এ অধ্যায়ের পাঠ্যসূচী
- অংশায়ন পদ্ধতিতে যোগজীকরণ (Integration by parts)
- অংশায়ন সূত্র (Participation formula)
- \(\int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
- \(LIATE\) পদ্ধতি (\(LIATE\) Method)
- \(\int{\ln{|x|}dx}\)
- \(\int{\log{|x|}dx}\)
- \(\int{\sin^{-1}{x}dx}\)
- \(\int{\cos^{-1}{x}dx}\)
- \(\int{\tan^{-1}{x}dx}\)
- \(\int{cosec^{-1}{x}dx}\)
- \(\int{\sec^{-1}{x}dx}\)
- \(\int{\cot^{-1}{x}dx}\)
- \(\int{e^x\sin{x}dx}\)
- \(\int{e^{ax}\sin{bx}dx}\)
- \(\int{e^{ax}\cos{bx}dx}\)
- \(\int{\sqrt{x^2-a^2}dx}\)
- \(\int{\sqrt{x^2+a^2}dx}\)
- \(\int{e^x\{f(x)+f^{\prime}(x)\}dx}\)
- \(\int{e^{ax}\{af(x)+f^{\prime}(x)\}dx}\)
- অধ্যায় \(x.E\)-এর উদাহরণসমুহ
- অধ্যায় \(x.E\) / \(Q.1\)-এর সংক্ষিপ্ত প্রশ্নসমূহ
- অধ্যায় \(x.E\) / \(Q.2\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(x.E\) / \(Q.3\)-এর বর্ণনামূলক প্রশ্নসমূহ
- অধ্যায় \(x.E\) / \(Q.4\)-এর বর্ণনামূলক প্রশ্নসমূহ
অংশায়ন পদ্ধতিতে যোগজীকরণ
Integration by parts
যখন প্রতিস্থাপন সহ আর সকল কৌশল যোগজফল নির্ণয়ে ব্যার্থ ঠিক সেই ক্ষেত্রে এটি একটি বিশেষ পদ্ধতি। ইহার সাহায্যে দুইটি ফাংশন-এর গুনফলের যোগজীকরণ করা হয়।
অংশায়ন সূত্র
Participation formula
যোজ্যরাশিকে দুইটি ফাংশনে বিভক্ত করে যোগজ নির্ণয় করা হয় বলে এ পদ্ধতিকে যোগজীকরণের অংশায়ন সূত্র বলে। দুইটি ফাংশন-এর গুণনের অন্তরীকরণ নির্ণয়ের ক্ষেত্রে যে কোনো একটিকে \(u\) এবং অপরটিকে \(v\) ধরে অন্তরজ নির্ণয় করা যায়। কিন্তু যোগজীকরণের ক্ষেত্রে নির্দিষ্ট একটি ফাংশনকে \(u\) এবং অপরটিকে \(v\) বিবেচনা করতে হয়।
\((uv)\) এর যোগজীকরণ
Integration of \((uv)\)
যদি \(u\) এবং \(v\) উভয়েই \(x\) এর ফাংশন হয় তবে,
\(\int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(LIATE\) পদ্ধতি
\(LIATE\) Method
