+\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=\frac{\pi}{2}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-1+2x-x^2}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-(x^2-2x+1)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1-(x^2-2.x.1+1^2)}}}\)
\(=\int_{0}^{1}{\frac{dx}{\sqrt{1^2-(x-1)^2}}}\)
\(=\left[\sin^{-1}{\left(\frac{x-1}{1}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\sin^{-1}{(x-1)}\right]_{0}^{1}\)
\(=\sin^{-1}{(1-1)}-\sin^{-1}{(0-1)}\)
\(=\sin^{-1}{(0)}-\sin^{-1}{(-1)}\)
\(=\sin^{-1}{\sin{0}}+\sin^{-1}{(1)}\)
\(=0+\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=\frac{\pi}{2}\)
\(Q.2.(Lv)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\sqrt{4-\sin^2{x}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{6}\)
উত্তরঃ \(\frac{\pi}{6}\)
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=\sin{x}\)
\(\Rightarrow t=\sin{(0)}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\sqrt{4-\sin^2{x}}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{\sqrt{4-\sin^2{x}}}.\cos{x}dx}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{4-t^2}}dt}\)
\(=\int_{0}^{1}{\frac{1}{\sqrt{2^2-t^2}}dt}\)
\(=\left[\sin^{-1}{\left(\frac{t}{2}\right)}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\sin^{-1}{\left(\frac{1}{2}\right)}-\sin^{-1}{\left(\frac{0}{2}\right)}\)
\(=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}-\sin^{-1}{(0)}\)
\(=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}-\sin^{-1}{\sin{0}}\)
\(=\frac{\pi}{6}-0\)
\(=\frac{\pi}{6}\)
\(Q.2.(Lvi)\) \(\int_{2}^{3}{\frac{dx}{9x^2-16}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
উত্তরঃ \(\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
সমাধানঃ
\(\int_{2}^{3}{\frac{dx}{9x^2-16}}\)
\(=\int_{2}^{3}{\frac{dx}{9\left(x^2-\frac{16}{9}\right)}}\)
\(=\frac{1}{9}\int_{2}^{3}{\frac{dx}{x^2-\frac{16}{9}}}\)
\(=\frac{1}{9}\int_{2}^{3}{\frac{dx}{x^2-\left(\frac{4}{3}\right)^2}}\)
\(=\frac{1}{9}\left[\frac{1}{2.\frac{4}{3}}\ln{\left|\frac{x-\frac{4}{3}}{x+\frac{4}{3}}\right|}\right]_{2}^{3}\) ➜ \(\because \int{\frac{dx}{x^2-a^2}}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\)
\(=\frac{1}{9}\left[\frac{1}{\frac{8}{3}}\ln{\left|\frac{3x-4}{3x+4}\right|}\right]_{2}^{3}\) ➜ লব ও হরের সহিত \(3\) গুণ করে।
\(=\frac{1}{9}\times{\frac{3}{8}}\left[\ln{\left|\frac{3x-4}{3x+4}\right|}\right]_{2}^{3}\)
\(=\frac{1}{24}\left[\ln{\left|\frac{3.3-4}{3.3+4}\right|}-\ln{\left|\frac{3.2-4}{3.2+4}\right|}\right]\)
\(=\frac{1}{24}\left[\ln{\left|\frac{9-4}{9+4}\right|}-\ln{\left|\frac{6-4}{6+4}\right|}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\right)}-\ln{\left(\frac{2}{10}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\right)}-\ln{\left(\frac{1}{5}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{\frac{5}{13}}{\frac{1}{5}}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\times{\frac{5}{1}}\right)}\right]\)
\(=\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
\(=\int_{2}^{3}{\frac{dx}{9\left(x^2-\frac{16}{9}\right)}}\)
\(=\frac{1}{9}\int_{2}^{3}{\frac{dx}{x^2-\frac{16}{9}}}\)
\(=\frac{1}{9}\int_{2}^{3}{\frac{dx}{x^2-\left(\frac{4}{3}\right)^2}}\)
\(=\frac{1}{9}\left[\frac{1}{2.\frac{4}{3}}\ln{\left|\frac{x-\frac{4}{3}}{x+\frac{4}{3}}\right|}\right]_{2}^{3}\) ➜ \(\because \int{\frac{dx}{x^2-a^2}}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\)
\(=\frac{1}{9}\left[\frac{1}{\frac{8}{3}}\ln{\left|\frac{3x-4}{3x+4}\right|}\right]_{2}^{3}\) ➜ লব ও হরের সহিত \(3\) গুণ করে।
\(=\frac{1}{9}\times{\frac{3}{8}}\left[\ln{\left|\frac{3x-4}{3x+4}\right|}\right]_{2}^{3}\)
\(=\frac{1}{24}\left[\ln{\left|\frac{3.3-4}{3.3+4}\right|}-\ln{\left|\frac{3.2-4}{3.2+4}\right|}\right]\)
\(=\frac{1}{24}\left[\ln{\left|\frac{9-4}{9+4}\right|}-\ln{\left|\frac{6-4}{6+4}\right|}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\right)}-\ln{\left(\frac{2}{10}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\right)}-\ln{\left(\frac{1}{5}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{\frac{5}{13}}{\frac{1}{5}}\right)}\right]\)
\(=\frac{1}{24}\left[\ln{\left(\frac{5}{13}\times{\frac{5}{1}}\right)}\right]\)
\(=\frac{1}{24}\ln{\left(\frac{25}{13}\right)}\)
অনুশীলনী \(10.G / Q.3\)-এর বর্ণনামূলক প্রশ্নসমুহ
নিচের যোগজগুলির মান নির্ণয় করঃ
\(Q.3.(i)\) \(\int_{0}^{4}{\sqrt{16-x^2}dx}\)
উত্তরঃ \(4\pi\)
[ সিঃ২০১৩,২০১১,২০০৯,২০০৬,২০০৩; কুঃ২০১১,২০০৩; রাঃ২০০৯,২০০৬,২০০৩; যঃ২০০৮; বঃ২০০৭; চঃ২০০৬ ]
\(Q.3.(ii)\) \(\int_{0}^{5}{\sqrt{25-x^2}dx}\)
উত্তরঃ \(\frac{25\pi}{4}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(iii)\) \(\int_{-a}^{a}{\sqrt{a^2-x^2}dx}\)
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(iv)\) \(\int_{-1}^{1}{x^2\sqrt{4-x^2}dx}\)
উত্তরঃ \(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)
[ ঢাঃ২০০৮; চঃ২০০৩; যঃ২০০৯,২০০৫;বঃ২০০৮ ]
\(Q.3.(v)\) \(\int_{0}^{2a}{\sqrt{2ax-x^2}dx}\)
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(vi)\) \(\int_{-1}^{1}{\sqrt{\frac{1-x}{1+x}}dx}\)
উত্তরঃ \(\pi\)
\(Q.3.(vii)\) \(\int_{0}^{a}{\sqrt{\frac{a+x}{a-x}}dx}\)
উত্তরঃ \(\frac{a(\pi+2)}{2}\)
\(Q.3.(viii)\) \(\int_{1}^{4}{\ln{|x|}dx}\)
উত্তরঃ \(8\ln{(2)}-3\)
[ কুঃ২০০৪ ]
\(Q.3.(ix)\) \(\int_{0}^{1}{\ln{(x^2+1)}dx}\)
উত্তরঃ \(\ln{2}+\frac{\pi}{2}-2\)
[ ঢাঃ২০০৭ ]
\(Q.3.(x)\) \(\int_{1}^{4}{\frac{\ln{|x|}}{\sqrt{x}}dx}\)
উত্তরঃ \(\ln{2}-4\)
[ ঢাঃ২০১২ ]
\(Q.3.(xi)\) \(\int_{1}^{e}{\ln{|x|}dx}\)
উত্তরঃ \(1\)
[ বুয়েটঃ২০০৫-২০০৬ ]
\(Q.3.(xii)\) \(\int_{1}^{\sqrt{e}}{x\ln{|x|}dx}\)
উত্তরঃ \(\frac{1}{4}\)
[ রাঃ২০১৪; যঃ২০০৪ ]
\(Q.3.(xiii)\) \(\int_{2}^{4}{\ln{|2x|}dx}\)
উত্তরঃ \(8\ln{2}-2\)
[ বঃ২০০৯ ]
\(Q.3.(xiv)\) \(\int_{0}^{1}{\ln{|x|}dx}\)
উত্তরঃ \(-1\)
[ রাঃ২০১৭ ]
\(Q.3.(xv)\) \(\int_{0}^{3}{\frac{xe^x}{3(x+1)^2}dx}\)
উত্তরঃ \(\frac{1}{3}\left(\frac{e^2}{4}-1\right)\)
[ রাঃ২০১৭ ]
\(Q.3.(xvi)\) \(\int_{0}^{1}{\frac{xe^x}{(x+1)^2}dx}\)
উত্তরঃ \(\frac{e}{2}-1\)
[ যঃ২০১৭ ]
\(Q.3.(xvii)\) \(\int_{0}^{1}{2x^3e^{-x^2}dx}\)
উত্তরঃ \(1-\frac{2}{e}\)
\(Q.3.(xviii)\) \(\int_{0}^{1}{xe^{-3x}dx}\)
উত্তরঃ \(\frac{1}{9}(1-4e^{-3})\)
[ দিঃ২০১০; সিঃ২০০৩]
\(Q.3.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{x^2\cos{x}dx}\)
উত্তরঃ \(\frac{\pi^2}{4}-2\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৪ ]
\(Q.3.(xx)\) \(\int_{0}^{\pi}{x^2\sin{x}dx}\)
উত্তরঃ \(\pi^2+4\)
\(Q.3.(xxi)\) \(\int_{0}^{1}{\sin^{-1}{x}dx}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
\(Q.3.(xxii)\) \(\int_{1}^{\sqrt{3}}{x\tan^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{12}(5\pi-6\sqrt{3}+6)\)
[ বঃ২০১৩; রাঃ,চঃ২০১২,২০০৮; দিঃ২০১২; যঃ২০১১ ]
\(Q.3.(xxiii)\) \(\int_{0}^{2}{(x-2)\tan^{-1}{(x-2)}dx}\)
উত্তরঃ \(\frac{5}{2}\tan^{-1}{(2)}-1\)
[ সিঃ২০১৭ ]
\(Q.3.(xxiv)\) \(\int_{1}^{\sqrt{3}}{x\cot^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{12}(\pi+6\sqrt{3}-6)\)
[ বুয়েটঃ২০০৯ ]
\(Q.3.(xxv)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{dx}{1+\sin{x}-\cos{x}}}\)
উত্তরঃ \(\ln{\left(\frac{\sqrt{3}+1}{2}\right)}\)
[ বুয়েটঃ২০১১-২০১২ ]
\(Q.3.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\sin{x}+\cos{x}}}\)
উত্তরঃ \(-\sqrt{2}\ln{(\sqrt{2}-1)}\)
[ কুয়েটঃ২০১৩-২০১৪; বুটেক্সঃ২০০১-২০০২ ]
\(Q.3.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{x}\sqrt{\cos{x}}dx}\)
উত্তরঃ \(\frac{8}{21}\)
[ চঃ২০১৩,২০০৯; বঃ,সিঃ২০১৩; যঃ২০১০; রাঃ২০০৮ ]
\(Q.3.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}\sqrt{\sin{x}}dx}\)
উত্তরঃ \(\frac{8}{21}\)
[ ঢাঃ২০১৪,২০০৬,২০০৪; রাঃ২০১৪; কুঃ২০১৩,২০১১,২০০৭; দিঃ২০১১; চঃ২০০৭,২০০৩; সিঃ২০১১,২০০৮,২০০৪; যঃ২০১২; বঃ২০১১,২০০৯,২০০৭,২০০৫ ]
\(Q.3.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{\theta}\sqrt[3]{\sin{\theta}}d\theta}\)
উত্তরঃ \(\frac{9}{20}\)
[ বঃ২০১৭ ]
\(Q.3.(xxx)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos^3{\theta}}{\sqrt{\sin{\theta}}}d\theta}\)
উত্তরঃ \(\frac{8}{5}\)
[ রাঃ২০১২; চঃ,বঃ২০১০ ]
উত্তরঃ \(4\pi\)
[ সিঃ২০১৩,২০১১,২০০৯,২০০৬,২০০৩; কুঃ২০১১,২০০৩; রাঃ২০০৯,২০০৬,২০০৩; যঃ২০০৮; বঃ২০০৭; চঃ২০০৬ ]
\(Q.3.(ii)\) \(\int_{0}^{5}{\sqrt{25-x^2}dx}\)
উত্তরঃ \(\frac{25\pi}{4}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(iii)\) \(\int_{-a}^{a}{\sqrt{a^2-x^2}dx}\)
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(iv)\) \(\int_{-1}^{1}{x^2\sqrt{4-x^2}dx}\)
উত্তরঃ \(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)
[ ঢাঃ২০০৮; চঃ২০০৩; যঃ২০০৯,২০০৫;বঃ২০০৮ ]
\(Q.3.(v)\) \(\int_{0}^{2a}{\sqrt{2ax-x^2}dx}\)
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
\(Q.3.(vi)\) \(\int_{-1}^{1}{\sqrt{\frac{1-x}{1+x}}dx}\)
উত্তরঃ \(\pi\)
\(Q.3.(vii)\) \(\int_{0}^{a}{\sqrt{\frac{a+x}{a-x}}dx}\)
উত্তরঃ \(\frac{a(\pi+2)}{2}\)
\(Q.3.(viii)\) \(\int_{1}^{4}{\ln{|x|}dx}\)
উত্তরঃ \(8\ln{(2)}-3\)
[ কুঃ২০০৪ ]
\(Q.3.(ix)\) \(\int_{0}^{1}{\ln{(x^2+1)}dx}\)
উত্তরঃ \(\ln{2}+\frac{\pi}{2}-2\)
[ ঢাঃ২০০৭ ]
\(Q.3.(x)\) \(\int_{1}^{4}{\frac{\ln{|x|}}{\sqrt{x}}dx}\)
উত্তরঃ \(\ln{2}-4\)
[ ঢাঃ২০১২ ]
\(Q.3.(xi)\) \(\int_{1}^{e}{\ln{|x|}dx}\)
উত্তরঃ \(1\)
[ বুয়েটঃ২০০৫-২০০৬ ]
\(Q.3.(xii)\) \(\int_{1}^{\sqrt{e}}{x\ln{|x|}dx}\)
উত্তরঃ \(\frac{1}{4}\)
[ রাঃ২০১৪; যঃ২০০৪ ]
\(Q.3.(xiii)\) \(\int_{2}^{4}{\ln{|2x|}dx}\)
উত্তরঃ \(8\ln{2}-2\)
[ বঃ২০০৯ ]
\(Q.3.(xiv)\) \(\int_{0}^{1}{\ln{|x|}dx}\)
উত্তরঃ \(-1\)
[ রাঃ২০১৭ ]
\(Q.3.(xv)\) \(\int_{0}^{3}{\frac{xe^x}{3(x+1)^2}dx}\)
উত্তরঃ \(\frac{1}{3}\left(\frac{e^2}{4}-1\right)\)
[ রাঃ২০১৭ ]
\(Q.3.(xvi)\) \(\int_{0}^{1}{\frac{xe^x}{(x+1)^2}dx}\)
উত্তরঃ \(\frac{e}{2}-1\)
[ যঃ২০১৭ ]
\(Q.3.(xvii)\) \(\int_{0}^{1}{2x^3e^{-x^2}dx}\)
উত্তরঃ \(1-\frac{2}{e}\)
\(Q.3.(xviii)\) \(\int_{0}^{1}{xe^{-3x}dx}\)
উত্তরঃ \(\frac{1}{9}(1-4e^{-3})\)
[ দিঃ২০১০; সিঃ২০০৩]
\(Q.3.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{x^2\cos{x}dx}\)
উত্তরঃ \(\frac{\pi^2}{4}-2\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৪ ]
\(Q.3.(xx)\) \(\int_{0}^{\pi}{x^2\sin{x}dx}\)
উত্তরঃ \(\pi^2+4\)
\(Q.3.(xxi)\) \(\int_{0}^{1}{\sin^{-1}{x}dx}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
\(Q.3.(xxii)\) \(\int_{1}^{\sqrt{3}}{x\tan^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{12}(5\pi-6\sqrt{3}+6)\)
[ বঃ২০১৩; রাঃ,চঃ২০১২,২০০৮; দিঃ২০১২; যঃ২০১১ ]
\(Q.3.(xxiii)\) \(\int_{0}^{2}{(x-2)\tan^{-1}{(x-2)}dx}\)
উত্তরঃ \(\frac{5}{2}\tan^{-1}{(2)}-1\)
[ সিঃ২০১৭ ]
\(Q.3.(xxiv)\) \(\int_{1}^{\sqrt{3}}{x\cot^{-1}{x}dx}\)
উত্তরঃ \(\frac{1}{12}(\pi+6\sqrt{3}-6)\)
[ বুয়েটঃ২০০৯ ]
\(Q.3.(xxv)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{dx}{1+\sin{x}-\cos{x}}}\)
উত্তরঃ \(\ln{\left(\frac{\sqrt{3}+1}{2}\right)}\)
[ বুয়েটঃ২০১১-২০১২ ]
\(Q.3.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\sin{x}+\cos{x}}}\)
উত্তরঃ \(-\sqrt{2}\ln{(\sqrt{2}-1)}\)
[ কুয়েটঃ২০১৩-২০১৪; বুটেক্সঃ২০০১-২০০২ ]
\(Q.3.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{x}\sqrt{\cos{x}}dx}\)
উত্তরঃ \(\frac{8}{21}\)
[ চঃ২০১৩,২০০৯; বঃ,সিঃ২০১৩; যঃ২০১০; রাঃ২০০৮ ]
\(Q.3.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}\sqrt{\sin{x}}dx}\)
উত্তরঃ \(\frac{8}{21}\)
[ ঢাঃ২০১৪,২০০৬,২০০৪; রাঃ২০১৪; কুঃ২০১৩,২০১১,২০০৭; দিঃ২০১১; চঃ২০০৭,২০০৩; সিঃ২০১১,২০০৮,২০০৪; যঃ২০১২; বঃ২০১১,২০০৯,২০০৭,২০০৫ ]
\(Q.3.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{\theta}\sqrt[3]{\sin{\theta}}d\theta}\)
উত্তরঃ \(\frac{9}{20}\)
[ বঃ২০১৭ ]
\(Q.3.(xxx)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos^3{\theta}}{\sqrt{\sin{\theta}}}d\theta}\)
উত্তরঃ \(\frac{8}{5}\)
[ রাঃ২০১২; চঃ,বঃ২০১০ ]
\(Q.3.(xxxi)\) \(\int_{0}^{\pi}{\frac{xdx}{1+\sin{x}}}\)
উত্তরঃ \(\pi\)
\(Q.3.(xxxii)\) \(\int_{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}}\)
উত্তরঃ \(\frac{1}{2}\ln{\left(\frac{5}{3}\right)}\)
\(Q.3.(xxxiii)\) \(\int_{\frac{1}{2}}^{1}{\frac{dx}{x\sqrt{4x^2-1}}}\)
উত্তরঃ \(\frac{\pi}{3}\)
[ বুয়েটঃ২০০৪ ]
\(Q.3.(xxxiv)\) \(\int_{1}^{2}{\frac{dx}{x^2\sqrt{4-x^2}}}\)
উত্তরঃ \(\frac{\sqrt{3}}{4}\)
[ বুয়েটঃ২০০৪ ]
\(Q.3.(xxxv)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{dx}{1-\tan^2{x}}}\)
উত্তরঃ \(\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
[ বুয়েটঃ২০০৮; বঃ২০১৫ ]
\(Q.3.(xxxvi)\) \(\int_{0}^{\infty}{e^{-2x}\cos{4x}dx}\)
উত্তরঃ \(\frac{1}{10}\)
[ কুয়েটঃ২০১৩-২০১৪ ]
\(Q.3.(xxxvii)\) \(\int_{2}^{3}{\frac{dx}{2(x-1)\sqrt{x^2-2x}}}\)
উত্তরঃ \(\frac{\pi}{6}\)
[ বুয়েটঃ২০০১-২০০২,২০০৩-২০০৪ ]
\(Q.3.(xxxviii)\) \(\int_{2}^{e}{\left\{\frac{1}{\ln{|x|}}-\frac{1}{(\ln{|x|})^2}\right\}}\)
উত্তরঃ \(e-\frac{2}{\ln{2}}\)
[ বুয়েটঃ২০০৩-২০০৪; বিআইটিঃ১৯৯৮-১৯৯৫ ]
\(Q.3.(xxxix)\) \(\int_{0}^{a}{\frac{a^2-x^2}{(a^2+x^2)^2}dx}\)
উত্তরঃ \(\frac{1}{2a}\)
[ বুয়েটঃ২০০০-২০০১ ]
\(Q.3.(xL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})(2+\sin{x})}dx}\)
উত্তরঃ \(\ln{\left(\frac{4}{3}\right)}\)
[ কুয়েটঃ২০০৯-২০১০; বিআইটিঃ১৯৯৭-১৯৯৮ ]
\(Q.3.(xLi)\) \(\int_{0}^{\pi}{x\sin^2{x}dx}\)
উত্তরঃ \(\frac{\pi^2}{4}\)
[ বুয়েটঃ২০০৫ ]
\(Q.3.(xLii)\) \(\int_{0}^{a}{\sqrt{a^2-x^2}dx}\)
উত্তরঃ \(\frac{1}{4}\pi{a^2}\)
[ ঢাঃ২০১১; চঃ২০১২ ]
\(Q.3.(xLiii)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin{x}+\cos{x})^2dx}\)
উত্তরঃ \(\pi\)
\(Q.3.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
উত্তরঃ \(2\)
[ চঃ২০০৪ ]
\(Q.3.(xLv)\) \(\int_{\frac{\pi}{2}}^{\pi}{(1+\sin{2\theta})d\theta}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
\(Q.3.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})^3}dx}\)
উত্তরঃ \(\frac{3}{8}\)
[ দিঃ২০১৪ ]
\(Q.3.(xLvii)\) \(\int_{0}^{\frac{1}{2}}{\frac{dx}{(1-2x^2)\sqrt{1-x^2}}}\)
উত্তরঃ \(\frac{1}{2}\ln{(2+\sqrt{3})}\)
\(Q.3.(xLviii)\) \(\int_{0}^{\frac{\pi}{4}}{\sec{x}\sqrt{\frac{1-\sin{x}}{1+\sin{x}}}dx}\)
উত্তরঃ \(2-\sqrt{2}\)
\(Q.3.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.3.(L)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\tan{x}}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.3.(Li)\) \(\int_{8}^{27}{\frac{dx}{x-x^{\frac{1}{3}}}}\)
উত্তরঃ \(\frac{3}{2}\ln{\left(\frac{8}{3}\right)}\)
[ যঃ২০১৪ ]
\(Q.3.(Lii)\) \(\int_{0}^{4}{f(x)dx}=5\) হয় তবে \(\int_{1}^{5}{f(x-1)dx}\) এর মাণ কত?
উত্তরঃ \(5\)
[ ঢাঃবিঃ২০০৭-২০০৮ ]
\(Q.3.(Liii)\) \(\int_{0}^{1}{\frac{1-x}{1+x}dx}\)
উত্তরঃ \(\ln{\left(\frac{4}{e}\right)}\)
[ যঃ২০০৪; বঃ২০০৪,২০০৩; মাঃ২০১০ ]
\(Q.3.(Liv)\) \(\int_{0}^{16}{\frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}dx}\)
উত্তরঃ \(4\left(\frac{2}{3}+\tan^{-1}{(2)}\right)\)
\(Q.3.(Lv)\) \(\int_{1}^{3}{\frac{x-3}{x^3+x^2}dx}\)
উত্তরঃ \(4\ln{\left(\frac{3}{2}\right)}-2\)
\(Q.3.(Lvi)\) \(\int_{0}^{2}{\frac{x^4+1}{x^2+1}dx}\)
উত্তরঃ \(\frac{2}{3}+2\tan^{-1}{(2)}\)
\(Q.3.(Lvii)\) \(\int_{0}^{3}{\frac{dx}{(2+x^2)^{\frac{3}{2}}}}\)
উত্তরঃ \(\frac{3\sqrt{11}}{22}\)
\(Q.3.(Lviii)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a^2\sin^2{\theta}+b^2\cos^2{\theta}}}=\frac{\pi}{2ab}\)
[ রাঃ ২০১১ ]
\(Q.3.(Lix)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{\theta}+b\sin^2{\theta})d\theta}=\frac{1}{4}(a+b)\pi\)
[ চঃ ২০০৩ ]
\(Q.3.(Lx)\) \(\int_{0}^{\sqrt{2}}{\frac{x^2}{(4-x^2)^{\frac{3}{2}}}dx}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
\(Q.3.(Lxi)\) \(\int_{1}^{15}{\frac{x+2}{(x+1)(x+3)}dx}\)
উত্তরঃ \(\ln{(6)}\)
\(Q.3.(Lxii)\) \(\int_{0}^{\frac{a}{2}}{\frac{1}{a^2-x^2}dx}\)
উত্তরঃ \(\frac{1}{2a}\ln{3}\)
উত্তরঃ \(\pi\)
\(Q.3.(xxxii)\) \(\int_{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}}\)
উত্তরঃ \(\frac{1}{2}\ln{\left(\frac{5}{3}\right)}\)
\(Q.3.(xxxiii)\) \(\int_{\frac{1}{2}}^{1}{\frac{dx}{x\sqrt{4x^2-1}}}\)
উত্তরঃ \(\frac{\pi}{3}\)
[ বুয়েটঃ২০০৪ ]
\(Q.3.(xxxiv)\) \(\int_{1}^{2}{\frac{dx}{x^2\sqrt{4-x^2}}}\)
উত্তরঃ \(\frac{\sqrt{3}}{4}\)
[ বুয়েটঃ২০০৪ ]
\(Q.3.(xxxv)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{dx}{1-\tan^2{x}}}\)
উত্তরঃ \(\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
[ বুয়েটঃ২০০৮; বঃ২০১৫ ]
\(Q.3.(xxxvi)\) \(\int_{0}^{\infty}{e^{-2x}\cos{4x}dx}\)
উত্তরঃ \(\frac{1}{10}\)
[ কুয়েটঃ২০১৩-২০১৪ ]
\(Q.3.(xxxvii)\) \(\int_{2}^{3}{\frac{dx}{2(x-1)\sqrt{x^2-2x}}}\)
উত্তরঃ \(\frac{\pi}{6}\)
[ বুয়েটঃ২০০১-২০০২,২০০৩-২০০৪ ]
\(Q.3.(xxxviii)\) \(\int_{2}^{e}{\left\{\frac{1}{\ln{|x|}}-\frac{1}{(\ln{|x|})^2}\right\}}\)
উত্তরঃ \(e-\frac{2}{\ln{2}}\)
[ বুয়েটঃ২০০৩-২০০৪; বিআইটিঃ১৯৯৮-১৯৯৫ ]
\(Q.3.(xxxix)\) \(\int_{0}^{a}{\frac{a^2-x^2}{(a^2+x^2)^2}dx}\)
উত্তরঃ \(\frac{1}{2a}\)
[ বুয়েটঃ২০০০-২০০১ ]
\(Q.3.(xL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})(2+\sin{x})}dx}\)
উত্তরঃ \(\ln{\left(\frac{4}{3}\right)}\)
[ কুয়েটঃ২০০৯-২০১০; বিআইটিঃ১৯৯৭-১৯৯৮ ]
\(Q.3.(xLi)\) \(\int_{0}^{\pi}{x\sin^2{x}dx}\)
উত্তরঃ \(\frac{\pi^2}{4}\)
[ বুয়েটঃ২০০৫ ]
\(Q.3.(xLii)\) \(\int_{0}^{a}{\sqrt{a^2-x^2}dx}\)
উত্তরঃ \(\frac{1}{4}\pi{a^2}\)
[ ঢাঃ২০১১; চঃ২০১২ ]
\(Q.3.(xLiii)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin{x}+\cos{x})^2dx}\)
উত্তরঃ \(\pi\)
\(Q.3.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
উত্তরঃ \(2\)
[ চঃ২০০৪ ]
\(Q.3.(xLv)\) \(\int_{\frac{\pi}{2}}^{\pi}{(1+\sin{2\theta})d\theta}\)
উত্তরঃ \(\frac{\pi}{2}-1\)
\(Q.3.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})^3}dx}\)
উত্তরঃ \(\frac{3}{8}\)
[ দিঃ২০১৪ ]
\(Q.3.(xLvii)\) \(\int_{0}^{\frac{1}{2}}{\frac{dx}{(1-2x^2)\sqrt{1-x^2}}}\)
উত্তরঃ \(\frac{1}{2}\ln{(2+\sqrt{3})}\)
\(Q.3.(xLviii)\) \(\int_{0}^{\frac{\pi}{4}}{\sec{x}\sqrt{\frac{1-\sin{x}}{1+\sin{x}}}dx}\)
উত্তরঃ \(2-\sqrt{2}\)
\(Q.3.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.3.(L)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\tan{x}}}\)
উত্তরঃ \(\frac{\pi}{4}\)
\(Q.3.(Li)\) \(\int_{8}^{27}{\frac{dx}{x-x^{\frac{1}{3}}}}\)
উত্তরঃ \(\frac{3}{2}\ln{\left(\frac{8}{3}\right)}\)
[ যঃ২০১৪ ]
\(Q.3.(Lii)\) \(\int_{0}^{4}{f(x)dx}=5\) হয় তবে \(\int_{1}^{5}{f(x-1)dx}\) এর মাণ কত?
