এ অধ্যায়ে আমরা যে বিষয়গুলি আলোচনা করব।
- কতিপয় স্মরণীয় আকারের যোগজ
- \(\int{f\{g(x)\}g^{\prime}(x)dx}=\int{f(t)dz}\)
- \(\int{\{f(x)\}^{n}f^{\prime}(x)dx}=\frac{\{f(x)\}^{n+1}}{n+1}+c\)
- \(\int{e^{f(x)}f^{\prime}(x)dx}=e^{f(x)}+c\)
- \(\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}=2\sqrt{f(x)}+c\)
- \(\int{\frac{f^{\prime}(x)}{f(x)}dx}=\ln{|f(x)|}+c\)
- \(\int{\tan{x}dx}=\ln{|\sec{x}|}+c\)
- \(\int{\cot{x}dx}=\ln{|\sin{x}|}+c\)
- \(\int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}+c\)\(=\ln{|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}|}+c\)
- \(\int{cosec \ {x}dx}=\ln{|cosec \ {x}-\cot{x}|}+c\)\(=\ln{|\tan{\frac{x}{2}}|}+c\)
- \(\int{\tan{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}|}+c\)
- \(\int{\cot{(ax+b)}dx}=\frac{1}{a}\ln{|\sin{(ax+b)}|}+c\)
- \(\int{\sec{(ax+b)}dx}\)\(=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c\)\(=\frac{1}{a}\ln{|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}|}+c\)
- \(\int{cosec \ {(ax+b)}dx}\)\(=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c\)\(=\frac{1}{a}\ln{|\tan{\frac{(ax+b)}{2}}|}+c\)
- বিশেষ আকারের যোগজ
- \(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকারের যোগজ
- \(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকারের যোগজ
- \(\int{\sin^{n}{x}dx}\) আকারের যোগজ
- \(\int{\cos^{n}{x}dx}\) আকারের যোগজ
- সমাধানকৃত উদাহরণমালা
- অতি সংক্ষিপ্ত প্রশ্ন-উত্তর
- সংক্ষিপ্ত প্রশ্ন-উত্তর
- বর্ণনামূলক প্রশ্ন-উত্তর
কতিপয় স্মরণীয় আকারের যোগজ
বিশেষ আকারের যোগজ
\(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকারের যোগজের ক্ষেত্রে।
\(1.\) যদি, \(m\) অথবা \(n\) বিজোড় সংখ্যা হয় তবে,\(1.(i)\) \(m\) বিজোড় সংখ্যা হলে, \(\cos{x}=t\)
এবং
\(1.(ii)\) \(n\) বিজোড় সংখ্যা হলে, \(\sin{x}=t\)
ধরে সরলীকরণ করার পর যোগজ নির্ণয় করতে হয়।
\(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকারের যোগজের ক্ষেত্রে।
\(2.\) যদি, \(m\)এবং \(n\) উভয়ে বিজোড় সংখ্যা হয় তবে,\(\sin{x}=t\)
ধরে সরলীকরণ করার পর যোগজ নির্ণয় করতে হয়।
\(\int{\sin^{m}{x}\cos^{n}{x}dx}\) আকারের যোগজের ক্ষেত্রে।
\(3.\) যদি, \(m\)এবং \(n\) উভয়ে জোড় সংখ্যা হয় তবে,এই ক্ষেত্রটি উচ্চমাধ্যমিক গণিতে আলোচনা করা হয়নি। পরবর্তি উচ্চতর শ্রেণীতে এর যথেষ্ট আলোচনা আছে।
\(\int{\sin^{n}{x}dx}\) আকারের যোগজের ক্ষেত্রে।
\(4.(i)\) যদি, \(n\) বিজোড় সংখ্যা হয় তবে,\(\cos{x}=t\) ধরে সরলীকরণ করার পর যোগজ নির্ণয় করতে হয়।
\(4.(ii)\) যদি, \(n\) জোড় সংখ্যা হয় তবে,
ইন্টিগ্র্যান্ডকে গুণিতক কোণে প্রকাশ করার পর যোগজীকরণ করতে হয়।
\(\int{\cos^{n}{x}dx}\) আকারের যোগজের ক্ষেত্রে।
\(5.(i)\) যদি, \(n\) বিজোড় সংখ্যা হয় তবে,\(\sin{x}=t\) ধরে সরলীকরণ করার পর যোগজ নির্ণয় করতে হয়।
\(5.(ii)\) যদি, \(n\) জোড় সংখ্যা হয় তবে,
ইন্টিগ্র্যান্ডকে গুণিতক কোণে প্রকাশ করার পর যোগজীকরণ করতে হয়।
\((1.)\) প্রমাণ কর যে, \(\int{f\{g(x)\}g^{\prime}(x)dx}=\int{f(t)dz}\)
Proof:
ধরি,
\(g(x)=t\)
\(\Rightarrow \frac{d}{dx}(g(x))=\frac{d}{dx}(t)\)
\(\Rightarrow g^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore g^{\prime}(x)dx=dt\)
\(L.H=\int{f\{g(x)\}g^{\prime}(x)dx}\)\(g(x)=t\)
\(\Rightarrow \frac{d}{dx}(g(x))=\frac{d}{dx}(t)\)
\(\Rightarrow g^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore g^{\prime}(x)dx=dt\)
\(=\int{f(t)dt}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((2.)\) প্রমাণ কর যে, \(\int{\{f(x)\}^{n}f^{\prime}(x)dx}=\frac{\{f(x)\}^{n+1}}{n+1}+c\)
Proof:
ধরি,
\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(L.H=\int{\{f(x)\}^{n}f^{\prime}(x)dx}\)\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(=\int{t^{n}dt}\)
\(=\frac{t^{n+1}}{n+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{\{f(x)\}^{n+1}}{n+1}+c\) ➜ \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((3.)\) প্রমাণ কর যে, \(\int{e^{f(x)}f^{\prime}(x)dx}=e^{f(x)}+c\)
Proof:
ধরি,
\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(L.H=\int{e^{f(x)}f^{\prime}(x)dx}\)\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{f(x)}+c\) ➜ \(\because t=f(x)\)
\(=R.H\) \(L.H=R.H\) (Proved)
\((4.)\) প্রমাণ কর যে, \(\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}=2\sqrt{f(x)}+c\)
Proof:
ধরি,
\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(L.H=\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}\)\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(=\int{\frac{1}{\sqrt{f(x)}}f^{\prime}(x)dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{f(x)}+c\) ➜ \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((5.)\) প্রমাণ কর যে, \(\int{\frac{f^{\prime}(x)}{f(x)}dx}=\ln{|f(x)|}+c\)
Proof:
ধরি,
\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(L.H=\int{\frac{f^{\prime}(x)}{f(x)}dx}\)\(f(x)=t\)
\(\Rightarrow \frac{d}{dx}(f(x))=\frac{d}{dx}(t)\)
\(\Rightarrow f^{\prime}(x)=\frac{dt}{dx}\)
\(\therefore f^{\prime}(x)dx=dt\)
\(=\int{\frac{1}{f(x)}f^{\prime}(x)dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|f(x)|}+c\) ➜ \(\because t=f(x)\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((6.)\) প্রমাণ কর যে, \(\int{\tan{x}dx}=\ln{|\sec{x}|}+c\)
Proof:
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(L.H=\int{\tan{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\frac{\sin{x}}{\cos{x}}dx}\)
\(=\int{\frac{1}{\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{|\cos{x}|}+c\) ➜ \(\because t=\cos{x}\)
\(=\ln{|(\cos{x})^{-1}|}+c\) ➜ \(\because n\ln{|x|}=\ln{x^n}\)
\(=\ln{|\frac{1}{\cos{x}}|}+c\)
\(=\ln{|\sec{x}|}+c\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((7.)\) প্রমাণ কর যে, \(\int{\cot{x}dx}=\ln{|\sin{x}|}+c\)
Proof:
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}dx=dt\)
\(\therefore \cos{x}dx=dt\)
\(L.H=\int{\cot{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}dx=dt\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\frac{\cos{x}}{\sin{x}}dx}\)
\(=\int{\frac{1}{\sin{x}}\cos{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sin{x}|}+c\) ➜ \(\because t=\sin{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((8.)\) প্রমাণ কর যে, \(\int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}+c=\ln{|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}|}+c\)
Proof:
ধরি,
\(\sec{x}+\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec{x}+\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{x}\tan{x}+\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}(\tan{x}+\sec{x})=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}(\sec{x}+\tan{x})=\frac{dt}{dx}\)
\(\therefore \sec{x}(\sec{x}+\tan{x})dx=dt\)
\(L.H=\int{\sec{x}dx}\)\(\sec{x}+\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec{x}+\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{x}\tan{x}+\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}(\tan{x}+\sec{x})=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}(\sec{x}+\tan{x})=\frac{dt}{dx}\)
\(\therefore \sec{x}(\sec{x}+\tan{x})dx=dt\)
\(=\int{\frac{\sec{x}(\sec{x}+\tan{x})}{(\sec{x}+\tan{x})}dx}\)
\(=\int{\frac{1}{(\sec{x}+\tan{x})}\sec{x}(\sec{x}+\tan{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\sec{x}+\tan{x}|}+c\) ➜ \(\because t=\sec{x}+\tan{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\ln{|\sec{x}+\tan{x}|}+c\)
\(=\ln{\left|\frac{1}{\cos{x}}+\frac{\sin{x}}{\cos{x}}\right|}+c\)
\(=\ln{\left|\frac{1+\sin{x}}{\cos{x}}\right|}+c\)
\(=\ln{\left|\frac{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}\right|}+c\) ➜ \(\because \sin^2{\frac{A}{2}}+\cos^2{\frac{A}{2}}=1, 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=\sin{A}\), \(\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}=\cos{A}\)
\(=\ln{\left|\frac{(\sin{\frac{x}{2}}+\cos{\frac{x}{2}})^2}{(\cos{\frac{x}{2}}+\sin{\frac{x}{2}})(\cos{\frac{x}{2}}-\sin{\frac{x}{2}})}\right|}+c\) ➜ \(\because a^2+b^2+2ab=(a+b)^2, (a+b)(a-b)=a^2-b^2\)
\(=\ln{\left|\frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\cos{\frac{x}{2}}-\sin{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\cos{\frac{x}{2}}+\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}-\sin{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\cos{\frac{x}{2}}\left(1+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right)}{\cos{\frac{x}{2}}\left(1-\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right)}\right|}+c\)
\(=\ln{\left|\frac{1+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}}{1-\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}}\right|}+c\)
\(=\ln{\left|\frac{1+\tan{\frac{x}{2}}}{1-\tan{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\frac{\tan{\frac{\pi}{4}}+\tan{\frac{x}{2}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{x}{2}}}\right|}+c\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=\ln{\left|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right|}+c\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\therefore \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}+c=\ln{\left|\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right|}+c\)
(proved)
\((9.)\) প্রমাণ কর যে, \(\int{cosec \ {x}dx}=\ln{|cosec \ {x}-\cot{x}|}+c=\ln{|\tan{\frac{x}{2}}|}+c\)
Proof:
ধরি,
\(cosec \ {x}-\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(cosec \ {x}-\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}\cot{x}+cosec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow cosec^2{x}-cosec \ {x}\cot{x}=\frac{dt}{dx}\)
\(\Rightarrow cosec \ {x}(cosec \ {x}-\cot{x})=\frac{dt}{dx}\)
\(\therefore cosec \ {x}(cosec \ {x}-\cot{x})dx=dt\)
\(L.H=\int{cosec \ {x}dx}\)\(cosec \ {x}-\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(cosec \ {x}-\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}\cot{x}+cosec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow cosec^2{x}-cosec \ {x}\cot{x}=\frac{dt}{dx}\)
\(\Rightarrow cosec \ {x}(cosec \ {x}-\cot{x})=\frac{dt}{dx}\)
\(\therefore cosec \ {x}(cosec \ {x}-\cot{x})dx=dt\)
\(=\int{\frac{cosec \ {x}(cosec \ {x}-\cot{x})}{(cosec \ {x}-\cot{x})}dx}\)
\(=\int{\frac{1}{(cosec \ {x}-\cot{x})}cosec \ {x}(cosec \ {x}-\cot{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|cosec \ {x}-\cot{x}|}+c\) ➜ \(\because t=cosec \ {x}-\cot{x}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\ln{|cosec \ {x}-\cot{x}|}+c\)
\(=\ln{\left|\frac{1}{\sin{x}}-\frac{\cos{x}}{\sin{x}}\right|}+c\)
\(=\ln{\left|\frac{1-\cos{x}}{\sin{x}}\right|}+c\)
\(=\ln{\left|\frac{2\sin^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\right|}+c\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\ln{\left|\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\right|}+c\)
\(=\ln{\left|\tan{\frac{x}{2}}\right|}+c\)
\(\therefore \int{cosec \ {x}dx}=\ln{|cosec \ {x}-\cot{x}|}+c=\ln{\left|\tan{\frac{x}{2}}\right|}+c\)
(proved)
\((10.)\) প্রমাণ কর যে, \(\int{\tan{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}|}+c\)
Proof:
ধরি,
\(\cos{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\cos{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{(ax+b)}.\frac{d}{dx}(ax+b)=\frac{dt}{dx}\)
\(\Rightarrow -\sin{(ax+b)}.(a+0)=\frac{dt}{dx}\)
\(\Rightarrow -a\sin{(ax+b)}=\frac{dt}{dx}\)
\(\Rightarrow -a\sin{(ax+b)}dx=dt\)
\(\therefore \sin{(ax+b)}dx=-\frac{1}{a}dt\)
\(L.H=\int{\tan{(ax+b)}dx}\)\(\cos{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\cos{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{(ax+b)}.\frac{d}{dx}(ax+b)=\frac{dt}{dx}\)
\(\Rightarrow -\sin{(ax+b)}.(a+0)=\frac{dt}{dx}\)
\(\Rightarrow -a\sin{(ax+b)}=\frac{dt}{dx}\)