দুইটি ফাংশন-এর গুণনের অন্তরীকরণ নির্ণয়ের ক্ষেত্রে যে কোনো একটিকে \(u\) এবং অপরটিকে \(v\) ধরে অন্তরজ নির্ণয় করা যায়। কিন্তু যোগজীকরণের ক্ষেত্রে নির্দিষ্ট একটি ফাংশনকে \(u\) এবং অপরটিকে \(v\) বিবেচনা করতে হয়। এ ক্ষেত্রে যোজ্যরাশির ফাংশণ দুইটিকে \(LIATE\) শব্দের অক্ষরগুলির ক্রমানুযায়ী সাজিয়ে প্রথমটিকে \(u\) এবং দ্বিতীয়টিকে \(v\) বিবেচনা করা সহজ হয়।
\(LIATE\) শব্দের অক্ষরগুলির বিশ্লেষণ নিম্নরূপঃ\(L\rightarrow Logarithm \ function \ \ ( \ \ln{|x|},\ \log{|x|} ... )\)
\(I\rightarrow Inverse \ trigonometric \ function \ \ ( \ \sin^{-1}{x},\) \(\cos^{-1}{x}, \ \tan^{-1}{x} ... )\)
\(A\rightarrow Algebric \ function \ \ ( \ x^3, \ x^2+3x \ ... )\)
\(T\rightarrow trigonometric \ function \ \ ( \ \sin{x}, \ \cos{x} \ ... )\)
\(E\rightarrow Exponential \ function \ \ ( \ e^{x}, \ a^{x} \ ... )\)
\(\ln{|x|}\) এবং \(log{|x|}\) এর যোগজীকরণ
Integration of \(\ln{|x|}\) and \(log{|x|}\)
\(\ln{|x|}\) এর যোগজীকরণ
\(log{|x|}\) এর যোগজীকরণ
\(\sin^{-1}{x}\) এবং \(\cos^{-1}{x}\) এর যোগজীকরণ
Integration of \(\sin^{-1}{x}\) and \(\cos^{-1}{x}\)
\(\sin^{-1}{x}\) এর যোগজীকরণ
\(\cos^{-1}{x}\) এর যোগজীকরণ
\(\tan^{-1}{x}\) এবং \(cosec^{-1}{x}\) এর যোগজীকরণ
Integration of \(\tan^{-1}{x}\) and \(cosec^{-1}{x}\)
\(\tan^{-1}{x}\) এর যোগজীকরণ
\(cosec^{-1}{x}\) এর যোগজীকরণ
\(\sec^{-1}{x}\) এবং \(\cot^{-1}{x}\) এর যোগজীকরণ
Integration of \(\sec^{-1}{x}\) and \(\cot^{-1}{x}\)
\(\sec^{-1}{x}\) এর যোগজীকরণ
\(\cot^{-1}{x}\) এর যোগজীকরণ
\(\int{(uv)dx}\) এর যোগজীকরণে পুনরাবৃত্তি
Iteration in integration of \(\int{(uv)dx}\)
\(\int{(uv)dx}\) সূত্র প্রয়োগ করে যোগজীকরণ করতে গিয়ে যদি প্রদত্ত যোজ্যরাশির পুনরাবৃত্তি ঘটে, সে ক্ষেত্রে যোগজটিকে \(I\) ধরে সমাধান করতে হয়।
যেমনঃ
\(\int{e^x\sin{x}dx}\) এর যোজিত ফল নির্ণয় কর। \(e^{ax}\sin{bx}\) এর যোগজীকরণ
Integration of \(e^{ax}\sin{bx}\).