উত্তরঃ \(5\)
[ ঢাঃবিঃ২০০৭-২০০৮ ]
\(Q.3.(Liii)\) \(\int_{0}^{1}{\frac{1-x}{1+x}dx}\)
উত্তরঃ \(\ln{\left(\frac{4}{e}\right)}\)
[ যঃ২০০৪; বঃ২০০৪,২০০৩; মাঃ২০১০ ]
\(Q.3.(Liv)\) \(\int_{0}^{16}{\frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}dx}\)
উত্তরঃ \(4\left(\frac{2}{3}+\tan^{-1}{(2)}\right)\)
\(Q.3.(Lv)\) \(\int_{1}^{3}{\frac{x-3}{x^3+x^2}dx}\)
উত্তরঃ \(4\ln{\left(\frac{3}{2}\right)}-2\)
\(Q.3.(Lvi)\) \(\int_{0}^{2}{\frac{x^4+1}{x^2+1}dx}\)
উত্তরঃ \(\frac{2}{3}+2\tan^{-1}{(2)}\)
\(Q.3.(Lvii)\) \(\int_{0}^{3}{\frac{dx}{(2+x^2)^{\frac{3}{2}}}}\)
উত্তরঃ \(\frac{3\sqrt{11}}{22}\)
\(Q.3.(Lviii)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a^2\sin^2{\theta}+b^2\cos^2{\theta}}}=\frac{\pi}{2ab}\)
[ রাঃ ২০১১ ]
\(Q.3.(Lix)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{\theta}+b\sin^2{\theta})d\theta}=\frac{1}{4}(a+b)\pi\)
[ চঃ ২০০৩ ]
\(Q.3.(Lx)\) \(\int_{0}^{\sqrt{2}}{\frac{x^2}{(4-x^2)^{\frac{3}{2}}}dx}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
\(Q.3.(Lxi)\) \(\int_{1}^{15}{\frac{x+2}{(x+1)(x+3)}dx}\)
উত্তরঃ \(\ln{(6)}\)
\(Q.3.(Lxii)\) \(\int_{0}^{\frac{a}{2}}{\frac{1}{a^2-x^2}dx}\)
উত্তরঃ \(\frac{1}{2a}\ln{3}\)
\(Q.3.(i)\) \(\int_{0}^{4}{\sqrt{16-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\pi\)
[ সিঃ২০১৩,২০১১,২০০৯,২০০৬,২০০৩; কুঃ২০১১,২০০৩; রাঃ২০০৯,২০০৬,২০০৩; যঃ২০০৮; বঃ২০০৭; চঃ২০০৬ ]
উত্তরঃ \(4\pi\)
[ সিঃ২০১৩,২০১১,২০০৯,২০০৬,২০০৩; কুঃ২০১১,২০০৩; রাঃ২০০৯,২০০৬,২০০৩; যঃ২০০৮; বঃ২০০৭; চঃ২০০৬ ]
সমাধানঃ
\(\int_{0}^{4}{\sqrt{16-x^2}dx}\)
\(=\int_{0}^{4}{\sqrt{4^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{4^2-x^2}}{2}+\frac{4^2}{2}\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{16-x^2}}{2}+\frac{16}{2}\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\)
\(=\left[\frac{x\sqrt{16-x^2}}{2}+8\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\)
\(=\frac{4\sqrt{16-4^2}}{2}+8\sin^{-1}{\left(\frac{4}{4}\right)}-\frac{0.\sqrt{16-0^2}}{2}-8\sin^{-1}{\left(\frac{0}{4}\right)}\)
\(=\frac{4\sqrt{0}}{2}+8\sin^{-1}{1}-\frac{0}{2}-8\sin^{-1}{0}\)
\(=\frac{4.0}{2}+8\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-8\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+8\times{\frac{\pi}{2}}-8.0\)
\(=0+4\pi-0\)
\(=4\pi\)
\(=\int_{0}^{4}{\sqrt{4^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{4^2-x^2}}{2}+\frac{4^2}{2}\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{16-x^2}}{2}+\frac{16}{2}\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\)
\(=\left[\frac{x\sqrt{16-x^2}}{2}+8\sin^{-1}{\left(\frac{x}{4}\right)}\right]_{0}^{4}\)
\(=\frac{4\sqrt{16-4^2}}{2}+8\sin^{-1}{\left(\frac{4}{4}\right)}-\frac{0.\sqrt{16-0^2}}{2}-8\sin^{-1}{\left(\frac{0}{4}\right)}\)
\(=\frac{4\sqrt{0}}{2}+8\sin^{-1}{1}-\frac{0}{2}-8\sin^{-1}{0}\)
\(=\frac{4.0}{2}+8\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-8\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+8\times{\frac{\pi}{2}}-8.0\)
\(=0+4\pi-0\)
\(=4\pi\)
\(Q.3.(ii)\) \(\int_{0}^{5}{\sqrt{25-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{25\pi}{4}\)
[ রাঃ২০০১১,২০০৩ ]
উত্তরঃ \(\frac{25\pi}{4}\)
[ রাঃ২০০১১,২০০৩ ]
সমাধানঃ
\(\int_{0}^{5}{\sqrt{25-x^2}dx}\)
\(=\int_{0}^{5}{\sqrt{5^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{5^2-x^2}}{2}+\frac{5^2}{2}\sin^{-1}{\left(\frac{x}{5}\right)}\right]_{0}^{5}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{25-x^2}}{2}+\frac{25}{2}\sin^{-1}{\left(\frac{x}{5}\right)}\right]_{0}^{5}\)
\(=\frac{5\sqrt{25-5^2}}{2}+\frac{25}{2}\sin^{-1}{\left(\frac{5}{5}\right)}-\frac{0.\sqrt{25-0^2}}{2}-\frac{25}{2}\sin^{-1}{\left(\frac{0}{5}\right)}\)
\(=\frac{5\sqrt{0}}{2}+\frac{25}{2}\sin^{-1}{1}-\frac{0}{2}-\frac{25}{2}\sin^{-1}{0}\)
\(=\frac{5.0}{2}+\frac{25}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{25}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{25}{2}.\frac{\pi}{2}-\frac{25}{2}.0\)
\(=0+\frac{25\pi}{4}-0\)
\(=\frac{25\pi}{4}\)
\(=\int_{0}^{5}{\sqrt{5^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{5^2-x^2}}{2}+\frac{5^2}{2}\sin^{-1}{\left(\frac{x}{5}\right)}\right]_{0}^{5}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\left[\frac{x\sqrt{25-x^2}}{2}+\frac{25}{2}\sin^{-1}{\left(\frac{x}{5}\right)}\right]_{0}^{5}\)
\(=\frac{5\sqrt{25-5^2}}{2}+\frac{25}{2}\sin^{-1}{\left(\frac{5}{5}\right)}-\frac{0.\sqrt{25-0^2}}{2}-\frac{25}{2}\sin^{-1}{\left(\frac{0}{5}\right)}\)
\(=\frac{5\sqrt{0}}{2}+\frac{25}{2}\sin^{-1}{1}-\frac{0}{2}-\frac{25}{2}\sin^{-1}{0}\)
\(=\frac{5.0}{2}+\frac{25}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{25}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{25}{2}.\frac{\pi}{2}-\frac{25}{2}.0\)
\(=0+\frac{25\pi}{4}-0\)
\(=\frac{25\pi}{4}\)
\(Q.3.(iii)\) \(\int_{-a}^{a}{\sqrt{a^2-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
সমাধানঃ
এখানে,
\(f(x)=\sqrt{a^2-x^2}\)
\(\Rightarrow f(-x)=\sqrt{a^2-(-x)^2}\)
\(=\sqrt{a^2-x^2}\)
\(=f(x)\)
\(\therefore f(x)=f(-x)\)
\(\therefore f(x)\) একটি জোড় ফাংশন।
\(\int_{-a}^{a}{\sqrt{a^2-x^2}dx}\)\(f(x)=\sqrt{a^2-x^2}\)
\(\Rightarrow f(-x)=\sqrt{a^2-(-x)^2}\)
\(=\sqrt{a^2-x^2}\)
\(=f(x)\)
\(\therefore f(x)=f(-x)\)
\(\therefore f(x)\) একটি জোড় ফাংশন।
\(=2\int_{0}^{a}{\sqrt{a^2-x^2}dx}\) ➜ \(\because \int_{-a}^{a}{f(x)dx}=2\int_{0}^{a}{f(x)dx}\) যেখানে \(f(x)\) একটি জোড় ফাংশন।
\(=2\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=2\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\)
\(=2\left[\frac{a\sqrt{a^2-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{a}{a}\right)}-\frac{0.\sqrt{a^2-0^2}}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{0}{a}\right)}\right]\)
\(=2\left[\frac{a\sqrt{0}}{2}+\frac{a^2}{2}\sin^{-1}{1}-\frac{0}{2}-\frac{a^2}{2}\sin^{-1}{0}\right]\)
\(=2\left[\frac{a.0}{2}+\frac{a^2}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{a^2}{2}\sin^{-1}{\sin{0}}\right]\)
\(=2\left[\frac{0}{2}+\frac{a^2}{2}.\frac{\pi}{2}-\frac{a^2}{2}.0\right]\)
\(=2\left[0+\frac{a^2\pi}{4}-0\right]\)
\(=2\times{\frac{a^2\pi}{4}}\)
\(=\frac{a^2\pi}{2}\)
\(Q.3.(iv)\) \(\int_{-1}^{1}{x^2\sqrt{4-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)
[ ঢাঃ২০০৮; চঃ২০০৩; যঃ২০০৯,২০০৫;বঃ২০০৮ ]
উত্তরঃ \(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)
[ ঢাঃ২০০৮; চঃ২০০৩; যঃ২০০৯,২০০৫;বঃ২০০৮ ]
সমাধানঃ
এখানে,
\(f(x)=x^2\sqrt{4-x^2}\)
\(\Rightarrow f(-x)=(-x)^2\sqrt{4-(-x)^2}\)
\(=x^2\sqrt{4-x^2}\)
\(=f(x)\)
\(\therefore f(x)=f(-x)\)
\(\therefore f(x)\) একটি জোড় ফাংশন।
ধরি,
\(x=2\sin{\theta}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{\theta})\)
\(\Rightarrow 1=2\cos{\theta}\frac{d\theta}{dx}\)
\(\therefore dx=2\cos{\theta}d\theta\)
➜ \(\because 2\sin{\theta}=x\)
\(\Rightarrow \sin{\theta}=\frac{x}{2}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{x}{2}}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{0}{2}}\), যখন \(x=0\)
\(\Rightarrow \theta=\sin^{-1}{0}\)
\(\Rightarrow \theta=\sin^{-1}{\sin{0}}\)
\(\therefore \theta=0\)
আবার,
\(\theta=\sin^{-1}{\frac{x}{2}}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{1}{2}}\), যখন \(x=1\)
\(\Rightarrow \theta=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore \theta=\frac{\pi}{6}\)
\(\int_{-1}^{1}{x^2\sqrt{4-x^2}dx}\)\(f(x)=x^2\sqrt{4-x^2}\)
\(\Rightarrow f(-x)=(-x)^2\sqrt{4-(-x)^2}\)
\(=x^2\sqrt{4-x^2}\)
\(=f(x)\)
\(\therefore f(x)=f(-x)\)
\(\therefore f(x)\) একটি জোড় ফাংশন।
ধরি,
\(x=2\sin{\theta}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{\theta})\)
\(\Rightarrow 1=2\cos{\theta}\frac{d\theta}{dx}\)
\(\therefore dx=2\cos{\theta}d\theta\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(\frac{\pi}{6}\) |
\(\Rightarrow \theta=\sin^{-1}{\frac{x}{2}}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{0}{2}}\), যখন \(x=0\)
\(\Rightarrow \theta=\sin^{-1}{0}\)
\(\Rightarrow \theta=\sin^{-1}{\sin{0}}\)
\(\therefore \theta=0\)
আবার,
\(\theta=\sin^{-1}{\frac{x}{2}}\)
\(\Rightarrow \theta=\sin^{-1}{\frac{1}{2}}\), যখন \(x=1\)
\(\Rightarrow \theta=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore \theta=\frac{\pi}{6}\)
\(=2\int_{0}^{1}{x^2\sqrt{4-x^2}dx}\) ➜ \(\because \int_{-a}^{a}{f(x)dx}=2\int_{0}^{a}{f(x)dx}\) যেখানে \(f(x)\) একটি জোড় ফাংশন।
\(=2\int_{0}^{\frac{\pi}{6}}{4\sin^2{\theta}\sqrt{4-4\sin^2{\theta}}.2\cos{\theta}d\theta}\)
\(=2\int_{0}^{\frac{\pi}{6}}{8\sin^2{\theta}.2\sqrt{1-\sin^2{\theta}}.\cos{\theta}d\theta}\)
\(=2\int_{0}^{\frac{\pi}{6}}{16\sin^2{\theta}.\cos{\theta}.\cos{\theta}d\theta}\) ➜ \(\because \sqrt{1-\sin^2{A}}=\cos{A}\)
\(=32\int_{0}^{\frac{\pi}{6}}{\sin^2{\theta}\cos^2{\theta}d\theta}\)
\(=8\int_{0}^{\frac{\pi}{6}}{(2\sin{\theta}\cos{\theta})^2d\theta}\)
\(=8\int_{0}^{\frac{\pi}{6}}{(\sin{2\theta})^2d\theta}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(=8\int_{0}^{\frac{\pi}{6}}{\sin^2{2\theta}d\theta}\)
\(=4\int_{0}^{\frac{\pi}{6}}{2\sin^2{2\theta}d\theta}\)
\(=4\int_{0}^{\frac{\pi}{6}}{(1-\cos{4\theta})d\theta}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=4\int_{0}^{\frac{\pi}{6}}{1d\theta}-4\int_{0}^{\frac{\pi}{6}}{\cos{4\theta}d\theta}\)
\(=4\left[\theta\right]_{0}^{\frac{\pi}{6}}-4\left[\frac{1}{4}\sin{4\theta}\right]_{0}^{\frac{\pi}{6}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=4\left[\theta\right]_{0}^{\frac{\pi}{6}}-\left[\sin{4\theta}\right]_{0}^{\frac{\pi}{6}}\)
\(=4\left[\frac{\pi}{6}-0\right]-\left[\sin{4\left(\frac{\pi}{6}\right)}-\sin{4.0}\right]\)
\(=\frac{2\pi}{3}-\left[\sin{\left(\frac{2\pi}{3}\right)}-\sin{0}\right]\)
\(=\frac{2\pi}{3}-\left[\sin{\left(\pi-\frac{\pi}{3}\right)}-0\right]\)
\(=\frac{2\pi}{3}-\sin{\left(\frac{\pi}{3}\right)}\) ➜ \(\because \sin{\left(\pi-\frac{\pi}{3}\right)}=\sin{\left(\frac{\pi}{3}\right)}\)
\(=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)
\(Q.3.(v)\) \(\int_{0}^{2a}{\sqrt{2ax-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
উত্তরঃ \(\frac{a^2\pi}{2}\)
[ রাঃ২০০১১,২০০৩ ]
সমাধানঃ
\(\int_{0}^{2a}{\sqrt{2ax-x^2}dx}\)
\(=\int_{0}^{2a}{\sqrt{a^2-a^2+2ax-x^2}dx}\)
\(=\int_{0}^{2a}{\sqrt{a^2-(x^2-2ax+a^2)}dx}\)
\(=\int_{0}^{2a}{\sqrt{a^2-(x-a)^2}dx}\)
\(=\left[\frac{x\sqrt{a^2-(x-a)^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x-a}{a}\right)}\right]_{0}^{2a}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{2a\sqrt{a^2-(2a-a)^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{2a-a}{a}\right)}-\frac{0.\sqrt{a^2-(0-a)^2}}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{0-a}{a}\right)}\)
\(=\frac{2a\sqrt{a^2-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{a}{a}\right)}-\frac{0}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{-a}{a}\right)}\)
\(=\frac{2a\sqrt{0}}{2}+\frac{a^2}{2}\sin^{-1}{\left(1\right)}-0+\frac{a^2}{2}\sin^{-1}{\left(1\right)}\)
\(=\frac{2a.0}{2}+2\frac{a^2}{2}\sin^{-1}{\left(1\right)}\)
\(=\frac{0}{2}+a^2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=0+a^2\frac{\pi}{2}\)
\(=\frac{a^2\pi}{2}\)
\(=\int_{0}^{2a}{\sqrt{a^2-a^2+2ax-x^2}dx}\)
\(=\int_{0}^{2a}{\sqrt{a^2-(x^2-2ax+a^2)}dx}\)
\(=\int_{0}^{2a}{\sqrt{a^2-(x-a)^2}dx}\)
\(=\left[\frac{x\sqrt{a^2-(x-a)^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x-a}{a}\right)}\right]_{0}^{2a}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{2a\sqrt{a^2-(2a-a)^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{2a-a}{a}\right)}-\frac{0.\sqrt{a^2-(0-a)^2}}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{0-a}{a}\right)}\)
\(=\frac{2a\sqrt{a^2-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{a}{a}\right)}-\frac{0}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{-a}{a}\right)}\)
\(=\frac{2a\sqrt{0}}{2}+\frac{a^2}{2}\sin^{-1}{\left(1\right)}-0+\frac{a^2}{2}\sin^{-1}{\left(1\right)}\)
\(=\frac{2a.0}{2}+2\frac{a^2}{2}\sin^{-1}{\left(1\right)}\)
\(=\frac{0}{2}+a^2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=0+a^2\frac{\pi}{2}\)
\(=\frac{a^2\pi}{2}\)
\(Q.3.(vi)\) \(\int_{-1}^{1}{\sqrt{\frac{1-x}{1+x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi\)
উত্তরঃ \(\pi\)
সমাধানঃ
ধরি,
\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow x=-t\frac{dt}{dx}\)
\(\therefore xdx=-tdt\)
➜ \(\because t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-(-1)^2}\), যখন \(x=-1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(\int_{-1}^{1}{\sqrt{\frac{1-x}{1+x}}dx}\)\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow x=-t\frac{dt}{dx}\)
\(\therefore xdx=-tdt\)
| \(x\) | \(-1\) | \(1\) |
| \(t\) | \(0\) | \(0\) |
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(=\int_{-1}^{1}{\frac{\sqrt{1-x}}{\sqrt{1+x}}dx}\)
\(=\int_{-1}^{1}{\frac{\sqrt{1-x}\sqrt{1-x}}{\sqrt{1+x}\sqrt{1-x}}dx}\) ➜ লব ও হরের সহিত \(\sqrt{1-x}\) গুণ করে।
\(=\int_{-1}^{1}{\frac{(\sqrt{1-x})^2}{\sqrt{(1+x)(1-x)}}dx}\)
\(=\int_{-1}^{1}{\frac{1-x}{\sqrt{1-x^2}}dx}\)
\(=\int_{-1}^{1}{\left(\frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right)dx}\)
\(=\int_{-1}^{1}{\frac{1}{\sqrt{1-x^2}}dx}-\int_{-1}^{1}{\frac{1}{\sqrt{1-x^2}}xdx}\)
\(=\left[\sin^{-1}{x}\right]_{-1}^{1}-\int_{0}^{0}{\frac{1}{t}\times{-tdt}}\) ➜ \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\)
\(=\sin^{-1}{(1)}-\sin^{-1}{(-1)}+0\) ➜ \(\because \int_{0}^{0}{f(x)dx}=0\)
\(=\sin^{-1}{(1)}+\sin^{-1}{(1)}\)
\(=2\sin^{-1}{(1)}\)
\(=2\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(=2\times{\frac{\pi}{2}}\)
\(=\pi\)
\(Q.3.(vii)\) \(\int_{0}^{a}{\sqrt{\frac{a+x}{a-x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{a(\pi+2)}{2}\)
উত্তরঃ \(\frac{a(\pi+2)}{2}\)
সমাধানঃ
ধরি,
\(\sqrt{a^2-x^2}=t\)
\(\Rightarrow a^2-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(a^2-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow x=-t\frac{dt}{dx}\)
\(\therefore xdx=-tdt\)
➜ \(\because t=\sqrt{a^2-x^2}\)
\(\Rightarrow t=\sqrt{a^2-(0)^2}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{a^2-0}\)
\(\Rightarrow t=\sqrt{a^2}\)
\(\therefore t=a\)
আবার,
\(t=\sqrt{a^2-x^2}\)
\(\Rightarrow t=\sqrt{a^2-a^2}\), যখন \(x=a\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(\int_{0}^{a}{\sqrt{\frac{a+x}{a-x}}dx}\)\(\sqrt{a^2-x^2}=t\)
\(\Rightarrow a^2-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(a^2-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow x=-t\frac{dt}{dx}\)
\(\therefore xdx=-tdt\)
| \(x\) | \(0\) | \(a\) |
| \(t\) | \(a\) | \(0\) |
\(\Rightarrow t=\sqrt{a^2-0}\)
\(\Rightarrow t=\sqrt{a^2}\)
\(\therefore t=a\)
আবার,
\(t=\sqrt{a^2-x^2}\)
\(\Rightarrow t=\sqrt{a^2-a^2}\), যখন \(x=a\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(=\int_{0}^{a}{\frac{\sqrt{a+x}}{\sqrt{a-x}}dx}\)
\(=\int_{0}^{a}{\frac{\sqrt{a+x}\sqrt{a+x}}{\sqrt{a+x}\sqrt{a-x}}dx}\) ➜ লব ও হরের সহিত \(\sqrt{a+x}\) গুণ করে।
\(=\int_{0}^{a}{\frac{(\sqrt{a+x})^2}{\sqrt{(a+x)(a-x)}}dx}\)
\(=\int_{0}^{a}{\frac{a+x}{\sqrt{a^2-x^2}}dx}\)
\(=\int_{0}^{a}{\left(\frac{a}{\sqrt{a^2-x^2}}+\frac{x}{\sqrt{a^2-x^2}}\right)dx}\)
\(=a\int_{0}^{a}{\frac{1}{\sqrt{a^2-x^2}}dx}+\int_{0}^{a}{\frac{1}{\sqrt{a^2-x^2}}xdx}\)
\(=a\left[\sin^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}+\int_{a}^{0}{\frac{1}{t}\times{-tdt}}\) ➜ \(\because \int{\frac{1}{\sqrt{a^2-x^2}}dx}=\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=a\left[\sin^{-1}{\left(\frac{a}{a}\right)}-\sin^{-1}{\left(\frac{0}{a}\right)}\right]-\int_{a}^{0}{1dt}\)
\(=a\left[\sin^{-1}{(1)}-\sin^{-1}{(0)}\right]-\left[t\right]_{a}^{0}\) ➜ \(\because \int{1dx}=x\)
\(=a\left[\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-\sin^{-1}{\sin{(0)}}\right]-\left[0-a\right]\)
\(=a\left[\frac{\pi}{2}-0\right]+a\)
\(=\frac{a\pi}{2}+a\)
\(=\frac{a\pi+2a}{2}\)
\(=\frac{a(\pi+2)}{2}\)
\(Q.3.(viii)\) \(\int_{1}^{4}{\ln{|x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(8\ln{(2)}-3\)
[ কুঃ২০০৪ ]
উত্তরঃ \(8\ln{(2)}-3\)
[ কুঃ২০০৪ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{1}^{4}{\ln{|x|}dx}\)\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{4}{\ln{|x|}.1dx}\)
\(=\left[\ln{|x|}.x\right]_{1}^{4}-\int_{1}^{4}{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[\ln{(4)}.4-\ln{(1)}.1\right]-\int_{1}^{4}{\frac{1}{x}.xdx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=4\ln{(4)}-0.1-\int_{1}^{4}{1dx}\) ➜ \(\because \ln{(1)}=0\)
\(=4\ln{(2)^2}-\left[x\right]_{1}^{4}\) ➜ \(\because \int{1dx}=x\)
\(=4.2\ln{(2)}-\left[4-1\right]\)
\(=8\ln{(2)}-3\)
\(Q.3.(ix)\) \(\int_{0}^{1}{\ln{(x^2+1)}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{2}+\frac{\pi}{2}-2\)
[ ঢাঃ২০০৭ ]
উত্তরঃ \(\ln{2}+\frac{\pi}{2}-2\)
[ ঢাঃ২০০৭ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{(x^2+1)}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{(x^2+1)}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{\ln{(x^2+1)}dx}\)\(A=1, \ \ L=\ln{(x^2+1)}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{(x^2+1)}\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{1}{\ln{(x^2+1)}.1dx}\)
\(=\left[\ln{(x^2+1)}.x\right]_{0}^{1}-\int_{0}^{1}{\left\{\frac{d}{dx}\ln{(x^2+1)}\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[\ln{(1^2+1)}.1-\ln{(0^2+1)}.0\right]-\int_{0}^{1}{\frac{1}{x^2+1}.2x.xdx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=\ln{(1+1)}-0-2\int_{0}^{1}{\frac{x^2}{x^2+1}dx}\)
\(=\ln{(2)}-2\int_{0}^{1}{\frac{x^2+1-1}{x^2+1}dx}\)
\(=\ln{(2)}-2\int_{0}^{1}{\left(\frac{x^2+1}{x^2+1}-\frac{1}{x^2+1}\right)dx}\)
\(=\ln{(2)}-2\int_{0}^{1}{\left(1-\frac{1}{x^2+1}\right)dx}\)
\(=\ln{(2)}-2\int_{0}^{1}{1dx}+2\int_{0}^{1}{\frac{1}{1+x^2}dx}\)
\(=\ln{(2)}-2\left[x\right]_{0}^{1}+2\left[\tan^{-1}{x}\right]_{0}^{1}\) ➜ \(\because \int{1dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\ln{(2)}-2\left[1-0\right]+2\left[\tan^{-1}{(1)}-\tan^{-1}{(0)}\right]\)
\(=\ln{(2)}-2+2\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\ln{(2)}-2+2\left[\frac{\pi}{4}-0\right]\)
\(=\ln{(2)}-2+\frac{\pi}{2}\)
\(=\ln{(2)}+\frac{\pi}{2}-2\)
\(Q.3.(x)\) \(\int_{1}^{4}{\frac{\ln{|x|}}{\sqrt{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{2}-4\)
[ ঢাঃ২০১২ ]
উত্তরঃ \(\ln{2}-4\)
[ ঢাঃ২০১২ ]
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow x=t^2\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 1=2t\frac{dt}{dx}\)
\(\therefore dx=2tdt\)
➜ \(\because t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{1}\), যখন \(x=1\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{4}\), যখন \(x=4\)
\(\therefore t=2\)
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|t|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|t|}\) কে \(u\) ধরা হয়েছে।
\(\int_{1}^{4}{\frac{\ln{|x|}}{\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow x=t^2\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 1=2t\frac{dt}{dx}\)
\(\therefore dx=2tdt\)
| \(x\) | \(1\) | \(4\) |
| \(t\) | \(1\) | \(2\) |
\(\therefore t=1\)
আবার,
\(t=\sqrt{x}\)
\(\Rightarrow t=\sqrt{4}\), যখন \(x=4\)
\(\therefore t=2\)