\(\Rightarrow -a\sin{(ax+b)}dx=dt\)
\(\therefore \sin{(ax+b)}dx=-\frac{1}{a}dt\)
\(=\int{\frac{\sin{(ax+b)}}{\cos{(ax+b)}}dx}\)
\(=\int{\frac{1}{\cos{(ax+b)}}\sin{(ax+b)}dx}\)
\(=\int{\frac{1}{t}\times{-\frac{1}{a}dt}}\)
\(=-\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{a}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{a}\ln{|\cos{(ax+b)}|}+c\) ➜ \(\because t=\cos{(ax+b)}\)
\(=\frac{1}{a}\ln{|(\cos{(ax+b)})^{-1}|}+c\) ➜ \(\because n\ln{|x|}=\ln{x^n}\)
\(=\frac{1}{a}\ln{|\frac{1}{\cos{(ax+b)}}|}+c\)
\(=\frac{1}{a}\ln{|\sec{(ax+b)}|}+c\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((11.)\) প্রমাণ কর যে, \(\int{\cot{(ax+b)}dx}=\frac{1}{a}\ln{|\sin{(ax+b)}|}+c\)
Proof:
ধরি,
\(\sin{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\sin{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{(ax+b)}.\frac{d}{dx}(ax+b)=\frac{dt}{dx}\)
\(\Rightarrow \cos{(ax+b)}.(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a\cos{(ax+b)}dx=dt\)
\(\therefore \cos{(ax+b)}dx=\frac{1}{a}dt\)
\(L.H=\int{\cot{(ax+b)}dx}\)\(\sin{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\sin{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{(ax+b)}.\frac{d}{dx}(ax+b)=\frac{dt}{dx}\)
\(\Rightarrow \cos{(ax+b)}.(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a\cos{(ax+b)}dx=dt\)
\(\therefore \cos{(ax+b)}dx=\frac{1}{a}dt\)
\(=\int{\frac{\cos{(ax+b)}}{\sin{(ax+b)}}dx}\)
\(=\int{\frac{1}{\sin{(ax+b)}}\cos{(ax+b)}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|\sin{(ax+b)}|}+c\) ➜ \(\because t=\sin{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
\((12.)\) প্রমাণ কর যে, \(\int{\sec{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c=\frac{1}{a}\ln{|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}|}+c\)
Proof:
ধরি,
\(\sec{(ax+b)}+\tan{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\sec{(ax+b)}+\tan{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{(ax+b)}\tan{(ax+b)}(a+0)+\sec^2{(ax+b)}(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a\sec{(ax+b)}\{\tan{(ax+b)}+\sec{(ax+b)}\}=\frac{dt}{dx}\)
\(\Rightarrow a\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}=\frac{dt}{dx}\)
\(\therefore \sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}dx=\frac{1}{a}dt\)
\(L.H=\int{\sec{(ax+b)}dx}\)\(\sec{(ax+b)}+\tan{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{\sec{(ax+b)}+\tan{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{(ax+b)}\tan{(ax+b)}(a+0)+\sec^2{(ax+b)}(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a\sec{(ax+b)}\{\tan{(ax+b)}+\sec{(ax+b)}\}=\frac{dt}{dx}\)
\(\Rightarrow a\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}=\frac{dt}{dx}\)
\(\therefore \sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}dx=\frac{1}{a}dt\)
\(=\int{\frac{\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}}{\sec{(ax+b)}+\tan{(ax+b)}}dx}\)
\(=\int{\frac{1}{\sec{(ax+b)}+\tan{(ax+b)}}\sec{(ax+b)}\{\sec{(ax+b)}+\tan{(ax+b)}\}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c\) ➜ \(\because t=\sec{(ax+b)}+\tan{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1}{\cos{(ax+b)}}+\frac{\sin{(ax+b)}}{\cos{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\sin{(ax+b)}}{\cos{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\sin^2{\frac{(ax+b)}{2}}+\cos^2{\frac{(ax+b)}{2}}+2\sin{\frac{(ax+b)}{2}}\cos{\frac{(ax+b)}{2}}}{\cos^2{\frac{(ax+b)}{2}}-\sin^2{\frac{(ax+b)}{2}}}\right|}+c\) ➜ \(\because \sin^2{\frac{A}{2}}+\cos^2{\frac{A}{2}}=1, 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}=\sin{A}\), \(\cos^2{\frac{A}{2}}-\sin^2{\frac{A}{2}}=\cos{A}\)
\(=\frac{1}{a}\ln{\left|\frac{(\sin{\frac{(ax+b)}{2}}+\cos{\frac{(ax+b)}{2}})^2}{(\cos{\frac{(ax+b)}{2}}+\sin{\frac{(ax+b)}{2}})(\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}})}\right|}+c\) ➜ \(\because a^2+b^2+2ab=(a+b)^2, (a+b)(a-b)=a^2-b^2\)
\(=\frac{1}{a}\ln{\left|\frac{\sin{\frac{(ax+b)}{2}}+\cos{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\cos{\frac{(ax+b)}{2}}+\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}-\sin{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\cos{\frac{(ax+b)}{2}}\left(1+\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right)}{\cos{\frac{(ax+b)}{2}}\left(1-\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right)}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}}{1-\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1+\tan{\frac{(ax+b)}{2}}}{1-\tan{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{\tan{\frac{\pi}{4}}+\tan{\frac{(ax+b)}{2}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{(ax+b)}{2}}}\right|}+c\) ➜ \(\because \tan{\frac{\pi}{4}}=1\)
\(=\frac{1}{a}\ln{\left|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}\right|}+c\) ➜ \(\because \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\tan{(A+B)}\)
\(\therefore \int{\sec{(ax+b)}dx}=\frac{1}{a}\ln{|\sec{(ax+b)}+\tan{(ax+b)}|}+c=\frac{1}{a}\ln{\left|\tan{\left\{\frac{\pi}{4}+\frac{(ax+b)}{2}\right\}}\right|}+c\)
(proved)
\((13.)\) প্রমাণ কর যে, \(\int{cosec \ {(ax+b)}dx}=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c=\frac{1}{a}\ln{|\tan{\frac{(ax+b)}{2}}|}+c\)
Proof:
ধরি,
\(cosec \ {(ax+b)}-\cot{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {(ax+b)}\cot{(ax+b)}(a+0)+cosec^2{(ax+b)}(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a \ cosec^2{(ax+b)}-a \ cosec \ {(ax+b)}\cot{(ax+b)}=\frac{dt}{dx}\)
\(\Rightarrow a \ cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}=\frac{dt}{dx}\)
\(\therefore cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}dx=\frac{1}{a}dt\)
\(L.H=\int{cosec \ {(ax+b)}dx}\)\(cosec \ {(ax+b)}-\cot{(ax+b)}=t\)
\(\Rightarrow \frac{d}{dx}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {(ax+b)}\cot{(ax+b)}(a+0)+cosec^2{(ax+b)}(a+0)=\frac{dt}{dx}\)
\(\Rightarrow a \ cosec^2{(ax+b)}-a \ cosec \ {(ax+b)}\cot{(ax+b)}=\frac{dt}{dx}\)
\(\Rightarrow a \ cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}=\frac{dt}{dx}\)
\(\therefore cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}dx=\frac{1}{a}dt\)
\(=\int{\frac{cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}}{cosec \ {(ax+b)}-\cot{(ax+b)}}dx}\)
\(=\int{\frac{1}{cosec \ {(ax+b)}-\cot{(ax+b)}}cosec \ {(ax+b)}\{cosec \ {(ax+b)}-\cot{(ax+b)}\}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c\) ➜ \(\because t=cosec \ {(ax+b)}-\cot{(ax+b)}\)
\(=R.H\)
\(L.H=R.H\)
(Proved)
আবার,
\(\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1}{\sin{(ax+b)}}-\frac{\cos{(ax+b)}}{\sin{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{1-\cos{(ax+b)}}{\sin{(ax+b)}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\frac{2\sin^2{\frac{(ax+b)}{2}}}{2\sin{\frac{(ax+b)}{2}}\cos{\frac{(ax+b)}{2}}}\right|}+c\) ➜ \(\because 1-\cos{A}=2\sin^2{\frac{A}{2}}, \sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}\)
\(=\frac{1}{a}\ln{\left|\frac{\sin{\frac{(ax+b)}{2}}}{\cos{\frac{(ax+b)}{2}}}\right|}+c\)
\(=\frac{1}{a}\ln{\left|\tan{\frac{(ax+b)}{2}}\right|}+c\)
\(\therefore \int{cosec \ {(ax+b)}dx}=\frac{1}{a}\ln{|cosec \ {(ax+b)}-\cot{(ax+b)}|}+c=\frac{1}{a}\ln{\left|\tan{\frac{(ax+b)}{2}}\right|}+c\)
(proved)
\(1.\) \(\int{\sin^{5}{x}\cos^{4}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধান
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\sin^{5}{x}\cos^{4}{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\sin^{4}{x}\cos^{4}{x}\sin{x}dx}\)
\(=\int{(\sin^{2}{x})^2\cos^{4}{x}\sin{x}dx}\)
\(=\int{(1-\cos^{2}{x})^2\cos^{4}{x}\sin{x}dx}\)
\(=\int{(1-t^2)^2t^4(-dt)}\)
\(=-\int{(1-2t^2+t^4)t^4dt}\)
\(=-\int{(t^4-2t^6+t^8)dt}\)
\(=-\int{t^4dt}+2\int{t^6dt}-\int{t^8dt}\)
\(=-\frac{t^{4+1}}{4+1}+2\frac{t^{6+1}}{6+1}-\frac{t^{8+1}}{8+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{5}}{5}+2\frac{t^{7}}{7}-\frac{t^{9}}{9}+c\)
\(=-\frac{1}{5}t^{5}+2.\frac{1}{7}t^{7}-\frac{1}{9}t^{9}+c\)
\(=-\frac{1}{5}\cos^{5}{x}+\frac{2}{7}\cos^{7}{x}-\frac{1}{9}\cos^{9}{x}+c\) ➜ \(\because t=\cos{x}\)
\(2.\) \(\int{\sin^{5}{x}\cos^{3}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধান
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\sin^{5}{x}\cos^{3}{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\sin^{5}{x}\cos^{2}{x}\cos{x}dx}\)
\(=\int{\sin^{5}{x}(1-\sin^{2}{x})\cos{x}dx}\)
\(=\int{t^{5}(1-t^{2})dt}\)
\(=\int{(t^{5}-t^{7})dt}\)
\(=\int{t^{5}dt}-\int{t^{7}dt}\)
\(=\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{6}}{6}-\frac{t^{8}}{8}+c\)
\(=\frac{1}{6}t^{6}-\frac{1}{8}t^{8}+c\)
\(=\frac{1}{6}\sin^{6}{x}-\frac{1}{8}\sin^{8}{x}+c\) ➜ \(\because t=\sin{x}\)
\(4.(i)\) \(\int{\sin^{7}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধান
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\sin^{7}{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\sin^{6}{x}\sin{x}dx}\)
\(=\int{(\sin^{2}{x})^3\sin{x}dx}\)
\(=\int{(1-\cos^{2}{x})^3\sin{x}dx}\)
\(=\int{(1-t^{2})^3(-dt)}\)
\(=-\int{(1-3t^{2}+3t^4-t^6)dt}\) ➜ \(\because (a-b)^3=a^3-3a^2b+3ab^2-b^3\)
\(=-\int{dt}+3\int{t^{2}dt}-3\int{t^4dt}+\int{t^6dt}\)
\(=-t+3\frac{t^{3+1}}{3+1}-3\frac{t^{4+1}}{4+1}+\frac{t^{6+1}}{6+1}+c\) ➜ \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-t+3\frac{t^{4}}{4}-3\frac{t^{5}}{5}+\frac{t^{7}}{7}+c\)
\(=-t+\frac{3}{4}t^{4}-\frac{3}{5}t^{5}+\frac{1}{7}t^{7}+c\)
\(=-\cos{x}+\frac{3}{4}\cos^{4}{x}-\frac{3}{5}\cos^{5}{x}+\frac{1}{7}cos^{7}{x}+c\) ➜ \(\because t=\cos{x}\)
\(4.(ii)\) \(\int{\sin^{6}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধান
\(\int{\sin^{6}{x}dx}\)
\(=\int{(\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(2\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(1-\cos{2x})^3dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{8}\int{(1-3\cos{2x}+3\cos^2{2x}-\cos^3{2x})dx}\) ➜ \(\because (a-b)^3=a^3-3a^2b+3ab^2-b^3\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{8}\int{\cos^2{2x}dx}-\frac{1}{8}\int{\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{2\cos^2{2x}dx}-\frac{1}{32}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{(1+\cos{4x})dx}\)\(-\frac{1}{32}\int{(\cos{6x}+3\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{dx}+\frac{3}{16}\int{\cos{4x}dx}\)\(-\frac{1}{32}\int{\cos{6x}dx}-\frac{3}{32}\int{\cos{2x}dx}\)
\(=\frac{1}{8}x-\frac{3}{8}.\frac{1}{2}\sin{2x}+\frac{3}{16}x+\frac{3}{16}.\frac{1}{4}\sin{4x}\)\(-\frac{1}{32}.\frac{1}{6}\sin{6x}-\frac{3}{32}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{8}x-\frac{3}{16}\sin{2x}+\frac{3}{16}x+\frac{3}{64}\sin{4x}\)\(-\frac{1}{192}\sin{6x}-\frac{3}{64}\sin{2x}+c\)
\(=\frac{1}{8}x+\frac{3}{16}x-\frac{3}{16}\sin{2x}-\frac{3}{64}\sin{2x}\)\(+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{2+3}{16}x+\frac{-12-3}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x+\frac{-15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x-\frac{15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{1}{192}(60x-45\sin{2x}+9\sin{4x}-\sin{6x})+c\)
\(=\int{(\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(2\sin^{2}{x})^3dx}\)
\(=\frac{1}{8}\int{(1-\cos{2x})^3dx}\) ➜ \(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{8}\int{(1-3\cos{2x}+3\cos^2{2x}-\cos^3{2x})dx}\) ➜ \(\because (a-b)^3=a^3-3a^2b+3ab^2-b^3\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{8}\int{\cos^2{2x}dx}-\frac{1}{8}\int{\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{2\cos^2{2x}dx}-\frac{1}{32}\int{4\cos^3{2x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{(1+\cos{4x})dx}\)\(-\frac{1}{32}\int{(\cos{6x}+3\cos{2x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}, 4\cos^3{A}=\cos{3A}+3\cos{A}\)