যোগজীকরণের সূত্র
\(\int{e^{ax}\sin{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+c\)
\(\int{e^{ax}\sin{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+c\)
\(e^{ax}\cos{bx}\) এর যোগজীকরণ
Integration of \(e^{ax}\cos{bx}\)
যোগজীকরণের সূত্র
\(\int{e^{ax}\cos{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+c\)
\(\int{e^{ax}\cos{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+c\)
\(\sqrt{x^2-a^2}\) এর যোগজীকরণ
Integration of \(\sqrt{x^2-a^2}\)
যোগজীকরণের সূত্র
\(\int{\sqrt{x^2-a^2}dx}=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{|x+\sqrt{x^2-a^2}|}+c\)
\(\int{\sqrt{x^2-a^2}dx}=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{|x+\sqrt{x^2-a^2}|}+c\)
\(\sqrt{x^2+a^2}\) এর যোগজীকরণ
Integration of \(\sqrt{x^2+a^2}\)
যোগজীকরণের সূত্র
\(\int{\sqrt{x^2+a^2}dx}=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{|x+\sqrt{x^2+a^2}|}+c\)
\(\int{\sqrt{x^2+a^2}dx}=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{|x+\sqrt{x^2+a^2}|}+c\)
\(e^x\{f(x)+f^{\prime}(x)\}\) এর যোগজীকরণ
Integration of \(e^x\{f(x)+f^{\prime}(x)\}\)
যোগজীকরণের সূত্র
\(\int{e^x\{f(x)+f^{\prime}(x)\}dx}=e^xf(x)+c\)
\(\int{e^x\{f(x)+f^{\prime}(x)\}dx}=e^xf(x)+c\)
\(e^{ax}\{af(x)+f^{\prime}(x)\}\) এর যোগজীকরণ
Integration of \(e^{ax}\{af(x)+f^{\prime}(x)\}\)
যোগজীকরণের সূত্র
\(\int{e^{ax}\{af(x)+f^{\prime}(x)\}dx}=e^{ax}f(x)+c\)
\(\int{e^{ax}\{af(x)+f^{\prime}(x)\}dx}=e^{ax}f(x)+c\)
×
যদি \(u\) এবং \(v\) উভয়েই \(x\) এর ফাংশন হয় তবে, \(\int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
Proof:
শর্তমতে,
\(u=u(x)\) এবং \(w=w(x)\)
ধরি,
\(v=\frac{d}{dx}(w)\)
\(\Rightarrow \int{vdx}=\int{\frac{d}{dx}(w)dx}\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{vdx}=\int{dw}\)
\(\Rightarrow \int{vdx}=w\) ➜ যোগজীকরণ ধ্রুবক \((c)\) উপেক্ষা করে ।
\(\Rightarrow u\int{vdx}=uw\) ➜ উভয় পার্শে \(u\) গুণ করে ।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=\frac{d}{dx}(uw)\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=u\frac{d}{dx}(w)+w\frac{d}{dx}(u)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\int{vdx}.\frac{d}{dx}(u)\) ➜ \(\because \frac{d}{dx}(w)=v, w=\int{vdx}\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\frac{d}{dx}(u)\int{vdx}\)
\(\Rightarrow \int{\frac{d}{dx}\left(u\int{vdx}\right)dx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{d\left(u\int{vdx}\right)}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) \(\Rightarrow u\int{vdx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜ \(\because \int{dx}=x\)
\(\Rightarrow \int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}=u\int{vdx}\)
\(\therefore \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(u=u(x)\) এবং \(w=w(x)\)
ধরি,
\(v=\frac{d}{dx}(w)\)
\(\Rightarrow \int{vdx}=\int{\frac{d}{dx}(w)dx}\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{vdx}=\int{dw}\)
\(\Rightarrow \int{vdx}=w\) ➜ যোগজীকরণ ধ্রুবক \((c)\) উপেক্ষা করে ।
\(\Rightarrow u\int{vdx}=uw\) ➜ উভয় পার্শে \(u\) গুণ করে ।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=\frac{d}{dx}(uw)\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=u\frac{d}{dx}(w)+w\frac{d}{dx}(u)\) ➜ \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\int{vdx}.\frac{d}{dx}(u)\) ➜ \(\because \frac{d}{dx}(w)=v, w=\int{vdx}\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\frac{d}{dx}(u)\int{vdx}\)
\(\Rightarrow \int{\frac{d}{dx}\left(u\int{vdx}\right)dx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜ উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{d\left(u\int{vdx}\right)}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) \(\Rightarrow u\int{vdx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜ \(\because \int{dx}=x\)