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|t|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|t|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{4}{\frac{\ln{(\sqrt{x})^2}}{\sqrt{x}}dx}\)
\(=\int_{1}^{4}{\frac{2\ln{\sqrt{x}}}{\sqrt{x}}dx}\)
\(=2\int_{1}^{2}{\frac{\ln{|t|}}{t}\times{2tdt}}\)
\(=4\int_{1}^{2}{\ln{|t|}.1dt}\)
\(=4\left[\ln{|t|}.t\right]_{1}^{2}-4\int_{1}^{2}{\left\{\frac{d}{dt}(\ln{|t|})\int{1dt}\right\}dt}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=4\left[\ln{(2)}.2-\ln{(1)}.1\right]-4\int_{1}^{2}{\frac{1}{t}.tdt}\)
\(=4\left[2\ln{(2)}-0.1\right]-4\int_{1}^{2}{1dt}\)
\(=4\left[2\ln{(2)}-0\right]-4\left[t\right]_{1}^{2}\) ➜ \(\because \int{1dx}=x\)
\(=8\ln{(2)}-4\left[2-1\right]\)
\(=8\ln{(2)}-4.1\)
\(=8\ln{(2)}-4\)
\(Q.3.(xi)\) \(\int_{1}^{e}{\ln{|x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1\)
[ বুয়েটঃ২০০৫-২০০৬ ]
উত্তরঃ \(1\)
[ বুয়েটঃ২০০৫-২০০৬ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{1}^{e}{\ln{|x|}dx}\)\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{e}{\ln{|x|}.1dx}\)
\(=\left[\ln{|x|}.x\right]_{1}^{e}-\int_{1}^{e}{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[\ln{(e)}.e-\ln{(1)}.1\right]-\int_{1}^{e}{\frac{1}{x}.xdx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=1.e-0.1-\int_{1}^{e}{1dx}\) ➜ \(\because \ln{(e)}=1, \ln{(1)}=0\)
\(=e-0-\left[x\right]_{1}^{e}\) ➜ \(\because \int{1dx}=x\)
\(=e-\left[e-1\right]\)
\(=e-e+1\)
\(=1\)
\(Q.3.(xii)\) \(\int_{1}^{\sqrt{e}}{x\ln{|x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\)
[ রাঃ২০১৪; যঃ২০০৪ ]
উত্তরঃ \(\frac{1}{4}\)
[ রাঃ২০১৪; যঃ২০০৪ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=x, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{1}^{\sqrt{e}}{x\ln{|x|}dx}\)\(A=x, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{\sqrt{e}}{\ln{|x|}.xdx}\)
\(=\left[\ln{|x|}.\frac{x^2}{2}\right]_{1}^{\sqrt{e}}-\int_{1}^{\sqrt{e}}{\left\{\frac{d}{dx}(\ln{|x|})\int{xdx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[x^2\ln{|x|}\right]_{1}^{\sqrt{e}}-\int_{1}^{\sqrt{e}}{\frac{1}{x}.\frac{x^2}{2}dx}\) ➜ \(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}, \int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[(\sqrt{e})^2\ln{(\sqrt{e})}-1^2\ln{(1)}\right]-\frac{1}{2}\int_{1}^{\sqrt{e}}{xdx}\)
\(=\frac{1}{2}\left[e\ln{e^{\frac{1}{2}}}-1.0\right]-\frac{1}{2}\left[\frac{x^2}{2}\right]_{1}^{\sqrt{e}}\) ➜ \(\because \ln{(1)}=0, \int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[e.\frac{1}{2}\ln{e}-0\right]-\frac{1}{4}\left[x^2\right]_{1}^{\sqrt{e}}\)
\(=\frac{1}{2}\times{\frac{e}{2}}-\frac{1}{4}\left[(\sqrt{e})^2-1^2\right]\)
\(=\frac{e}{4}-\frac{1}{4}\left[e-1\right]\)
\(=\frac{e}{4}-\frac{e}{4}+\frac{1}{4}\)
\(=\frac{1}{4}\)
\(Q.3.(xiii)\) \(\int_{2}^{4}{\ln{|2x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(8\ln{2}-2\)
[ বঃ২০০৯ ]
উত্তরঃ \(8\ln{2}-2\)
[ বঃ২০০৯ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|2x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|2x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{2}^{4}{\ln{|2x|}dx}\)\(A=1, \ \ L=\ln{|2x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|2x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{2}^{4}{\ln{|2x|}.1dx}\)
\(=\left[\ln{|2x|}.x\right]_{2}^{4}-\int_{2}^{4}{\left\{\frac{d}{dx}(\ln{|2x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[x\ln{|2x|}\right]_{2}^{4}-\int_{2}^{4}{\frac{1}{2x}.2.xdx}\) ➜ \(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}, \int{1dx}=x\)
\(=4\ln{(2.4)}-2\ln{(2.2)}-\int_{2}^{4}{1dx}\)
\(=4\ln{(8)}-2\ln{(4)}-\left[x\right]_{2}^{4}\) ➜ \(\because \int{1dx}=x\)
\(=4\ln{(2)^3}-2\ln{(2)^2}-\left[4-2\right]\)
\(=4.3\ln{(2)}-2.2\ln{(2)}-2\)
\(=12\ln{(2)}-4\ln{(2)}-2\)
\(=8\ln{(2)}-2\)
\(Q.3.(xiv)\) \(\int_{0}^{1}{\ln{|x|}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(-1\)
[ রাঃ২০১৭ ]
উত্তরঃ \(-1\)
[ রাঃ২০১৭ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{\ln{|x|}dx}\)\(A=1, \ \ L=\ln{|x|}\)
যেহেতু L, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\ln{|x|}\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{1}{\ln{|x|}.1dx}\)
\(=\left[\ln{|x|}.x\right]_{0}^{1}-\int_{0}^{1}{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[\ln{(1)}.1-\ln{(0)}.0\right]-\int_{0}^{1}{\frac{1}{x}.xdx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=0.1-0-\int_{0}^{1}{1dx}\) ➜ \(\because \ln{(e)}=1, \ln{(1)}=0\)
\(=0-\left[x\right]_{0}^{1}\) ➜ \(\because \int{1dx}=x\)
\(=-\left[1-0\right]\)
\(=-1\)
\(Q.3.(xv)\) \(\int_{0}^{3}{\frac{xe^x}{3(x+1)^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\left(\frac{e^2}{4}-1\right)\)
[ রাঃ২০১৭ ]
উত্তরঃ \(\frac{1}{3}\left(\frac{e^2}{4}-1\right)\)
[ রাঃ২০১৭ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=\frac{1}{x+1}, \ \ E=e^x\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\frac{1}{x+1}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{3}{\frac{xe^x}{3(x+1)^2}dx}\)\(A=\frac{1}{x+1}, \ \ E=e^x\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\frac{1}{x+1}\) কে \(u\) ধরা হয়েছে।
\(=\frac{1}{3}\int_{0}^{3}{\frac{xe^x}{(x+1)^2}dx}\)
\(=\frac{1}{3}\int_{0}^{3}{\frac{e^x(x+1)-e^x}{(x+1)^2}dx}\)
\(=\frac{1}{3}\int_{0}^{3}{\left\{\frac{e^x(x+1)}{(x+1)^2}-\frac{e^x}{(x+1)^2}\right\}dx}\)
\(=\frac{1}{3}\int_{0}^{3}{\left\{\frac{e^x}{x+1}-\frac{e^x}{(x+1)^2}\right\}dx}\)
\(=\frac{1}{3}\int_{0}^{3}{\frac{e^x}{x+1}dx}-\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}\)
\(=\frac{1}{3}\int_{0}^{3}{\frac{1}{x+1}e^xdx}-\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}\)
\(=\frac{1}{3}\left[\frac{1}{x+1}.e^x\right]_{0}^{3}-\frac{1}{3}\int_{0}^{3}{\left\{\frac{d}{dx}\left(\frac{1}{x+1}\right)\int{e^xdx}\right\}dx}-\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{e^xdx}=e^x\)
\(=\frac{1}{3}\left[\frac{1}{3+1}.e^3-\frac{1}{0+1}.e^0\right]-\frac{1}{3}\int_{0}^{3}{\left\{-\frac{1}{(x+1)^2}\right\}e^xdx}-\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}\) ➜ \(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\), \(\int{e^xdx}=e^x\)
\(=\frac{1}{3}\left[\frac{1}{3+1}.e^3-\frac{1}{0+1}.e^0\right]+\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}-\frac{1}{3}\int_{0}^{3}{\frac{e^x}{(x+1)^2}dx}\)
\(=\frac{1}{3}\left[\frac{e^3}{4}-1.1\right]\)
\(=\frac{1}{3}\left(\frac{e^3}{4}-1\right)\)
\(Q.3.(xvi)\) \(\int_{0}^{1}{\frac{xe^x}{(x+1)^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{e}{2}-1\)
[ যঃ২০১৭ ]
উত্তরঃ \(\frac{e}{2}-1\)
[ যঃ২০১৭ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=\frac{1}{x+1}, \ \ E=e^x\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\frac{1}{x+1}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{\frac{xe^x}{(x+1)^2}dx}\)\(A=\frac{1}{x+1}, \ \ E=e^x\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(\frac{1}{x+1}\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{1}{\frac{xe^x}{(x+1)^2}dx}\)
\(=\int_{0}^{1}{\frac{e^x(x+1)-e^x}{(x+1)^2}dx}\)
\(=\int_{0}^{1}{\left\{\frac{e^x(x+1)}{(x+1)^2}-\frac{e^x}{(x+1)^2}\right\}dx}\)
\(=\int_{0}^{1}{\left\{\frac{e^x}{x+1}-\frac{e^x}{(x+1)^2}\right\}dx}\)
\(=\int_{0}^{1}{\frac{e^x}{x+1}dx}-\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}\)
\(=\int_{0}^{1}{\frac{1}{x+1}e^xdx}-\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}\)
\(=\left[\frac{1}{x+1}.e^x\right]_{0}^{1}-\int_{0}^{1}{\left\{\frac{d}{dx}\left(\frac{1}{x+1}\right)\int{e^xdx}\right\}dx}-\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{e^xdx}=e^x\)
\(=\frac{1}{1+1}.e^1-\frac{1}{0+1}.e^0-\int_{0}^{1}{\left\{-\frac{1}{(x+1)^2}\right\}e^xdx}-\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}\) ➜ \(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\), \(\int{e^xdx}=e^x\)
\(=\frac{1}{1+1}.e^1-\frac{1}{0+1}.e^0+\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}-\int_{0}^{1}{\frac{e^x}{(x+1)^2}dx}\)
\(=\frac{e}{2}-1.1\) ➜ \(\because e^0=1\)
\(=\frac{e}{2}-1\)
\(Q.3.(xvii)\) \(\int_{0}^{1}{2x^3e^{-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1-\frac{2}{e}\)
উত্তরঃ \(1-\frac{2}{e}\)
সমাধানঃ
ধরি,
\(-x^2=t\)
\(\Rightarrow \frac{d}{dx}(-x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow -2x=\frac{dt}{dx}\)
\(\therefore -2xdx=dt\)
➜ \(\because t=-x^2\)
\(\Rightarrow t=-0^2\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=-x^2\)
\(\Rightarrow t=-1^2\), যখন \(x=1\)
\(\therefore t=-1\)
এখানে, LIATE শব্দের
\(A=t, \ \ E=e^t\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(t\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{2x^3e^{-x^2}dx}\)\(-x^2=t\)
\(\Rightarrow \frac{d}{dx}(-x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow -2x=\frac{dt}{dx}\)
\(\therefore -2xdx=dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(-1\) |
\(\therefore t=0\)
আবার,
\(t=-x^2\)
\(\Rightarrow t=-1^2\), যখন \(x=1\)
\(\therefore t=-1\)
এখানে, LIATE শব্দের
\(A=t, \ \ E=e^t\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(t\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{1}{-x^2e^{-x^2}\times{-2xdx}}\)
\(=\int_{0}^{-1}{te^{t}dt}\)
\(=\left[t.e^t\right]_{0}^{-1}-\int_{0}^{-1}{\left\{\frac{d}{dt}(t)\int{e^tdt}\right\}dt}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{e^xdx}=e^x\)
\(=-1.e^{-1}-0.e^{0}-\int_{0}^{-1}{1.e^tdt}\) ➜ \(\because \frac{d}{dx}(x)=1\), \(\int{e^xdx}=e^x\)
\(=-\frac{1}{e}-0-\int_{0}^{-1}{e^tdt}\)
\(=-\frac{1}{e}-\left[e^t\right]_{0}^{-1}\)
\(=-\frac{1}{e}-\left[e^{-1}-e^0\right]\)
\(=-\frac{1}{e}-\left[\frac{1}{e}-1\right]\) ➜ \(\because e^0=1\)
\(=-\frac{1}{e}-\frac{1}{e}+1\)
\(=-2\frac{1}{e}+1\)
\(=1-\frac{2}{e}\)
\(Q.3.(xviii)\) \(\int_{0}^{1}{xe^{-3x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{9}(1-4e^{-3})\)
[ দিঃ২০১০; সিঃ২০০৩]
উত্তরঃ \(\frac{1}{9}(1-4e^{-3})\)
[ দিঃ২০১০; সিঃ২০০৩]
সমাধানঃ
ধরি,
\(-3x=t\)
\(\Rightarrow x=-\frac{1}{3}(t)\)
\(\Rightarrow \frac{d}{dx}(x)=-\frac{1}{3}\frac{d}{dx}(t)\)
\(\Rightarrow 1=-\frac{1}{3}\frac{dt}{dx}\)
\(\therefore dx=-\frac{1}{3}dt\)
➜ \(\because t=-3x\)
\(\Rightarrow t=-3.0\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=-3x\)
\(\Rightarrow t=-3.1\), যখন \(x=1\)
\(\therefore t=-3\)
এখানে, LIATE শব্দের
\(A=t, \ \ E=e^t\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(t\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{1}{xe^{-3x}dx}\)\(-3x=t\)
\(\Rightarrow x=-\frac{1}{3}(t)\)
\(\Rightarrow \frac{d}{dx}(x)=-\frac{1}{3}\frac{d}{dx}(t)\)
\(\Rightarrow 1=-\frac{1}{3}\frac{dt}{dx}\)
\(\therefore dx=-\frac{1}{3}dt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(0\) | \(-3\) |
\(\therefore t=0\)
আবার,
\(t=-3x\)
\(\Rightarrow t=-3.1\), যখন \(x=1\)
\(\therefore t=-3\)
এখানে, LIATE শব্দের
\(A=t, \ \ E=e^t\)
যেহেতু A, শব্দের মধ্যে E এর পূর্বে আছে, তাই \(t\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{-3}{-\frac{1}{3}te^{t}\times{-\frac{1}{3}}dt}\)
\(=\frac{1}{9}\int_{0}^{-3}{te^{t}dt}\)
\(=\frac{1}{9}\left[t.e^t\right]_{0}^{-3}-\frac{1}{9}\int_{0}^{-3}{\left\{\frac{d}{dt}(t)\int{e^tdt}\right\}dt}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{e^xdx}=e^x\)
\(=\frac{1}{9}\left[-3.e^{-3}-0.e^{0}\right]-\frac{1}{9}\int_{0}^{-3}{1.e^tdt}\) ➜ \(\because \frac{d}{dx}(x)=1\), \(\int{e^xdx}=e^x\)
\(=\frac{1}{9}\left[-\frac{3}{e^3}-0\right]-\frac{1}{9}\int_{0}^{-3}{e^tdt}\)
\(=-\frac{1}{3e^3}-\frac{1}{9}\left[e^t\right]_{0}^{-3}\)
\(=-\frac{1}{3e^3}-\frac{1}{9}\left[e^{-3}-e^0\right]\)
\(=-\frac{1}{3e^3}-\frac{1}{9}\left[\frac{1}{e^3}-1\right]\) ➜ \(\because e^0=1\)
\(=-\frac{1}{3e^3}-\frac{1}{9e^3}+\frac{1}{9}\)
\(=\frac{1}{9}-\frac{1}{3e^3}-\frac{1}{9e^3}\)
\(=\frac{1}{9}-\left(\frac{1}{3e^3}+\frac{1}{9e^3}\right)\)
\(=\frac{1}{9}-\frac{3+1}{9e^3}\)
\(=\frac{1}{9}-\frac{4}{9e^3}\)
\(=\frac{1}{9}\left(1-\frac{4}{e^3}\right)\)
\(=\frac{1}{9}\left(1-4e^{-3}\right)\)
\(Q.3.(xix)\) \(\int_{0}^{\frac{\pi}{2}}{x^2\cos{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^2}{4}-2\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৪ ]
উত্তরঃ \(\frac{\pi^2}{4}-2\)
[ চঃ২০০৭; ঢাঃ২০০৬; কুঃ২০০৪ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=x^2, \ \ T=\cos{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x^2\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\frac{\pi}{2}}{x^2\cos{x}dx}\)\(A=x^2, \ \ T=\cos{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x^2\) কে \(u\) ধরা হয়েছে।
\(=\left[x^2\sin{x}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}{\left\{\frac{d}{dx}(x^2)\int{\cos{x}dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\cos{x}dx}=\sin{x}\)
\(=\left(\frac{\pi}{2}\right)^2\sin{\left(\frac{\pi}{2}\right)}-0^2\sin{0}-\int_{0}^{\frac{\pi}{2}}{2x\sin{x}dx}\) ➜ \(\because \frac{d}{dx}(x^2)=2x,\) \(\int{\cos{x}dx}=\sin{x}\)
\(=\frac{\pi^2}{4}.1-0-2\int_{0}^{\frac{\pi}{2}}{x\sin{x}dx}\)
\(=\frac{\pi^2}{4}-2\left[-x\cos{x}\right]_{0}^{\frac{\pi}{2}}+2\int_{0}^{\frac{\pi}{2}}{\left\{\frac{d}{dx}(x)\int{\sin{x}dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\sin{x}dx}=-\cos{x}\)
\(=\frac{\pi^2}{4}+2\left[\frac{\pi}{2}\cos{\left(\frac{\pi}{2}\right)}-0.\cos{0}\right]+2\int_{0}^{\frac{\pi}{2}}{1.(-\cos{x})dx}\) ➜ \(\because \frac{d}{dx}(x)=1,\) \(\int{\sin{x}dx}=-\cos{x}\)
\(=\frac{\pi^2}{4}+2\left[\frac{\pi}{2}.0-0.1\right]-2\int_{0}^{\frac{\pi}{2}}{\cos{x}dx}\)
\(=\frac{\pi^2}{4}+2\left[0-0\right]-2\left[\sin{x}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=\frac{\pi^2}{4}-2\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{0}\right]\)
\(=\frac{\pi^2}{4}-2\left[1-0\right]\)
\(=\frac{\pi^2}{4}-2\)
\(Q.3.(xx)\) \(\int_{0}^{\pi}{x^2\sin{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi^2+4\)
উত্তরঃ \(\pi^2+4\)
সমাধানঃ
এখানে, LIATE শব্দের
\(A=x^2, \ \ T=\sin{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x^2\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\pi}{x^2\sin{x}dx}\)\(A=x^2, \ \ T=\sin{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x^2\) কে \(u\) ধরা হয়েছে।
\(=\left[-x^2\cos{x}\right]_{0}^{\pi}-\int_{0}^{\pi}{\left\{\frac{d}{dx}(x^2)\int{\sin{x}dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\sin{x}dx}=-\cos{x}\)
\(=-\left[(\pi)^2\cos{\pi}-0^2\cos{0}\right]-\int_{0}^{\pi}{\left\{-2x\cos{x}\right\}dx}\) ➜ \(\because \frac{d}{dx}(x^2)=2x,\) \(\int{\sin{x}dx}=-\cos{x}\)
\(=-\left[(\pi)^2(-1)-0.1\right]+2\int_{0}^{\pi}{x\cos{x}dx}\)
\(=-\left[-\pi^2-0\right]+2\left[x\sin{x}\right]_{0}^{\pi}-2\int_{0}^{\pi}{\left\{\frac{d}{dx}(x)\int{\cos{x}dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\cos{x}dx}=\sin{x}\)
\(=\pi^2+2\left[\pi\sin{(\pi)}-0.\sin{0}\right]+2\int_{0}^{\pi}{1.\sin{x}dx}\) ➜ \(\because \frac{d}{dx}(x)=1,\) \(\int{\cos{x}dx}=\sin{x}\)
\(=\pi^2+2\left[\pi.0-0\right]+2\int_{0}^{\pi}{\sin{x}dx}\)
\(=\pi^2+2\left[0-0\right]+2\left[-\cos{x}\right]_{0}^{\pi}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\)
\(=\pi^2-2\left[\cos{(\pi)}-\cos{0}\right]\)
\(=\pi^2-2\left[-1-1\right]\)
\(=\pi^2-2\left[-2\right]\)
\(=\pi^2+4\)
\(Q.3.(xxi)\) \(\int_{0}^{1}{\sin^{-1}{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}-1\)
উত্তরঃ \(\frac{\pi}{2}-1\)
সমাধানঃ
এখানে, LIATE শব্দের
\(I=\sin^{-1}{x}, \ \ A=1\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\sin^{-1}{x}\) কে \(u\) ধরা হয়েছে।
ধরি,
\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\therefore xdx=-tdt\)
➜ \(\because t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-0^2}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1-0}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\therefore t=0\)
\(\int_{0}^{1}{\sin^{-1}{x}dx}\)\(I=\sin^{-1}{x}, \ \ A=1\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\sin^{-1}{x}\) কে \(u\) ধরা হয়েছে।
ধরি,
\(\sqrt{1-x^2}=t\)
\(\Rightarrow 1-x^2=t^2\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t^2)\)
\(\Rightarrow 0-2x=2t\frac{dt}{dx}\)
\(\Rightarrow -2xdx=2tdt\)
\(\therefore xdx=-tdt\)
| \(x\) | \(0\) | \(1\) |
| \(t\) | \(1\) | \(0\) |
\(\Rightarrow t=\sqrt{1-0}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{1-x^2}\)
\(\Rightarrow t=\sqrt{1-1^2}\), যখন \(x=1\)
\(\Rightarrow t=\sqrt{1-1}\)
\(\therefore t=0\)
\(=\int_{0}^{1}{\sin^{-1}{x}.1dx}\)
\(=\left[x\sin^{-1}{x}\right]_{0}^{1}-\int_{0}^{1}{\left\{\frac{d}{dx}(\sin^{-1}{x})\int{.1dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=\left[1\sin^{-1}{(1)}-0.\sin^{-1}{(0)}\right]-\int_{0}^{1}{\frac{1}{\sqrt{1-x^2}}.xdx}\) ➜ \(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}},\) \(\int{1dx}=x\)
\(=\left[\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0\right]-\int_{1}^{0}{\frac{1}{t}\times{-tdt}}\)
\(=\frac{\pi}{2}+\int_{1}^{0}{1dt}\)
\(=\frac{\pi}{2}+\left[t\right]_{1}^{0}\) ➜ \(\because \int{1dx}=x\)
\(=\frac{\pi}{2}+\left[0-1\right]\)
\(=\frac{\pi}{2}-1\)
\(Q.3.(xxii)\) \(\int_{1}^{\sqrt{3}}{x\tan^{-1}{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(5\pi-6\sqrt{3}+6)\)
[ বঃ২০১৩; রাঃ,চঃ২০১২,২০০৮; দিঃ২০১২; যঃ২০১১ ]
উত্তরঃ \(\frac{1}{12}(5\pi-6\sqrt{3}+6)\)
[ বঃ২০১৩; রাঃ,চঃ২০১২,২০০৮; দিঃ২০১২; যঃ২০১১ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(I=\tan^{-1}{x}, \ \ A=x\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{x}\) কে \(u\) ধরা হয়েছে।
\(\int_{1}^{\sqrt{3}}{x\tan^{-1}{x}dx}\)\(I=\tan^{-1}{x}, \ \ A=x\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{x}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{\sqrt{3}}{\tan^{-1}{x}.xdx}\)
\(=\left[\frac{x^2}{2}\tan^{-1}{x}\right]_{1}^{\sqrt{3}}-\int_{1}^{\sqrt{3}}{\left\{\frac{d}{dx}(\tan^{-1}{x})\int{.xdx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[x^2\tan^{-1}{x}\right]_{1}^{\sqrt{3}}-\int_{1}^{\sqrt{3}}{\frac{1}{1+x^2}.\frac{x^2}{2}dx}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[(\sqrt{3})^2\tan^{-1}{(\sqrt{3})}-1^2.\tan^{-1}{(1)}\right]-\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{x^2}{1+x^2}dx}\)
\(=\frac{1}{2}\left[3\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}-1.\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\right]-\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{1+x^2-1}{1+x^2}dx}\)
\(=\frac{1}{2}\left[3.\frac{\pi}{3}-\frac{\pi}{4}\right]-\frac{1}{2}\int_{1}^{\sqrt{3}}{\left(\frac{1+x^2}{1+x^2}-\frac{1}{1+x^2}\right)dx}\)
\(=\frac{1}{2}\left[\pi-\frac{\pi}{4}\right]-\frac{1}{2}\int_{1}^{\sqrt{3}}{\left(1-\frac{1}{1+x^2}\right)dx}\)
\(=\frac{1}{2}.\frac{4\pi-\pi}{4}-\frac{1}{2}\int_{1}^{\sqrt{3}}{1dx}+\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{1}{1+x^2}dx}\)
\(=\frac{3\pi}{8}-\frac{1}{2}\left[x\right]_{1}^{\sqrt{3}}+\frac{1}{2}\left[\tan^{-1}{x}\right]_{1}^{\sqrt{3}}\) ➜ \(\because \int{1dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\frac{3\pi}{8}-\frac{1}{2}\left[\sqrt{3}-1\right]+\frac{1}{2}\left[\tan^{-1}{(\sqrt{3})}-\tan^{-1}{(1)}\right]\)
\(=\frac{3\pi}{8}-\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{2}\left[\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}-\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}\right]\)