\(=\frac{1}{8}\int{dx}-\frac{3}{8}\int{\cos{2x}dx}+\frac{3}{16}\int{dx}+\frac{3}{16}\int{\cos{4x}dx}\)\(-\frac{1}{32}\int{\cos{6x}dx}-\frac{3}{32}\int{\cos{2x}dx}\)
\(=\frac{1}{8}x-\frac{3}{8}.\frac{1}{2}\sin{2x}+\frac{3}{16}x+\frac{3}{16}.\frac{1}{4}\sin{4x}\)\(-\frac{1}{32}.\frac{1}{6}\sin{6x}-\frac{3}{32}.\frac{1}{2}\sin{2x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{8}x-\frac{3}{16}\sin{2x}+\frac{3}{16}x+\frac{3}{64}\sin{4x}\)\(-\frac{1}{192}\sin{6x}-\frac{3}{64}\sin{2x}+c\)
\(=\frac{1}{8}x+\frac{3}{16}x-\frac{3}{16}\sin{2x}-\frac{3}{64}\sin{2x}\)\(+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{2+3}{16}x+\frac{-12-3}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x+\frac{-15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{5}{16}x-\frac{15}{64}\sin{2x}+\frac{3}{64}\sin{4x}-\frac{1}{192}\sin{6x}+c\)
\(=\frac{1}{192}(60x-45\sin{2x}+9\sin{4x}-\sin{6x})+c\)
\(5.(i)\) এর যোজিত ফল নির্ণয় কর।
সমাধান
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\cos^{5}{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\cos^{4}{x}\cos{x}dx}\)
\(=\int{(\cos^{2}{x})^2\cos{x}dx}\)
\(=\int{(1-\sin^{2}{x})^2\cos{x}dx}\)
\(=\int{(1-t^{2})^2dt}\)
\(=\int{(1-2t^{2}+t^4)dt}\) ➜ \(\because (a-b)^2=a^2-2ab+b^2\)
\(=\int{dt}-2\int{t^{2}dt}+\int{t^4dt}\)
\(=t-2\frac{t^{2+1}}{2+1}+\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=t-2\frac{t^{3}}{3}+\frac{t^{5}}{5}+c\)
\(=t-\frac{2}{3}t^{3}+\frac{1}{5}t^{5}+c\)
\(=\sin{x}-\frac{2}{3}\sin^{3}{x}+\frac{1}{5}\sin^{5}{x}+c\) ➜ \(\because t=\sin{x}\)
\(5.(ii)\) \(\int{\cos^{4}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
সমাধান
\(\int{\cos^{4}{x}dx}\)
\(=\int{(\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(2\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x+\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
\(=\int{(\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(2\cos^{2}{x})^2dx}\)
\(=\frac{1}{4}\int{(1+\cos{2x})^2dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{(1+2\cos{2x}+\cos^2{2x})dx}\) ➜ \(\because (a+b)^2=a^2+2ab+b^2\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{4}\int{\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{(1+\cos{4x})dx}\) ➜ \(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{4}\int{dx}+\frac{1}{2}\int{\cos{2x}dx}+\frac{1}{8}\int{dx}+\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{4}x+\frac{1}{2}.\frac{1}{2}\sin{2x}+\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}\sin{4x}+c\) ➜ \(\because \int{dx}=x, \int{\cos{ax}dx}=\frac{1}{a}\sin{ax}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}x+\frac{1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{2+1}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{3}{8}x+\frac{1}{4}\sin{2x}+\frac{1}{32}\sin{4x}+c\)
\(=\frac{1}{32}(12x+8\sin{2x}+\sin{4x})+c\)
অনুশীলনী \(10.C\) উদাহরণ সমুহ
যোজিত ফল নির্ণয় করঃ
\((1.)\) \(\int{x^2(3-2x^3)^4dx}\)
উত্তরঃ \(-\frac{1}{30}(3-2x^3)^{5}+c\)
\((2.)\) \(\int{cosec^2 \ {x}e^{\cot{x}}dx}\)
উত্তরঃ \(-e^{\cot{x}}+c\)
\((3.)\) \(\int{\sec{x}\tan{x}e^{\sec{x}}dx}\)
উত্তরঃ \(e^{\sec{x}}+c\)
\((4.)\) \(\int{cosec \ {x}\cot{x}e^{cosec \ {x}}dx}\)
উত্তরঃ \(-e^{cosec \ {x}}+c\)
\((5.)\) \(\int{\frac{\sec{x}}{\ln{|\sec{x}+\tan{x}|}}dx}\)
উত্তরঃ \(\ln{(\ln{|\sec{x}+\tan{x}|})}+c\)
\((6.)\) \(\int{\frac{cosec \ {x}}{\ln{|cosec \ {x}-\cot{x}|}}dx}\)
উত্তরঃ \(\ln{(\ln{|cosec \ {x}-\cot{x}|})}+c\)
\((7.)\) \(\int{\frac{xdx}{\sqrt{1-x}}}\)
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\)
\((8.)\) \(\int{\tan^3{x}dx}\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}-\ln{|\sec{x}|}+c\)
\((9.)\) \(\int{\frac{\sin{2x}}{a+b\sin^2{x}}dx}\)
উত্তরঃ \(\frac{1}{b}\ln{|a+b\sin^2{x}|}+c\)
উত্তরঃ \(-\frac{1}{30}(3-2x^3)^{5}+c\)
\((2.)\) \(\int{cosec^2 \ {x}e^{\cot{x}}dx}\)
উত্তরঃ \(-e^{\cot{x}}+c\)
\((3.)\) \(\int{\sec{x}\tan{x}e^{\sec{x}}dx}\)
উত্তরঃ \(e^{\sec{x}}+c\)
\((4.)\) \(\int{cosec \ {x}\cot{x}e^{cosec \ {x}}dx}\)
উত্তরঃ \(-e^{cosec \ {x}}+c\)
\((5.)\) \(\int{\frac{\sec{x}}{\ln{|\sec{x}+\tan{x}|}}dx}\)
উত্তরঃ \(\ln{(\ln{|\sec{x}+\tan{x}|})}+c\)
\((6.)\) \(\int{\frac{cosec \ {x}}{\ln{|cosec \ {x}-\cot{x}|}}dx}\)
উত্তরঃ \(\ln{(\ln{|cosec \ {x}-\cot{x}|})}+c\)
\((7.)\) \(\int{\frac{xdx}{\sqrt{1-x}}}\)
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\)
\((8.)\) \(\int{\tan^3{x}dx}\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}-\ln{|\sec{x}|}+c\)
\((9.)\) \(\int{\frac{\sin{2x}}{a+b\sin^2{x}}dx}\)
উত্তরঃ \(\frac{1}{b}\ln{|a+b\sin^2{x}|}+c\)
\((10.)\) \(\int{\frac{\sin{(\tan^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(-\frac{1}{\sqrt{1+x^2}}+c\)
\((11.)\) \(\int{\frac{\cos{(\tan^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(\frac{x}{\sqrt{1+x^2}}+c\)
\((12.)\) \(\int{\frac{\sec{(\cot^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(-\ln{\left|\frac{x^2+2}{x}\right|}+c\)
\((13.)\) \(\int{\frac{2x\tan^{-1}{x^2}}{1+x^4}dx}\)
উত্তরঃ \(\frac{1}{2}\left(\tan^{-1}{x^2}\right)^2+c\)
\((14.)\) \(\int{\frac{\tan^{-1}{\sqrt{x}}}{\sqrt{x}(1+x)}dx}\)
উত্তরঃ \((\tan^{-1}{\sqrt{x}})^2+c\)
\((15.)\) \(\int{\frac{\tan^{-1}{\sqrt[3]{x}}}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}dx}\)
উত্তরঃ \(\frac{3}{2}(\tan^{-1}{\sqrt[3]{x}})^2+c\)
\((16.)\) \(\int{\frac{\sec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)
উত্তরঃ \(\{\sec^{-1}{(\sqrt{x})}\}^2+c\)
\((17.)\) \(\int{\frac{cosec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)
উত্তরঃ \(-\{cosec^{-1}{(\sqrt{x})}\}^2+c\)
উত্তরঃ \(-\frac{1}{\sqrt{1+x^2}}+c\)
\((11.)\) \(\int{\frac{\cos{(\tan^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(\frac{x}{\sqrt{1+x^2}}+c\)
\((12.)\) \(\int{\frac{\sec{(\cot^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(-\ln{\left|\frac{x^2+2}{x}\right|}+c\)
\((13.)\) \(\int{\frac{2x\tan^{-1}{x^2}}{1+x^4}dx}\)
উত্তরঃ \(\frac{1}{2}\left(\tan^{-1}{x^2}\right)^2+c\)
\((14.)\) \(\int{\frac{\tan^{-1}{\sqrt{x}}}{\sqrt{x}(1+x)}dx}\)
উত্তরঃ \((\tan^{-1}{\sqrt{x}})^2+c\)
\((15.)\) \(\int{\frac{\tan^{-1}{\sqrt[3]{x}}}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}dx}\)
উত্তরঃ \(\frac{3}{2}(\tan^{-1}{\sqrt[3]{x}})^2+c\)
\((16.)\) \(\int{\frac{\sec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)
উত্তরঃ \(\{\sec^{-1}{(\sqrt{x})}\}^2+c\)
\((17.)\) \(\int{\frac{cosec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)
উত্তরঃ \(-\{cosec^{-1}{(\sqrt{x})}\}^2+c\)
\((1.)\) \(\int{x^2(3-2x^3)^4dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{30}(3-2x^3)^{5}+c\)
উত্তরঃ \(-\frac{1}{30}(3-2x^3)^{5}+c\)
সমাধানঃ
ধরি,
\(3-2x^3=t\)
\(\Rightarrow \frac{d}{dx}(3-2x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2.3x^2=\frac{dt}{dx}\)
\(\Rightarrow -6x^2=\frac{dt}{dx}\)
\(\therefore x^2dx=-\frac{1}{6}dt\)
\(\int{x^2(3-2x^3)^4dx}\)\(3-2x^3=t\)
\(\Rightarrow \frac{d}{dx}(3-2x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2.3x^2=\frac{dt}{dx}\)
\(\Rightarrow -6x^2=\frac{dt}{dx}\)
\(\therefore x^2dx=-\frac{1}{6}dt\)
\(=\int{(3-2x^3)^4x^2dx}\)
\(=\int{t^4.\left(-\frac{1}{6}dt\right)}\)
\(=-\frac{1}{6}\int{t^4dt}\)
\(=-\frac{1}{6}\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{6}\frac{t^{5}}{5}+c\) \(=-\frac{1}{30}(3-2x^3)^{5}+c\) ➜ \(\because t=3-2x^3\)
\((2.)\) \(\int{cosec^2 \ {x}e^{\cot{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-e^{\cot{x}}+c\)
উত্তরঃ \(-e^{\cot{x}}+c\)
সমাধানঃ
ধরি,
\(\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec^2 \ {x}=\frac{dt}{dx}\)
\(\Rightarrow -cosec^2 \ {x}dx=dt\)
\(\therefore cosec^2 \ {x}dx=-dt\)
\(\int{cosec^2 \ {x}e^{\cot{x}}dx}\)\(\cot{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec^2 \ {x}=\frac{dt}{dx}\)
\(\Rightarrow -cosec^2 \ {x}dx=dt\)
\(\therefore cosec^2 \ {x}dx=-dt\)
\(=\int{e^{\cot{x}}cosec^2 \ {x}dx}\)
\(=\int{e^{t}(-dt)}\)
\(=-\int{e^{t}dt}\)
\(=-e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-e^{\cot{x}}+c\) ➜ \(\because t=\cot{x}\)
\((3.)\) \(\int{\sec{x}\tan{x}e^{\sec{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{\sec{x}}+c\)
উত্তরঃ \(e^{\sec{x}}+c\)
সমাধানঃ
ধরি,
\(\sec{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{x}\tan{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}\tan{x}dx=dt\)
\(\int{\sec{x}\tan{x}e^{\sec{x}}dx}\)\(\sec{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec{x}\tan{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}\tan{x}dx=dt\)
\(=\int{e^{\sec{x}}\sec{x}\tan{x}dx}\)
\(=\int{e^{t}dt}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{\sec{x}}+c\) ➜ \(\because t=\sec{x}\)
\((4.)\) \(\int{cosec \ {x}\cot{x}e^{cosec \ {x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-e^{cosec \ {x}}+c\)
উত্তরঃ \(-e^{cosec \ {x}}+c\)
সমাধানঃ
ধরি,
\(cosec \ {x}=t\)
\(\Rightarrow \frac{d}{dx}(cosec \ {x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}\cot{x}=\frac{dt}{dx}\)
\(\Rightarrow -cosec \ {x}\cot{x}dx=\frac{dt}{dx}\)
\(\therefore cosec \ {x}\cot{x}dx=(-dt)\)
\(\int{cosec \ {x}\cot{x}e^{cosec \ {x}}dx}\)\(cosec \ {x}=t\)
\(\Rightarrow \frac{d}{dx}(cosec \ {x})=\frac{d}{dx}(t)\)
\(\Rightarrow -cosec \ {x}\cot{x}=\frac{dt}{dx}\)
\(\Rightarrow -cosec \ {x}\cot{x}dx=\frac{dt}{dx}\)
\(\therefore cosec \ {x}\cot{x}dx=(-dt)\)
\(=\int{e^{cosec \ {x}}cosec \ {x}\cot{x}dx}\)
\(=\int{e^{t}(-dt)}\)
\(=-\int{e^{t}dt}\)
\(=-e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-e^{cosec \ {x}}+c\) ➜ \(\because t=cosec \ {x}\)
\((5.)\) \(\int{\frac{\sec{x}}{\ln{|\sec{x}+\tan{x}|}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(\ln{|\sec{x}+\tan{x}|})}+c\)
উত্তরঃ \(\ln{(\ln{|\sec{x}+\tan{x}|})}+c\)
সমাধানঃ
ধরি,
\(\ln{|\sec{x}+\tan{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|\sec{x}+\tan{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}(\sec{x}\tan{x}+\sec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}\{\sec{x}(\sec{x}+\tan{x})\}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}dx=dt\)
\(\int{\frac{\sec{x}}{\ln{|\sec{x}+\tan{x}|}}dx}\)\(\ln{|\sec{x}+\tan{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|\sec{x}+\tan{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}(\sec{x}\tan{x}+\sec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}\{\sec{x}(\sec{x}+\tan{x})\}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}dx=dt\)
\(=\int{\frac{1}{\ln{|\sec{x}+\tan{x}|}}\sec{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{t}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{(\ln{|\sec{x}+\tan{x}|})}+c\) ➜ \(\because t=\ln{|\sec{x}+\tan{x}|}\)
\((6.)\) \(\int{\frac{cosec \ {x}}{\ln{|cosec \ {x}-\cot{x}|}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(\ln{|cosec \ {x}-\cot{x}|})}+c\)
উত্তরঃ \(\ln{(\ln{|cosec \ {x}-\cot{x}|})}+c\)
সমাধানঃ
ধরি,
\(\ln{|cosec \ {x}-\cot{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|cosec \ {x}-\cot{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{cosec \ {x}-\cot{x}}(-cosec \ {x}\cot{x}+cosec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{cosec \ {x}-\cot{x}}\{cosec \ {x}(cosec \ {x}-\cot{x})\}=\frac{dt}{dx}\)
\(\Rightarrow cosec \ {x}=\frac{dt}{dx}\)