\(\Rightarrow \int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}=u\int{vdx}\)
\(\therefore \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
×
\(\int{\ln{|x|}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\ln{|x|}.1dx}\)
\(=\ln{|x|}\int{1dx}-\int{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\ln{|x|}.x-\int{\left\{\frac{1}{x}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=x\ln{|x|}-\int{1dx}\)
\(=x\ln{|x|}-x+c\) ➜ \(\because \int{1dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x(\ln{|x|}-1)+c\)
×
\(\int{\log{|x|}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\log_ae\times\log_e{|x|}dx}\) ➜ \(\because \log_p{|x|}=\log_pe\times\log_e{|x|}\)
\(=\log_ae\int{\ln{|x|}dx}\)
\(=\log_ae\int{\ln{|x|}.1dx}\)
\(=\log_ae\left[\ln{|x|}\int{1dx}-\int{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\right]\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\log_ae\left[\ln{|x|}.x-\int{\left\{\frac{1}{x}.x\right\}dx}\right]\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=\log_ae\left(x\ln{|x|}-\int{1dx}\right)\)
\(=\log_ae\left(x\ln{|x|}-x\right)+c\) ➜ \(\because \int{1dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\log_a{e}\times x(\ln{|x|}-1)+c\)
×
\(\int{\sin^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\sin^{-1}{x}.1dx}\)
\(=\sin^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\sin^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\sin^{-1}{x}.x-\int{\left\{\frac{1}{\sqrt{1-x^2}}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(=x\sin^{-1}{x}-\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=x\sin^{-1}{x}+\frac{1}{2}\int{\frac{-2xdx}{\sqrt{1-x^2}}}\)
\(=x\sin^{-1}{x}+\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}\)
\(=x\sin^{-1}{x}+\frac{1}{2}.2\sqrt{t}+c\) ➜ \(\because \int{\frac{dx}{\sqrt{x}}}=2\sqrt{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\sin^{-1}{x}+\sqrt{t}+c\)
\(=x\sin^{-1}{x}+\sqrt{1-x^2}+c\) ➜ \(\because t=1-x^2\)
×
\(\int{\cos^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\cos^{-1}{x}.1dx}\)
\(=\cos^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\cos^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\cos^{-1}{x}.x-\int{\left\{-\frac{1}{\sqrt{1-x^2}}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(=x\cos^{-1}{x}+\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=x\cos^{-1}{x}-\frac{1}{2}\int{\frac{-2xdx}{\sqrt{1-x^2}}}\)
\(=x\cos^{-1}{x}-\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}\)
\(=x\cos^{-1}{x}-\frac{1}{2}.2\sqrt{t}+c\) ➜ \(\because \int{\frac{dx}{\sqrt{x}}}=2\sqrt{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\cos^{-1}{x}-\sqrt{t}+c\)
\(=x\cos^{-1}{x}-\sqrt{1-x^2}+c\) ➜ \(\because t=1-x^2\)
×
\(\int{\tan^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\tan^{-1}{x}.1dx}\)
\(=\tan^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\tan^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\tan^{-1}{x}.x-\int{\left\{\frac{1}{1+x^2}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(=x\tan^{-1}{x}-\int{\frac{1}{1+x^2}xdx}\)
\(=x\tan^{-1}{x}-\int{\frac{1}{t}\times{\frac{1}{2}}dt}\)
\(=x\tan^{-1}{x}-\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=x\tan^{-1}{x}-\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\tan^{-1}{x}-\frac{1}{2}\ln{|1+x^2|}+c\) ➜ \(\because t=1+x^2\)
×
\(\int{cosec^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{cosec^{-1}{x}.1dx}\)