\(=\frac{3\pi}{8}-\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{4}\right]\)
\(=\frac{3\pi}{8}-\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{2}.\frac{4\pi-3\pi}{12}\)
\(=\frac{3\pi}{8}-\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{\pi}{24}\)
\(=\frac{3\pi}{8}+\frac{\pi}{24}-\frac{\sqrt{3}}{2}+\frac{1}{2}\)
\(=\frac{9\pi+\pi}{24}-\frac{\sqrt{3}}{2}+\frac{1}{2}\)
\(=\frac{10\pi}{24}-\frac{\sqrt{3}}{2}+\frac{1}{2}\)
\(=\frac{5\pi}{12}-\frac{\sqrt{3}}{2}+\frac{1}{2}\)
\(=\frac{1}{12}(5\pi-6\sqrt{3}+6)\)
\(Q.3.(xxiii)\) \(\int_{0}^{2}{(x-2)\tan^{-1}{(x-2)}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{5}{2}\tan^{-1}{(2)}-1\)
[ সিঃ২০১৭ ]
উত্তরঃ \(\frac{5}{2}\tan^{-1}{(2)}-1\)
[ সিঃ২০১৭ ]
সমাধানঃ
ধরি,
\(x-2=t\)
\(\Rightarrow \frac{d}{dx}(x-2)=\frac{d}{dx}(t)\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
➜ \(\because t=x-2\)
\(\Rightarrow t=0-2\), যখন \(x=0\)
\(\therefore t=-2\)
আবার,
\(t=x-2\)
\(\Rightarrow t=2-2\), যখন \(x=2\)v \(\therefore t=0\)
এখানে, LIATE শব্দের
\(I=\tan^{-1}{t}, \ \ A=t\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{t}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{2}{(x-2)\tan^{-1}{(x-2)}dx}\)\(x-2=t\)
\(\Rightarrow \frac{d}{dx}(x-2)=\frac{d}{dx}(t)\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
| \(x\) | \(0\) | \(2\) |
| \(t\) | \(-2\) | \(0\) |
\(\therefore t=-2\)
আবার,
\(t=x-2\)
\(\Rightarrow t=2-2\), যখন \(x=2\)v \(\therefore t=0\)
এখানে, LIATE শব্দের
\(I=\tan^{-1}{t}, \ \ A=t\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{t}\) কে \(u\) ধরা হয়েছে।
\(=\int_{-2}^{0}{t\tan^{-1}{t}dt}\)
\(=\left[\frac{t^2}{2}\tan^{-1}{t}\right]_{-2}^{0}-\int_{-2}^{0}{\left\{\frac{d}{dt}(\tan^{-1}{t})\int{tdt}\right\}dt}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[t^2\tan^{-1}{t}\right]_{-2}^{0}-\int_{-2}^{0}{\frac{1}{1+t^2}.\frac{t^2}{2}dx}\) ➜ \(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[0^2\tan^{-1}{(0)}-(-2)^2.\tan^{-1}{(-2)}\right]-\frac{1}{2}\int_{-2}^{0}{\frac{t^2}{1+t^2}dt}\)
\(=\frac{1}{2}\left[0\tan^{-1}{(0)}+4\tan^{-1}{(2)}\right]-\frac{1}{2}\int_{-2}^{0}{\frac{1+t^2-1}{1+t^2}dt}\)
\(=\frac{1}{2}\left[0+4\tan^{-1}{(2)}\right]-\frac{1}{2}\int+4\tan^{-1}{(2)}{\left(\frac{1+t^2}{1+t^2}-\frac{1}{1+t^2}\right)dt}\)
\(=2\tan^{-1}{(2)}-\frac{1}{2}\int_{-2}^{0}{\left(1-\frac{1}{1+t^2}\right)dt}\)
\(=2\tan^{-1}{(2)}-\frac{1}{2}\int_{-2}^{0}{1dt}+\frac{1}{2}\int_{-2}^{0}{\frac{1}{1+t^2}dt}\)
\(=2\tan^{-1}{(2)}-\frac{1}{2}\left[t\right]_{-2}^{0}+\frac{1}{2}\left[\tan^{-1}{t}\right]_{-2}^{0}\) ➜ \(\because \int{1dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=2\tan^{-1}{(2)}-\frac{1}{2}\left[0+2\right]+\frac{1}{2}\left[\tan^{-1}{(0)}-\tan^{-1}{(-2)}\right]\)
\(=2\tan^{-1}{(2)}-1+\frac{1}{2}\left[\tan^{-1}{\tan{0}}+\tan^{-1}{(2)}\right]\)
\(=2\tan^{-1}{(2)}-1+\frac{1}{2}\left[0+\tan^{-1}{(2)}\right]\)
\(=2\tan^{-1}{(2)}-1+\frac{1}{2}\tan^{-1}{(2)}\)
\(=2\tan^{-1}{(2)}+\frac{1}{2}\tan^{-1}{(2)}-1\)
\(=\frac{4+1}{2}\tan^{-1}{(2)}-1\)
\(=\frac{5}{2}\tan^{-1}{(2)}-1\)
\(Q.3.(xxiv)\) \(\int_{1}^{\sqrt{3}}{x\cot^{-1}{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{12}(\pi+6\sqrt{3}-6)\)
[ বুয়েটঃ২০০৯ ]
উত্তরঃ \(\frac{1}{12}(\pi+6\sqrt{3}-6)\)
[ বুয়েটঃ২০০৯ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(I=\cot^{-1}{x}, \ \ A=x\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{x}\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\sqrt{3}}{x\cot^{-1}{x}dx}\)\(I=\cot^{-1}{x}, \ \ A=x\)
যেহেতু I, শব্দের মধ্যে A এর পূর্বে আছে, তাই \(\tan^{-1}{x}\) কে \(u\) ধরা হয়েছে।
\(=\int_{1}^{\sqrt{3}}{\cot^{-1}{x}.xdx}\)
\(=\left[\frac{x^2}{2}\cot^{-1}{x}\right]_{1}^{\sqrt{3}}-\int_{1}^{\sqrt{3}}{\left\{\frac{d}{dx}(\cot^{-1}{x})\int{.xdx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[x^2\cot^{-1}{x}\right]_{1}^{\sqrt{3}}-\int_{1}^{\sqrt{3}}{\left(-\frac{1}{1+x^2}\right)\frac{x^2}{2}dx}\) ➜ \(\because \frac{d}{dx}(\cot^{-1}{x})=-\frac{1}{1+x^2},\) \(\int{xdx}=\frac{x^2}{2}\)
\(=\frac{1}{2}\left[(\sqrt{3})^2\cot^{-1}{(\sqrt{3})}-1^2.\cot^{-1}{(1)}\right]+\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{x^2}{1+x^2}dx}\)
\(=\frac{1}{2}\left[3\cot^{-1}{\cot{\left(\frac{\pi}{6}\right)}}-1.\cot^{-1}{\cot{\left(\frac{\pi}{4}\right)}}\right]+\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{1+x^2-1}{1+x^2}dx}\)
\(=\frac{1}{2}\left[3.\frac{\pi}{6}-\frac{\pi}{4}\right]+\frac{1}{2}\int_{1}^{\sqrt{3}}{\left(\frac{1+x^2}{1+x^2}-\frac{1}{1+x^2}\right)dx}\)
\(=\frac{1}{2}\left[\frac{\pi}{2}-\frac{\pi}{4}\right]+\frac{1}{2}\int_{1}^{\sqrt{3}}{\left(1-\frac{1}{1+x^2}\right)dx}\)
\(=\frac{1}{2}.\frac{2\pi-\pi}{4}+\frac{1}{2}\int_{1}^{\sqrt{3}}{1dx}-\frac{1}{2}\int_{1}^{\sqrt{3}}{\frac{1}{1+x^2}dx}\) \(=\frac{\pi}{8}+\frac{1}{2}\left[x\right]_{1}^{\sqrt{3}}-\frac{1}{2}\left[\tan^{-1}{x}\right]_{1}^{\sqrt{3}}\) ➜ \(\because \int{1dx}=x, \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\frac{\pi}{8}+\frac{1}{2}\left[\sqrt{3}-1\right]-\frac{1}{2}\left[\tan^{-1}{(\sqrt{3})}-\tan^{-1}{(1)}\right]\)
\(=\frac{\pi}{8}+\frac{\sqrt{3}}{2}-\frac{1}{2}-\frac{1}{2}\left[\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}-\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}\right]\)
\(=\frac{\pi}{8}+\frac{\sqrt{3}}{2}-\frac{1}{2}-\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{4}\right]\)
\(=\frac{\pi}{8}+\frac{\sqrt{3}}{2}-\frac{1}{2}-\frac{1}{2}.\frac{4\pi-3\pi}{12}\)
\(=\frac{\pi}{8}+\frac{\sqrt{3}}{2}-\frac{1}{2}-\frac{\pi}{24}\)
\(=\frac{\pi}{8}-\frac{\pi}{24}+\frac{\sqrt{3}}{2}-\frac{1}{2}\)
\(=\frac{3\pi-\pi}{24}+\frac{\sqrt{3}}{2}-\frac{1}{2}\)
\(=\frac{2\pi}{24}+\frac{\sqrt{3}}{2}-\frac{1}{2}\)
\(=\frac{\pi}{12}+\frac{\sqrt{3}}{2}-\frac{1}{2}\)
\(=\frac{1}{12}(\pi+6\sqrt{3}-6)\)
\(Q.3.(xxv)\) \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{dx}{1+\sin{x}-\cos{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{\left(\frac{\sqrt{3}+1}{2}\right)}\)
[ বুয়েটঃ২০১১-২০১২ ]
উত্তরঃ \(\ln{\left(\frac{\sqrt{3}+1}{2}\right)}\)
[ বুয়েটঃ২০১১-২০১২ ]
সমাধানঃ
ধরি,
\(\tan{\left(\frac{x}{2}\right)}=t\)
\(\Rightarrow \frac{d}{dx}\tan{\left(\frac{x}{2}\right)}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}=2\frac{dt}{dx}\)
\(\therefore \sec^2{\left(\frac{x}{2}\right)}dx=2dt\)
➜ \(\because t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{3}}\right)}\), যখন \(x=\frac{\pi}{3}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{6}\right)}\)
\(\therefore t=\frac{1}{\sqrt{3}}\)
আবার,
\(t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{4}\right)}\)
\(\therefore t=1\)
\(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{dx}{1+\sin{x}-\cos{x}}}\)\(\tan{\left(\frac{x}{2}\right)}=t\)
\(\Rightarrow \frac{d}{dx}\tan{\left(\frac{x}{2}\right)}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}=2\frac{dt}{dx}\)
\(\therefore \sec^2{\left(\frac{x}{2}\right)}dx=2dt\)
| \(x\) | \(\frac{\pi}{3}\) | \(\frac{\pi}{2}\) |
| \(t\) | \(\frac{1}{\sqrt{3}}\) | \(1\) |
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{3}}\right)}\), যখন \(x=\frac{\pi}{3}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{6}\right)}\)
\(\therefore t=\frac{1}{\sqrt{3}}\)
আবার,
\(t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{4}\right)}\)
\(\therefore t=1\)
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{dx}{1+\frac{2\tan{\left(\frac{x}{2}\right)}}{1+\tan^2{\left(\frac{x}{2}\right)}}-\frac{1-\tan^2{\left(\frac{x}{2}\right)}}{1+\tan^2{\left(\frac{x}{2}\right)}}}}\) ➜ \(\because \sin{A}=\frac{2\tan{\left(\frac{A}{2}\right)}}{1+\tan^2{\left(\frac{A}{2}\right)}},\) \(\cos{A}=\frac{1-\tan^2{\left(\frac{A}{2}\right)}}{1+\tan^2{\left(\frac{A}{2}\right)}}\)
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{1+\tan^2{\left(\frac{x}{2}\right)}}{1+\tan^2{\left(\frac{x}{2}\right)}+2\tan{\left(\frac{x}{2}\right)}-1+\tan^2{\left(\frac{x}{2}\right)}}dx}\) ➜ লব ও হরের সহিত \(1+\tan^2{\left(\frac{x}{2}\right)}\) গুণ করে।
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{\sec^2{\left(\frac{x}{2}\right)}}{2\tan{\left(\frac{x}{2}\right)}+2\tan^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\tan^2{A}=sec^2{A}\)
\(=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{1}{2\tan{\left(\frac{x}{2}\right)}+2\tan^2{\left(\frac{x}{2}\right)}}.\sec^2{\left(\frac{x}{2}\right)}dx}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1}{2t+2t^2}.2dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1}{2t(1+t)}.2dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1}{t(1+t)}dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1+t-t}{t(1+t)}dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\left\{\frac{1+t}{t(1+t)}-\frac{t}{t(1+t)}\right\}dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\left\{\frac{1}{t}-\frac{1}{1+t}\right\}dt}\)
\(=\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1}{t}dt}-\int_{\frac{1}{\sqrt{3}}}^{1}{\frac{1}{1+t}dt}\)
\(=\left[\ln{|t|}\right]_{\frac{1}{\sqrt{3}}}^{1}-\left[\ln{|1+t|}\right]_{\frac{1}{\sqrt{3}}}^{1}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\ln{(1)}-\ln{\left(\frac{1}{\sqrt{3}}\right)}-\left[\ln{(1+1)}-\ln{\left(1+\frac{1}{\sqrt{3}}\right)}\right]\)
\(=0-\ln{\left(\frac{1}{\sqrt{3}}\right)}-\left[\ln{(2)}-\ln{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}\right]\)
\(=0-\ln{\left(\frac{1}{\sqrt{3}}\right)}-\ln{(2)}+\ln{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}\)
\(=\ln{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}-\ln{\left(\frac{1}{\sqrt{3}}\right)}-\ln{(2)}\)
\(=\ln{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}-\ln{(\sqrt{3})^{-1}}-\ln{(2)}\)
\(=\ln{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}+\ln{(\sqrt{3})}-\ln{(2)}\)
\(=\ln{\left\{\frac{(\sqrt{3}+1)\sqrt{3}}{2\sqrt{3}}\right\}}\) ➜ \(\because \ln{M}+\ln{N}-\ln{p}=\ln{\left(\frac{MN}{p}\right)}\)
\(=\ln{\left(\frac{\sqrt{3}+1}{2}\right)}\)
\(Q.3.(xxvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\sin{x}+\cos{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(-\sqrt{2}\ln{(\sqrt{2}-1)}\)
[ কুয়েটঃ২০১৩-২০১৪; বুটেক্সঃ২০০১-২০০২ ]
উত্তরঃ \(-\sqrt{2}\ln{(\sqrt{2}-1)}\)
[ কুয়েটঃ২০১৩-২০১৪; বুটেক্সঃ২০০১-২০০২ ]
সমাধানঃ
ধরি,
\(\tan{\left(\frac{x}{2}\right)}=t\)
\(\Rightarrow \frac{d}{dx}\tan{\left(\frac{x}{2}\right)}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}=2\frac{dt}{dx}\)
\(\therefore \sec^2{\left(\frac{x}{2}\right)}dx=2dt\)
➜ \(\because t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{0}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan{(0)}\)
\(\therefore t=0\)
আবার,
\(t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{4}\right)}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\sin{x}+\cos{x}}}\)\(\tan{\left(\frac{x}{2}\right)}=t\)
\(\Rightarrow \frac{d}{dx}\tan{\left(\frac{x}{2}\right)}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{\left(\frac{x}{2}\right)}=2\frac{dt}{dx}\)
\(\therefore \sec^2{\left(\frac{x}{2}\right)}dx=2dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{0}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan{(0)}\)
\(\therefore t=0\)
আবার,
\(t=\tan{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}x\right)}\)
\(\Rightarrow t=\tan{\left(\frac{1}{2}\times{\frac{\pi}{2}}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{4}\right)}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{dx}{\frac{2\tan{\left(\frac{x}{2}\right)}}{1+\tan^2{\left(\frac{x}{2}\right)}}+\frac{1-\tan^2{\left(\frac{x}{2}\right)}}{1+\tan^2{\left(\frac{x}{2}\right)}}}}\) ➜ \(\because \sin{A}=\frac{2\tan{\left(\frac{A}{2}\right)}}{1+\tan^2{\left(\frac{A}{2}\right)}},\) \(\cos{A}=\frac{1-\tan^2{\left(\frac{A}{2}\right)}}{1+\tan^2{\left(\frac{A}{2}\right)}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1+\tan^2{\left(\frac{x}{2}\right)}}{2\tan{\left(\frac{x}{2}\right)}+1+\tan^2{\left(\frac{x}{2}\right)}}dx}\) ➜ লব ও হরের সহিত \(1+\tan^2{\left(\frac{x}{2}\right)}\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{\left(\frac{x}{2}\right)}}{2\tan{\left(\frac{x}{2}\right)}+1-\tan^2{\left(\frac{x}{2}\right)}}dx}\) ➜ \(\because 1+\tan^2{A}=sec^2{A}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{2\tan{\left(\frac{x}{2}\right)}+1-\tan^2{\left(\frac{x}{2}\right)}}.\sec^2{\left(\frac{x}{2}\right)}dx}\)
\(=\int_{0}^{1}{\frac{1}{2t+1-t^2}.2dt}\)
\(=2\int_{0}^{1}{\frac{1}{2-t^2+2t-1}dt}\)
\(=2\int_{0}^{1}{\frac{1}{2-(t^2-2t+1)}dt}\)
\(=2\int_{0}^{1}{\frac{1}{(\sqrt{2})^2-(t-1)^2}dt}\)
\(=2\left[\frac{1}{2\sqrt{2}}\ln{\left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\frac{1}{\sqrt{2}}\left[\ln{\left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|}\right]_{0}^{1}\)
\(=\frac{1}{\sqrt{2}}\left[\ln{\left|\frac{\sqrt{2}+(1-1)}{\sqrt{2}-(1-1)}\right|}-\ln{\left|\frac{\sqrt{2}+(0-1)}{\sqrt{2}-(0-1)}\right|}\right]\)
\(=\frac{1}{\sqrt{2}}\left[\ln{\left|\frac{\sqrt{2}+0}{\sqrt{2}-0}\right|}-\ln{\left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right|}\right]\)
\(=\frac{1}{\sqrt{2}}\left[\ln{\left|\frac{\sqrt{2}}{\sqrt{2}}\right|}-\ln{\left|\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}\right|}\right]\) ➜ লব ও হরের সহিত \((\sqrt{2}-1)\) গুণ করে।
\(=\frac{1}{\sqrt{2}}\left[\ln{(1)}-\ln{\left|\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-1^2}\right|}\right]\)
\(=\frac{1}{\sqrt{2}}\left[0-\ln{\left|\frac{(\sqrt{2}-1)^2}{2-1}\right|}\right]\) ➜ \(\because \ln{(1)}=0\)
\(=\frac{1}{\sqrt{2}}\times{-\ln{(\sqrt{2}-1)^2}}\)
\(=\frac{1}{\sqrt{2}}\times{-2\ln{(\sqrt{2}-1)}}\)
\(=-\frac{2}{\sqrt{2}}\ln{(\sqrt{2}-1)}\)
\(=-\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}}\ln{(\sqrt{2}-1)}\)
\(=-\sqrt{2}\ln{(\sqrt{2}-1)}\)
\(Q.3.(xxvii)\) \(\int_{0}^{\frac{\pi}{2}}{\sin^3{x}\sqrt{\cos{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{8}{21}\)
[ চঃ২০১৩,২০০৯; বঃ,সিঃ২০১৩; যঃ২০১০; রাঃ২০০৮ ]
উত্তরঃ \(\frac{8}{21}\)
[ চঃ২০১৩,২০০৯; বঃ,সিঃ২০১৩; যঃ২০১০; রাঃ২০০৮ ]
সমাধানঃ
ধরি,
\(\sqrt{\cos{x}}=t\)
\(\Rightarrow \cos{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow -\sin{x}=2t\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-2t\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-2tdt\)
➜ \(\because t=\sqrt{\cos{x}}\)
\(\Rightarrow t=\sqrt{\cos{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{\cos{x}}\)
\(\Rightarrow t=\sqrt{\cos{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(\int_{0}^{\frac{\pi}{2}}{\sin^3{x}\sqrt{\cos{x}}dx}\)\(\sqrt{\cos{x}}=t\)
\(\Rightarrow \cos{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow -\sin{x}=2t\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=-2t\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-2tdt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(0\) |
\(\Rightarrow t=\sqrt{\cos{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
আবার,
\(t=\sqrt{\cos{x}}\)
\(\Rightarrow t=\sqrt{\cos{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin^2{x}\sqrt{\cos{x}}.\sin{x}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{(1-\cos^2{x})\sqrt{\cos{x}}.\sin{x}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\{1-(\sqrt{\cos{x}})^4\}\sqrt{\cos{x}}.\sin{x}dx}\)
\(=\int_{1}^{0}{\{1-t^4\}t\times{-2t}dt}\)
\(=\int_{1}^{0}{\{-(t^4-1)\times{-2t^2}\}dt}\)
\(=2\int_{1}^{0}{t^2(t^4-1)dt}\)
\(=2\int_{1}^{0}{(t^6-t^2)dt}\)
\(=2\left[\frac{t^7}{7}-\frac{t^3}{3}\right]_{1}^{0}\) ➜ \(\because \int{x^ndx}=\frac{n^{n+1}}{n+1}\)
\(=2\left[\left(\frac{0^7}{7}-\frac{0^3}{3}\right)-\left(\frac{1^7}{7}-\frac{1^3}{3}\right)\right]\)
\(=2\left[\left(\frac{0}{7}-\frac{0}{3}\right)-\left(\frac{1}{7}-\frac{1}{3}\right)\right]\)
\(=2\left[\left(0-0\right)-\left(\frac{3-7}{21}\right)\right]\)
\(=2\left[0-\left(\frac{-4}{21}\right)\right]\)
\(=2\times{\frac{4}{21}}\)
\(=\frac{8}{21}\)
\(Q.3.(xxviii)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}\sqrt{\sin{x}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{8}{21}\)
[ ঢাঃ২০১৪,২০০৬,২০০৪; রাঃ২০১৪; কুঃ২০১৩,২০১১,২০০৭; দিঃ২০১১; চঃ২০০৭,২০০৩; সিঃ২০১১,২০০৮,২০০৪; যঃ২০১২; বঃ২০১১,২০০৯,২০০৭,২০০৫ ]
উত্তরঃ \(\frac{8}{21}\)
[ ঢাঃ২০১৪,২০০৬,২০০৪; রাঃ২০১৪; কুঃ২০১৩,২০১১,২০০৭; দিঃ২০১১; চঃ২০০৭,২০০৩; সিঃ২০১১,২০০৮,২০০৪; যঃ২০১২; বঃ২০১১,২০০৯,২০০৭,২০০৫ ]
সমাধানঃ
ধরি,
\(\sqrt{\sin{x}}=t\)
\(\Rightarrow \sin{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow \cos{x}=2t\frac{dt}{dx}\)
\(\therefore \cos{x}dx=2tdt\)
➜ \(\because t=\sqrt{\sin{x}}\)
\(\Rightarrow t=\sqrt{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{\sin{x}}\)
\(\Rightarrow t=\sqrt{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\cos^3{x}\sqrt{\sin{x}}dx}\)\(\sqrt{\sin{x}}=t\)
\(\Rightarrow \sin{x}=t^2\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t^2)\)
\(\Rightarrow \cos{x}=2t\frac{dt}{dx}\)
\(\therefore \cos{x}dx=2tdt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\Rightarrow t=\sqrt{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{\sin{x}}\)
\(\Rightarrow t=\sqrt{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\cos^2{x}\sqrt{\sin{x}}.\cos{x}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{(1-\sin^2{x})\sqrt{\sin{x}}.\cos{x}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\{1-(\sqrt{\sin{x}})^4\}\sqrt{\sin{x}}.\cos{x}dx}\)
\(=\int_{0}^{1}{\{1-t^4\}t\times{2t}dt}\)
\(=2\int_{0}^{1}{t^2(1-t^4)dt}\)
\(=2\int_{0}^{1}{(t^2-t^6)dt}\)
\(=2\left[\frac{t^3}{3}-\frac{t^7}{7}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{n^{n+1}}{n+1}\)
\(=2\left[\left(\frac{1^3}{3}-\frac{1^7}{7}\right)-\left(\frac{0^3}{3}-\frac{0^7}{7}\right)\right]\)
\(=2\left[\left(\frac{1}{3}-\frac{1}{7}\right)-\left(\frac{0}{3}-\frac{0}{7}\right)\right]\)
\(=2\left[\left(\frac{7-3}{21}\right)-\left(0-0\right)\right]\)
\(=2\left[\left(\frac{4}{21}\right)-0\right]\)
\(=2\times{\frac{4}{21}}\)
\(=\frac{8}{21}\)
\(Q.3.(xxix)\) \(\int_{0}^{\frac{\pi}{2}}{\cos^3{\theta}\sqrt[3]{\sin{\theta}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{9}{20}\)
[ বঃ২০১৭ ]
উত্তরঃ \(\frac{9}{20}\)
[ বঃ২০১৭ ]
সমাধানঃ
ধরি,
\(\sqrt[3]{\sin{\theta}}=t\)
\(\Rightarrow \sin{\theta}=t^3\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t^3)\)
\(\Rightarrow \cos{\theta}=3t^2\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=3t^2dt\)
➜ \(\because t=\sqrt[3]{\sin{\theta}}\)
\(\Rightarrow t=\sqrt[3]{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt[3]{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt[3]{\sin{\theta}}\)
\(\Rightarrow t=\sqrt[3]{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt[3]{1}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\cos^3{\theta}\sqrt[3]{\sin{\theta}}d\theta}\)\(\sqrt[3]{\sin{\theta}}=t\)
\(\Rightarrow \sin{\theta}=t^3\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t^3)\)