\(\therefore cosec \ {x}dx=dt\)
\(\int{\frac{cosec \ {x}}{\ln{|cosec \ {x}-\cot{x}|}}dx}\)\(\ln{|cosec \ {x}-\cot{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|cosec \ {x}-\cot{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{cosec \ {x}-\cot{x}}(-cosec \ {x}\cot{x}+cosec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{cosec \ {x}-\cot{x}}\{cosec \ {x}(cosec \ {x}-\cot{x})\}=\frac{dt}{dx}\)
\(\Rightarrow cosec \ {x}=\frac{dt}{dx}\)
\(\therefore cosec \ {x}dx=dt\)
\(=\int{\frac{1}{\ln{|cosec \ {x}-\cot{x}|}}cosec \ {x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{t}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{(\ln{|cosec \ {x}-\cot{x}|})}+c\) ➜ \(\because t=\ln{|cosec \ {x}-\cot{x}|}\)
\((7.)\) \(\int{\frac{xdx}{\sqrt{1-x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\)
উত্তরঃ \(\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\)
সমাধানঃ
ধরি,
\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(\int{\frac{xdx}{\sqrt{1-x}}}\)\(1-x=t\)
\(\Rightarrow \frac{d}{dx}(1-x)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-1=\frac{dt}{dx}\)
\(\Rightarrow -1=\frac{dt}{dx}\)
\(\Rightarrow -dx=dt\)
\(\therefore dx=-dt\)
\(=\int{\frac{1-(1-x)}{\sqrt{1-x}}dx}\)
\(=\int{\left\{\frac{1}{\sqrt{1-x}}-\frac{(1-x)}{\sqrt{1-x}}\right\}dx}\)
\(=\int{\left\{\frac{1}{\sqrt{1-x}}-\frac{\sqrt{1-x}\sqrt{1-x}}{\sqrt{1-x}}\right\}dx}\)
\(=\int{\left\{\frac{1}{\sqrt{1-x}}-\sqrt{1-x}\right\}dx}\)
\(=\int{\left\{\frac{1}{(1-x)^{\frac{1}{2}}}-(1-x)^{\frac{1}{2}}\right\}dx}\)
\(=\int{\left\{(1-x)^{-\frac{1}{2}}-(1-x)^{\frac{1}{2}}\right\}dx}\)
\(=\int{t^{-\frac{1}{2}}(-dt)}-\int{t^{\frac{1}{2}}(-dt)}\)
\(=-\int{t^{-\frac{1}{2}}dt}+\int{t^{\frac{1}{2}}dt}\)
\(=-\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{1}{-1}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-2t^{\frac{1}{2}}+\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-2t^{\frac{1}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-2\sqrt{t}+c\)
\(=\frac{2}{3}(1-x)^{\frac{3}{2}}-2\sqrt{1-x}+c\) ➜ \(\because t=1-x\)
\((8.)\) \(\int{\tan^3{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\tan^2{x}-\ln{|\sec{x}|}+c\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}-\ln{|\sec{x}|}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\tan^3{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{\tan{x}\tan^2{x}dx}\)
\(=\int{\tan{x}(\sec^2{x}-1)dx}\)
\(=\int{(\tan{x}\sec^2{x}-\tan{x})dx}\)
\(=\int{\tan{x}\sec^2{x}dx}-\int{\tan{x}dx}\)
\(=\int{tdt}-\int{\tan{x}dx}\)
\(=\frac{t^2}{2}-\ln{|\sec{x}|}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\tan{x}dx}=\ln{|\sec{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^2}{2}-\ln{|\sec{x}|}+c\)
\(=\frac{1}{2}t^2-\ln{|\sec{x}|}+c\)
\(=\frac{1}{2}\tan^2{x}-\ln{|\sec{x}|}+c\) ➜ \(\because t=\tan{x}\)
\((9.)\) \(\int{\frac{\sin{2x}}{a+b\sin^2{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{b}\ln{|a+b\sin^2{x}|}+c\)
উত্তরঃ \(\frac{1}{b}\ln{|a+b\sin^2{x}|}+c\)
সমাধানঃ
ধরি,
\(a+b\sin^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(a+b\sin^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+b.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow b\sin{2x}=\frac{dt}{dx}\)
\(\Rightarrow b\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=\frac{1}{b}dt\)
\(\int{\frac{\sin{2x}}{a+b\sin^2{x}}dx}\)\(a+b\sin^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(a+b\sin^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+b.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow b\sin{2x}=\frac{dt}{dx}\)
\(\Rightarrow b\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=\frac{1}{b}dt\)
\(=\int{\frac{1}{a+b\sin^2{x}}\sin{2x}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{b}dt}\)
\(=\frac{1}{b}\int{\frac{1}{t}dt}\)
\(=\frac{1}{b}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{b}\ln{|a+b\sin^2{x}|}+c\) ➜ \(\because t=a+b\sin^2{x}\)
\((10.)\) \(\int{\frac{\sin{(\tan^{-1}{x})}}{1+x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{\sqrt{1+x^2}}+c\)
উত্তরঃ \(-\frac{1}{\sqrt{1+x^2}}+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(\int{\frac{\sin{(\tan^{-1}{x})}}{1+x^2}dx}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(=\int{\sin{(\tan^{-1}{x})}\frac{1}{1+x^2}dx}\)
\(=\int{\sin{t}dt}\)
\(=-\cos{t}+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\cos{(\tan^{-1}{x})}+c\) ➜ \(\because t=\tan^{-1}{x}\)
\(=-\cos{\left(\cos^{-1}{\frac{1}{\sqrt{1+x^2}}}\right)}+c\) ➜ \(\because \tan^{-1}{A}=\cos^{-1}{\left(\frac{1}{\sqrt{1+A^2}}\right)}\)
\(=-\frac{1}{\sqrt{1+x^2}}+c\)
\((11.)\) \(\int{\frac{\cos{(\tan^{-1}{x})}}{1+x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{x}{\sqrt{1+x^2}}+c\)
উত্তরঃ \(\frac{x}{\sqrt{1+x^2}}+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(\int{\frac{\cos{(\tan^{-1}{x})}}{1+x^2}dx}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(=\int{\cos{(\tan^{-1}{x})}\frac{1}{1+x^2}dx}\)
\(=\int{\cos{t}dt}\)
\(=\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin{(\tan^{-1}{x})}+c\) ➜ \(\because t=\tan^{-1}{x}\)
\(=\sin{\left(\sin^{-1}{\frac{x}{\sqrt{1+x^2}}}\right)}+c\) ➜ \(\because \tan^{-1}{A}=\sin^{-1}{\left(\frac{A}{\sqrt{1+A^2}}\right)}\)
\(=\frac{x}{\sqrt{1+x^2}}+c\)
\((12.)\) \(\int{\frac{\sec{(\cot^{-1}{x})}}{1+x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\ln{\left|\frac{x^2+2}{x}\right|}+c\)
উত্তরঃ \(-\ln{\left|\frac{x^2+2}{x}\right|}+c\)
সমাধানঃ
ধরি,
\(\cot^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=-dt\)
\(\int{\frac{\sec{(\cot^{-1}{x})}}{1+x^2}dx}\)\(\cot^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=-dt\)
\(=\int{\sec{(\cot^{-1}{x})}\frac{1}{1+x^2}dx}\)
\(=\int{\sec{t}(-dt)}\)
\(=-\int{\sec{t}dt}\)
\(=-\ln{|\sec{t}+\tan{t}|}+c\) ➜ \(\because \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{|\sec{(\cot^{-1}{x})}+\tan{(\cot^{-1}{x})}|}+c\) ➜ \(\because t=\cot^{-1}{x}\)
\(=-\ln{\left|\sec{\left(\sec^{-1}{\frac{1+x^2}{x}}\right)}+\tan{\left(\tan^{-1}{\frac{1}{x}}\right)}\right|}+c\) ➜ \(\because \cot^{-1}{A}=\sec^{-1}{\left(\frac{\sqrt{1+A^2}}{A}\right)}, \cot^{-1}{A}=\tan^{-1}{\left(\frac{1}{A}\right)}\)
\(=-\ln{\left|\frac{1+x^2}{x}+\frac{1}{x}\right|}+c\)
\(=-\ln{\left|\frac{1+x^2+1}{x}\right|}+c\)
\(=-\ln{\left|\frac{x^2+2}{x}\right|}+c\)
\((13.)\) \(\int{\frac{2x\tan^{-1}{x^2}}{1+x^4}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\left(\tan^{-1}{x^2}\right)^2+c\)
উত্তরঃ \(\frac{1}{2}\left(\tan^{-1}{x^2}\right)^2+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^2)^2}\frac{d}{dx}(x^2)=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x^4}2x=\frac{dt}{dx}\)
\(\therefore \frac{2x}{1+x^4}dx=dt\)
\(\int{\frac{2x\tan^{-1}{x^2}}{1+x^4}dx}\)\(\tan^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^2)^2}\frac{d}{dx}(x^2)=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x^4}2x=\frac{dt}{dx}\)
\(\therefore \frac{2x}{1+x^4}dx=dt\)
\(=\int{\tan^{-1}{x^2}\times{\frac{2x}{1+x^4}}dx}\)
\(=\int{tdt}\)
\(=\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}t^2+c\)
\(=\frac{1}{2}\left(\tan^{-1}{x^2}\right)^2+c\) ➜ \(\because t=\tan^{-1}{x^2}\)
\((14.)\) \(\int{\frac{\tan^{-1}{\sqrt{x}}}{\sqrt{x}(1+x)}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \((\tan^{-1}{\sqrt{x}})^2+c\)
উত্তরঃ \((\tan^{-1}{\sqrt{x}})^2+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{\sqrt{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{\sqrt{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(\sqrt{x})^2}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}(1+x)}=\frac{dt}{dx}\)
\(\Rightarrow \frac{dx}{2\sqrt{x}(1+x)}=dt\)
\(\therefore \frac{dx}{\sqrt{x}(1+x)}=2dt\)
\(\int{\frac{\tan^{-1}{\sqrt{x}}}{\sqrt{x}(1+x)}dx}\)\(\tan^{-1}{\sqrt{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{\sqrt{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(\sqrt{x})^2}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+x}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}(1+x)}=\frac{dt}{dx}\)
\(\Rightarrow \frac{dx}{2\sqrt{x}(1+x)}=dt\)
\(\therefore \frac{dx}{\sqrt{x}(1+x)}=2dt\)
\(=\int{\tan^{-1}{\sqrt{x}}\times{\frac{dx}{\sqrt{x}(1+x)}}}\)
\(=\int{t.2dt}\)
\(=2\int{tdt}\)
\(=2\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=t^2+c\)
\(=(\tan^{-1}{\sqrt{x}})^2+c\) ➜ \(\because t=\tan^{-1}{\sqrt{x}}\)
\((15.)\) \(\int{\frac{\tan^{-1}{\sqrt[3]{x}}}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{3}{2}(\tan^{-1}{\sqrt[3]{x}})^2+c\)
উত্তরঃ \(\frac{3}{2}(\tan^{-1}{\sqrt[3]{x}})^2+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{\sqrt[3]{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{\sqrt[3]{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(\sqrt[3]{x})^2}\frac{d}{dx}(\sqrt[3]{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}\frac{d}{dx}(x^{\frac{1}{3}})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}.\frac{1}{3}x^{\frac{1}{3}-1}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}x^{\frac{1-3}{3}}=3\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}x^{\frac{-2}{3}}=3\frac{dt}{dx}\)
\(\Rightarrow \frac{dx}{x^{\frac{2}{3}}(1+\sqrt[3]{x^2})}=3dt\)
\(\therefore \frac{dx}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}=3dt\)
\(\int{\frac{\tan^{-1}{\sqrt[3]{x}}}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}dx}\)\(\tan^{-1}{\sqrt[3]{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{\sqrt[3]{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(\sqrt[3]{x})^2}\frac{d}{dx}(\sqrt[3]{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}\frac{d}{dx}(x^{\frac{1}{3}})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}.\frac{1}{3}x^{\frac{1}{3}-1}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}x^{\frac{1-3}{3}}=3\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{1+\sqrt[3]{x^2}}x^{\frac{-2}{3}}=3\frac{dt}{dx}\)
\(\Rightarrow \frac{dx}{x^{\frac{2}{3}}(1+\sqrt[3]{x^2})}=3dt\)
\(\therefore \frac{dx}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}=3dt\)
\(=\int{\tan^{-1}{\sqrt[3]{x}}\times{\frac{dx}{\sqrt[3]{x^2}(1+\sqrt[3]{x^2})}}}\)
\(=\int{t.3dt}\)
\(=3\int{tdt}\)
\(=3\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{2}t^2+c\)
\(=\frac{3}{2}(\tan^{-1}{\sqrt[3]{x}})^2+c\) ➜ \(\because t=\tan^{-1}{\sqrt[3]{x}}\)
\((16.)\) \(\int{\frac{\sec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\{\sec^{-1}{(\sqrt{x})}\}^2+c\)
উত্তরঃ \(\{\sec^{-1}{(\sqrt{x})}\}^2+c\)
সমাধানঃ
ধরি,
\(\sec^{-1}{\sqrt{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\sec^{-1}{\sqrt{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{x}\sqrt{(\sqrt{x})^2-1}}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{x}\sqrt{x-1}}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2x\sqrt{x-1}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x\sqrt{x-1}}dx=2dt\)
\(\int{\frac{\sec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)\(\sec^{-1}{\sqrt{x}}=t\)
\(\Rightarrow \frac{d}{dx}(\sec^{-1}{\sqrt{x}})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{x}\sqrt{(\sqrt{x})^2-1}}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sqrt{x}\sqrt{x-1}}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2x\sqrt{x-1}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x\sqrt{x-1}}dx=2dt\)
\(=\int{\sec^{-1}{(\sqrt{x})}.\frac{1}{x\sqrt{x-1}}dx}\)
\(=\int{t.2dt}\)
\(=2\int{tdt}\)