\(=cosec^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(cosec^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=cosec^{-1}{x}.x-\int{\left\{-\frac{1}{x\sqrt{x^2-1}}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(cosec^{-1}{x})=-\frac{1}{x\sqrt{x^2-1}}\)
\(=x \ cosec^{-1}{x}+\int{\frac{1}{\sqrt{x^2-1}}dx}\)
\(=x \ cosec^{-1}{x}+\int{\frac{1}{\sqrt{x^2-1^2}}dx}\)
\(=x \ cosec^{-1}{x}+\ln{|x+\sqrt{x^2-1^2}|}+c\) ➜ \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|x+\sqrt{x^2-a^2}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x \ cosec^{-1}{x}+\ln{|x+\sqrt{x^2-1}|}+c\)
×
\(\int{\sec^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\sec^{-1}{x}.1dx}\)
\(=\sec^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\sec^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\sec^{-1}{x}.x-\int{\left\{\frac{1}{x\sqrt{x^2-1}}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\sec^{-1}{x})=\frac{1}{x\sqrt{x^2-1}}\)
\(=x\sec^{-1}{x}-\int{\frac{1}{\sqrt{x^2-1}}dx}\)
\(=x\sec^{-1}{x}-\int{\frac{1}{\sqrt{x^2-1^2}}dx}\)
\(=x\sec^{-1}{x}-\ln{|x+\sqrt{x^2-1^2}|}+c\) ➜ \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|x+\sqrt{x^2-a^2}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\sec^{-1}{x}-\ln{|x+\sqrt{x^2-1}|}+c\)
×
\(\int{\cot^{-1}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(=\int{\cot^{-1}{x}.1dx}\)
\(=\cot^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\cot^{-1}{x})\int{1dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\cot^{-1}{x}.x-\int{\left\{-\frac{1}{1+x^2}.x\right\}dx}\) ➜ \(\because \int{1dx}=x, \frac{d}{dx}(\cot^{-1}{x})=-\frac{1}{1+x^2}\)
\(=x\cot^{-1}{x}+\int{\frac{1}{1+x^2}xdx}\)
\(=x\cot^{-1}{x}+\int{\frac{1}{t}\times{\frac{1}{2}}dt}\)
\(=x\cot^{-1}{x}+\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=x\cot^{-1}{x}+\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\cot^{-1}{x}+\frac{1}{2}\ln{|1+x^2|}+c\) ➜ \(\because t=1+x^2\)
×
\(\int{e^x\sin{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধানঃ
\(I=\int{e^x\sin{x}dx}\)
\(\Rightarrow I=\int{\sin{x}.e^xdx}\)
\(\Rightarrow I=\sin{x}\int{e^xdx}-\int{\left\{\frac{d}{dx}(\sin{x})\int{e^xdx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sin{x}.e^x-\int{\left\{\cos{x}.e^x\right\}dx}\) ➜ \(\because \int{e^xdx}=e^x, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=e^x\sin{x}-\int{\cos{x}.e^xdx}\)
\(\Rightarrow I=e^x\sin{x}-\left[\cos{x}\int{e^xdx}-\int{\left\{\frac{d}{dx}(\cos{x})\int{e^xdx}\right\}dx}\right]\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=e^x\sin{x}-\left[\cos{x}.e^x-\int{(-\sin{x}).e^xdx}\right]\) ➜ \(\because \int{e^xdx}=e^x, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=e^x\sin{x}-\left[e^x\cos{x}+\int{e^x\sin{x}dx}\right]\)
\(\Rightarrow I=e^x\sin{x}-\left[e^x\cos{x}+I\right]\) ➜ \(\because \int{e^x\sin{x}dx}=I\)
\(\Rightarrow I=e^x\sin{x}-e^x\cos{x}-I\)
\(\Rightarrow I+I=e^x\sin{x}-e^x\cos{x}\)
\(\Rightarrow 2I=e^x\sin{x}-e^x\cos{x}\)
\(\Rightarrow I=\frac{1}{2}(e^x\sin{x}-e^x\cos{x})+c\) ➜\(c\) যোগজীকরণ ধ্রুবক।
\(\therefore \int{e^x\sin{x}dx}=\frac{1}{2}(e^x\sin{x}-e^x\cos{x})+c\)
×
\(\int{e^{ax}\sin{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+c\)
Proof:
\(\Rightarrow I=\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\sin{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\sin{bx})\int{e^{ax}dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sin{bx}.\frac{e^{ax}}{a}-\int{\left\{\cos{bx}.\frac{d}{dx}(bx).\frac{e^{ax}}{a}\right\}dx}\) ➜ \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\int{\left\{\cos{bx}.b.\frac{e^{ax}}{a}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\cos{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\cos{bx})\int{e^{ax}dx}\right\}dx}\right]\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\cos{bx}.\frac{e^{ax}}{a}-\int{(-\sin{bx}).\frac{d}{dx}(bx).\frac{e^{ax}}{a}dx}\right]\) ➜ \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\frac{1}{a}e^{ax}\cos{bx}+\int{\sin{bx}.b.\frac{e^{ax}}{a}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\int{\sin{bx}.e^{ax}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}-\frac{b^2}{a^2}\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}-\frac{b^2}{a^2}I\) ➜ \(\because \int{\sin{bx}.e^{ax}dx}=I\)
\(\Rightarrow I+\frac{b^2}{a^2}I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}\)
\(\Rightarrow I\left(1+\frac{b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\sin{bx}-b\cos{bx})\)
\(\Rightarrow I\left(\frac{a^2+b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\sin{bx}-b\cos{bx})\)
\(\Rightarrow I=\frac{a^2}{a^2+b^2}\times{\frac{1}{a^2}}e^{ax}(a\sin{bx}-b\cos{bx})+c\) ➜\(c\) যোগজীকরণ ধ্রুবক।
\(\Rightarrow I=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+c\)
\(\Rightarrow I=R.S\)
\(\therefore L.S=R.S\)
(Proved)
×
\(\int{e^{ax}\cos{bx}dx}=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+c\)
Proof:
\(\Rightarrow I=\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\cos{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\cos{bx})\int{e^{ax}dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\cos{bx}.\frac{e^{ax}}{a}-\int{\left\{(-\sin{bx}).\frac{d}{dx}(bx).\frac{e^{ax}}{a}\right\}dx}\) ➜ \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\int{\left\{\sin{bx}.b.\frac{e^{ax}}{a}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\sin{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\sin{bx})\int{e^{ax}dx}\right\}dx}\right]\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\sin{bx}.\frac{e^{ax}}{a}-\int{\cos{bx}.\frac{d}{dx}(bx).\frac{e^{ax}}{a}dx}\right]\) ➜ \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\frac{1}{a}e^{ax}\sin{bx}-\int{\cos{bx}.b.\frac{e^{ax}}{a}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\int{\cos{bx}.e^{ax}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}-\frac{b^2}{a^2}\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}-\frac{b^2}{a^2}I\) ➜ \(\because \int{\cos{bx}.e^{ax}dx}=I\)
\(\Rightarrow I+\frac{b^2}{a^2}I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}\)
\(\Rightarrow I\left(1+\frac{b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\cos{bx}+b\sin{bx})\)
\(\Rightarrow I\left(\frac{a^2+b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\cos{bx}+b\sin{bx})\)
\(\Rightarrow I=\frac{a^2}{a^2+b^2}\times{\frac{1}{a^2}}e^{ax}(a\cos{bx}+b\sin{bx})+c\) ➜\(c\) যোগজীকরণ ধ্রুবক।
\(\Rightarrow I=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+c\)
\(\Rightarrow I=R.S\)
\(\therefore L.S=R.S\)
(Proved)
×
\(\int{\sqrt{x^2-a^2}dx}=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{|x+\sqrt{x^2-a^2}|}+c\)
Proof:
\(=\int{\sqrt{a^2\sec^2{\theta}-a^2}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\sqrt{a^2(\sec^2{\theta}-1)}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{a\sqrt{\tan^2{\theta}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=a^2\int{\tan{\theta}.\sec{\theta}\tan{\theta}d\theta}\)
\(=a^2\left[\tan{\theta}\int{\sec{\theta}\tan{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\tan{\theta})\int{\sec{\theta}\tan{\theta}d\theta}\right\}d\theta}\right]\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=a^2\left[\tan{\theta}\sec{\theta}-\int{\sec^2{\theta}.\sec{\theta}d\theta}\right]\)