\(\Rightarrow \cos{\theta}=3t^2\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=3t^2dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\Rightarrow t=\sqrt[3]{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt[3]{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt[3]{\sin{\theta}}\)
\(\Rightarrow t=\sqrt[3]{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt[3]{1}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\cos^2{\theta}\sqrt[3]{\sin{\theta}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{(1-\sin^2{\theta})\sqrt[3]{\sin{\theta}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\{1-(\sqrt[3]{\sin{\theta}})^6\}\sqrt[3]{\sin{\theta}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{1}{\{1-t^6\}t.3t^2dt}\)
\(=3\int_{0}^{1}{t^3(1-t^6)dt}\)
\(=3\int_{0}^{1}{(t^3-t^9)dt}\)
\(=3\left[\frac{t^4}{4}-\frac{t^{10}}{10}\right]_{0}^{1}\) ➜ \(\because \int{x^ndx}=\frac{n^{n+1}}{n+1}\)
\(=3\left[\left(\frac{1^4}{4}-\frac{1^{10}}{10}\right)-\left(\frac{0^4}{4}-\frac{0^{10}}{10}\right)\right]\)
\(=3\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{0}{4}-\frac{0}{10}\right)\right]\)
\(=3\left[\left(\frac{5-2}{20}\right)-\left(0-0\right)\right]\)
\(=3\left[\left(\frac{3}{20}\right)-0\right]\)
\(=3\times{\frac{3}{20}}\)
\(=\frac{9}{20}\)
\(Q.3.(xxx)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos^3{\theta}}{\sqrt{\sin{\theta}}}d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{8}{5}\)
[ রাঃ২০১২; চঃ,বঃ২০১০ ]
উত্তরঃ \(\frac{8}{5}\)
[ রাঃ২০১২; চঃ,বঃ২০১০ ]
সমাধানঃ
ধরি,
\(\sqrt{\sin{\theta}}=t\)
\(\Rightarrow \sin{\theta}=t^2\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t^2)\)
\(\Rightarrow \cos{\theta}=2t\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=2tdt\)
➜ \(\because t=\sqrt{\sin{\theta}}\)
\(\Rightarrow t=\sqrt{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{\sin{\theta}}\)
\(\Rightarrow t=\sqrt{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos^3{\theta}}{\sqrt{\sin{\theta}}}d\theta}\)\(\sqrt{\sin{\theta}}=t\)
\(\Rightarrow \sin{\theta}=t^2\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t^2)\)
\(\Rightarrow \cos{\theta}=2t\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=2tdt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\Rightarrow t=\sqrt{\sin{0}}\), যখন \(x=0\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{\sin{\theta}}\)
\(\Rightarrow t=\sqrt{\sin{\left(\frac{\pi}{2}\right)}}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=\sqrt{1}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\cos^2{\theta}}{\sqrt{\sin{\theta}}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1-\sin^2{\theta}}{\sqrt{\sin{\theta}}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1-(\sqrt{\sin{\theta}})^4}{\sqrt{\sin{\theta}}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1-(\sqrt{\sin{\theta}})^4}{\sqrt{\sin{\theta}}}.\cos{\theta}d\theta}\)
\(=\int_{0}^{1}{\frac{1-t^4}{t}\times{2tdt}}\)
\(=2\int_{0}^{1}{(1-t^4)dt}\)
\(=2\left[t-\frac{t^5}{5}\right]_{0}^{1}\) ➜ \(\because \int{1dx}=x, \int{x^ndx}=\frac{n^{n+1}}{n+1}\)
\(=2\left[\left(1-\frac{1^5}{5}\right)-\left(0-\frac{0^5}{5}\right)\right]\)
\(=2\left[\left(1-\frac{1}{5}\right)-\left(0-\frac{0}{5}\right)\right]\)
\(=2\left[\frac{5-1}{5}-\left(0-0\right)\right]\)
\(=2\left[\frac{4}{5}-0\right]\)
\(=2\times{\frac{4}{5}}\)
\(=\frac{8}{5}\)
\(Q.3.(xxxi)\) \(\int_{0}^{\pi}{\frac{xdx}{1+\sin{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi\)
উত্তরঃ \(\pi\)
সমাধানঃ
প্রথম ক্ষেত্রে,
এখানে, LIATE শব্দের
\(A=x, \ \ T=\sec^2{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
দ্বিতীয় ক্ষেত্রে,
এখানে, LIATE শব্দের
\(A=x, \ \ T=\sec{x}\tan{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\pi}{\frac{xdx}{1+\sin{x}}}\)এখানে, LIATE শব্দের
\(A=x, \ \ T=\sec^2{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
দ্বিতীয় ক্ষেত্রে,
এখানে, LIATE শব্দের
\(A=x, \ \ T=\sec{x}\tan{x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
\(=\int_{0}^{\pi}{\frac{x(1-\sin{x})}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \((1-\sin{x})\) গুণ করে।
\(=\int_{0}^{\pi}{\frac{x-x\sin{x}}{1-\sin^2{x}}dx}\)
\(=\int_{0}^{\pi}{\frac{x-x\sin{x}}{\cos^2{x}}dx}\)
\(=\int_{0}^{\pi}{\left(\frac{x}{\cos^2{x}}-\frac{x\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\pi}{\left(x\frac{1}{\cos^2{x}}-x\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int_{0}^{\pi}{\left(x\sec^2{x}-x\sec{x}\tan{x}\right)dx}\)
\(=\int_{0}^{\pi}{x\sec^2{x}dx}-\int_{0}^{\pi}{x\sec{x}\tan{x}dx}\)
\(=\left[x\tan{x}\right]_{0}^{\pi}-\int_{0}^{\pi}{\left\{\frac{d}{dx}(x)\int{\sec^2{x}dx}\right\}dx}-\left[x\sec{x}\right]_{0}^{\pi}+\int_{0}^{\pi}{\left\{\frac{d}{dx}(x)\int{\sec{x}\tan{x}dx}\right\}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\pi\tan{(\pi)}-0\tan{(0)}\right]-\int_{0}^{\pi}{1.\tan{x}dx}-\left[\pi\sec{(\pi)}-0\sec{(0)}\right]+\int_{0}^{\pi}{1.\sec{x}dx}\) ➜ \(\because \frac{d}{dx}(x)=1,\) \(\int{\sec^2{x}dx}=\tan{x},\) \(\int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\left[\pi.0-0\right]-\int_{0}^{\pi}{\tan{x}dx}-\left[\pi(-1)-0\right]+\int_{0}^{\pi}{\sec{x}dx}\)
\(=\left[0-0\right]-\left[\ln{|\sec{x}|}\right]_{0}^{\pi}-\left[-\pi-0\right]+\left[\ln{|\sec{x}+\tan{x}|}\right]_{0}^{\pi}\) ➜ \(\because \int{\tan{x}dx}=\ln{|\sec{x}|},\) \(\int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}\)
\(=0-\left[\ln{|\sec{\pi}|}-\ln{|\sec{0}|}\right]+\pi-\left[\ln{|\sec{(\pi)}+\tan{(\pi)}|}+\ln{|\sec{(0)}+\tan{(0)}|}\right]\)
\(=-\left[\ln{|-1|}-\ln{|1|}\right]+\pi+\left[\ln{|-1+0|}-\ln{|1+0|}\right]\)
\(=-\left[\ln{(1)}-\ln{(1)}\right]+\pi+\left[\ln{(1)}-\ln{(1)}\right]\)
\(=-0+\pi+0\)
\(=\pi\)
\(Q.3.(xxxii)\) \(\int_{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\ln{\left(\frac{5}{3}\right)}\)
উত্তরঃ \(\frac{1}{2}\ln{\left(\frac{5}{3}\right)}\)
সমাধানঃ
ধরি,
\(\sqrt{x+1}=t\)
\(\Rightarrow x+1=t^2\)
\(\Rightarrow x=t^2-1\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^2-1)\)
\(\Rightarrow 1=2t\frac{dt}{dx}-0\)
\(\Rightarrow 1=2t\frac{dt}{dx}\)
\(\therefore dx=2tdt\)
➜ \(\because t=\sqrt{x+1}\)
\(\Rightarrow t=\sqrt{8+1}\), যখন \(x=8\)
\(\Rightarrow t=\sqrt{9}\)
\(\therefore t=3\)
আবার,
\(t=\sqrt{x+1}\)
\(\Rightarrow t=\sqrt{15+1}\), যখন \(x=15\)
\(\Rightarrow t=\sqrt{16}\)
\(\therefore t=4\)
\(\int_{8}^{15}{\frac{dx}{(x-3)\sqrt{x+1}}}\)\(\sqrt{x+1}=t\)
\(\Rightarrow x+1=t^2\)
\(\Rightarrow x=t^2-1\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^2-1)\)
\(\Rightarrow 1=2t\frac{dt}{dx}-0\)
\(\Rightarrow 1=2t\frac{dt}{dx}\)
\(\therefore dx=2tdt\)
| \(x\) | \(8\) | \(15\) |
| \(t\) | \(3\) | \(4\) |
\(\Rightarrow t=\sqrt{8+1}\), যখন \(x=8\)
\(\Rightarrow t=\sqrt{9}\)
\(\therefore t=3\)
আবার,
\(t=\sqrt{x+1}\)
\(\Rightarrow t=\sqrt{15+1}\), যখন \(x=15\)
\(\Rightarrow t=\sqrt{16}\)
\(\therefore t=4\)
\(=\int_{8}^{15}{\frac{1}{(x-3)\sqrt{x+1}}dx}\)
\(=\int_{3}^{4}{\frac{1}{(t^2-1-3)t}\times{2tdt}}\)
\(=2\int_{3}^{4}{\frac{1}{t^2-4}dt}\)
\(=2\int_{3}^{4}{\frac{1}{t^2-2^2}dt}\)
\(=2\left[\frac{1}{2.2}\ln{\left|\frac{t-2}{t+2}\right|}\right]_{3}^{4}\) ➜ \(\because \int{\frac{1}{x^2-a^2}dx}=\frac{1}{2a}\ln{\left|\frac{x-a}{x+a}\right|}\)
\(=2\left[\frac{1}{4}\ln{\left|\frac{t-2}{t+2}\right|}\right]_{3}^{4}\)
\(=\frac{1}{2}\left[\ln{\left|\frac{t-2}{t+2}\right|}\right]_{3}^{4}\)
\(=\frac{1}{2}\left[\ln{\left|\frac{4-2}{4+2}\right|}-\ln{\left|\frac{3-2}{3+2}\right|}\right]\)
\(=\frac{1}{2}\left[\ln{\left(\frac{2}{6}\right)}-\ln{\left(\frac{1}{5}\right)}\right]\)
\(=\frac{1}{2}\left[\ln{\left(\frac{1}{3}\right)}-\ln{\left(\frac{1}{5}\right)}\right]\)
\(=\frac{1}{2}\left[\ln{\left(\frac{\frac{1}{3}}{\frac{1}{5}}\right)}\right]\)
\(=\frac{1}{2}\ln{\left(\frac{5}{3}\right)}\)
\(Q.3.(xxxiii)\) \(\int_{\frac{1}{2}}^{1}{\frac{dx}{x\sqrt{4x^2-1}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{3}\)
[ বুয়েটঃ২০০৪ ]
উত্তরঃ \(\frac{\pi}{3}\)
[ বুয়েটঃ২০০৪ ]
সমাধানঃ
\(\int_{\frac{1}{2}}^{1}{\frac{dx}{x\sqrt{4x^2-1}}}\)
\(=\int_{\frac{1}{2}}^{1}{\frac{2dx}{2x\sqrt{(2x)^2-1}}}\)
\(=\int_{\frac{1}{2}}^{1}{\frac{d(2x)}{2x\sqrt{(2x)^2-1}}}\)
\(=\left[\sec^{-1}{(2x)}\right]_{\frac{1}{2}}^{1}\) ➜ \(\because \int{\frac{dx}{x\sqrt{x^2-1}}}=\sec^{-1}{x}\)
\(=\sec^{-1}{(2.1)}-\sec^{-1}{\left(2.\frac{1}{2}\right)}\)
\(=\sec^{-1}{(2)}-\sec^{-1}{(1)}\)
\(=\sec^{-1}{\sec{\left(\frac{\pi}{3}\right)}}-\sec^{-1}{\sec{0}}\)
\(=\frac{\pi}{3}-0\)
\(=\frac{\pi}{3}\)
\(=\int_{\frac{1}{2}}^{1}{\frac{2dx}{2x\sqrt{(2x)^2-1}}}\)
\(=\int_{\frac{1}{2}}^{1}{\frac{d(2x)}{2x\sqrt{(2x)^2-1}}}\)
\(=\left[\sec^{-1}{(2x)}\right]_{\frac{1}{2}}^{1}\) ➜ \(\because \int{\frac{dx}{x\sqrt{x^2-1}}}=\sec^{-1}{x}\)
\(=\sec^{-1}{(2.1)}-\sec^{-1}{\left(2.\frac{1}{2}\right)}\)
\(=\sec^{-1}{(2)}-\sec^{-1}{(1)}\)
\(=\sec^{-1}{\sec{\left(\frac{\pi}{3}\right)}}-\sec^{-1}{\sec{0}}\)
\(=\frac{\pi}{3}-0\)
\(=\frac{\pi}{3}\)
\(Q.3.(xxxiv)\) \(\int_{1}^{2}{\frac{dx}{x^2\sqrt{4-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\sqrt{3}}{4}\)
[ বুয়েটঃ২০০৪ ]
উত্তরঃ \(\frac{\sqrt{3}}{4}\)
[ বুয়েটঃ২০০৪ ]
সমাধানঃ
ধরি,
\(x=2\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=2\cos{t}\frac{dt}{dx}\)
\(\therefore dx=2\cos{t}dt\)
➜ \(\because 2\sin{t}=x\)
\(\Rightarrow \sin{t}=\frac{x}{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{2}\right)}\), যখন \(x=1\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore t=\frac{\pi}{6}\)
আবার,
\(t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{2}{2}\right)}\), যখন \(x=2\)
\(\Rightarrow t=\sin^{-1}{1}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(\int_{1}^{2}{\frac{dx}{x^2\sqrt{4-x^2}}}\)\(x=2\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=2\cos{t}\frac{dt}{dx}\)
\(\therefore dx=2\cos{t}dt\)
| \(x\) | \(1\) | \(2\) |
| \(t\) | \(\frac{\pi}{6}\) | \(\frac{\pi}{2}\) |
\(\Rightarrow \sin{t}=\frac{x}{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{2}\right)}\), যখন \(x=1\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore t=\frac{\pi}{6}\)
আবার,
\(t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{2}{2}\right)}\), যখন \(x=2\)
\(\Rightarrow t=\sin^{-1}{1}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}\)
\(\therefore t=\frac{\pi}{2}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{2\cos{t}dt}{4\sin^2{t}\sqrt{4-4\sin^2{t}}}}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos{t}dt}{2\sin^2{t}\sqrt{4(1-\sin^2{t})}}}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos{t}dt}{2\sin^2{t}.2\sqrt{1-\sin^2{t}}}}\)
\(=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{\cos{t}dt}{4\sin^2{t}\cos{t}}}\) ➜ \(\because \sqrt{1-\sin^2{A}}=\cos{A}\)
\(=\frac{1}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\frac{1}{\sin^2{t}}dt}\)
\(=\frac{1}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\ cosec^{2}{t}dt}\) ➜ \(\because \frac{1}{\sin^2{A}}=\ cosec \ {A}\)
\(=\frac{1}{4}\left[-\cot{t}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\) ➜ \(\because \int{cosec^2{x}dx}=-\cot{x}\)
\(=-\frac{1}{4}\left[\cot{\left(\frac{\pi}{2}\right)}-\cot{\left(\frac{\pi}{6}\right)}\right]\)
\(=-\frac{1}{4}\left[0-\sqrt{3}\right]\)
\(=\frac{\sqrt{3}}{4}\)
\(Q.3.(xxxv)\) \(\int_{0}^{\frac{\pi}{6}}{\frac{dx}{1-\tan^2{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
[ বুয়েটঃ২০০৮; বঃ২০১৫ ]
উত্তরঃ \(\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
[ বুয়েটঃ২০০৮; বঃ২০১৫ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{6}}{\frac{dx}{1-\tan^2{x}}}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{1-\frac{\sin^2{x}}{\cos^2{x}}}dx}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{\cos^2{x}}{\cos^2{x}-\sin^2{x}}dx}\) ➜ লব ও হরের সহিত \(\cos^2{x}\) গুণ করে।
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\frac{2\cos^2{x}}{\cos{2x}}dx}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\frac{1+\cos{2x}}{\cos{2x}}dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\left(\frac{1}{\cos{2x}}+\frac{\cos{2x}}{\cos{2x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{(\sec{2x}+1)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\sec{2x}dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{1dx}\)
\(=\frac{1}{2}\left[\frac{1}{2}\ln{|\sec{2x}+\tan{2x}|}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{6}}\) ➜ \(\because \int{\sec{ax}dx}=\frac{1}{a}\ln{|\sec{ax}+\tan{ax}|}\)
\(=\frac{1}{4}\left[\ln{|\sec{2x}+\tan{2x}|}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}\left[\frac{\pi}{6}-0\right]\)
\(=\frac{1}{4}\left[\ln{|\sec{\left(2.\frac{\pi}{6}\right)}+\tan{\left(2.\frac{\pi}{6}\right)}|}-\ln{|\sec{2.0}+\tan{2.0}|}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{|\sec{\left(\frac{\pi}{3}\right)}+\tan{\left(\frac{\pi}{3}\right)}|}-\ln{|\sec{0}+\tan{0}|}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-\ln{(1+0)}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-\ln{(1)}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-0\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\ln{(2+\sqrt{3})}+\frac{\pi}{12}\)
\(=\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{1-\frac{\sin^2{x}}{\cos^2{x}}}dx}\) ➜ \(\because \tan{A}=\frac{\sin{A}}{\cos{A}}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{\cos^2{x}}{\cos^2{x}-\sin^2{x}}dx}\) ➜ লব ও হরের সহিত \(\cos^2{x}\) গুণ করে।
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\frac{2\cos^2{x}}{\cos{2x}}dx}\) ➜ \(\because \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\frac{1+\cos{2x}}{\cos{2x}}dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\left(\frac{1}{\cos{2x}}+\frac{\cos{2x}}{\cos{2x}}\right)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{(\sec{2x}+1)dx}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{\sec{2x}dx}+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}{1dx}\)
\(=\frac{1}{2}\left[\frac{1}{2}\ln{|\sec{2x}+\tan{2x}|}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{6}}\) ➜ \(\because \int{\sec{ax}dx}=\frac{1}{a}\ln{|\sec{ax}+\tan{ax}|}\)
\(=\frac{1}{4}\left[\ln{|\sec{2x}+\tan{2x}|}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}\left[\frac{\pi}{6}-0\right]\)
\(=\frac{1}{4}\left[\ln{|\sec{\left(2.\frac{\pi}{6}\right)}+\tan{\left(2.\frac{\pi}{6}\right)}|}-\ln{|\sec{2.0}+\tan{2.0}|}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{|\sec{\left(\frac{\pi}{3}\right)}+\tan{\left(\frac{\pi}{3}\right)}|}-\ln{|\sec{0}+\tan{0}|}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-\ln{(1+0)}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-\ln{(1)}\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\left[\ln{(2+\sqrt{3})}-0\right]+\frac{\pi}{12}\)
\(=\frac{1}{4}\ln{(2+\sqrt{3})}+\frac{\pi}{12}\)
\(=\frac{\pi}{12}+\frac{1}{4}\ln{(2+\sqrt{3})}\)
\(Q.3.(xxxvi)\) \(\int_{0}^{\infty}{e^{-2x}\cos{4x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{10}\)
[ কুয়েটঃ২০১৩-২০১৪ ]
উত্তরঃ \(\frac{1}{10}\)
[ কুয়েটঃ২০১৩-২০১৪ ]
সমাধানঃ
\(\int_{0}^{\infty}{e^{-2x}\cos{4x}dx}\)
\(=\left[\frac{e^{-2x}}{(-2)^2+4^2}(-2\cos{4x}+4\sin{4x})\right]_{0}^{\infty}\) ➜ \(\because \int{e^{ax}\cos{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})\)
\(=\left[\frac{e^{-2x}}{4+16}(-2\cos{4x}+4\sin{4x})\right]_{0}^{\infty}\)
\(=\left[\frac{e^{-2x}}{20}\times{2}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\left[\frac{e^{-2x}}{10}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\frac{1}{10}\left[e^{-2x}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\frac{1}{10}\left[e^{-2\times{\infty}}\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-e^{-2\times{0}}\{2\sin{(4\times{0})}-\cos{(4\times{0})}\}\right]\)
\(=\frac{1}{10}\left[e^{-\infty}\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-e^{0}\{2\sin{(0)}-\cos{(0)}\}\right]\)
\(=\frac{1}{10}\left[0.\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-1(2.0-1)\right]\) ➜ \(\because e^{-\infty}=0, e^{0}=1\)
\(=\frac{1}{10}\left[0-1(0-1)\right]\)
\(=\frac{1}{10}\left[-1(-1)\right]\)
\(=\frac{1}{10}\times{1}\)
\(=\frac{1}{10}\)
\(=\left[\frac{e^{-2x}}{(-2)^2+4^2}(-2\cos{4x}+4\sin{4x})\right]_{0}^{\infty}\) ➜ \(\because \int{e^{ax}\cos{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})\)
\(=\left[\frac{e^{-2x}}{4+16}(-2\cos{4x}+4\sin{4x})\right]_{0}^{\infty}\)
\(=\left[\frac{e^{-2x}}{20}\times{2}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\left[\frac{e^{-2x}}{10}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\frac{1}{10}\left[e^{-2x}(2\sin{4x}-\cos{4x})\right]_{0}^{\infty}\)
\(=\frac{1}{10}\left[e^{-2\times{\infty}}\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-e^{-2\times{0}}\{2\sin{(4\times{0})}-\cos{(4\times{0})}\}\right]\)
\(=\frac{1}{10}\left[e^{-\infty}\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-e^{0}\{2\sin{(0)}-\cos{(0)}\}\right]\)
\(=\frac{1}{10}\left[0.\{2\sin{(4\times{\infty})}-\cos{(4\times{\infty})}\}-1(2.0-1)\right]\) ➜ \(\because e^{-\infty}=0, e^{0}=1\)
\(=\frac{1}{10}\left[0-1(0-1)\right]\)
\(=\frac{1}{10}\left[-1(-1)\right]\)
\(=\frac{1}{10}\times{1}\)
\(=\frac{1}{10}\)
\(Q.3.(xxxvii)\) \(\int_{2}^{3}{\frac{dx}{2(x-1)\sqrt{x^2-2x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{6}\)
[ বুয়েটঃ২০০১-২০০২,২০০৩-২০০৪ ]
উত্তরঃ \(\frac{\pi}{6}\)
[ বুয়েটঃ২০০১-২০০২,২০০৩-২০০৪ ]
সমাধানঃ
ধরি,
\(\sqrt{x^2-2x}=t\)
\(\Rightarrow x^2-2x=t^2\)
\(\Rightarrow x^2-2x+1=t^2+1\)
\(\Rightarrow (x-1)^2=t^2+1\)
\(\Rightarrow \frac{d}{dx}(x-1)^2=\frac{dt}{dx}(t^2+1)\)
\(\Rightarrow 2(x-1)=2t\frac{dt}{dx}+0\)
\(\Rightarrow (x-1)=t\frac{dt}{dx}\)
\(\therefore (x-1)dx=tdt\)
➜ \(\because t=\sqrt{x^2-2x}\)
\(\Rightarrow t=\sqrt{2^2-2.2}\), যখন \(x=2\)
\(\Rightarrow t=\sqrt{4-4}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{x^2-2x}\)
\(\Rightarrow t=\sqrt{3^2-2.3}\), যখন \(x=3\)
\(\Rightarrow t=\sqrt{9-6}\)
\(\therefore t=\sqrt{3}\)
\(\int_{2}^{3}{\frac{dx}{2(x-1)\sqrt{x^2-2x}}}\)\(\sqrt{x^2-2x}=t\)
\(\Rightarrow x^2-2x=t^2\)
\(\Rightarrow x^2-2x+1=t^2+1\)
\(\Rightarrow (x-1)^2=t^2+1\)
\(\Rightarrow \frac{d}{dx}(x-1)^2=\frac{dt}{dx}(t^2+1)\)
\(\Rightarrow 2(x-1)=2t\frac{dt}{dx}+0\)
\(\Rightarrow (x-1)=t\frac{dt}{dx}\)
\(\therefore (x-1)dx=tdt\)
| \(x\) | \(2\) | \(3\) |
| \(t\) | \(0\) | \(\sqrt{3}\) |
\(\Rightarrow t=\sqrt{2^2-2.2}\), যখন \(x=2\)
\(\Rightarrow t=\sqrt{4-4}\)
\(\Rightarrow t=\sqrt{0}\)
\(\therefore t=0\)
আবার,
\(t=\sqrt{x^2-2x}\)
\(\Rightarrow t=\sqrt{3^2-2.3}\), যখন \(x=3\)
\(\Rightarrow t=\sqrt{9-6}\)
\(\therefore t=\sqrt{3}\)
\(=\int_{2}^{3}{\frac{(x-1)dx}{2(x-1)^2\sqrt{x^2-2x}}}\)
\(=\int_{0}^{\sqrt{3}}{\frac{tdt}{2(t^2+1)t}}\)
\(=\frac{1}{2}\int_{0}^{\sqrt{3}}{\frac{dt}{t^2+1}}\)
\(=\frac{1}{2}\int_{0}^{\sqrt{3}}{\frac{1}{1+t^2}dt}\)
\(=\frac{1}{2}\left[\tan^{-1}{t}\right]_{0}^{\sqrt{3}}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\frac{1}{2}\left[\tan^{-1}{(\sqrt{3})}-\tan^{-1}{(0)}\right]\)