\(=2\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=t^2+c\)
\(=\{\sec^{-1}{(\sqrt{x})}\}^2+c\) ➜ \(\because t=\sec^{-1}{(\sqrt{x})}\)
\((17.)\) \(\int{\frac{cosec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\{cosec^{-1}{(\sqrt{x})}\}^2+c\)
উত্তরঃ \(-\{cosec^{-1}{(\sqrt{x})}\}^2+c\)
সমাধানঃ
ধরি,
\(cosec^{-1}{(\sqrt{x})}=t\)
\(\Rightarrow \frac{d}{dx}\{cosec^{-1}{(\sqrt{x})}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{\sqrt{x}\sqrt{(\sqrt{x})^2-1}}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{\sqrt{x}\sqrt{x-1}}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{2x\sqrt{x-1}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x\sqrt{x-1}}dx=-2dt\)
\(\int{\frac{cosec^{-1}{(\sqrt{x})}}{x\sqrt{x-1}}dx}\)\(cosec^{-1}{(\sqrt{x})}=t\)
\(\Rightarrow \frac{d}{dx}\{cosec^{-1}{(\sqrt{x})}\}=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{\sqrt{x}\sqrt{(\sqrt{x})^2-1}}\frac{d}{dx}(\sqrt{x})=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{\sqrt{x}\sqrt{x-1}}\frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{2x\sqrt{x-1}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x\sqrt{x-1}}dx=-2dt\)
\(=\int{cosec^{-1}{(\sqrt{x})}.\frac{1}{x\sqrt{x-1}}dx}\)
\(=\int{t.(-2dt)}\)
\(=-2\int{tdt}\)
\(=-2\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-t^2+c\)
\(=-\{cosec^{-1}{(\sqrt{x})}\}^2+c\) ➜ \(\because t=cosec^{-1}{(\sqrt{x})}\)
অনুশীলনী \(10.C / Q.1\)-এর অতি সংক্ষিপ্ত প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(\int{f\{g(x)\}g^{\prime}(x)dx}\) ও \(\int{\{f(x)\}^nf^{\prime}(x)dx}\) আকার।
\(\int{f\{g(x)\}g^{\prime}(x)dx}\) ও \(\int{\{f(x)\}^nf^{\prime}(x)dx}\) আকার।
\(Q.1.(i)\) \(\int{x\sin{x^2}dx}\)
উত্তরঃ \(-\frac{1}{2}\cos{x^2}+c\)
\(Q.1.(ii)\) \(\int{x^2\cos{x^3}dx}\)
উত্তরঃ \(\frac{1}{3}\sin{x^3}+c\)
\(Q.1.(iii)\) \(\int{\frac{\sin{\frac{1}{x}}}{x^2}dx}\)
উত্তরঃ \(\cos{\frac{1}{x}}+c\)
[ যঃ২০০৭; ঢাঃ২০০৪ ]
\(Q.1.(iv)\) \(\int{\frac{\sec{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2\ln{|\sec{\sqrt{x}}+\tan{\sqrt{x}}|}+c\)
\(Q.1.(v)\) \(\int{\frac{\cos{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2\sin{\sqrt{x}}+c\)
\(Q.1.(vi)\) \(\int{e^x\cos{e^x}dx}\)
উত্তরঃ \(\sin{e^x}+c\)
\(Q.1.(vii)\) \(\int{e^x\tan{e^x}\sec{e^x}dx}\)
উত্তরঃ \(\sec{e^x}+c\)
[ কুঃ২০০৩ ]
\(Q.1.(viii)\) \(\int{e^{2x}\tan{e^{2x}}\sec{e^{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}\sec{e^{2x}}+c\)
[ চঃ২০০৭ ]
\(Q.1.(ix)\) \(\int{\frac{e^{x}(1+x)dx}{\cos^2{(xe^x)}}}\)
উত্তরঃ \(\tan{(xe^{x})}+c\)
[ রুয়েটঃ২০১১-২০১২]
\(Q.1.(x)\) \(\int{\frac{\cos{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\sin{(\ln{|x|})}+c\)
\(Q.1.(xi)\) \(\int{\frac{\tan^2{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\tan{(\ln{|x|})}-\ln{|x|}+c\)
[ বঃ২০০২ ]
\(Q.1.(xii)\) \(\int{\frac{\tan{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\ln{\{\sec{(\ln{|x|})}\}}+c\)
\(Q.1.(xiii)\) \(\int{\frac{\tan{(\sin^{-1}{x})}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\ln{\{|\sec{(\sin^{-1}{x})}|\}}+c\)
[ বঃ২০১১,২০০৭; যঃ২০০৯; কুঃ২০০৭,২০০৪; সিঃ২০০৬; রাঃ২০০৫; ঢাঃ২০০৩; মাঃ২০১১,২০০৭ ]
\(Q.1.(xiv)\) \(\int{(2x+3)\sqrt{x^2+3x}dx}\)
উত্তরঃ \(\frac{2}{3}(x^2+3x)^{\frac{3}{2}}+c\)
\(Q.1.(xv)\) \(\int{x^2\sqrt(1-x^3)dx}\)
উত্তরঃ \(-\frac{2}{9}(1-x^3)^{\frac{3}{2}}+c\)
\(Q.1.(xvi)\) \(\int{\frac{e^{-x}dx}{(5+e^{-x})^2}}\)
উত্তরঃ \(\frac{1}{5+e^{-x}}+c\)
\(Q.1.(xvii)\) \(\int{\left(e^x+\frac{1}{x}\right)(e^x+\ln{|x|})dx}\)
উত্তরঃ \(\frac{1}{2}(e^x+\ln{|x|})^2+c\)
\(Q.1.(xviii)\) \(\int{\frac{(\ln{|x|})^2}{x}dx}\)
উত্তরঃ \(\frac{1}{3}(\ln{|x|})^3+c\)
\(Q.1.(xix)\) \(\int{\frac{\ln{|x|}}{x}dx}\)
উত্তরঃ \(\frac{1}{2}(\ln{|x|})^2+c\)
\(Q.1.(xx)\) \(\int{\frac{1}{x(\ln{|x|})^2}dx}\)
উত্তরঃ \(-\frac{1}{\ln{|x|}}+c\)
উত্তরঃ \(-\frac{1}{2}\cos{x^2}+c\)
\(Q.1.(ii)\) \(\int{x^2\cos{x^3}dx}\)
উত্তরঃ \(\frac{1}{3}\sin{x^3}+c\)
\(Q.1.(iii)\) \(\int{\frac{\sin{\frac{1}{x}}}{x^2}dx}\)
উত্তরঃ \(\cos{\frac{1}{x}}+c\)
[ যঃ২০০৭; ঢাঃ২০০৪ ]
\(Q.1.(iv)\) \(\int{\frac{\sec{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2\ln{|\sec{\sqrt{x}}+\tan{\sqrt{x}}|}+c\)
\(Q.1.(v)\) \(\int{\frac{\cos{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(2\sin{\sqrt{x}}+c\)
\(Q.1.(vi)\) \(\int{e^x\cos{e^x}dx}\)
উত্তরঃ \(\sin{e^x}+c\)
\(Q.1.(vii)\) \(\int{e^x\tan{e^x}\sec{e^x}dx}\)
উত্তরঃ \(\sec{e^x}+c\)
[ কুঃ২০০৩ ]
\(Q.1.(viii)\) \(\int{e^{2x}\tan{e^{2x}}\sec{e^{2x}}dx}\)
উত্তরঃ \(\frac{1}{2}\sec{e^{2x}}+c\)
[ চঃ২০০৭ ]
\(Q.1.(ix)\) \(\int{\frac{e^{x}(1+x)dx}{\cos^2{(xe^x)}}}\)
উত্তরঃ \(\tan{(xe^{x})}+c\)
[ রুয়েটঃ২০১১-২০১২]
\(Q.1.(x)\) \(\int{\frac{\cos{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\sin{(\ln{|x|})}+c\)
\(Q.1.(xi)\) \(\int{\frac{\tan^2{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\tan{(\ln{|x|})}-\ln{|x|}+c\)
[ বঃ২০০২ ]
\(Q.1.(xii)\) \(\int{\frac{\tan{(\ln{|x|})}}{x}dx}\)
উত্তরঃ \(\ln{\{\sec{(\ln{|x|})}\}}+c\)
\(Q.1.(xiii)\) \(\int{\frac{\tan{(\sin^{-1}{x})}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\ln{\{|\sec{(\sin^{-1}{x})}|\}}+c\)
[ বঃ২০১১,২০০৭; যঃ২০০৯; কুঃ২০০৭,২০০৪; সিঃ২০০৬; রাঃ২০০৫; ঢাঃ২০০৩; মাঃ২০১১,২০০৭ ]
\(Q.1.(xiv)\) \(\int{(2x+3)\sqrt{x^2+3x}dx}\)
উত্তরঃ \(\frac{2}{3}(x^2+3x)^{\frac{3}{2}}+c\)
\(Q.1.(xv)\) \(\int{x^2\sqrt(1-x^3)dx}\)
উত্তরঃ \(-\frac{2}{9}(1-x^3)^{\frac{3}{2}}+c\)
\(Q.1.(xvi)\) \(\int{\frac{e^{-x}dx}{(5+e^{-x})^2}}\)
উত্তরঃ \(\frac{1}{5+e^{-x}}+c\)
\(Q.1.(xvii)\) \(\int{\left(e^x+\frac{1}{x}\right)(e^x+\ln{|x|})dx}\)
উত্তরঃ \(\frac{1}{2}(e^x+\ln{|x|})^2+c\)
\(Q.1.(xviii)\) \(\int{\frac{(\ln{|x|})^2}{x}dx}\)
উত্তরঃ \(\frac{1}{3}(\ln{|x|})^3+c\)
\(Q.1.(xix)\) \(\int{\frac{\ln{|x|}}{x}dx}\)
উত্তরঃ \(\frac{1}{2}(\ln{|x|})^2+c\)
\(Q.1.(xx)\) \(\int{\frac{1}{x(\ln{|x|})^2}dx}\)
উত্তরঃ \(-\frac{1}{\ln{|x|}}+c\)
\(Q.1.(xxi)\) \(\int{\frac{\sqrt{1+\ln{|x|}}}{x}dx}\)
উত্তরঃ \(\frac{2}{3}(1+\ln{|x|})^{\frac{3}{2}}+c\)
\(Q.1.(xxii)\) \(\int{\frac{1}{x(1+\ln{|x|})^3}dx}\)
উত্তরঃ \(-\frac{1}{2(1+\ln{|x|})^2}+c\)
[ ঢাঃ২০০১]
\(Q.1.(xxiii)\) \(\int{(1+\cos{x})^3\sin{x}dx}\)
উত্তরঃ \(-\frac{1}{4}(1+\cos{x})^4+c\)
[ কুঃ২০০৩]
\(Q.1.(xxiv)\) \(\int{\sqrt{1-\sin{x}}\cos{x}dx}\)
উত্তরঃ \(-\frac{2}{3}(1-\sin{x})^{\frac{3}{2}}+c\)
[ সিঃ২০০১]
\(Q.1.(xxv)\) \(\int{\frac{\cos{x}dx}{(1-\sin{x})^2}}\)
উত্তরঃ \(\frac{1}{1-\sin{x}}+c\)
[ কুঃ২০০৬; বঃ২০১১ ]
\(Q.1.(xxvi)\) \(\int{\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{4}\tan^4{x}+c\)
\(Q.1.(xxvii)\) \(\int{\frac{1+\tan^2{x}}{(1+\tan{x})^2}dx}\)
উত্তরঃ \(-\frac{1}{1+\tan{x}}+c\)
[ কুয়েটঃ ২০১৩-২০১৪ ]
\(Q.1.(xxviii)\) \(\int{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x})^2+c\)
\(Q.1.(xxix)\) \(\int{\frac{2x\sin^{-1}{x^2}}{\sqrt{1-x^4}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x^2})^2+c\)
\(Q.1.(xxx)\) \(\int{\frac{x^2\tan^{-1}{x^3}}{1+x^6}dx}\)
উত্তরঃ \(\frac{1}{6}(\tan^{-1}{x^3})^2+c\)
[ কুঃ২০১৫,২০০৮; যঃ২০০৬; চঃ২০০৭; মাঃ২০০৮ ]
\(Q.1.(xxxi)\) \(\int{x7^{x^2}dx}\)
উত্তরঃ \(\frac{7^{x^2}}{2\ln{7}}+c\)
\(Q.1.(xxxii)\) \(\int{\frac{\tan{\sqrt{x}}\sec^2{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(\tan^2{\sqrt{x}}+c\)
\(Q.1.(xxxiii)\) \(\int{\tan^2{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{3}\tan^3{x}+c\)
\(Q.1.(xxxiv)\) \(\int{\tan^4{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{5}\tan^5{x}+c\)
\(Q.1.(xxxv)\) \(\int{\cos{x}\cos{(\sin{x})}dx}\)
উত্তরঃ \(\sin{(\sin{x})}+c\)
\(Q.1.(xxxvi)\) \(\int{\sin{x}\sin{(\cos{x})}dx}\)
উত্তরঃ \(\cos{(\cos{x})}+c\)
\(Q.1.(xxxvii)\) \(\int{\frac{(\sec^{-1}{x})^4}{x\sqrt{x^2-1}}dx}\)
উত্তরঃ \(\frac{1}{5}(\sec^{-1}{x})^5+c\)
\(Q.1.(xxxviii)\) \(\int{\frac{\cos{2x}}{(\sqrt{2+\sin{2x}})^3}dx}\)
উত্তরঃ \(-\frac{1}{\sqrt{2+\sin{2x}}}+c\)
\(Q.1.(xxxix)\) \(\int{\sqrt{1+\sec{x}}dx}\)
উত্তরঃ \(2\sin^{-1}{\left(\sqrt{2}\sin{\frac{x}{2}}\right)}+c\)
\(Q.1.(xL)\) \(\int{\frac{\sin{(2+3\ln{|x|})}}{x}dx}\)
উত্তরঃ \(-\frac{1}{3}\cos{(2+3\ln{|x|})}+c\)
উত্তরঃ \(\frac{2}{3}(1+\ln{|x|})^{\frac{3}{2}}+c\)
\(Q.1.(xxii)\) \(\int{\frac{1}{x(1+\ln{|x|})^3}dx}\)
উত্তরঃ \(-\frac{1}{2(1+\ln{|x|})^2}+c\)
[ ঢাঃ২০০১]
\(Q.1.(xxiii)\) \(\int{(1+\cos{x})^3\sin{x}dx}\)
উত্তরঃ \(-\frac{1}{4}(1+\cos{x})^4+c\)
[ কুঃ২০০৩]
\(Q.1.(xxiv)\) \(\int{\sqrt{1-\sin{x}}\cos{x}dx}\)
উত্তরঃ \(-\frac{2}{3}(1-\sin{x})^{\frac{3}{2}}+c\)
[ সিঃ২০০১]
\(Q.1.(xxv)\) \(\int{\frac{\cos{x}dx}{(1-\sin{x})^2}}\)
উত্তরঃ \(\frac{1}{1-\sin{x}}+c\)
[ কুঃ২০০৬; বঃ২০১১ ]
\(Q.1.(xxvi)\) \(\int{\tan^3{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{4}\tan^4{x}+c\)
\(Q.1.(xxvii)\) \(\int{\frac{1+\tan^2{x}}{(1+\tan{x})^2}dx}\)
উত্তরঃ \(-\frac{1}{1+\tan{x}}+c\)
[ কুয়েটঃ ২০১৩-২০১৪ ]
\(Q.1.(xxviii)\) \(\int{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x})^2+c\)
\(Q.1.(xxix)\) \(\int{\frac{2x\sin^{-1}{x^2}}{\sqrt{1-x^4}}dx}\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x^2})^2+c\)
\(Q.1.(xxx)\) \(\int{\frac{x^2\tan^{-1}{x^3}}{1+x^6}dx}\)
উত্তরঃ \(\frac{1}{6}(\tan^{-1}{x^3})^2+c\)
[ কুঃ২০১৫,২০০৮; যঃ২০০৬; চঃ২০০৭; মাঃ২০০৮ ]
\(Q.1.(xxxi)\) \(\int{x7^{x^2}dx}\)
উত্তরঃ \(\frac{7^{x^2}}{2\ln{7}}+c\)
\(Q.1.(xxxii)\) \(\int{\frac{\tan{\sqrt{x}}\sec^2{\sqrt{x}}}{\sqrt{x}}dx}\)
উত্তরঃ \(\tan^2{\sqrt{x}}+c\)
\(Q.1.(xxxiii)\) \(\int{\tan^2{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{3}\tan^3{x}+c\)
\(Q.1.(xxxiv)\) \(\int{\tan^4{x}\sec^2{x}dx}\)
উত্তরঃ \(\frac{1}{5}\tan^5{x}+c\)
\(Q.1.(xxxv)\) \(\int{\cos{x}\cos{(\sin{x})}dx}\)
উত্তরঃ \(\sin{(\sin{x})}+c\)
\(Q.1.(xxxvi)\) \(\int{\sin{x}\sin{(\cos{x})}dx}\)
উত্তরঃ \(\cos{(\cos{x})}+c\)
\(Q.1.(xxxvii)\) \(\int{\frac{(\sec^{-1}{x})^4}{x\sqrt{x^2-1}}dx}\)
উত্তরঃ \(\frac{1}{5}(\sec^{-1}{x})^5+c\)
\(Q.1.(xxxviii)\) \(\int{\frac{\cos{2x}}{(\sqrt{2+\sin{2x}})^3}dx}\)
উত্তরঃ \(-\frac{1}{\sqrt{2+\sin{2x}}}+c\)
\(Q.1.(xxxix)\) \(\int{\sqrt{1+\sec{x}}dx}\)
উত্তরঃ \(2\sin^{-1}{\left(\sqrt{2}\sin{\frac{x}{2}}\right)}+c\)
\(Q.1.(xL)\) \(\int{\frac{\sin{(2+3\ln{|x|})}}{x}dx}\)
উত্তরঃ \(-\frac{1}{3}\cos{(2+3\ln{|x|})}+c\)
\(Q.1.(i)\) \(\int{x\sin{x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{2}\cos{x^2}+c\)
উত্তরঃ \(-\frac{1}{2}\cos{x^2}+c\)
সমাধানঃ
ধরি,
\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
\(\int{x\sin{x^2}dx}\)\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
\(=\int{\sin{x^2}xdx}\)
\(=\int{\sin{t}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\sin{t}dt}\)
\(=\frac{1}{2}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\cos{t}+c\)
\(=-\frac{1}{2}\cos{x^2}+c\) ➜\(\because t=x^2\)
\(Q.1.(ii)\) \(\int{x^2\cos{x^3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\sin{x^3}+c\)
উত্তরঃ \(\frac{1}{3}\sin{x^3}+c\)
সমাধানঃ
ধরি,
\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
\(\int{x^2\cos{x^3}dx}\)\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
\(=\int{\cos{x^3}x^2dx}\)
\(=\int{\cos{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\cos{t}dt}\)
\(=\frac{1}{3}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\sin{x^3}+c\) ➜\(\because t=x^3\)
\(Q.1.(iii)\) \(\int{\frac{\sin{\frac{1}{x}}}{x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\cos{\frac{1}{x}}+c\)
[ যঃ২০০৭; ঢাঃ২০০৪ ]
উত্তরঃ \(\cos{\frac{1}{x}}+c\)
[ যঃ২০০৭; ঢাঃ২০০৪ ]
সমাধানঃ
ধরি,
\(\frac{1}{x}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{x^2}dx=dt\)
\(\therefore \frac{1}{x^2}dx=-dt\)
\(\int{\frac{\sin{\frac{1}{x}}}{x^2}dx}\)\(\frac{1}{x}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{1}{x}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{x^2}dx=dt\)