\(=a^2\tan{\theta}\sec{\theta}-a^2\int{\sec^3{\theta}d\theta}\)
\(=a^2\tan{\theta}\sec{\theta}-a^2I\)
ধরি,
\(I=\int{\sec^3{\theta}d\theta}\)
\(\Rightarrow I=\int{\sec^2{\theta}.\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\int{\sec^2{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\sec{\theta})\int{\sec^2{\theta}d\theta}\right\}d\theta}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan{\theta}.\tan{\theta}d\theta}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \frac{d}{dx}(sec{x})=sec{x}\tan{x}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan^2{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}(\sec^2{\theta}-1)d\theta}\) ➜ \(\because tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec^3{\theta}d\theta}+\int{\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-I+\int{\sec{\theta}d\theta}\) ➜ \(\because \int{\sec^3{\theta}d\theta}=I\)
\(\Rightarrow I+I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow 2I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow I=\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\) ➜ \(\because \int{\sec{\theta}d\theta}=\ln{\left|\sec{x}+\tan{x}\right|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
এখন,
\(L.S=a^2\tan{\theta}\sec{\theta}-a^2I\)
\(=a^2\tan{\theta}\sec{\theta}-a^2\left\{\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\right\}\)
\(=a^2\tan{\theta}\sec{\theta}-\frac{a^2}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)-a^2c_{1}\)
\(=a^2\tan{\theta}\sec{\theta}-\frac{a^2}{2}\sec{\theta}\tan{\theta}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\tan{\theta}\right|}-a^2c_{1}\)
\(=\frac{2a^2-a^2}{2}\sec{\theta}\tan{\theta}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\tan{\theta}\right|}-a^2c_{1}\)
\(=\frac{a^2}{2}\sec{\theta}\sqrt{\sec^2{\theta}-1}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\sqrt{\sec^2{\theta}-1}\right|}-a^2c_{1}\) ➜ \(\because \tan{x}=\sqrt{\sec^2{x}-1}\)
\(=\frac{a^2}{2}\times{\frac{x}{a}}\sqrt{\left(\frac{x}{a}\right)^2-1}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}\right|}-a^2c_{1}\) ➜ \(\because \sec{\theta}=\frac{x}{a}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2}{a^2}-1}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2-a^2}{a^2}}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2-a^2}{a^2}}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\frac{\sqrt{x^2-a^2}}{a}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\frac{\left|x+\sqrt{x^2-a^2}\right|}{|a|}}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\left(\ln{\left|x+\sqrt{x^2-a^2}\right|}-\ln{|a|}\right)-a^2c_{1}\) ➜ \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|x+\sqrt{x^2-a^2}\right|}+\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|x+\sqrt{x^2-a^2}\right|}+c\) ➜ \(\because c=\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)
×
\(\int{\sqrt{x^2+a^2}dx}=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{|x+\sqrt{x^2+a^2}|}+c\)
Proof:
\(=\int{\sqrt{a^2\tan^2{\theta}+a^2}.a\sec^2{\theta}d\theta}\)
\(=\int{\sqrt{a^2(\tan^2{\theta}+1)}.a\sec^2{\theta}d\theta}\)
\(=\int{a\sqrt{\sec^2{\theta}}.a\sec^2{\theta}d\theta}\)
\(=a^2\int{\sec{\theta}.\sec^2{\theta}d\theta}\)
\(=a^2\int{\sec^3{\theta}d\theta}\)
\(=a^2I\)
ধরি,
\(I=\int{\sec^3{\theta}d\theta}\)