\(=\frac{1}{2}\left[\tan^{-1}{\tan{\left(\frac{\pi}{3}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{2}\left[\frac{\pi}{3}-0\right]\)
\(=\frac{\pi}{6}\)
\(Q.3.(xxxviii)\) \(\int_{2}^{e}{\left\{\frac{1}{\ln{|x|}}-\frac{1}{(\ln{|x|})^2}\right\}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(e-\frac{2}{\ln{2}}\)
[ বুয়েটঃ২০০৩-২০০৪; বিআইটিঃ১৯৯৮-১৯৯৫ ]
উত্তরঃ \(e-\frac{2}{\ln{2}}\)
[ বুয়েটঃ২০০৩-২০০৪; বিআইটিঃ১৯৯৮-১৯৯৫ ]
সমাধানঃ
\(\int_{2}^{e}{\left\{\frac{1}{\ln{|x|}}-\frac{1}{(\ln{|x|})^2}\right\}dx}\)
\(=\int_{2}^{e}{\frac{1}{\ln{|x|}}.1dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\)
\(=\left[x\frac{1}{\ln{|x|}}\right]_{2}^{e}-\int_{2}^{e}{\left\{\frac{d}{dx}\left(\frac{1}{\ln{|x|}}\right)\int{1dx}\right\}dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=e\frac{1}{\ln{(e)}}-2\frac{1}{\ln{(2)}}-\int_{2}^{e}{\left(-\frac{1}{(\ln{|x|})^2}\right).\frac{1}{x}.xdx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\) ➜ \(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},\) \(\int{1dx}=x\)
\(=e\frac{1}{1}-\frac{2}{\ln{(2)}}+\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\)
\(=e.1-\frac{2}{\ln{(2)}}\)
\(=e-\frac{2}{\ln{(2)}}\)
\(=\int_{2}^{e}{\frac{1}{\ln{|x|}}.1dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\)
\(=\left[x\frac{1}{\ln{|x|}}\right]_{2}^{e}-\int_{2}^{e}{\left\{\frac{d}{dx}\left(\frac{1}{\ln{|x|}}\right)\int{1dx}\right\}dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\) ➜ \(\because \int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{1dx}=x\)
\(=e\frac{1}{\ln{(e)}}-2\frac{1}{\ln{(2)}}-\int_{2}^{e}{\left(-\frac{1}{(\ln{|x|})^2}\right).\frac{1}{x}.xdx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\) ➜ \(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2},\) \(\int{1dx}=x\)
\(=e\frac{1}{1}-\frac{2}{\ln{(2)}}+\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}-\int_{2}^{e}{\frac{1}{(\ln{|x|})^2}dx}\)
\(=e.1-\frac{2}{\ln{(2)}}\)
\(=e-\frac{2}{\ln{(2)}}\)
\(Q.3.(xxxix)\) \(\int_{0}^{a}{\frac{a^2-x^2}{(a^2+x^2)^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2a}\)
[ বুয়েটঃ২০০০-২০০১ ]
উত্তরঃ \(\frac{1}{2a}\)
[ বুয়েটঃ২০০০-২০০১ ]
সমাধানঃ
ধরি,
\(x=a\tan{t}\)
\(\Rightarrow \frac{d}{dx}(x)=a\frac{d}{dx}(\tan{t})\)
\(\Rightarrow 1=a\sec^2{t}\frac{dt}{dx}\)
\(\therefore dx=a\sec^2{t}dt\)
➜ \(\because 2\sin{t}=x\)
\(\Rightarrow \tan{t}=\frac{x}{a}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{0}{a}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{(0)}\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{a}{a}\right)}\), যখন \(x=a\)
\(\Rightarrow t=\tan^{-1}{(1)}\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(\int_{0}^{a}{\frac{a^2-x^2}{(a^2+x^2)^2}dx}\)\(x=a\tan{t}\)
\(\Rightarrow \frac{d}{dx}(x)=a\frac{d}{dx}(\tan{t})\)
\(\Rightarrow 1=a\sec^2{t}\frac{dt}{dx}\)
\(\therefore dx=a\sec^2{t}dt\)
| \(x\) | \(0\) | \(a\) |
| \(t\) | \(0\) | \(\frac{\pi}{4}\) |
\(\Rightarrow \tan{t}=\frac{x}{a}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{0}{a}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{(0)}\)
\(\Rightarrow t=\tan^{-1}{\tan{0}}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{a}{a}\right)}\), যখন \(x=a\)
\(\Rightarrow t=\tan^{-1}{(1)}\)
\(\Rightarrow t=\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(=\int_{0}^{a}{\frac{2a^2-a^2-x^2}{(a^2+x^2)^2}dx}\)
\(=\int_{0}^{a}{\frac{2a^2-(a^2+x^2)}{(a^2+x^2)^2}dx}\)
\(=\int_{0}^{a}{\left\{\frac{2a^2}{(a^2+x^2)^2}-\frac{a^2+x^2}{(a^2+x^2)^2}\right\}dx}\)
\(=\int_{0}^{a}{\left\{\frac{2a^2}{(a^2+x^2)^2}-\frac{1}{a^2+x^2}\right\}dx}\)
\(=\int_{0}^{a}{\frac{2a^2}{(a^2+x^2)^2}dx}-\int_{0}^{a}{\frac{1}{a^2+x^2}dx}\)
\(=2a^2\int_{0}^{a}{\frac{1}{(a^2+x^2)^2}dx}-\left[\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=2a^2\int_{0}^{\frac{\pi}{4}}{\frac{1}{(a^2+a^2\tan^2{t})^2}a\sec^2{t}dt}-\frac{1}{a}\left[\tan^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\)
\(=2a^2\int_{0}^{\frac{\pi}{4}}{\frac{1}{a^4(1+\tan^2{t})^2}a\sec^2{t}dt}-\frac{1}{a}\left[\tan^{-1}{\left(\frac{a}{a}\right)}-\tan^{-1}{\left(\frac{0}{a}\right)}\right]\)
\(=\frac{2}{a}\int_{0}^{\frac{\pi}{4}}{\frac{1}{(\sec^2{t})^2}\sec^2{t}dt}-\frac{1}{a}\left[\tan^{-1}{(1)}-\tan^{-1}{(0)}\right]\) ➜ \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\frac{2}{a}\int_{0}^{\frac{\pi}{4}}{\frac{1}{\sec^2{t}}dt}-\frac{1}{a}\left[\tan^{-1}{\tan{\left(\frac{\pi}{4}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{a}\int_{0}^{\frac{\pi}{4}}{2\cos^2{t}dt}-\frac{1}{a}\left[\frac{\pi}{4}-0\right]\) ➜ \(\because \frac{1}{\sec^2{A}}=\cos^2{A}\)
\(=\frac{1}{a}\int_{0}^{\frac{\pi}{4}}{(1+\cos{2t})dt}-\frac{\pi}{4a}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{a}\left[t+\frac{1}{2}\sin{2t}\right]_{0}^{\frac{\pi}{4}}-\frac{\pi}{4a}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{a}\left[\frac{\pi}{4}+\frac{1}{2}\sin{\left(2\times{\frac{\pi}{4}}\right)}-0-\frac{1}{2}\sin{(2\times{0})}\right]-\frac{\pi}{4a}\)
\(=\frac{1}{a}\left[\frac{\pi}{4}+\frac{1}{2}\sin{\frac{\pi}{2}}-\frac{1}{2}\sin{0}\right]-\frac{\pi}{4a}\)
\(=\frac{1}{a}\left[\frac{\pi}{4}+\frac{1}{2}.1-\frac{1}{2}.0\right]-\frac{\pi}{4a}\)
\(=\frac{1}{a}\left[\frac{\pi}{4}+\frac{1}{2}-0\right]-\frac{\pi}{4a}\)
\(=\frac{\pi}{4a}+\frac{1}{2a}-\frac{\pi}{4a}\)
\(=\frac{1}{2a}\)
\(Q.3.(xL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})(2+\sin{x})}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{\left(\frac{4}{3}\right)}\)
[ কুয়েটঃ২০০৯-২০১০; বিআইটিঃ১৯৯৭-১৯৯৮ ]
উত্তরঃ \(\ln{\left(\frac{4}{3}\right)}\)
[ কুয়েটঃ২০০৯-২০১০; বিআইটিঃ১৯৯৭-১৯৯৮ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=\sin{x}\)
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})(2+\sin{x})}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(1\) |
\(\Rightarrow t=\sin{0}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=\sin{x}\)
\(\Rightarrow t=\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\therefore t=1\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{(1+\sin{x})(2+\sin{x})}.\cos{x}dx}\)
\(=\int_{0}^{1}{\frac{1}{(1+t)(2+t)}dt}\)
\(=\int_{0}^{1}{\frac{(2+t)-(1+t)}{(1+t)(2+t)}dt}\)
\(=\int_{0}^{1}{\left(\frac{(2+t)}{(1+t)(2+t)}-\frac{(1+t)}{(1+t)(2+t)}\right)dt}\)
\(=\int_{0}^{1}{\left(\frac{1}{1+t}-\frac{1}{2+t}\right)dt}\)
\(=\int_{0}^{1}{\frac{1}{1+t}dt}-\int_{0}^{1}{\frac{1}{2+t}dt}\)
\(=\left[\ln{|1+t|}\right]_{0}^{1}-\left[\ln{|2+t|}\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\left[\ln{(1+1)}-\ln{(1+0)}\right]-\left[\ln{(2+1)}-\ln{(2+0)}\right]\)
\(=\left[\ln{(2)}-\ln{(1)}\right]-\left[\ln{(3)}-\ln{(2)}\right]\)
\(=\left[\ln{(2)}-0\right]-\left[\ln{(3)}-\ln{(2)}\right]\) ➜ \(\because \ln{(1)}=0\)
\(=\ln{(2)}-\ln{(3)}+\ln{(2)}\)
\(=2\ln{(2)}-\ln{(3)}\)
\(=\ln{(2^2)}-\ln{(3)}\)
\(=\ln{(4)}-\ln{(3)}\)
\(=\ln{\left(\frac{4}{3}\right)}\) ➜ \(\because \ln{(M)}-\ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(Q.3.(xLi)\) \(\int_{0}^{\pi}{x\sin^2{x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi^2}{4}\)
[ বুয়েটঃ২০০৫ ]
উত্তরঃ \(\frac{\pi^2}{4}\)
[ বুয়েটঃ২০০৫ ]
সমাধানঃ
এখানে, LIATE শব্দের
\(A=x, \ \ T=\cos{2x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
\(\int_{0}^{\pi}{x\sin^2{x}dx}\)\(A=x, \ \ T=\cos{2x}\)
যেহেতু A, শব্দের মধ্যে T এর পূর্বে আছে, তাই \(x\) কে \(u\) ধরা হয়েছে।
\(=\frac{1}{2}\int_{0}^{\pi}{x.2\sin^2{x}dx}\)
\(=\frac{1}{2}\int_{0}^{\pi}{x(1-\cos{2x})dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\pi}{(x-x\cos{2x})dx}\)
\(=\frac{1}{2}\int_{0}^{\pi}{xdx}-\frac{1}{2}\int_{0}^{\pi}{x\cos{2x}dx}\)
\(=\frac{1}{2}\left[\frac{x^2}{2}\right]_{0}^{\pi}-\frac{1}{2}\left[x.\frac{1}{2}\sin{2x}\right]_{0}^{\pi}+\int_{0}^{\pi}{\left\{\frac{d}{dx}(x)\int{\cos{2x}dx}\right\}dx}\) ➜ \(\because \int{xdx}=\frac{x^2}{2},\) \(\int{uvdx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx},\) \(\int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{4}\left[x^2\right]_{0}^{\pi}-\frac{1}{4}\left[x\sin{2x}\right]_{0}^{\pi}+\int_{0}^{\pi}{1.\frac{1}{2}\sin{2x}dx}\) ➜ \(\because \frac{d}{dx}(x)=1,\) \(\int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{4}\left[(\pi)^2-0^2\right]-\frac{1}{4}\left[\pi\sin{(2\pi)}-0.\sin{(2.0)}\right]+\frac{1}{2}\int_{0}^{\pi}{\sin{2x}dx}\)
\(=\frac{1}{4}\left[(\pi)^2-0\right]-\frac{1}{4}\left[\pi.0-0\right]+\frac{1}{2}\left[-\frac{1}{2}\cos{2x}\right]_{0}^{\pi}\) ➜ \(\because \sin{(2\pi)}=0,\) \(\int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=\frac{\pi^2}{4}-\frac{1}{4}\left[0-0\right]-\frac{1}{4}\left[\cos{2x}\right]_{0}^{\pi}\)
\(=\frac{\pi^2}{4}-\frac{1}{4}.0-\frac{1}{4}\left[\cos{(2\pi)}-\cos{(2.0)}\right]\)
\(=\frac{\pi^2}{4}-\frac{1}{4}\left[1-\cos{(0)}\right]\)
\(=\frac{\pi^2}{4}-\frac{1}{4}\left[1-1\right]\)
\(=\frac{\pi^2}{4}-\frac{1}{4}.0\)
\(=\frac{\pi^2}{4}\)
\(Q.3.(xLii)\) \(\int_{0}^{a}{\sqrt{a^2-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\pi{a^2}\)
[ ঢাঃ২০১১; চঃ২০১২ ]
উত্তরঃ \(\frac{1}{4}\pi{a^2}\)
[ ঢাঃ২০১১; চঃ২০১২ ]
সমাধানঃ
\(\int_{0}^{a}{\sqrt{a^2-x^2}dx}\)
\(=\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{a\sqrt{a^2-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{a}{a}\right)}-\frac{0.\sqrt{a^2-0^2}}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{0}{a}\right)}\)
\(=\frac{a\sqrt{0}}{2}+\frac{a^2}{2}\sin^{-1}{(1)}-\frac{0}{2}-\frac{a^2}{2}\sin^{-1}{(0)}\)
\(=\frac{a.0}{2}+\frac{a^2}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{a^2}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{a^2}{2}\times{\frac{\pi}{2}}-\frac{a^2}{2}.0\)
\(=0+\frac{1}{4}\pi{a^2}-0\)
\(=\frac{1}{4}\pi{a^2}\)
\(=\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\right]_{0}^{a}\) ➜ \(\because \int{\sqrt{a^2-x^2}dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{a\sqrt{a^2-a^2}}{2}+\frac{a^2}{2}\sin^{-1}{\left(\frac{a}{a}\right)}-\frac{0.\sqrt{a^2-0^2}}{2}-\frac{a^2}{2}\sin^{-1}{\left(\frac{0}{a}\right)}\)
\(=\frac{a\sqrt{0}}{2}+\frac{a^2}{2}\sin^{-1}{(1)}-\frac{0}{2}-\frac{a^2}{2}\sin^{-1}{(0)}\)
\(=\frac{a.0}{2}+\frac{a^2}{2}\sin^{-1}{\sin{\left(\frac{\pi}{2}\right)}}-0-\frac{a^2}{2}\sin^{-1}{\sin{0}}\)
\(=\frac{0}{2}+\frac{a^2}{2}\times{\frac{\pi}{2}}-\frac{a^2}{2}.0\)
\(=0+\frac{1}{4}\pi{a^2}-0\)
\(=\frac{1}{4}\pi{a^2}\)
\(Q.3.(xLiii)\) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin{x}+\cos{x})^2dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\pi\)
উত্তরঃ \(\pi\)
সমাধানঃ
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin{x}+\cos{x})^2dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin^2{x}+2\sin{x}\cos{x}+\cos^2{x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin^2{x}+\cos^2{x}+2\sin{x}\cos{x})dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(1+\sin{2x})dx}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{1dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sin{2x})dx}\)
\(=\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[-\frac{1}{2}\cos{2x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)-\frac{1}{2}\left[\cos{2x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}+\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\left(\frac{\pi}{2}\right)}-\cos{2\left(-\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi+\pi}{2}-\frac{1}{2}\left[\cos{(\pi)}-\cos{(-\pi)}\right]\)
\(=\frac{2\pi}{2}-\frac{1}{2}\left[\cos{(\pi)}-\cos{(\pi)}\right]\) ➜ \(\because \cos{(-\theta)}=\cos{\theta}\)
\(=\pi-\frac{1}{2}\times{0}\)
\(=\pi-0\)
\(=\pi\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin^2{x}+2\sin{x}\cos{x}+\cos^2{x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(\sin^2{x}+\cos^2{x}+2\sin{x}\cos{x})dx}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{(1+\sin{2x})dx}\) ➜ \(\because \sin^2{A}+\cos^2{A}=1, 2\sin{A}\cos{A}=\sin{2A}\)
\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{1dx}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sin{2x})dx}\)
\(=\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[-\frac{1}{2}\cos{2x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)-\frac{1}{2}\left[\cos{2x}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)
\(=\frac{\pi}{2}+\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\left(\frac{\pi}{2}\right)}-\cos{2\left(-\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi+\pi}{2}-\frac{1}{2}\left[\cos{(\pi)}-\cos{(-\pi)}\right]\)
\(=\frac{2\pi}{2}-\frac{1}{2}\left[\cos{(\pi)}-\cos{(\pi)}\right]\) ➜ \(\because \cos{(-\theta)}=\cos{\theta}\)
\(=\pi-\frac{1}{2}\times{0}\)
\(=\pi-0\)
\(=\pi\)
\(Q.3.(xLiv)\) \(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2\)
[ চঃ২০০৪ ]
উত্তরঃ \(2\)
[ চঃ২০০৪ ]
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{(\sin{\theta}+\cos{\theta})d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}\)
\(=\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\)
\(=-\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{(0)}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{(0)}\right]\)
\(=-\left[0-1\right]+\left[1-0\right]\)
\(=1+1\)
\(=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\sin{\theta}d\theta}+\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}\)
\(=\left[-\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}, \int{\cos{x}dx}=\sin{x}\)
\(=-\left[\cos{\theta}\right]_{0}^{\frac{\pi}{2}}+\left[\sin{\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=-\left[\cos{\left(\frac{\pi}{2}\right)}-\cos{(0)}\right]+\left[\sin{\left(\frac{\pi}{2}\right)}-\sin{(0)}\right]\)
\(=-\left[0-1\right]+\left[1-0\right]\)
\(=1+1\)
\(=2\)
\(Q.3.(xLv)\) \(\int_{\frac{\pi}{2}}^{\pi}{(1+\sin{2\theta})d\theta}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{2}-1\)
উত্তরঃ \(\frac{\pi}{2}-1\)
সমাধানঃ
\(\int_{\frac{\pi}{2}}^{\pi}{(1+\sin{2\theta})d\theta}\)
\(=\left[\theta\right]_{\frac{\pi}{2}}^{\pi}+\left[-\frac{1}{2}\cos{2\theta}\right]_{\frac{\pi}{2}}^{\pi}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=\pi-\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\theta}\right]_{\frac{\pi}{2}}^{\pi}\)
\(=\frac{2\pi-\pi}{2}-\frac{1}{2}\left[\cos{2\pi}-\cos{2\left(\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\pi}-\cos{(\pi)}\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[1-(-1)\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[1+1\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\times{2}\)
\(=\frac{\pi}{2}-1\)
\(=\left[\theta\right]_{\frac{\pi}{2}}^{\pi}+\left[-\frac{1}{2}\cos{2\theta}\right]_{\frac{\pi}{2}}^{\pi}\) ➜ \(\because \int{\sin{ax}dx}=-\frac{1}{a}\cos{ax}\)
\(=\pi-\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\theta}\right]_{\frac{\pi}{2}}^{\pi}\)
\(=\frac{2\pi-\pi}{2}-\frac{1}{2}\left[\cos{2\pi}-\cos{2\left(\frac{\pi}{2}\right)}\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[\cos{2\pi}-\cos{(\pi)}\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[1-(-1)\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\left[1+1\right]\)
\(=\frac{\pi}{2}-\frac{1}{2}\times{2}\)
\(=\frac{\pi}{2}-1\)
\(Q.3.(xLvi)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})^3}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{8}\)
[ দিঃ২০১৪ ]
উত্তরঃ \(\frac{3}{8}\)
[ দিঃ২০১৪ ]
সমাধানঃ
ধরি,
\(1+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
➜ \(\because t=1+\sin{x}\)
\(\Rightarrow t=1+\sin{0}\), যখন \(x=0\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\sin{x}\)
\(\Rightarrow t=1+\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
\(\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{(1+\sin{x})^3}dx}\)\(1+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(1\) | \(2\) |
\(\Rightarrow t=1+\sin{0}\), যখন \(x=0\)
\(\Rightarrow t=1+0\)
\(\therefore t=1\)
আবার,
\(t=1+\sin{x}\)
\(\Rightarrow t=1+\sin{\left(\frac{\pi}{2}\right)}\), যখন \(x=\frac{\pi}{2}\)
\(\Rightarrow t=1+1\)
\(\therefore t=2\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{(1+\sin{x})^3}.\cos{x}dx}\)
\(=\int_{1}^{2}{\frac{1}{t^3}dt}\)
\(=\int_{1}^{2}{t^{-3}dt}\)
\(=\left[\frac{t^{-3+1}}{-3+1}\right]_{1}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\)
\(=\left[\frac{t^{-2}}{-2}\right]_{1}^{2}\)
\(=\frac{1}{-2}\left[t^{-2}\right]_{1}^{2}\)
\(=-\frac{1}{2}\left[\frac{1}{t^2}\right]_{1}^{2}\)
\(=-\frac{1}{2}\left[\frac{1}{2^2}-\frac{1}{1^2}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{4}-\frac{1}{1}\right]\)
\(=-\frac{1}{2}\left[\frac{1}{4}-1\right]\)
\(=-\frac{1}{2}\times{\frac{1-4}{4}}\)
\(=-\frac{1}{2}\times{\frac{-3}{4}}\)
\(=\frac{3}{8}\)
\(Q.3.(xLvii)\) \(\int_{0}^{\frac{1}{2}}{\frac{dx}{(1-2x^2)\sqrt{1-x^2}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\ln{(2+\sqrt{3})}\)
উত্তরঃ \(\frac{1}{2}\ln{(2+\sqrt{3})}\)
সমাধানঃ
ধরি,
\(x=\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=\cos{t}\frac{dt}{dx}\)
\(\therefore dx=\cos{t}dt\)
➜ \(\because \sin{t}=x\)
\(\Rightarrow t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{2}\right)}\), যখন \(x=\frac{1}{2}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore t=\frac{\pi}{6}\)
\(\int_{0}^{\frac{1}{2}}{\frac{dx}{(1-2x^2)\sqrt{1-x^2}}}\)\(x=\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=\cos{t}\frac{dt}{dx}\)
\(\therefore dx=\cos{t}dt\)
| \(x\) | \(0\) | \(\frac{1}{2}\) |
| \(t\) | \(0\) | \(2\) |
\(\Rightarrow t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{0}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{x}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{2}\right)}\), যখন \(x=\frac{1}{2}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{6}\right)}}\)
\(\therefore t=\frac{\pi}{6}\)
\(=\int_{0}^{\frac{1}{2}}{\frac{1}{(1-2x^2)\sqrt{1-x^2}}dx}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{(1-2\sin^2{t})\sqrt{1-\sin^2{t}}}\cos{t}dt}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{\cos{2t}\cos{t}}\cos{t}dt}\) ➜ \(\because \sqrt{1-\sin^2{A}}=\cos{A}, 1-2\sin^2{A}=\cos{2A}\)
\(=\int_{0}^{\frac{\pi}{6}}{\frac{1}{\cos{2t}}dt}\)
\(=\int_{0}^{\frac{\pi}{6}}{\sec{2t}dt}\) ➜ \(\because \frac{1}{\cos{2A}}=\sec{2A}\)
\(=\left[\frac{1}{2}\ln{|\sec{2t}+\tan{2t}|}\right]_{0}^{\frac{\pi}{6}}\) ➜ \(\because \int{\sec{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}\)
\(=\frac{1}{2}\left[\ln{|\sec{2t}+\tan{2t}|}\right]_{0}^{\frac{\pi}{6}}\)
\(=\frac{1}{2}\left[\ln{|\sec{2\left(\frac{\pi}{6}\right)}+\tan{2\left(\frac{\pi}{6}\right)}|}-\ln{|\sec{2.0}+\tan{2.0}|}\right]\)
\(=\frac{1}{2}\left[\ln{|\sec{\left(\frac{\pi}{3}\right)}+\tan{\left(\frac{\pi}{3}\right)}|}-\ln{|\sec{0}+\tan{0}|}\right]\)
\(=\frac{1}{2}\left[\ln{(2+\sqrt{3})}-\ln{(1)}\right]\) ➜ \(\because \sec{\left(\frac{\pi}{3}\right)}=2,\) \(\tan{\left(\frac{\pi}{3}\right)}=\sqrt{3},\) \(\sec{0}=1,\) \(\tan{0}=0\)
\(=\frac{1}{2}\left[\ln{(2+\sqrt{3})}-0\right]\) ➜ \(\because \ln{(1)}=0\)
\(=\frac{1}{2}\ln{(2+\sqrt{3})}\)