\(\therefore \frac{1}{x^2}dx=-dt\)
\(=\int{\sin{\left(\frac{1}{x}\right)}\frac{1}{x^2}dx}\)
\(=\int{\sin{t}(-dt)}\)
\(=-\int{\sin{t}dt}\)
\(=-(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\cos{t}+c\)
\(=\cos{\frac{1}{x}}+c\) ➜\(\because t=\frac{1}{x}\)
\(Q.1.(iv)\) \(\int{\frac{\sec{\sqrt{x}}}{\sqrt{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\ln{|\sec{\sqrt{x}}+\tan{\sqrt{x}}|}+c\)
উত্তরঃ \(2\ln{|\sec{\sqrt{x}}+\tan{\sqrt{x}}|}+c\)
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(\int{\frac{\sec{\sqrt{x}}}{\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(=\int{\sec{\sqrt{x}}\frac{1}{\sqrt{x}}dx}\)
\(=\int{\sec{t}.2dt}\)
\(=2\int{\sec{t}dt}\)
\(=2\ln{|\sec{t}+\tan{t}|}+c\) ➜ \(\because \int{\sec{x}dx}=\ln{|\sec{x}+\tan{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{|\sec{\sqrt{x}}+\tan{\sqrt{x}}|}+c\) ➜\(\because t=\sqrt{x}\)
\(Q.1.(v)\) \(\int{\frac{\cos{\sqrt{x}}}{\sqrt{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sin{\sqrt{x}}+c\)
উত্তরঃ \(2\sin{\sqrt{x}}+c\)
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(\int{\frac{\cos{\sqrt{x}}}{\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(=\int{\cos{\sqrt{x}}\frac{1}{\sqrt{x}}dx}\)
\(=\int{\cos{t}.2dt}\)
\(=2\int{\cos{t}dt}\)
\(=2\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sin{\sqrt{x}}+c\) ➜\(\because t=\sqrt{x}\)
\(Q.1.(vi)\) \(\int{e^x\cos{e^x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\sin{e^x}+c\)
উত্তরঃ \(\sin{e^x}+c\)
সমাধানঃ
ধরি,
\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(\int{e^x\cos{e^x}dx}\)\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(=\int{\cos{e^x}e^xdx}\)
\(=\int{\cos{t}dt}\)
\(=\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin{e^x}+c\) ➜\(\because t=e^x\)
\(Q.1.(vii)\) \(\int{e^x\tan{e^x}\sec{e^x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\sec{e^x}+c\) [ কুঃ২০০৩ ]
উত্তরঃ \(\sec{e^x}+c\) [ কুঃ২০০৩ ]
সমাধানঃ
ধরি,
\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(\int{e^x\tan{e^x}\sec{e^x}dx}\)\(e^x=t\)
\(\Rightarrow \frac{d}{dx}(e^x)=\frac{d}{dx}(t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(=\int{\tan{e^x}\sec{e^x}e^xdx}\)
\(=\int{\tan{t}\sec{t}dt}\)
\(=\int{\sec{t}\tan{t}dt}\)
\(=\sec{t}+c\) ➜ \(\because \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sec{e^x}+c\) ➜\(\because t=e^x\)
\(Q.1.(viii)\) \(\int{e^{2x}\tan{e^{2x}}\sec{e^{2x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\sec{e^{2x}}+c\)
[ চঃ২০০৭ ]
উত্তরঃ \(\frac{1}{2}\sec{e^{2x}}+c\)
[ চঃ২০০৭ ]
সমাধানঃ
ধরি,
\(e^{2x}=t\)
\(\Rightarrow \frac{d}{dx}(e^{2x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{2x}.2=\frac{dt}{dx}\)
\(\Rightarrow 2e^{2x}dx=dt\)
\(\therefore e^{2x}dx=\frac{1}{2}dt\)
\(\int{e^{2x}\tan{e^{2x}}\sec{e^{2x}}dx}\)\(e^{2x}=t\)
\(\Rightarrow \frac{d}{dx}(e^{2x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{2x}.2=\frac{dt}{dx}\)
\(\Rightarrow 2e^{2x}dx=dt\)
\(\therefore e^{2x}dx=\frac{1}{2}dt\)
\(=\int{\tan{e^{2x}}\sec{e^{2x}}e^{2x}dx}\)
\(=\int{\tan{t}\sec{t}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\tan{t}\sec{t}dt}\)
\(=\frac{1}{2}\int{\sec{t}\tan{t}dt}\)
\(=\frac{1}{2}\sec{t}+c\) ➜ \(\because \int{\sec{x}\tan{x}dx}=\sec{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\sec{e^{2x}}+c\) ➜\(\because t=e^{2x}\)
\(Q.1.(ix)\) \(\int{\frac{e^{x}(1+x)dx}{\cos^2{(xe^x)}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan{(xe^{x})}+c\)
[ রুয়েটঃ২০১১-২০১২]
উত্তরঃ \(\tan{(xe^{x})}+c\)
[ রুয়েটঃ২০১১-২০১২]
সমাধানঃ
ধরি,
\(xe^x=t\)
\(\Rightarrow \frac{d}{dx}(xe^x)=\frac{d}{dx}(t)\)
\(\Rightarrow xe^x+e^x=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow e^x(x+1)=\frac{dt}{dx}\)
\(\therefore e^x(1+x)dx=dt\)
\(\int{\frac{e^{x}(1+x)dx}{\cos^2{(xe^x)}}}\)\(xe^x=t\)
\(\Rightarrow \frac{d}{dx}(xe^x)=\frac{d}{dx}(t)\)
\(\Rightarrow xe^x+e^x=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow e^x(x+1)=\frac{dt}{dx}\)
\(\therefore e^x(1+x)dx=dt\)
\(=\int{\frac{1}{\cos^2{(xe^x)}}.e^{x}(1+x)dx}\)
\(=\int{\frac{1}{\cos^2{t}}dt}\)
\(=\int{\sec^2{t}dt}\)
\(=\tan{t}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan{(xe^x)}+c\) ➜\(\because t=xe^x\)
\(Q.1.(x)\) \(\int{\frac{\cos{(\ln{|x|})}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\sin{(\ln{|x|})}+c\)
উত্তরঃ \(\sin{(\ln{|x|})}+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{\cos{(\ln{|x|})}}{x}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\cos{(\ln{|x|})}\frac{1}{x}dx}\)
\(=\int{\cos{t}dt}\)
\(=\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin{(\ln{|x|})}+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xi)\) \(\int{\frac{\tan^2{(\ln{|x|})}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan{(\ln{|x|})}-\ln{|x|}+c\)
[ বঃ২০০২ ]
উত্তরঃ \(\tan{(\ln{|x|})}-\ln{|x|}+c\)
[ বঃ২০০২ ]
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{\tan^2{(\ln{|x|})}}{x}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\tan^2{(\ln{|x|})}\frac{1}{x}dx}\)
\(=\int{\tan^2{t}dt}\)
\(=\int{(\sec^2{t}-1)dt}\)
\(=\int{\sec^2{t}dt}-\int{dt}\)
\(=\tan{t}-t+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}, \int{dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\tan{(\ln{|x|})}-\ln{|x|}+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xii)\) \(\int{\frac{\tan{(\ln{|x|})}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{\{\sec{(\ln{|x|})}\}}+c\)
উত্তরঃ \(\ln{\{\sec{(\ln{|x|})}\}}+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{\tan{(\ln{|x|})}}{x}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\tan{(\ln{|x|})}\frac{1}{x}dx}\)
\(=\int{\tan{t}dt}\)
\(=\ln{\sec{t}}+c\) ➜ \(\because \int{\tan{x}dx}=\ln{|\sec{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\sec{\ln{|x|}}}+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xiii)\) \(\int{\frac{\tan{(\sin^{-1}{x})}}{\sqrt{1-x^2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{\{|\sec{(\sin^{-1}{x})}|\}}+c\)
[ বঃ২০১১,২০০৭; যঃ২০০৯; কুঃ২০০৭,২০০৪; সিঃ২০০৬; রাঃ২০০৫; ঢাঃ২০০৩; মাঃ২০১১,২০০৭ ]
উত্তরঃ \(\ln{\{|\sec{(\sin^{-1}{x})}|\}}+c\)
[ বঃ২০১১,২০০৭; যঃ২০০৯; কুঃ২০০৭,২০০৪; সিঃ২০০৬; রাঃ২০০৫; ঢাঃ২০০৩; মাঃ২০১১,২০০৭ ]
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\int{\frac{\tan{(\sin^{-1}{x})}}{\sqrt{1-x^2}}dx}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(=\int{\tan{(\sin^{-1}{x})}\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{\tan{t}dt}\)
\(=\ln{\sec{t}}+c\) ➜ \(\because \int{\tan{x}dx}=\ln{|\sec{x}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\{|\sec{(\sin^{-1}{x})}|\}}+c\) ➜\(\because t=\sin^{-1}{x}\)
\(Q.1.(xiv)\) \(\int{(2x+3)\sqrt{x^2+3x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}(x^2+3x)^{\frac{3}{2}}+c\)
উত্তরঃ \(\frac{2}{3}(x^2+3x)^{\frac{3}{2}}+c\)
সমাধানঃ
ধরি,
\(x^2+3x=t\)
\(\Rightarrow \frac{d}{dx}(x^2+3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x+3=\frac{dt}{dx}\)
\(\therefore (2x+3)dx=dt\)
\(\int{(2x+3)\sqrt{x^2+3x}dx}\)\(x^2+3x=t\)
\(\Rightarrow \frac{d}{dx}(x^2+3x)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x+3=\frac{dt}{dx}\)
\(\therefore (2x+3)dx=dt\)
\(=\int{\sqrt{x^2+3x}(2x+3)dx}\)
\(=\int{\sqrt{t}dt}\)
\(=\int{t^{\frac{1}{2}}dt}\)
\(=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{1+3}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{2}{3}(x^2+3x)^{\frac{3}{2}}+c\) ➜\(\because t=x^2+3x\)
\(Q.1.(xv)\) \(\int{x^2\sqrt(1-x^3)dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{9}(1-x^3)^{\frac{3}{2}}+c\)
উত্তরঃ \(-\frac{2}{9}(1-x^3)^{\frac{3}{2}}+c\)
সমাধানঃ
ধরি,
\(1-x^3=t\)
\(\Rightarrow \frac{d}{dx}(1-x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow -3x^2=\frac{dt}{dx}\)
\(\Rightarrow -3x^2dx=\frac{dt}{dx}\)
\(\therefore x^2dx=-\frac{1}{3}dt\)
\(\int{x^2\sqrt{1-x^3}dx}\)\(1-x^3=t\)
\(\Rightarrow \frac{d}{dx}(1-x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow -3x^2=\frac{dt}{dx}\)
\(\Rightarrow -3x^2dx=\frac{dt}{dx}\)
\(\therefore x^2dx=-\frac{1}{3}dt\)
\(=\int{\sqrt{1-x^3}x^2dx}\)
\(=\int{\sqrt{t}\left(-\frac{1}{3}dt\right)}\)
\(=-\frac{1}{3}\int{\sqrt{t}dt}\)
\(=-\frac{1}{3}\int{t^{\frac{1}{2}}dt}\)
\(=-\frac{1}{3}\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\frac{t^{\frac{1+3}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{1}{3}.\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=-\frac{2}{9}(1-x^3)^{\frac{3}{2}}+c\) ➜\(\because t=1-x^3\)
\(Q.1.(xvi)\) \(\int{\frac{e^{-x}dx}{(5+e^{-x})^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{5+e^{-x}}+c\)
উত্তরঃ \(\frac{1}{5+e^{-x}}+c\)
সমাধানঃ
ধরি,
\(5+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(5+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}dx=dt\)
\(\therefore e^{-x}dx=-dt\)
\(\int{\frac{e^{-x}dx}{(5+e^{-x})^2}}\)\(5+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(5+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}dx=dt\)
\(\therefore e^{-x}dx=-dt\)
\(=\int{\frac{1}{(5+e^{-x})^2}e^{-x}dx}\)
\(=\int{\frac{1}{t^2}(-dt)}\)
\(=-\int{\frac{1}{t^2}dt}\)
\(=-\left(-\frac{1}{t}\right)+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{t}+c\)
\(=\frac{1}{5+e^{-x}}+c\) ➜\(\because t=5+e^{-x}\)
\(Q.1.(xvii)\) \(\int{\left(e^x+\frac{1}{x}\right)(e^x+\ln{|x|})dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(e^x+\ln{|x|})^2+c\)
উত্তরঃ \(\frac{1}{2}(e^x+\ln{|x|})^2+c\)
সমাধানঃ
ধরি,
\(e^x+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(e^x+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{x}+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \left(e^{x}+\frac{1}{x}\right)dx=dt\)
\(\int{\left(e^x+\frac{1}{x}\right)(e^x+\ln{|x|})dx}\)\(e^x+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(e^x+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{x}+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \left(e^{x}+\frac{1}{x}\right)dx=dt\)
\(=\int{(e^x+\ln{|x|})\left(e^x+\frac{1}{x}\right)dx}\)
\(=\int{tdt}\)
\(=\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^2}{2}+c\)
\(=\frac{1}{2}t^2+c\)
\(=\frac{1}{2}(e^x+\ln{|x|})^2+c\) ➜\(\because t=e^x+\ln{|x|}\)
\(Q.1.(xviii)\) \(\int{\frac{(\ln{|x|})^2}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}(\ln{|x|})^3+c\)
উত্তরঃ \(\frac{1}{3}(\ln{|x|})^3+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{(\ln{|x|})^2}{x}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{(\ln{|x|})^2\frac{1}{x}dx}\)
\(=\int{t^2dt}\)
\(=\frac{t^{2+1}}{2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{3}}{3}+c\)
\(=\frac{1}{3}t^{3}+c\)
\(=\frac{1}{3}(\ln{|x|})^{3}+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xix)\) \(\int{\frac{\ln{|x|}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\ln{|x|})^2+c\)
উত্তরঃ \(\frac{1}{2}(\ln{|x|})^2+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{\ln{|x|}}{x}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\ln{|x|}\frac{1}{x}dx}\)
\(=\int{tdt}\)
\(=\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^2}{2}+c\)
\(=\frac{1}{2}t^2+c\)
\(=\frac{1}{2}(\ln{|x|})^2+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xx)\) \(\int{\frac{1}{x(\ln{|x|})^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{\ln{|x|}}+c\)
উত্তরঃ \(-\frac{1}{\ln{|x|}}+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{1}{x(\ln{|x|})^2}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\frac{1}{(\ln{|x|})^2}\frac{1}{x}dx}\)