\(\Rightarrow I=\int{\sec^2{\theta}.\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\int{\sec^2{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\sec{\theta})\int{\sec^2{\theta}d\theta}\right\}d\theta}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan{\theta}.\tan{\theta}d\theta}\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \frac{d}{dx}(sec{x})=sec{x}\tan{x}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan^2{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}(\sec^2{\theta}-1)d\theta}\) ➜ \(\because tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec^3{\theta}d\theta}+\int{\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-I+\int{\sec{\theta}d\theta}\) ➜ \(\because \int{\sec^3{\theta}d\theta}=I\)
\(\Rightarrow I+I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow 2I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow I=\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\) ➜ \(\because \int{\sec{\theta}d\theta}=\ln{\left|\sec{x}+\tan{x}\right|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
এখন,
\(L.S=a^2I\)
\(=a^2\left\{\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\right\}\)
\(=\frac{a^2}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)-a^2c_{1}\)
\(=\frac{a^2}{2}\sec{\theta}\tan{\theta}+\frac{a^2}{2}\ln{\left|\tan{\theta}+\sec{\theta}\right|}-a^2c_{1}\)
\(=\frac{a^2}{2}\tan{\theta}\sqrt{\tan^2{\theta}+1}+\frac{a^2}{2}\ln{\left|\tan{\theta}+\sqrt{\tan^2{\theta}+1}\right|}-a^2c_{1}\) ➜ \(\because \sec{x}=\sqrt{\tan^2{x}+1}\)
\(=\frac{a^2}{2}\times{\frac{x}{a}}\sqrt{\left(\frac{x}{a}\right)^2+1}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1}\right|}-a^2c_{1}\) ➜ \(\because \tan{\theta}=\frac{x}{a}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2}{a^2}+1}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2+a^2}{a^2}}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2+a^2}{a^2}}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\frac{\sqrt{x^2+a^2}}{a}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\frac{\sqrt{x^2+a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|\frac{x+\sqrt{x^2+a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\frac{\left|x+\sqrt{x^2+a^2}\right|}{|a|}}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\left(\ln{\left|x+\sqrt{x^2+a^2}\right|}-\ln{|a|}\right)-a^2c_{1}\) ➜ \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|x+\sqrt{x^2+a^2}\right|}-\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|x+\sqrt{x^2+a^2}\right|}+c\) ➜ \(\because c=-\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)
×
\(\int{e^x\{f(x)+f^{\prime}(x)\}dx}=e^xf(x)+c\)
Proof:
\(=\int{\{e^xf(x)+e^xf^{\prime}(x)\}dx}\)
\(=\int{e^xf(x)dx}+\int{e^xf^{\prime}(x)dx}\)
\(=\int{e^xf(x)dx}+e^x\int{f^{\prime}(x)dx}-\int{\left\{\frac{d}{dx}(e^x)\int{f^{\prime}(x)dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\int{e^xf(x)dx}+e^xf(x)-\int{e^xf(x)dx}\) ➜ \(\because \int{f^{\prime}(x)dx}=f(x), \frac{d}{dx}(e^x)=e^x\)
\(=e^xf(x)+c\) ➜ \(c\) যোগজীকরণ ধ্রুবক।
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)
×
\(\int{e^{ax}\{af(x)+f^{\prime}(x)\}dx}=e^{ax}f(x)+c\)
Proof:
\(=\int{\{ae^{ax}f(x)+e^{ax}f^{\prime}(x)\}dx}\)
\(=\int{ae^{ax}f(x)dx}+\int{e^{ax}f^{\prime}(x)dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}\int{f^{\prime}(x)dx}-\int{\left\{\frac{d}{dx}(e^{ax})\int{f^{\prime}(x)dx}\right\}dx}\) ➜ \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{e^{ax}.\frac{d}{dx}(ax).f(x)dx}\) ➜ \(\because \int{f^{\prime}(x)dx}=f(x), \frac{d}{dx}(e^x)=e^x\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{e^{ax}.a.f(x)dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{ae^{ax}f(x)dx}\)
\(=e^{ax}f(x)+c\) ➜ \(c\) যোগজীকরণ ধ্রুবক।
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)
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