\(Q.3.(xLviii)\) \(\int_{0}^{\frac{\pi}{4}}{\sec{x}\sqrt{\frac{1-\sin{x}}{1+\sin{x}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(2-\sqrt{2}\)
উত্তরঃ \(2-\sqrt{2}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{4}}{\sec{x}\sqrt{\frac{1-\sin{x}}{1+\sin{x}}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin{x}}\times{\sqrt{1+\sin{x}}}}{\sqrt{1+\sin{x}}\times{\sqrt{1+\sin{x}}}}dx}\) ➜ লব ও হরের সহিত \(\sqrt{1+\sin{x}}\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin^2{x}}}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\cos{x}}{1+\sin{x}}dx}\) ➜ \(\because \sqrt{1-\sin^2{A}}=\cos{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{\cos{x}}\times{\frac{\cos{x}}{1+\sin{x}}}dx}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \((1-\sin{x})\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜\(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A},\) \(\frac{1}{\cos{A}}=\sec{A},\) \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[\sec{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\tan{\left(\frac{\pi}{4}\right)}-\tan{0}-\left[\sec{\left(\frac{\pi}{4}\right)}-\sec{0}\right]\)
\(=1-0-\left[\sqrt{2}-1\right]\)
\(=1-\sqrt{2}+1\)
\(=2-\sqrt{2}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin{x}}\times{\sqrt{1+\sin{x}}}}{\sqrt{1+\sin{x}}\times{\sqrt{1+\sin{x}}}}dx}\) ➜ লব ও হরের সহিত \(\sqrt{1+\sin{x}}\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\sqrt{1-\sin^2{x}}}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec{x}\frac{\cos{x}}{1+\sin{x}}dx}\) ➜ \(\because \sqrt{1-\sin^2{A}}=\cos{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{\cos{x}}\times{\frac{\cos{x}}{1+\sin{x}}}dx}\) ➜ \(\because \sec{A}=\frac{1}{\cos{A}}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1}{1+\sin{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{(1+\sin{x})(1-\sin{x})}dx}\) ➜ লব ও হরের সহিত \((1-\sin{x})\) গুণ করে।
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{1-\sin^2{x}}dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{1-\sin{x}}{\cos^2{x}}dx}\) ➜\(\because 1-\sin^2{A}=\cos^2{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{\sin{x}}{\cos^2{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\frac{1}{\cos^2{x}}-\frac{1}{\cos{x}}.\frac{\sin{x}}{\cos{x}}\right)dx}\)
\(=\int_{0}^{\frac{\pi}{4}}{\left(\sec^2{x}-\sec{x}\tan{x}\right)dx}\) ➜ \(\because \frac{1}{\cos^2{A}}=\sec^2{A},\) \(\frac{1}{\cos{A}}=\sec{A},\) \(\frac{\sin{A}}{\cos{A}}=\tan{A}\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{x}dx}-\int_{0}^{\frac{\pi}{4}}{\sec{x}\tan{x}dx}\)
\(=\left[\tan{x}\right]_{0}^{\frac{\pi}{4}}-\left[\sec{x}\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{\sec{x}\tan{x}dx}=\sec{x}\)
\(=\tan{\left(\frac{\pi}{4}\right)}-\tan{0}-\left[\sec{\left(\frac{\pi}{4}\right)}-\sec{0}\right]\)
\(=1-0-\left[\sqrt{2}-1\right]\)
\(=1-\sqrt{2}+1\)
\(=2-\sqrt{2}\)
\(Q.3.(iL)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
ধরি,
\(I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx} .....(1)\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{\left(\frac{\pi}{2}-x\right)}}}{\sqrt{\sin{\left(\frac{\pi}{2}-x\right)}}+\sqrt{\cos{\left(\frac{\pi}{2}-x\right)}}}dx}\) ➜ \(\because \int_{0}^{\frac{\pi}{2}}{f(x)dx}=\int_{0}^{\frac{\pi}{2}}{f\left(\frac{\pi}{2}-x\right)dx}\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\cos{x}}}{\sqrt{\cos{x}}+\sqrt{\sin{x}}}dx}\)
\(\therefore I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx} .........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(I+I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx}\)
\(\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}{1dx}\)
\(\Rightarrow 2I=\left[x\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x\)
\(\Rightarrow 2I=\frac{\pi}{2}-0\)
\(\Rightarrow 2I=\frac{\pi}{2}\)
\(\therefore I=\frac{\pi}{4}\)
\(I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx} .....(1)\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{\left(\frac{\pi}{2}-x\right)}}}{\sqrt{\sin{\left(\frac{\pi}{2}-x\right)}}+\sqrt{\cos{\left(\frac{\pi}{2}-x\right)}}}dx}\) ➜ \(\because \int_{0}^{\frac{\pi}{2}}{f(x)dx}=\int_{0}^{\frac{\pi}{2}}{f\left(\frac{\pi}{2}-x\right)dx}\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\cos{x}}}{\sqrt{\cos{x}}+\sqrt{\sin{x}}}dx}\)
\(\therefore I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx} .........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(I+I=\int_{0}^{\frac{\pi}{2}}{\frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}dx}\)
\(\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}{1dx}\)
\(\Rightarrow 2I=\left[x\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x\)
\(\Rightarrow 2I=\frac{\pi}{2}-0\)
\(\Rightarrow 2I=\frac{\pi}{2}\)
\(\therefore I=\frac{\pi}{4}\)
\(Q.3.(L)\) \(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\tan{x}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{\pi}{4}\)
উত্তরঃ \(\frac{\pi}{4}\)
সমাধানঃ
\(\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\tan{x}}}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\frac{\sin{x}}{\cos{x}}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\cos{x}+\sin{x}}dx}\) ➜ লব ও হরের সহিত \(\cos{x}\) গুণ করে।
ধরি,
\(I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\cos{x}+\sin{x}}dx} .....(1)\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\left(\frac{\pi}{2}-x\right)}}{\cos{\left(\frac{\pi}{2}-x\right)}+\sin{\left(\frac{\pi}{2}-x\right)}}dx}\) ➜ \(\because \int_{0}^{\frac{\pi}{2}}{f(x)dx}=\int_{0}^{\frac{\pi}{2}}{f\left(\frac{\pi}{2}-x\right)dx}\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(\therefore I=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\cos{x}+\sin{x}}dx} .........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(I+I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}+\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}{1dx}\)
\(\Rightarrow 2I=\left[x\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x\)
\(\Rightarrow 2I=\frac{\pi}{2}-0\)
\(\Rightarrow 2I=\frac{\pi}{2}\)
\(\therefore I=\frac{\pi}{4}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\frac{\sin{x}}{\cos{x}}}dx}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\cos{x}+\sin{x}}dx}\) ➜ লব ও হরের সহিত \(\cos{x}\) গুণ করে।
ধরি,
\(I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}}{\cos{x}+\sin{x}}dx} .....(1)\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{\left(\frac{\pi}{2}-x\right)}}{\cos{\left(\frac{\pi}{2}-x\right)}+\sin{\left(\frac{\pi}{2}-x\right)}}dx}\) ➜ \(\because \int_{0}^{\frac{\pi}{2}}{f(x)dx}=\int_{0}^{\frac{\pi}{2}}{f\left(\frac{\pi}{2}-x\right)dx}\)
\(\Rightarrow I=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
\(\therefore I=\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\cos{x}+\sin{x}}dx} .........(2)\)
\((1)\) ও \((2)\) যোগ করে।
\(I+I=\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}+\sin{x}}{\cos{x}+\sin{x}}dx}\)
\(\Rightarrow 2I=\int_{0}^{\frac{\pi}{2}}{1dx}\)
\(\Rightarrow 2I=\left[x\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x\)
\(\Rightarrow 2I=\frac{\pi}{2}-0\)
\(\Rightarrow 2I=\frac{\pi}{2}\)
\(\therefore I=\frac{\pi}{4}\)
\(Q.3.(Li)\) \(\int_{8}^{27}{\frac{dx}{x-x^{\frac{1}{3}}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3}{2}\ln{\left(\frac{8}{3}\right)}\)
[ যঃ২০১৪ ]
উত্তরঃ \(\frac{3}{2}\ln{\left(\frac{8}{3}\right)}\)
[ যঃ২০১৪ ]
সমাধানঃ
ধরি,
\(x^{\frac{1}{3}}=t\)
\(\Rightarrow x=t^3\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^3)\)
\(\Rightarrow 1=3t^2\frac{dt}{dx}\)
\(\therefore dx=3t^2dt\)
➜ \(\because t=x^{\frac{1}{3}}\)
\(\Rightarrow t=8^{\frac{1}{3}}\), যখন \(x=8\)
\(\Rightarrow t=(2^3)^{\frac{1}{3}}\)
\(\therefore t=2\)
আবার,
\(t=x^{\frac{1}{3}}\)
\(\Rightarrow t=(27)^{\frac{1}{3}}\), যখন \(x=27\)
\(\Rightarrow t=(3^3)^{\frac{1}{3}}\)
\(\therefore t=3\)
\(\int_{8}^{27}{\frac{dx}{x-x^{\frac{1}{3}}}}\)\(x^{\frac{1}{3}}=t\)
\(\Rightarrow x=t^3\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^3)\)
\(\Rightarrow 1=3t^2\frac{dt}{dx}\)
\(\therefore dx=3t^2dt\)
| \(x\) | \(8\) | \(27\) |
| \(t\) | \(2\) | \(3\) |
\(\Rightarrow t=8^{\frac{1}{3}}\), যখন \(x=8\)
\(\Rightarrow t=(2^3)^{\frac{1}{3}}\)
\(\therefore t=2\)
আবার,
\(t=x^{\frac{1}{3}}\)
\(\Rightarrow t=(27)^{\frac{1}{3}}\), যখন \(x=27\)
\(\Rightarrow t=(3^3)^{\frac{1}{3}}\)
\(\therefore t=3\)
\(=\int_{2}^{3}{\frac{1}{t^3-t}\times{3t^2dt}}\)
\(=3\int_{2}^{3}{\frac{t^2}{t^3-t}dt}\)
\(=3\int_{2}^{3}{\frac{t^2}{t(t^2-1)}dt}\)
\(=3\int_{2}^{3}{\frac{t}{t^2-1}dt}\)
\(=\frac{3}{2}\int_{2}^{3}{\frac{2t}{(t-1)(t+1)}dt}\)
\(=\frac{3}{2}\int_{2}^{3}{\frac{t+1+t-1}{(t-1)(t+1)}dt}\)
\(=\frac{3}{2}\int_{2}^{3}{\left\{\frac{t+1}{(t-1)(t+1)}+\frac{t-1}{(t-1)(t+1)}\right\}dt}\)
\(=\frac{3}{2}\int_{2}^{3}{\left\{\frac{1}{t-1}+\frac{1}{t+1}\right\}dt}\)
\(=\frac{3}{2}\int_{2}^{3}{\frac{1}{t-1}dt}+\frac{3}{2}\int_{2}^{3}{\frac{1}{t+1}dt}\)
\(=\frac{3}{2}\left[\ln{|t-1|}\right]_{2}^{3}+\frac{3}{2}\left[\ln{|t+1|}\right]_{2}^{3}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\frac{3}{2}\left[\ln{(3-1)}-\ln{(2-1)}\right]+\frac{3}{2}\left[\ln{(3+1)}-\ln{(2+1)}\right]\)
\(=\frac{3}{2}\left[\ln{(2)}-\ln{(1)}\right]+\frac{3}{2}\left[\ln{(4)}-\ln{(3)}\right]\)
\(=\frac{3}{2}\left[\ln{(2)}-0\right]+\frac{3}{2}\times{\ln{\left(\frac{4}{3}\right)}}\) ➜ \(\because \ln{1}=0, \ln{(M)}-\ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(=\frac{3}{2}\left\{\ln{(2)}+\ln{(4)}-\ln{(3)}\right\}\)
\(=\frac{3}{2}\ln{\left(\frac{2\times{4}}{3}\right)}\) ➜ \(\because \ln{(M)}+\ln{(N)}-\ln{(P)}=\ln{\left(\frac{MM}{P}\right)}\)
\(=\frac{3}{2}\ln{\frac{8}{3}}\)
\(Q.3.(Lii)\) \(\int_{0}^{4}{f(x)dx}=5\) হয় তবে \(\int_{1}^{5}{f(x-1)dx}\) এর মাণ কত?
উত্তরঃ \(5\)
[ ঢাঃবিঃ২০০৭-২০০৮ ]
উত্তরঃ \(5\)
[ ঢাঃবিঃ২০০৭-২০০৮ ]
সমাধানঃ
ধরি,
\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
➜ \(\because t=x-1\)
\(\Rightarrow t=1-1\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=x-1\)
\(\Rightarrow t=5-1\), যখন \(x=27\)
\(\therefore t=4\)
দেওয়া আছে,\(x-1=t\)
\(\Rightarrow \frac{d}{dx}(x-1)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-0=\frac{dt}{dx}\)
\(\Rightarrow 1=\frac{dt}{dx}\)
\(\therefore dx=dt\)
| \(x\) | \(1\) | \(5\) |
| \(t\) | \(0\) | \(4\) |
\(\Rightarrow t=1-1\), যখন \(x=1\)
\(\therefore t=0\)
আবার,
\(t=x-1\)
\(\Rightarrow t=5-1\), যখন \(x=27\)
\(\therefore t=4\)
\(\int_{0}^{4}{f(x)dx}=5 ........(1)\)
এখন,
\(\int_{1}^{5}{f(x-1)dx}\)
\(=\int_{0}^{4}{f(t)dt}\)
\(=\int_{0}^{4}{f(x)dx}\) ➜ \(\because \int_{a}^{b}{f(x)dx}=\int_{a}^{b}{f(z)dz}\)
\(=5\) ➜ \(\because \int_{0}^{4}{f(x)dx}=5 .....(1)\)
\(Q.3.(Liii)\) \(\int_{0}^{1}{\frac{1-x}{1+x}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{\left(\frac{4}{e}\right)}\)
[ যঃ২০০৪; বঃ২০০৪,২০০৩; মাঃ২০১০ ]
উত্তরঃ \(\ln{\left(\frac{4}{e}\right)}\)
[ যঃ২০০৪; বঃ২০০৪,২০০৩; মাঃ২০১০ ]
সমাধানঃ
\(\int_{0}^{1}{\frac{1-x}{1+x}dx}\)
\(=\int_{0}^{1}{\frac{2-(1+x)}{1+x}dx}\)
\(=\int_{0}^{1}{\left(\frac{2}{1+x}-\frac{1+x}{1+x}\right)dx}\)
\(=\int_{0}^{1}{\left(\frac{2}{1+x}-1\right)dx}\)
\(=\int_{0}^{1}{\frac{2}{1+x}dx}-\int_{0}^{1}{1dx}\)
\(=2\int_{0}^{1}{\frac{1}{1+x}dx}-\int_{0}^{1}{1dx}\)
\(=2\left[\ln{|1+x|}\right]_{0}^{1}-\left[x\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}, \int{1dx}=x\)
\(=2\left[\ln{(1+1)}-\ln{(1+0)}\right]-\left[1-0\right]\)
\(=2\left[\ln{(2)}-\ln{(1)}\right]-1\)
\(=2\left[\ln{(2)}-0\right]-1\) ➜ \(\because \ln{(1)}=0\)
\(=2\ln{(2)}-1\)
\(=\ln{(2^2)}-\ln{(e)}\) ➜ \(\because n\ln{(M)}=\ln{(M^n)}, 1=\ln{(e)}\)
\(=\ln{(4)}-\ln{(e)}\)
\(=\ln{\left(\frac{4}{e}\right)}\)
\(=\int_{0}^{1}{\frac{2-(1+x)}{1+x}dx}\)
\(=\int_{0}^{1}{\left(\frac{2}{1+x}-\frac{1+x}{1+x}\right)dx}\)
\(=\int_{0}^{1}{\left(\frac{2}{1+x}-1\right)dx}\)
\(=\int_{0}^{1}{\frac{2}{1+x}dx}-\int_{0}^{1}{1dx}\)
\(=2\int_{0}^{1}{\frac{1}{1+x}dx}-\int_{0}^{1}{1dx}\)
\(=2\left[\ln{|1+x|}\right]_{0}^{1}-\left[x\right]_{0}^{1}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}, \int{1dx}=x\)
\(=2\left[\ln{(1+1)}-\ln{(1+0)}\right]-\left[1-0\right]\)
\(=2\left[\ln{(2)}-\ln{(1)}\right]-1\)
\(=2\left[\ln{(2)}-0\right]-1\) ➜ \(\because \ln{(1)}=0\)
\(=2\ln{(2)}-1\)
\(=\ln{(2^2)}-\ln{(e)}\) ➜ \(\because n\ln{(M)}=\ln{(M^n)}, 1=\ln{(e)}\)
\(=\ln{(4)}-\ln{(e)}\)
\(=\ln{\left(\frac{4}{e}\right)}\)
\(Q.3.(Liv)\) \(\int_{0}^{16}{\frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\left(\frac{2}{3}+\tan^{-1}{(2)}\right)\)
উত্তরঃ \(4\left(\frac{2}{3}+\tan^{-1}{(2)}\right)\)
সমাধানঃ
ধরি,
\(x^{\frac{1}{4}}=t\)
\(\Rightarrow x=t^4\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^4)\)
\(\Rightarrow 1=4t^3\frac{dt}{dx}\)
\(\therefore dx=4t^3dt\)
➜ \(\because t=x^{\frac{1}{4}}\)
\(\Rightarrow t=0^{\frac{1}{4}}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=x^{\frac{1}{4}}\)
\(\Rightarrow t=(16)^{\frac{1}{4}}\), যখন \(x=16\)
\(\Rightarrow t=(2^4)^{\frac{1}{4}}\)
\(\therefore t=2\)
\(\int_{0}^{16}{\frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}dx}\)\(x^{\frac{1}{4}}=t\)
\(\Rightarrow x=t^4\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^4)\)
\(\Rightarrow 1=4t^3\frac{dt}{dx}\)
\(\therefore dx=4t^3dt\)
| \(x\) | \(0\) | \(16\) |
| \(t\) | \(0\) | \(2\) |
\(\Rightarrow t=0^{\frac{1}{4}}\), যখন \(x=0\)
\(\therefore t=0\)
আবার,
\(t=x^{\frac{1}{4}}\)
\(\Rightarrow t=(16)^{\frac{1}{4}}\), যখন \(x=16\)
\(\Rightarrow t=(2^4)^{\frac{1}{4}}\)
\(\therefore t=2\)
\(=\int_{0}^{16}{\frac{x^{\frac{1}{4}}}{1+\left(x^{\frac{1}{4}}\right)^2}dx}\)
\(=\int_{0}^{2}{\frac{t}{1+t^2}.4t^3dt}\)
\(=4\int_{0}^{2}{\frac{t^4}{1+t^2}dt}\)
\(=4\int_{0}^{2}{\frac{t^2(1+t^2)-(1+t^2)+1}{1+t^2}dt}\)
\(=4\int_{0}^{2}{\left\{\frac{t^2(1+t^2)}{1+t^2}-\frac{1+t^2}{1+t^2}+\frac{1}{1+t^2}\right\}dt}\)
\(=4\int_{0}^{2}{\left\{t^2-1+\frac{1}{1+t^2}\right\}dt}\)
\(=4\int_{0}^{2}{t^2dt}-4\int_{0}^{2}{1dt}+4\int_{0}^{2}{\frac{1}{1+t^2}dt}\)
\(=4\left[\frac{t^3}{3}\right]_{0}^{2}-4\left[t\right]_{0}^{2}+4\left[\tan^{-1}{t}\right]_{0}^{2}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{1dx}=x,\) \(\int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=4\frac{1}{3}\left[t^3\right]_{0}^{2}-4\left[2-0\right]+4\left[\tan^{-1}{2}-\tan^{-1}{0}\right]\)
\(=\frac{4}{3}\left[2^3-0^3\right]-8+4\left[\tan^{-1}{2}-\tan^{-1}{(\tan{0})}\right]\)
\(=\frac{4}{3}\left[8-0\right]-8+4\left[\tan^{-1}{2}-0\right]\)
\(=\frac{32}{3}-8+4\tan^{-1}{2}\)
\(=\frac{32-24}{3}+4\tan^{-1}{2}\)
\(=\frac{8}{3}+4\tan^{-1}{2}\)
\(=4\left(\frac{2}{3}+\tan^{-1}{2}\right)\)
\(Q.3.(Lv)\) \(\int_{1}^{3}{\frac{x-3}{x^3+x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(4\ln{\left(\frac{3}{2}\right)}-2\)
উত্তরঃ \(4\ln{\left(\frac{3}{2}\right)}-2\)
সমাধানঃ
\(\int_{1}^{3}{\frac{x-3}{x^3+x^2}dx}\)
\(=\int_{1}^{3}{\frac{x-3}{x^2(x+1)}dx}\)
ধরি,
\(\frac{x-3}{x^2(x+1)}\equiv{\frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1}} .......(i)\)
\(x-3\equiv{A(x+1)+Bx(x+1)+Cx^2} .......(ii)\) ➜ \((i)\) এর উভয় পার্শে \(x^2(x+1)\) গুন করে।
অর্থাৎ,
\(x=0\)
এখন, \(x=0\)
\((ii)\) নং সমীকরণে বসিয়ে,
\(0-3=A(0+1)+B.0(0+1)+C.0^2\)
\(\Rightarrow -3=A+0+0\)
\(\Rightarrow -3=A\)
\(\therefore A=-3\)
আবার,
\((ii)\) এর উভয় পার্শ হতে \(x^2\) এর সহগ সমীকৃত করে।
\(0=B+C\)
\(\Rightarrow 0=B-4\)
\(\Rightarrow B-4=0\)
\(\therefore B=4\)
\(A\), \(B\) ও \(C\) এর মান \((i)\) নং সমীকরণে বসিয়ে,
\(\frac{x-3}{x^2(x+1)}=\frac{-3}{x^2}+\frac{4}{x}+\frac{-4}{x+1}\)
\(\Rightarrow \frac{x-3}{x^2(x+1)}=-\frac{3}{x^2}+\frac{4}{x}-\frac{4}{x+1}\)
\(\therefore \frac{x-3}{x^2(x+1)}=\frac{4}{x}-\frac{3}{x^2}-\frac{4}{x+1}\)
প্রদত্ত যোগজটি দাঁড়ায়,
\(=\int_{1}^{3}{\left(\frac{4}{x}-\frac{3}{x^2}-\frac{4}{x+1}\right)dx}\)
\(=\int_{1}^{3}{\frac{4}{x}dx}-\int_{1}^{3}{\frac{3}{x^2}dx}-\int_{1}^{3}{\frac{4}{x+1}dx}\)
\(=4\int_{1}^{3}{\frac{1}{x}dx}-3\int_{1}^{3}{\frac{1}{x^2}dx}-4\int_{1}^{3}{\frac{1}{x+1}dx}\)
\(=4\left[\ln{|x|}\right]_{1}^{3}-3\left[-\frac{1}{x}\right]_{1}^{3}-4\left[\ln{|x+1|}\right]_{1}^{3}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}, \int{\frac{1}{x^2}dx}=-\frac{1}{x}\)
\(=4\left[\ln{(3)}-\ln{(1)}\right]+3\left[\frac{1}{x}\right]_{1}^{3}-4\left[\ln{(3+1)}-\ln{(1+1)}\right]\)
\(=4\left[\ln{(3)}-0\right]+3\left[\frac{1}{3}-\frac{1}{1}\right]-4\left[\ln{(4)}-\ln{(2)}\right]\) ➜ \(\because \ln{(1)}=0\)
\(=4\ln{(3)}+3\left[\frac{1}{3}-1\right]-4\left[\ln{(2^2)}-\ln{(2)}\right]\)
\(=4\ln{(3)}+1-3-4\left[2\ln{(2)}-\ln{(2)}\right]\) ➜ \(\because \ln{(M)^n}=n\ln{(M)}\)
\(=4\ln{(3)}-2-4\ln{(2)}\)
\(=4\ln{(3)}-4\ln{(2)}-2\)
\(=4\{\ln{(3)}-\ln{(2)}\}-2\)
\(=4\ln{\left(\frac{3}{2}\right)}-2\) ➜ \(\because \ln{(M)}-\ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(=\int_{1}^{3}{\frac{x-3}{x^2(x+1)}dx}\)
ধরি,
\(\frac{x-3}{x^2(x+1)}\equiv{\frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1}} .......(i)\)
\(x-3\equiv{A(x+1)+Bx(x+1)+Cx^2} .......(ii)\) ➜ \((i)\) এর উভয় পার্শে \(x^2(x+1)\) গুন করে।
আবার, \(x^2(x+1)\) উৎপাদকগুলির মধ্যে \((x+1)\) সমান শুন্য ধরি।
\(x+1=0\)
\(\therefore x=-1\)
এখন, \(x=-1\)
\((ii)\) নং সমীকরণে বসিয়ে,
\(-1-3=A.0+B.(-1).0+C.(-1)^2\)
\(\Rightarrow -4=0+0+C.1\)
\(\Rightarrow -4=C\)
\(\therefore C=-4\)
এখানে, \(x^2(x+1)\) উৎপাদকগুলির মধ্যে \(x\) সমান শুন্য ধরি।\(x+1=0\)
\(\therefore x=-1\)
এখন, \(x=-1\)
\((ii)\) নং সমীকরণে বসিয়ে,
\(-1-3=A.0+B.(-1).0+C.(-1)^2\)
\(\Rightarrow -4=0+0+C.1\)
\(\Rightarrow -4=C\)
\(\therefore C=-4\)
অর্থাৎ,
\(x=0\)
এখন, \(x=0\)
\((ii)\) নং সমীকরণে বসিয়ে,
\(0-3=A(0+1)+B.0(0+1)+C.0^2\)
\(\Rightarrow -3=A+0+0\)
\(\Rightarrow -3=A\)
\(\therefore A=-3\)
আবার,
\((ii)\) এর উভয় পার্শ হতে \(x^2\) এর সহগ সমীকৃত করে।
\(0=B+C\)
\(\Rightarrow 0=B-4\)
\(\Rightarrow B-4=0\)
\(\therefore B=4\)
\(A\), \(B\) ও \(C\) এর মান \((i)\) নং সমীকরণে বসিয়ে,
\(\frac{x-3}{x^2(x+1)}=\frac{-3}{x^2}+\frac{4}{x}+\frac{-4}{x+1}\)
\(\Rightarrow \frac{x-3}{x^2(x+1)}=-\frac{3}{x^2}+\frac{4}{x}-\frac{4}{x+1}\)
\(\therefore \frac{x-3}{x^2(x+1)}=\frac{4}{x}-\frac{3}{x^2}-\frac{4}{x+1}\)
প্রদত্ত যোগজটি দাঁড়ায়,
\(=\int_{1}^{3}{\left(\frac{4}{x}-\frac{3}{x^2}-\frac{4}{x+1}\right)dx}\)
\(=\int_{1}^{3}{\frac{4}{x}dx}-\int_{1}^{3}{\frac{3}{x^2}dx}-\int_{1}^{3}{\frac{4}{x+1}dx}\)
\(=4\int_{1}^{3}{\frac{1}{x}dx}-3\int_{1}^{3}{\frac{1}{x^2}dx}-4\int_{1}^{3}{\frac{1}{x+1}dx}\)
\(=4\left[\ln{|x|}\right]_{1}^{3}-3\left[-\frac{1}{x}\right]_{1}^{3}-4\left[\ln{|x+1|}\right]_{1}^{3}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}, \int{\frac{1}{x^2}dx}=-\frac{1}{x}\)
\(=4\left[\ln{(3)}-\ln{(1)}\right]+3\left[\frac{1}{x}\right]_{1}^{3}-4\left[\ln{(3+1)}-\ln{(1+1)}\right]\)
\(=4\left[\ln{(3)}-0\right]+3\left[\frac{1}{3}-\frac{1}{1}\right]-4\left[\ln{(4)}-\ln{(2)}\right]\) ➜ \(\because \ln{(1)}=0\)
\(=4\ln{(3)}+3\left[\frac{1}{3}-1\right]-4\left[\ln{(2^2)}-\ln{(2)}\right]\)
\(=4\ln{(3)}+1-3-4\left[2\ln{(2)}-\ln{(2)}\right]\) ➜ \(\because \ln{(M)^n}=n\ln{(M)}\)