\(=\int{\frac{1}{t^2}dt}\)
\(=-\frac{1}{t}+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{\ln{|x|}}+c\) ➜\(\because t=\ln{|x|}\)
\(Q.1.(xxi)\) \(\int{\frac{\sqrt{1+\ln{|x|}}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}(1+\ln{|x|})^{\frac{3}{2}}+c\)
উত্তরঃ \(\frac{2}{3}(1+\ln{|x|})^{\frac{3}{2}}+c\)
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{\sqrt{1+\ln{|x|}}}{x}dx}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\sqrt{1+\ln{|x|}}\frac{1}{x}dx}\)
\(=\int{\sqrt{t}dt}\)
\(=\int{t^{\frac{1}{2}}dt}\)
\(=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{1+3}{2}}}{\frac{1+2}{2}}+c\)
\(=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=\frac{2}{3}(1+\ln{|x|})^{\frac{3}{2}}+c\) ➜\(\because t=1+\ln{|x|}\)
\(Q.1.(xxii)\) \(\int{\frac{1}{x(1+\ln{|x|})^3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{2(1+\ln{|x|})^2}+c\)
[ ঢাঃ২০০১]
উত্তরঃ \(-\frac{1}{2(1+\ln{|x|})^2}+c\)
[ ঢাঃ২০০১]
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{1}{x(1+\ln{|x|})^3}dx}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\frac{1}{(1+\ln{|x|})^3}\frac{1}{x}dx}\)
\(=\int{\frac{1}{t^3}dt}\)
\(=\int{t^{-3}dt}\)
\(=\frac{t^{-3+1}}{-3+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{-2}}{-2}+c\)
\(=\frac{1}{-2}t^{-2}+c\)
\(=-\frac{1}{2}\frac{1}{t^{2}}+c\)
\(=-\frac{1}{2t^{2}}+c\)
\(=-\frac{1}{2(1+\ln{|x|})^{2}}+c\) ➜\(\because t=1+\ln{|x|}\)
\(Q.1.(xxiii)\) \(\int{(1+\cos{x})^3\sin{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4}(1+\cos{x})^4+c\)
[ কুঃ২০০৩]
উত্তরঃ \(-\frac{1}{4}(1+\cos{x})^4+c\)
[ কুঃ২০০৩]
সমাধানঃ
ধরি,
\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(\int{(1+\cos{x})^3\sin{x}dx}\)\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{t^3(-dt)}\)
\(=-\int{t^3dt}\)
\(=-\frac{t^{3+1}}{3+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{4}}{4}+c\)
\(=-\frac{1}{4}t^{4}+c\)
\(=-\frac{1}{4}(1+\cos{x})^{4}+c\) ➜\(\because t=1+\cos{x}\)
\(Q.1.(xxiv)\) \(\int{\sqrt{1-\sin{x}}\cos{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{2}{3}(1-\sin{x})^{\frac{3}{2}}+c\)
[ সিঃ২০০১]
উত্তরঃ \(-\frac{2}{3}(1-\sin{x})^{\frac{3}{2}}+c\)
[ সিঃ২০০১]
সমাধানঃ
ধরি,
\(1-\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1-\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -\cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=-dt\)
\(\int{\sqrt{1-\sin{x}}\cos{x}dx}\)\(1-\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1-\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -\cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=-dt\)
\(=\int{\sqrt{t}(-dt)}\)
\(=-\int{\sqrt{t}dt}\)
\(=-\int{t^{\frac{1}{2}}dt}\)
\(=-\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\) ➜ \(\because \int{x^{n}dx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{t^{\frac{1+3}{2}}}{\frac{1+2}{2}}+c\)
\(=-\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\)
\(=-\frac{2}{3}t^{\frac{3}{2}}+c\)
\(=-\frac{2}{3}(1-\sin{x})^{\frac{3}{2}}+c\) ➜\(\because t=1-\sin{x}\)
\(Q.1.(xxv)\) \(\int{\frac{\cos{x}dx}{(1-\sin{x})^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{1-\sin{x}}+c\)
[ কুঃ২০০৬; বঃ২০১১ ]
উত্তরঃ \(\frac{1}{1-\sin{x}}+c\)
[ কুঃ২০০৬; বঃ২০১১ ]
সমাধানঃ
ধরি,
\(1-\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1-\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -\cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=-dt\)
\(\int{\frac{\cos{x}dx}{(1-\sin{x})^2}}\)\(1-\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(1-\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -\cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=-dt\)
\(=\int{\frac{1}{(1-\sin{x})^2}\cos{x}dx}\)
\(=\int{\frac{1}{t^2}(-dt)}\)
\(=-\int{\frac{1}{t^2}dt}\)
\(=-\left(-\frac{1}{t}\right)+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{t}+c\)
\(=\frac{1}{1-\sin{x}}+c\) ➜\(\because t=1-\sin{x}\)
\(Q.1.(xxvi)\) \(\int{\tan^3{x}\sec^2{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\tan^4{x}+c\)
উত্তরঃ \(\frac{1}{4}\tan^4{x}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\tan^3{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{t^3dt}\)
\(=\frac{t^{3+1}}{3+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{4}}{4}+c\)
\(=\frac{1}{4}t^{4}+c\)
\(=\frac{1}{4}\tan^{4}{x}+c\) ➜\(\because t=\tan{x}\)
\(Q.1.(xxvii)\) \(\int{\frac{1+\tan^2{x}}{(1+\tan{x})^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{1+\tan{x}}+c\)
[ কুয়েটঃ ২০১৩-২০১৪ ]
উত্তরঃ \(-\frac{1}{1+\tan{x}}+c\)
[ কুয়েটঃ ২০১৩-২০১৪ ]
সমাধানঃ
ধরি,
\(1+\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\frac{1+\tan^2{x}}{(1+\tan{x})^2}dx}\)\(1+\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{\frac{sec^2{x}}{(1+\tan{x})^2}dx}\) ➜\(\because 1+\tan^2{x}=\sec^2{x}\)
\(=\int{\frac{1}{(1+\tan{x})^2}sec^2{x}dx}\)
\(=\int{\frac{1}{t^2}dt}\)
\(=-\frac{1}{t}+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{1+\tan{x}}+c\) ➜\(\because t=1+\tan{x}\)
\(Q.1.(xxviii)\) \(\int{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x})^2+c\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x})^2+c\)
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\int{\frac{\sin^{-1}{x}}{\sqrt{1-x^2}}dx}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(=\int{\sin^{-1}{x}\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{tdt}\)
\(=\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}t^2+c\)
\(=\frac{1}{2}(\sin^{-1}{x})^2+c\) ➜\(\because t=\sin^{-1}{x}\)
\(Q.1.(xxix)\) \(\int{\frac{2x\sin^{-1}{x^2}}{\sqrt{1-x^4}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x^2})^2+c\)
উত্তরঃ \(\frac{1}{2}(\sin^{-1}{x^2})^2+c\)
সমাধানঃ
ধরি,
\(\sin^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^4}}.2x=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{2x}{\sqrt{1-x^4}}dx=dt\)
\(\int{\frac{2x\sin^{-1}{x^2}}{\sqrt{1-x^4}}dx}\)\(\sin^{-1}{x^2}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x^2})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^4}}.2x=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \frac{2x}{\sqrt{1-x^4}}dx=dt\)
\(=\int{\sin^{-1}{x^2}\frac{2x}{\sqrt{1-x^4}}dx}\)
\(=\int{tdt}\)
\(=\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}t^2+c\)
\(=\frac{1}{2}(\sin^{-1}{x^2})^2+c\) ➜\(\because t=\sin^{-1}{x^2}\)
\(Q.1.(xxx)\) \(\int{\frac{x^2\tan^{-1}{x^3}}{1+x^6}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}(\tan^{-1}{x^3})^2+c\)
[ কুঃ২০১৫,২০০৮; যঃ২০০৬; চঃ২০০৭; মাঃ২০০৮ ]
উত্তরঃ \(\frac{1}{6}(\tan^{-1}{x^3})^2+c\)
[ কুঃ২০১৫,২০০৮; যঃ২০০৬; চঃ২০০৭; মাঃ২০০৮ ]
সমাধানঃ
ধরি,
\(\tan^{-1}{x^3}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^3})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^3)^2}.3x^2=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{3x^2}{1+x^6}dx=dt\)
\(\therefore \frac{x^2}{1+x^6}dx=\frac{1}{3}dt\)
\(\int{\frac{x^2\tan^{-1}{x^3}}{1+x^6}dx}\)\(\tan^{-1}{x^3}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x^3})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+(x^3)^2}.3x^2=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(\Rightarrow \frac{3x^2}{1+x^6}dx=dt\)
\(\therefore \frac{x^2}{1+x^6}dx=\frac{1}{3}dt\)
\(=\int{\tan^{-1}{x^3}\frac{x^2}{1+x^6}dx}\)
\(=\int{t\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{tdt}\)
\(=\frac{1}{3}\frac{t^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{6}t^2+c\)
\(=\frac{1}{6}(\tan^{-1}{x^3})^2+c\) ➜\(\because t=\tan^{-1}{x^3}\)
\(Q.1.(xxxi)\) \(\int{x7^{x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{7^{x^2}}{2\ln{7}}+c\)
উত্তরঃ \(\frac{7^{x^2}}{2\ln{7}}+c\)
সমাধানঃ
ধরি,
\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\therefore xdx=\frac{1}{2}dt\)
\(\int{x7^{x^2}dx}\)\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\therefore xdx=\frac{1}{2}dt\)
\(=\int{7^{x^2}xdx}\)
\(=\int{7^{t}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{7^{t}dt}\)
\(=\frac{1}{2}\frac{7^{t}}{\ln{7}}+c\) ➜ \(\because \int{a^xdx}=\frac{a^x}{ln{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{7^{t}}{2\ln{7}}+c\)
\(=\frac{7^{x^2}}{2\ln{7}}+c\) ➜\(\because t=x^2\)
\(Q.1.(xxxii)\) \(\int{\frac{\tan{\sqrt{x}}\sec^2{\sqrt{x}}}{\sqrt{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\tan^2{\sqrt{x}}+c\)
উত্তরঃ \(\tan^2{\sqrt{x}}+c\)
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
আবার,
ধরি,
\(\tan{t}=z\)
\(\Rightarrow \frac{d}{dt}(\tan{t})=\frac{d}{dt}(z)\)
\(\Rightarrow \sec^2{t}=\frac{dz}{dt}\)
\(\therefore \sec^2{t}dt=dz\)
\(\int{\frac{\tan{\sqrt{x}}\sec^2{\sqrt{x}}}{\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
আবার,
ধরি,
\(\tan{t}=z\)
\(\Rightarrow \frac{d}{dt}(\tan{t})=\frac{d}{dt}(z)\)
\(\Rightarrow \sec^2{t}=\frac{dz}{dt}\)
\(\therefore \sec^2{t}dt=dz\)
\(=\int{\tan{\sqrt{x}}\sec^2{\sqrt{x}}\frac{1}{\sqrt{x}}dx}\)
\(=\int{\tan{t}\sec^2{t}.2dt}\)
\(=2\int{\tan{t}\sec^2{t}dt}\)
\(=2\int{zdz}\)
\(=2\frac{z^2}{2}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=z^2+c\)
\(=\tan^2{t}+c\) ➜\(\because z=\tan{t}\)
\(=\tan^2{\sqrt{x}}+c\) ➜\(\because t=\sqrt{x}\)
\(Q.1.(xxxiii)\) \(\int{\tan^2{x}\sec^2{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\tan^3{x}+c\)
উত্তরঃ \(\frac{1}{3}\tan^3{x}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\tan^2{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{t^2dt}\)
\(=\frac{t^{2+1}}{2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{3}}{3}+c\)
\(=\frac{1}{3}t^3+c\)
\(=\frac{1}{3}tan^3{x}+c\) ➜\(\because t=\tan{x}\)
\(Q.1.(xxxiv)\) \(\int{\tan^4{x}\sec^2{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{5}\tan^5{x}+c\)
উত্তরঃ \(\frac{1}{5}\tan^5{x}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\tan^4{x}\sec^2{x}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{t^4dt}\)
\(=\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{5}}{5}+c\)
\(=\frac{1}{5}t^5+c\)
\(=\frac{1}{5}tan^5{x}+c\) ➜\(\because t=\tan{x}\)
\(Q.1.(xxxv)\) \(\int{\cos{x}\cos{(\sin{x})}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\sin{(\sin{x})}+c\)
উত্তরঃ \(\sin{(\sin{x})}+c\)
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\cos{x}\cos{(\sin{x})}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\cos{(\sin{x})}\cos{x}dx}\)
\(=\int{\cos{t}dt}\)
\(=\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sin{(\sin{x})}+c\) ➜\(\because t=\sin{x}\)
\(Q.1.(xxxvi)\) \(\int{\sin{x}\sin{(\cos{x})}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\cos{(\cos{x})}+c\)
উত্তরঃ \(\cos{(\cos{x})}+c\)
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\sin{x}\sin{(\cos{x})}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\sin{(\cos{x})}\sin{x}dx}\)
\(=\int{\sin{t}(-dt)}\)
\(=-\int{\sin{t}dt}\)
\(=-(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\cos{t}+c\)
\(=\cos{(\cos{x})}+c\) ➜\(\because t=\cos{x}\)
\(Q.1.(xxxvii)\) \(\int{\frac{(\sec^{-1}{x})^4}{x\sqrt{x^2-1}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{5}(\sec^{-1}{x})^5+c\)
উত্তরঃ \(\frac{1}{5}(\sec^{-1}{x})^5+c\)
সমাধানঃ
ধরি,
\(\sec^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x\sqrt{x^2-1}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sec^{-1}{x})=\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{1}{x\sqrt{x^2-1}}dx=dt\)
\(\int{\frac{(\sec^{-1}{x})^4}{x\sqrt{x^2-1}}dx}\)\(\sec^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sec^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x\sqrt{x^2-1}}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}(\sec^{-1}{x})=\frac{1}{x\sqrt{x^2-1}}\)