\(=4\ln{(3)}-2-4\ln{(2)}\)
\(=4\ln{(3)}-4\ln{(2)}-2\)
\(=4\{\ln{(3)}-\ln{(2)}\}-2\)
\(=4\ln{\left(\frac{3}{2}\right)}-2\) ➜ \(\because \ln{(M)}-\ln{(N)}=\ln{\left(\frac{M}{N}\right)}\)
\(Q.3.(Lvi)\) \(\int_{0}^{2}{\frac{x^4+1}{x^2+1}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}+2\tan^{-1}{(2)}\)
উত্তরঃ \(\frac{2}{3}+2\tan^{-1}{(2)}\)
সমাধানঃ
\(\int_{0}^{2}{\frac{x^4+1}{x^2+1}dx}\)
\(=\int_{0}^{2}{\frac{x^2(x^2+1)-(x^2+1)+2}{x^2+1}dx}\)
\(=\int_{0}^{2}{\left\{\frac{x^2(x^2+1)}{x^2+1}-\frac{(x^2+1)}{x^2+1}+\frac{2}{x^2+1}\right\}dx}\)
\(=\int_{0}^{2}{\left\{x^2-1+\frac{2}{x^2+1}\right\}dx}\)
\(=\int_{0}^{2}{x^2dx}-\int_{0}^{2}{1dx}+\int_{0}^{2}{\frac{2}{x^2+1}dx}\)
\(=\left[\frac{x^3}{3}\right]_{0}^{2}-\left[x\right]_{0}^{2}+2\int_{0}^{2}{\frac{1}{1+x^2}dx}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{1dx}=x\)
\(=\frac{1}{3}\left[x^3\right]_{0}^{2}-\left[2-0\right]+2\left[\tan^{-1}{x}\right]_{0}^{2}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\frac{1}{3}\left[2^3-0^3\right]-2+2\left[\tan^{-1}{(2)}-\tan^{-1}{(0)}\right]\)
\(=\frac{1}{3}\left[8-0\right]-2+2\left[\tan^{-1}{(2)}-\tan^{-1}{(\tan{0})}\right]\)
\(=\frac{8}{3}-2+2\left[\tan^{-1}{(2)}-0\right]\)
\(=\frac{8-6}{3}+2\tan^{-1}{(2)}\)
\(=\frac{2}{3}+2\tan^{-1}{(2)}\)
\(=\int_{0}^{2}{\frac{x^2(x^2+1)-(x^2+1)+2}{x^2+1}dx}\)
\(=\int_{0}^{2}{\left\{\frac{x^2(x^2+1)}{x^2+1}-\frac{(x^2+1)}{x^2+1}+\frac{2}{x^2+1}\right\}dx}\)
\(=\int_{0}^{2}{\left\{x^2-1+\frac{2}{x^2+1}\right\}dx}\)
\(=\int_{0}^{2}{x^2dx}-\int_{0}^{2}{1dx}+\int_{0}^{2}{\frac{2}{x^2+1}dx}\)
\(=\left[\frac{x^3}{3}\right]_{0}^{2}-\left[x\right]_{0}^{2}+2\int_{0}^{2}{\frac{1}{1+x^2}dx}\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{1dx}=x\)
\(=\frac{1}{3}\left[x^3\right]_{0}^{2}-\left[2-0\right]+2\left[\tan^{-1}{x}\right]_{0}^{2}\) ➜ \(\because \int{\frac{1}{1+x^2}dx}=\tan^{-1}{x}\)
\(=\frac{1}{3}\left[2^3-0^3\right]-2+2\left[\tan^{-1}{(2)}-\tan^{-1}{(0)}\right]\)
\(=\frac{1}{3}\left[8-0\right]-2+2\left[\tan^{-1}{(2)}-\tan^{-1}{(\tan{0})}\right]\)
\(=\frac{8}{3}-2+2\left[\tan^{-1}{(2)}-0\right]\)
\(=\frac{8-6}{3}+2\tan^{-1}{(2)}\)
\(=\frac{2}{3}+2\tan^{-1}{(2)}\)
\(Q.3.(Lvii)\) \(\int_{0}^{3}{\frac{dx}{(2+x^2)^{\frac{3}{2}}}}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{3\sqrt{11}}{22}\)
উত্তরঃ \(\frac{3\sqrt{11}}{22}\)
সমাধানঃ
ধরি,
\(x=\sqrt{2}\tan{t}\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(\sqrt{2}\tan{t})\)
\(\Rightarrow 1=\sqrt{2}\sec^2{t}\frac{dt}{dx}\)
\(\therefore dx=\sqrt{2}\sec^2{t}dt\)
➜ \(\because \sqrt{2}\tan{t}=x\)
\(\Rightarrow \tan{t}=\frac{x}{\sqrt{2}}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{x}{\sqrt{2}}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{0}{\sqrt{2}}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{(0)}\)
\(\Rightarrow t=\tan^{-1}{(\tan{0})}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{\left(\frac{x}{\sqrt{2}}\right)}\)
\(\therefore t=\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}\), যখন \(x=3\)
\(\int_{0}^{3}{\frac{dx}{(2+x^2)^{\frac{3}{2}}}}\)\(x=\sqrt{2}\tan{t}\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(\sqrt{2}\tan{t})\)
\(\Rightarrow 1=\sqrt{2}\sec^2{t}\frac{dt}{dx}\)
\(\therefore dx=\sqrt{2}\sec^2{t}dt\)
| \(x\) | \(0\) | \(3\) |
| \(t\) | \(0\) | \(\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}\) |
\(\Rightarrow \tan{t}=\frac{x}{\sqrt{2}}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{x}{\sqrt{2}}\right)}\)
\(\Rightarrow t=\tan^{-1}{\left(\frac{0}{\sqrt{2}}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\tan^{-1}{(0)}\)
\(\Rightarrow t=\tan^{-1}{(\tan{0})}\)
\(\therefore t=0\)
আবার,
\(t=\tan^{-1}{\left(\frac{x}{\sqrt{2}}\right)}\)
\(\therefore t=\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}\), যখন \(x=3\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{\{2+(\sqrt{2}\tan{t})^2\}^{\frac{3}{2}}}.\sqrt{2}\sec^2{t}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{\{2+2\tan^2{t}\}^{\frac{3}{2}}}.\sqrt{2}\sec^2{t}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{\{2(1+\tan^2{t})\}^{\frac{3}{2}}}.\sqrt{2}\sec^2{t}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{\{2\sec^2{t}\}^{\frac{3}{2}}}.\sqrt{2}\sec^2{t}dt}\) ➜ \(\because 1+\tan^2{A}=\sec^2{A}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{(2)^{\frac{3}{2}}\sec^3{t}}.\sqrt{2}\sec^2{t}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{(\sqrt{2})^3\sec^3{t}}.\sqrt{2}\sec^2{t}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{(\sqrt{2})^2\sec{t}}dt}\)
\(=\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{2\sec{t}}dt}\)
\(=\frac{1}{2}\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\frac{1}{\sec{t}}dt}\)
\(=\frac{1}{2}\int_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}{\cos{t}dt}\) ➜ \(\because \frac{1}{\sec{A}}=\cos{A}\)
\(=\frac{1}{2}\left[\sin{t}\right]_{0}^{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}}\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\)
\(=\frac{1}{2}\left[\sin{\left\{\tan^{-1}{\left(\frac{3}{\sqrt{2}}\right)}\right\}}-\sin{(0)}\right]\)
\(=\frac{1}{2}\left[\sin{\left\{\sin^{-1}{\left(\frac{3}{\sqrt{3^2+(\sqrt{2})^2}}\right)}\right\}}-0\right]\) ➜ \(\because \tan^{-1}{\left(\frac{a}{b}\right)}=\sin^{-1}{\left(\frac{a}{\sqrt{a^2+b^2}}\right)}\)
\(=\frac{1}{2}\times{\frac{3}{\sqrt{9+2}}}\)
\(=\frac{3}{2\sqrt{11}}\)
\(=\frac{3\sqrt{11}}{2\sqrt{11}.\sqrt{11}}\) ➜ লব ও হরের সহিত \(\sqrt{11}\) গুণ করে।
\(=\frac{3\sqrt{11}}{2\times{11}}\)
\(=\frac{3\sqrt{11}}{22}\)
\(Q.3.(Lviii)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a^2\sin^2{\theta}+b^2\cos^2{\theta}}}=\frac{\pi}{2ab}\)
[ রাঃ ২০১১ ]
[ রাঃ ২০১১ ]
সমাধানঃ
ধরি,
\(\tan{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\tan{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \sec^2{\theta}=\frac{dt}{d\theta}\)
\(\therefore \sec^2{\theta}d\theta=dt\)
➜ \(\because t=\tan{\theta}\)
\(\Rightarrow t=\tan{0}\), যখন \(\theta=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{\theta}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(\theta=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(L.S=\int_{0}^{\frac{\pi}{2}}{\frac{d\theta}{a^2\sin^2{\theta}+b^2\cos^2{\theta}}}\)\(\tan{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\tan{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \sec^2{\theta}=\frac{dt}{d\theta}\)
\(\therefore \sec^2{\theta}d\theta=dt\)
| \(x\) | \(0\) | \(\frac{\pi}{2}\) |
| \(t\) | \(0\) | \(\infty\) |
\(\Rightarrow t=\tan{0}\), যখন \(\theta=0\)
\(\therefore t=0\)
আবার,
\(t=\tan{\theta}\)
\(\Rightarrow t=\tan{\left(\frac{\pi}{2}\right)}\), যখন \(\theta=\frac{\pi}{2}\)
\(\therefore t=\infty\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{1}{a^2\sin^2{\theta}+b^2\cos^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\frac{1}{\cos^2{\theta}}}{a^2\frac{\sin^2{\theta}}{\cos^2{\theta}}+b^2}d\theta}\) ➜ লব ও হরের সহিত \(\cos^2{\theta}\) ভাগ করে।
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{\theta}}{b^2+a^2\tan^2{\theta}}d\theta}\)
\(=\int_{0}^{\frac{\pi}{2}}{\frac{\sec^2{\theta}}{a^2\left(\frac{b^2}{a^2}+\tan^2{\theta}\right)}d\theta}\)
\(=\frac{1}{a^2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{\frac{b^2}{a^2}+\tan^2{\theta}}\sec^2{\theta}d\theta}\)
\(=\frac{1}{a^2}\int_{0}^{\infty}{\frac{1}{\left(\frac{b}{a}\right)^2+t^2}dt}\)
\(=\frac{1}{a^2}\left[\frac{1}{\frac{b}{a}}\tan^{-1}{\left(\frac{t}{\frac{b}{a}}\right)}\right]_{0}^{\infty}\) ➜ \(\because \int{\frac{1}{a^2+x^2}dx}=\frac{1}{a}\tan^{-1}{\left(\frac{x}{a}\right)}\)
\(=\frac{1}{a^2}.\frac{a}{b}\left[\tan^{-1}{\left(\frac{at}{b}\right)}\right]_{0}^{\infty}\)
\(=\frac{1}{ab}\left[\tan^{-1}{\left(\frac{a\times{\infty}}{b}\right)}-\tan^{-1}{\left(\frac{a.0}{b}\right)}\right]\)
\(=\frac{1}{ab}\left[\tan^{-1}{\left(\infty\right)}-\tan^{-1}{0}\right]\)
\(=\frac{1}{ab}\left[\tan^{-1}{\tan{\left(\frac{\pi}{2}\right)}}-\tan^{-1}{\tan{0}}\right]\)
\(=\frac{1}{ab}\left[\frac{\pi}{2}-0\right]\)
\(=\frac{\pi}{2ab}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(proved)
\(Q.3.(Lix)\) দেখাও যে, \(\int_{0}^{\frac{\pi}{2}}{(a\cos^2{\theta}+b\sin^2{\theta})d\theta}=\frac{1}{4}(a+b)\pi\)
[ চঃ ২০০৩ ]
[ চঃ ২০০৩ ]
সমাধানঃ
\(L.S=\int_{0}^{\frac{\pi}{2}}{(a\cos^2{\theta}+b\sin^2{\theta})d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a.2\cos^2{\theta}+b.2\sin^2{\theta})d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a(1+\cos{2\theta})+b(1-\cos{2\theta})\}d\theta}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a+a\cos{2\theta}+b-b\cos{2\theta})\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{(a+b)+(a-b)\cos{2\theta}\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a+b)d\theta}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a-b)\cos{2\theta}d\theta}\)
\(=\frac{1}{2}(a+b)\int_{0}^{\frac{\pi}{2}}{1d\theta}+\frac{1}{2}(a-b)\int_{0}^{\frac{\pi}{2}}{\cos{2\theta}d\theta}\)
\(=\frac{1}{2}(a+b)\left[\theta\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}(a-b)\left[\frac{1}{2}\sin{2\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}(a+b)\left[\frac{\pi}{2}-0\right]+\frac{1}{4}(a-b)\left[\sin{2\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(2.\frac{\pi}{2}\right)}-\sin{(2.0)}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{(\pi)}-\sin{(0)}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[0-0\right]\) ➜ \(\because \sin{(\pi)}=0, \sin{(0)}=0\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b).0\)
\(=\frac{1}{4}(a+b)\pi+0\)
\(=\frac{1}{4}(a+b)\pi\)
\(=R.S\)
\(\therefore L.S=R.S\)
(proved)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a.2\cos^2{\theta}+b.2\sin^2{\theta})d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a(1+\cos{2\theta})+b(1-\cos{2\theta})\}d\theta}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{a+a\cos{2\theta}+b-b\cos{2\theta})\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\{(a+b)+(a-b)\cos{2\theta}\}d\theta}\)
\(=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a+b)d\theta}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{(a-b)\cos{2\theta}d\theta}\)
\(=\frac{1}{2}(a+b)\int_{0}^{\frac{\pi}{2}}{1d\theta}+\frac{1}{2}(a-b)\int_{0}^{\frac{\pi}{2}}{\cos{2\theta}d\theta}\)
\(=\frac{1}{2}(a+b)\left[\theta\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}(a-b)\left[\frac{1}{2}\sin{2\theta}\right]_{0}^{\frac{\pi}{2}}\) ➜ \(\because \int{1dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\)
\(=\frac{1}{2}(a+b)\left[\frac{\pi}{2}-0\right]+\frac{1}{4}(a-b)\left[\sin{2\theta}\right]_{0}^{\frac{\pi}{2}}\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{\left(2.\frac{\pi}{2}\right)}-\sin{(2.0)}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[\sin{(\pi)}-\sin{(0)}\right]\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b)\left[0-0\right]\) ➜ \(\because \sin{(\pi)}=0, \sin{(0)}=0\)
\(=\frac{1}{4}(a+b)\pi+\frac{1}{4}(a-b).0\)
\(=\frac{1}{4}(a+b)\pi+0\)
\(=\frac{1}{4}(a+b)\pi\)
\(=R.S\)
\(\therefore L.S=R.S\)
(proved)
\(Q.3.(Lx)\) \(\int_{0}^{\sqrt{2}}{\frac{x^2}{(4-x^2)^{\frac{3}{2}}}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(1-\frac{\pi}{4}\)
উত্তরঃ \(1-\frac{\pi}{4}\)
সমাধানঃ
ধরি,
\(x=2\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=2\cos{t}\frac{dt}{dx}\)
\(\therefore dx=2\cos{t}dt\)
➜ \(\because 2\sin{t}=x\)
\(\Rightarrow \sin{t}=\frac{x}{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{0}{2}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{(0)}\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{\sqrt{2}}{2}\right)}\), যখন \(x=\sqrt{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{\sqrt{2}}\right)}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(\int_{0}^{\sqrt{2}}{\frac{x^2}{(4-x^2)^{\frac{3}{2}}}dx}\)\(x=2\sin{t}\)
\(\Rightarrow \frac{d}{dx}(x)=2\frac{d}{dx}(\sin{t})\)
\(\Rightarrow 1=2\cos{t}\frac{dt}{dx}\)
\(\therefore dx=2\cos{t}dt\)
| \(x\) | \(0\) | \(\sqrt{2}\) |
| \(t\) | \(0\) | \(\frac{\pi}{4}\) |
\(\Rightarrow \sin{t}=\frac{x}{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{0}{2}\right)}\), যখন \(x=0\)
\(\Rightarrow t=\sin^{-1}{(0)}\)
\(\Rightarrow t=\sin^{-1}{\sin{0}}\)
\(\therefore t=0\)
আবার,
\(t=\sin^{-1}{\left(\frac{x}{2}\right)}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{\sqrt{2}}{2}\right)}\), যখন \(x=\sqrt{2}\)
\(\Rightarrow t=\sin^{-1}{\left(\frac{1}{\sqrt{2}}\right)}\)
\(\Rightarrow t=\sin^{-1}{\sin{\left(\frac{\pi}{4}\right)}}\)
\(\therefore t=\frac{\pi}{4}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{4\sin^2{t}}{(4-4\sin^2{t})^{\frac{3}{2}}}.2\cos{t}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{8\sin^2{t}}{4^{\frac{3}{2}}(1-\sin^2{t})^{\frac{3}{2}}}\cos{t}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{8\sin^2{t}}{(2^2)^{\frac{3}{2}}(\cos^2{t})^{\frac{3}{2}}}\cos{t}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{8\sin^2{t}}{(2)^3\cos^3{t}}\cos{t}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{8\sin^2{t}}{8\cos^2{t}}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\frac{\sin^2{t}}{\cos^2{t}}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{\tan^2{t}dt}\)
\(=\int_{0}^{\frac{\pi}{4}}{(\sec^2{t}-1)dt}\) ➜ \(\because \tan^2{A}=\sec^2{A}-1\)
\(=\int_{0}^{\frac{\pi}{4}}{\sec^2{t}dt}-\int_{0}^{\frac{\pi}{4}}{1dt}\)
\(=\left[\tan{t}\right]_{0}^{\frac{\pi}{4}}-\left[t\right]_{0}^{\frac{\pi}{4}}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{1dx}=x\)
\(=\tan{\left(\frac{\pi}{4}\right)}-\tan{(0)}-\left[\frac{\pi}{4}-0\right]\)
\(=1-0-\frac{\pi}{4}\)
\(=1-\frac{\pi}{4}\)
\(Q.3.(Lxi)\) \(\int_{1}^{15}{\frac{x+2}{(x+1)(x+3)}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\ln{(6)}\)
উত্তরঃ \(\ln{(6)}\)
সমাধানঃ
\(\int_{1}^{15}{\frac{x+2}{(x+1)(x+3)}dx}\)
এখন,
\(\frac{x+2}{(x+1)(x+3)}=\frac{-1+2}{(x+1)(-1+3)}+\frac{-3+2}{(-3+1)(x+3)}\) ➜ এখানে \((x+1)\) ব্যতীত \(\frac{x+2}{(x+1)(x+3)}\) এর সকল উৎপাদকে \(x=-1\) এবং \((x+3)\) ব্যতীত \(\frac{x+2}{(x+1)(x+3)}\) এর সকল উৎপাদকে \(x=-3\) বসানো হয়েছে।
\(=\frac{1}{(x+1).(2)}+\frac{-1}{(-2)(x+3)}\)
\(=\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\)
\(\therefore \frac{x+2}{(x+1)(x+3)}=\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\)
প্রদত্ত যোগজটি দাঁড়ায়,
\(\int_{1}^{15}{\left\{\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\right\}dx}\)
\(=\int_{1}^{15}{\frac{1}{2(x+1)}dx}+\int_{1}^{15}{\frac{1}{2(x+3)}dx}\)
\(=\frac{1}{2}\int_{1}^{15}{\frac{1}{x+1}dx}+\frac{1}{2}\int_{1}^{15}{\frac{1}{x+3}dx}\)
\(=\frac{1}{2}\left[\ln{|x+1|}\right]_{1}^{15}+\frac{1}{2}\left[\ln{|x+3|}\right]_{1}^{15}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\frac{1}{2}\left[\ln{(15+1)}-\ln{(1+1)}\right]+\frac{1}{2}\left[\ln{(15+3)}-\ln{(1+3)}\right]\)
\(=\frac{1}{2}\left[\ln{(16)}-\ln{(2)}\right]+\frac{1}{2}\left[\ln{(18)}-\ln{(4)}\right]\)
\(=\frac{1}{2}\left[\ln{\left(\frac{16}{2}\right)}\right]+\frac{1}{2}\ln{\left(\frac{18}{4}\right)}\)
\(=\frac{1}{2}\ln{(8)}+\frac{1}{2}\ln{\left(\frac{9}{2}\right)}\)
\(=\frac{1}{2}\{\ln{(8)}+\ln{\left(\frac{9}{2}\right)}\}\)
\(=\frac{1}{2}\ln{\left(8\times{\frac{9}{2}}\right)}\) ➜ \(\because \ln{(M)}+\ln{(N)}=\ln{(MN)}\)
\(=\frac{1}{2}\ln{\left(4\times{9}\right)}\)
\(=\frac{1}{2}\ln{(36)}\)
\(=\frac{1}{2}\ln{(6)^2}\)
\(=\frac{1}{2}\times{2}\ln{(6)}\) ➜ \(\because \ln{(M)^n}=n\ln{(M)}\)
\(=\ln{(6)}\)
এখন,
\(\frac{x+2}{(x+1)(x+3)}=\frac{-1+2}{(x+1)(-1+3)}+\frac{-3+2}{(-3+1)(x+3)}\) ➜ এখানে \((x+1)\) ব্যতীত \(\frac{x+2}{(x+1)(x+3)}\) এর সকল উৎপাদকে \(x=-1\) এবং \((x+3)\) ব্যতীত \(\frac{x+2}{(x+1)(x+3)}\) এর সকল উৎপাদকে \(x=-3\) বসানো হয়েছে।
\(=\frac{1}{(x+1).(2)}+\frac{-1}{(-2)(x+3)}\)
\(=\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\)
\(\therefore \frac{x+2}{(x+1)(x+3)}=\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\)
প্রদত্ত যোগজটি দাঁড়ায়,
\(\int_{1}^{15}{\left\{\frac{1}{2(x+1)}+\frac{1}{2(x+3)}\right\}dx}\)
\(=\int_{1}^{15}{\frac{1}{2(x+1)}dx}+\int_{1}^{15}{\frac{1}{2(x+3)}dx}\)
\(=\frac{1}{2}\int_{1}^{15}{\frac{1}{x+1}dx}+\frac{1}{2}\int_{1}^{15}{\frac{1}{x+3}dx}\)
\(=\frac{1}{2}\left[\ln{|x+1|}\right]_{1}^{15}+\frac{1}{2}\left[\ln{|x+3|}\right]_{1}^{15}\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\)
\(=\frac{1}{2}\left[\ln{(15+1)}-\ln{(1+1)}\right]+\frac{1}{2}\left[\ln{(15+3)}-\ln{(1+3)}\right]\)
\(=\frac{1}{2}\left[\ln{(16)}-\ln{(2)}\right]+\frac{1}{2}\left[\ln{(18)}-\ln{(4)}\right]\)
\(=\frac{1}{2}\left[\ln{\left(\frac{16}{2}\right)}\right]+\frac{1}{2}\ln{\left(\frac{18}{4}\right)}\)
\(=\frac{1}{2}\ln{(8)}+\frac{1}{2}\ln{\left(\frac{9}{2}\right)}\)
\(=\frac{1}{2}\{\ln{(8)}+\ln{\left(\frac{9}{2}\right)}\}\)
\(=\frac{1}{2}\ln{\left(8\times{\frac{9}{2}}\right)}\) ➜ \(\because \ln{(M)}+\ln{(N)}=\ln{(MN)}\)
\(=\frac{1}{2}\ln{\left(4\times{9}\right)}\)
\(=\frac{1}{2}\ln{(36)}\)
\(=\frac{1}{2}\ln{(6)^2}\)
\(=\frac{1}{2}\times{2}\ln{(6)}\) ➜ \(\because \ln{(M)^n}=n\ln{(M)}\)
\(=\ln{(6)}\)
\(Q.3.(Lxii)\) \(\int_{0}^{\frac{a}{2}}{\frac{1}{a^2-x^2}dx}\) যোগজটির মান নির্ণয় কর।
উত্তরঃ \(\frac{1}{2a}\ln{3}\)
উত্তরঃ \(\frac{1}{2a}\ln{3}\)
সমাধানঃ
\(\int_{0}^{\frac{a}{2}}{\frac{1}{a^2-x^2}dx}\)
\(=\left[\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\right]_{0}^{\frac{a}{2}}\) ➜ \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{a+x}{a-x}\right|}\right]_{0}^{\frac{a}{2}}\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{a+\frac{a}{2}}{a-\frac{a}{2}}\right|}-\ln{\left|\frac{a+0}{a-0}\right|}\right]\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{\frac{2a+a}{2}}{\frac{2a-a}{2}}\right|}-\ln{\left|\frac{a}{a}\right|}\right]\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{\frac{3a}{2}}{\frac{a}{2}}\right|}-\ln{(1)}\right]\)
\(=\frac{1}{2a}\left[\ln{\left(\frac{3a}{2}\times{\frac{2}{a}}\right)}-0\right]\) ➜ \(\because \ln{(1)}=0\)
\(=\frac{1}{2a}\ln{(3)}\)
\(=\left[\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\right]_{0}^{\frac{a}{2}}\) ➜ \(\because \int{\frac{1}{a^2-x^2}dx}=\frac{1}{2a}\ln{\left|\frac{a+x}{a-x}\right|}\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{a+x}{a-x}\right|}\right]_{0}^{\frac{a}{2}}\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{a+\frac{a}{2}}{a-\frac{a}{2}}\right|}-\ln{\left|\frac{a+0}{a-0}\right|}\right]\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{\frac{2a+a}{2}}{\frac{2a-a}{2}}\right|}-\ln{\left|\frac{a}{a}\right|}\right]\)
\(=\frac{1}{2a}\left[\ln{\left|\frac{\frac{3a}{2}}{\frac{a}{2}}\right|}-\ln{(1)}\right]\)
\(=\frac{1}{2a}\left[\ln{\left(\frac{3a}{2}\times{\frac{2}{a}}\right)}-0\right]\) ➜ \(\because \ln{(1)}=0\)
\(=\frac{1}{2a}\ln{(3)}\)
- About
- Author
-
Contact
Call me at +880-1715-651163Email: Golzarrahman1966@gmail.com
Visitors online: 000008