\(\therefore \frac{1}{x\sqrt{x^2-1}}dx=dt\)
\(=\int{(\sec^{-1}{x})^4\times{\frac{1}{x\sqrt{x^2-1}}}dx}\)
\(=\int{t^4dt}\)
\(=\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{5}}{5}+c\)
\(=\frac{1}{5}t^{5}+c\)
\(=\frac{1}{5}(\sec^{-1}{x})^{5}+c\) ➜\(\because t=\sec^{-1}{x}\)
\(Q.1.(xxxviii)\) \(\int{\frac{\cos{2x}}{(\sqrt{2+\sin{2x}})^3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{\sqrt{2+\sin{2x}}}+c\)
উত্তরঃ \(-\frac{1}{\sqrt{2+\sin{2x}}}+c\)
সমাধানঃ
ধরি,
\(2+\sin{2x}=t\)
\(\Rightarrow \frac{d}{dx}(2+\sin{2x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\cos{2x}.\frac{d}{dx}(2x)=\frac{dt}{dx}\)
\(\Rightarrow \cos{2x}.2=\frac{dt}{dx}\)
\(\Rightarrow 2\cos{2x}dx=dt\)
\(\therefore \cos{2x}dx=\frac{1}{2}dt\)
\(\int{\frac{\cos{2x}}{(\sqrt{2+\sin{2x}})^3}dx}\)\(2+\sin{2x}=t\)
\(\Rightarrow \frac{d}{dx}(2+\sin{2x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\cos{2x}.\frac{d}{dx}(2x)=\frac{dt}{dx}\)
\(\Rightarrow \cos{2x}.2=\frac{dt}{dx}\)
\(\Rightarrow 2\cos{2x}dx=dt\)
\(\therefore \cos{2x}dx=\frac{1}{2}dt\)
\(=\int{\frac{1}{(\sqrt{2+\sin{2x}})^3}\cos{2x}dx}\)
\(=\int{\frac{1}{(\sqrt{t})^3}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t^{\frac{3}{2}}}dt}\)
\(=\frac{1}{2}\int{t^{-\frac{3}{2}}dt}\)
\(=\frac{1}{2}\frac{t^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}\frac{t^{\frac{-3+2}{2}}}{\frac{-3+2}{2}}+c\)
\(=\frac{1}{2}\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+c\)
\(=-t^{\frac{-1}{2}}+c\)
\(=-\frac{1}{t^{\frac{1}{2}}}+c\)
\(=-\frac{1}{\sqrt{t}}+c\)
\(=-\frac{1}{\sqrt{2+\sin{2x}}}+c\) ➜\(\because t=2+\sin{2x}\)
\(Q.1.(xxxix)\) \(\int{\sqrt{1+\sec{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sin^{-1}{\left(\sqrt{2}\sin{\frac{x}{2}}\right)}+c\)
উত্তরঃ \(2\sin^{-1}{\left(\sqrt{2}\sin{\frac{x}{2}}\right)}+c\)
সমাধানঃ
ধরি,
\(\sqrt{2}\sin{\frac{x}{2}}=t\)
\(\Rightarrow \frac{d}{dx}\left(\sqrt{2}\sin{\frac{x}{2}}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \sqrt{2}\cos{\frac{x}{2}}\frac{d}{dx}\left(\frac{x}{2}\right)=\frac{dt}{dx}\)
\(\Rightarrow \sqrt{2}\cos{\frac{x}{2}}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\sqrt{2}}{2}\cos{\frac{x}{2}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\sqrt{2}}{\sqrt{2}.\sqrt{2}}\cos{\frac{x}{2}}dx=dt\)
\(\Rightarrow \frac{1}{\sqrt{2}}\cos{\frac{x}{2}}dx=dt\)
\(\therefore \cos{\frac{x}{2}}dx=\sqrt{2}dt\)
\(\int{\sqrt{1+\sec{x}}dx}\)\(\sqrt{2}\sin{\frac{x}{2}}=t\)
\(\Rightarrow \frac{d}{dx}\left(\sqrt{2}\sin{\frac{x}{2}}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \sqrt{2}\cos{\frac{x}{2}}\frac{d}{dx}\left(\frac{x}{2}\right)=\frac{dt}{dx}\)
\(\Rightarrow \sqrt{2}\cos{\frac{x}{2}}.\frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\sqrt{2}}{2}\cos{\frac{x}{2}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{\sqrt{2}}{\sqrt{2}.\sqrt{2}}\cos{\frac{x}{2}}dx=dt\)
\(\Rightarrow \frac{1}{\sqrt{2}}\cos{\frac{x}{2}}dx=dt\)
\(\therefore \cos{\frac{x}{2}}dx=\sqrt{2}dt\)
\(=\int{\sqrt{\left(1+\frac{1}{\cos{x}}\right)}dx}\)
\(=\int{\sqrt{\left(\frac{\cos{x}+1}{\cos{x}}\right)}dx}\)
\(=\int{\frac{\sqrt{2\cos^2{\frac{x}{2}}}}{\sqrt{1-2\sin^2{\frac{x}{2}}}}dx}\)
\(=\sqrt{2}\int{\frac{\cos{\frac{x}{2}}}{\sqrt{1-2\sin^2{\frac{x}{2}}}}dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-2\sin^2{\frac{x}{2}}}}\times{\cos{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-\left(\sqrt{2}\sin{\frac{x}{2}}\right)^2}}\times{\cos{\frac{x}{2}}}dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-t^2}}\times{\sqrt{2}}dt}\)
\(=2\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=2\sin^{-1}{t}+c\) ➜ \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\sin^{-1}{\left(\sqrt{2}\sin{\frac{x}{2}}\right)}+c\) ➜\(\because t=\sqrt{2}\sin{\frac{x}{2}}\)
\(Q.1.(xL)\) \(\int{\frac{\sin{(2+3\ln{|x|})}}{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{3}\cos{(2+3\ln{|x|})}+c\)
উত্তরঃ \(-\frac{1}{3}\cos{(2+3\ln{|x|})}+c\)
সমাধানঃ
ধরি,
\(2+3\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(2+3\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+3.\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow 3\frac{1}{x}dx=dt\)
\(\therefore \frac{1}{x}dx=\frac{1}{3}dt\)
\(\int{\frac{\sin{(2+3\ln{|x|})}}{x}dx}\)\(2+3\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(2+3\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+3.\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow 3\frac{1}{x}dx=dt\)
\(\therefore \frac{1}{x}dx=\frac{1}{3}dt\)
\(=\int{\sin{(2+3\ln{|x|})}.\frac{1}{x}dx}\)
\(=\int{\sin{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\sin{t}dt}\)
\(=\frac{1}{3}(-\cos{t})+c\) ➜ \(\because \int{\sin{x}dx}=-\cos{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{3}\cos{t}+c\)
\(=-\frac{1}{3}\cos{(2+3\ln{|x|})}+c\) ➜\(\because t=2+3\ln{|x|}\)
অনুশীলনী \(10.C / Q.2\)-এর সংক্ষিপ্ত প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(\int{\sin^m{x}\cos^n{x}dx}\), \(\int{\sin^m{x}dx}\), \(\int{\cos^n{x}dx}\) ও \(\int{e^{f(x)}f^{\prime}(x)dx}\) আকার।
\(\int{\sin^m{x}\cos^n{x}dx}\), \(\int{\sin^m{x}dx}\), \(\int{\cos^n{x}dx}\) ও \(\int{e^{f(x)}f^{\prime}(x)dx}\) আকার।
\(Q.2.(i)\) \(\int{\sin^3{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{4}\sin^4{x}+c\)
\(Q.2.(ii)\) \(\int{\sin^2{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{3}\sin^3{x}+c\)
[ ঢাঃ২০০২]
\(Q.2.(iii)\) \(\int{\sin^2{\frac{x}{2}}\cos{\frac{x}{2}}dx}\)
উত্তরঃ \(\frac{2}{3}\sin^3{\frac{x}{2}}+c\)
[ কুঃ২০০৩ ]
\(Q.2.(iv)\) \(\int{\sin^3{x}\cos^4{x}dx}\)
উত্তরঃ \(\frac{1}{7}\cos^7{x}-\frac{1}{5}\cos^5{x}+c\)
[ রাঃ২০০১ ]
\(Q.2.(v)\) \(\int{\sin^3{x}\cos^3{x}dx}\)
উত্তরঃ \(\frac{1}{4}\sin^4{x}-\frac{1}{6}\sin^6{x}+c\)
[ যঃ২০০৬ ]
\(Q.2.(vi)\) \(\int{\sin^5{x}dx}\)
উত্তরঃ \(-\cos{x}+\frac{2}{3}\cos^3{x}-\frac{1}{5}\cos^5{x}+c\)
[ রুয়েটঃ২০১০-২০১১ ]
\(Q.2.(vii)\) \(\int{xe^{x^2}dx}\)
উত্তরঃ \(\frac{1}{2}e^{x^2}+c\)
[ বঃ২০০৩ ]
\(Q.2.(viii)\) \(\int{x^2a^{x^3}dx}\)
উত্তরঃ \(\frac{a^{x^3}}{3\ln{a}}+c\)
[ মাঃ২০০৯ ]
\(Q.2.(ix)\) \(\int{\frac{e^{\sqrt{x}}}{5\sqrt{x}}dx}\)
উত্তরঃ \(\frac{2}{5}e^{\sqrt{x}}+c\)
\(Q.2.(x)\) \(\int{\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx}\)
উত্তরঃ \(e^{x+\frac{1}{x}}+c\)
[ বিআইটিঃ ২০০১-২০০২]
\(Q.2.(xi)\) \(\int{\cos{x}e^{\sin{x}}dx}\)
উত্তরঃ \(e^{\sin{x}}+c\)
[ ঢাঃ ২০১১; রাঃ২০০৮ ]
\(Q.2.(xii)\) \(\int{\sec^2{x}e^{\tan{x}}dx}\)
উত্তরঃ \(e^{\tan{x}}+c\)
\(Q.2.(xiii)\) \(\int{\frac{e^{\sin^{-1}{x}}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(e^{\sin^{-1}{x}}+c\)
[ বুয়েটঃ২০০৬; বুটেক্সঃ২০১০-২০১১; কুয়েটঃ২০০৬-২০০৭; চঃ২০১১ ]
উত্তরঃ \(\frac{1}{4}\sin^4{x}+c\)
\(Q.2.(ii)\) \(\int{\sin^2{x}\cos{x}dx}\)
উত্তরঃ \(\frac{1}{3}\sin^3{x}+c\)
[ ঢাঃ২০০২]
\(Q.2.(iii)\) \(\int{\sin^2{\frac{x}{2}}\cos{\frac{x}{2}}dx}\)
উত্তরঃ \(\frac{2}{3}\sin^3{\frac{x}{2}}+c\)
[ কুঃ২০০৩ ]
\(Q.2.(iv)\) \(\int{\sin^3{x}\cos^4{x}dx}\)
উত্তরঃ \(\frac{1}{7}\cos^7{x}-\frac{1}{5}\cos^5{x}+c\)
[ রাঃ২০০১ ]
\(Q.2.(v)\) \(\int{\sin^3{x}\cos^3{x}dx}\)
উত্তরঃ \(\frac{1}{4}\sin^4{x}-\frac{1}{6}\sin^6{x}+c\)
[ যঃ২০০৬ ]
\(Q.2.(vi)\) \(\int{\sin^5{x}dx}\)
উত্তরঃ \(-\cos{x}+\frac{2}{3}\cos^3{x}-\frac{1}{5}\cos^5{x}+c\)
[ রুয়েটঃ২০১০-২০১১ ]
\(Q.2.(vii)\) \(\int{xe^{x^2}dx}\)
উত্তরঃ \(\frac{1}{2}e^{x^2}+c\)
[ বঃ২০০৩ ]
\(Q.2.(viii)\) \(\int{x^2a^{x^3}dx}\)
উত্তরঃ \(\frac{a^{x^3}}{3\ln{a}}+c\)
[ মাঃ২০০৯ ]
\(Q.2.(ix)\) \(\int{\frac{e^{\sqrt{x}}}{5\sqrt{x}}dx}\)
উত্তরঃ \(\frac{2}{5}e^{\sqrt{x}}+c\)
\(Q.2.(x)\) \(\int{\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx}\)
উত্তরঃ \(e^{x+\frac{1}{x}}+c\)
[ বিআইটিঃ ২০০১-২০০২]
\(Q.2.(xi)\) \(\int{\cos{x}e^{\sin{x}}dx}\)
উত্তরঃ \(e^{\sin{x}}+c\)
[ ঢাঃ ২০১১; রাঃ২০০৮ ]
\(Q.2.(xii)\) \(\int{\sec^2{x}e^{\tan{x}}dx}\)
উত্তরঃ \(e^{\tan{x}}+c\)
\(Q.2.(xiii)\) \(\int{\frac{e^{\sin^{-1}{x}}}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(e^{\sin^{-1}{x}}+c\)
[ বুয়েটঃ২০০৬; বুটেক্সঃ২০১০-২০১১; কুয়েটঃ২০০৬-২০০৭; চঃ২০১১ ]
\(Q.2.(xiv)\) \(\int{e^{a\sin^{-1}{x}}.\frac{dx}{\sqrt{1-x^2}}}\)
উত্তরঃ \(\frac{1}{a}e^{a\sin^{-1}{x}}+c\)
\(Q.2.(xv)\) \(\int{e^{\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)
উত্তরঃ \(e^{\tan^{-1}{x}}+c\)
[ ঢাঃ২০০৯; মাঃ২০১২,২০১৪ ]
\(Q.2.(xvi)\) \(\int{e^{a\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)
উত্তরঃ \(\frac{1}{a}e^{a\tan^{-1}{x}}+c\)
\(Q.2.(xvii)\) \(\int{\sin^5{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{6}\sin^6{\theta}+c\)
\(Q.2.(xviii)\) \(\int{\cos^3{x}\sqrt{\sin{x}}dx}\)
উত্তরঃ \(\frac{2}{3}\sin^{\frac{3}{2}}{x}-\frac{2}{7}\sin^{\frac{7}{2}}{x}+c\)
\(Q.2.(xix)\) \(\int{\sin^2{x}\cos^2{x}dx}\)
উত্তরঃ \(\frac{1}{8}x-\frac{1}{8}\sin{4x}+c\)
\(Q.2.(xx)\) \(\int{\sin^2{x}\cos{2x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(Q.2.(xxi)\) \(\int{\sin^4{x}\cos^4{x}dx}\)
উত্তরঃ \(\frac{1}{1024}(24x-8\sin{4x}+\sin{8x})+c\)
\(Q.2.(xxii)\) \(\int{\frac{\sin{4x}}{\sin^4{x}+\cos^4{x}}dx}\)
উত্তরঃ \(-ln{|\sin^4{x}+\cos^4{x}|}+c\)
\(Q.2.(xxiii)\) \(\int{\frac{\tan{x}dx}{1+\cos^2{x}}}\)
উত্তরঃ \(\frac{1}{2}ln{|2+\tan^2{x}|}+c\)
\(Q.2.(xxiv)\) \(\int{\frac{\sin{x}}{\sin{(x+a)}}dx}\)
উত্তরঃ \(x\cos{a}-\sin{a}\ln{|\sin{(x+a)}|}+c\)
\(Q.2.(xxv)\) \(\int{(\sqrt{\tan{x}}+\sqrt{\cot{x}})dx}\)
উত্তরঃ \(\sqrt{2}\sin^{-1}{\sin{x}-\cos{x}}+c\)
\(Q.2.(xxvi)\) \(\int{\frac{dx}{\sin{x}\cos^3{x}}}\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}+\ln{|\tan{x}|}+c\)
\(Q.2.(xxvii)\) \(\int{\cos^2{x}\sin{2x}dx}\)
উত্তরঃ \(-\frac{1}{2}\cos^{4}{x}+c\)
উত্তরঃ \(\frac{1}{a}e^{a\sin^{-1}{x}}+c\)
\(Q.2.(xv)\) \(\int{e^{\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)
উত্তরঃ \(e^{\tan^{-1}{x}}+c\)
[ ঢাঃ২০০৯; মাঃ২০১২,২০১৪ ]
\(Q.2.(xvi)\) \(\int{e^{a\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)
উত্তরঃ \(\frac{1}{a}e^{a\tan^{-1}{x}}+c\)
\(Q.2.(xvii)\) \(\int{\sin^5{\theta}\cos{\theta}d\theta}\)
উত্তরঃ \(\frac{1}{6}\sin^6{\theta}+c\)
\(Q.2.(xviii)\) \(\int{\cos^3{x}\sqrt{\sin{x}}dx}\)
উত্তরঃ \(\frac{2}{3}\sin^{\frac{3}{2}}{x}-\frac{2}{7}\sin^{\frac{7}{2}}{x}+c\)
\(Q.2.(xix)\) \(\int{\sin^2{x}\cos^2{x}dx}\)
উত্তরঃ \(\frac{1}{8}x-\frac{1}{8}\sin{4x}+c\)
\(Q.2.(xx)\) \(\int{\sin^2{x}\cos{2x}dx}\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(Q.2.(xxi)\) \(\int{\sin^4{x}\cos^4{x}dx}\)
উত্তরঃ \(\frac{1}{1024}(24x-8\sin{4x}+\sin{8x})+c\)
\(Q.2.(xxii)\) \(\int{\frac{\sin{4x}}{\sin^4{x}+\cos^4{x}}dx}\)
উত্তরঃ \(-ln{|\sin^4{x}+\cos^4{x}|}+c\)
\(Q.2.(xxiii)\) \(\int{\frac{\tan{x}dx}{1+\cos^2{x}}}\)
উত্তরঃ \(\frac{1}{2}ln{|2+\tan^2{x}|}+c\)
\(Q.2.(xxiv)\) \(\int{\frac{\sin{x}}{\sin{(x+a)}}dx}\)
উত্তরঃ \(x\cos{a}-\sin{a}\ln{|\sin{(x+a)}|}+c\)
\(Q.2.(xxv)\) \(\int{(\sqrt{\tan{x}}+\sqrt{\cot{x}})dx}\)
উত্তরঃ \(\sqrt{2}\sin^{-1}{\sin{x}-\cos{x}}+c\)
\(Q.2.(xxvi)\) \(\int{\frac{dx}{\sin{x}\cos^3{x}}}\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}+\ln{|\tan{x}|}+c\)
\(Q.2.(xxvii)\) \(\int{\cos^2{x}\sin{2x}dx}\)
উত্তরঃ \(-\frac{1}{2}\cos^{4}{x}+c\)
\(Q.2.(i)\) \(\int{\sin^3{x}\cos{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\sin^4{x}+c\)
উত্তরঃ \(\frac{1}{4}\sin^4{x}+c\)
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\sin^3{x}\cos{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{t^3.dt}\)
\(=\frac{t^{3+1}}{3+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{4}}{4}+c\)
\(=\frac{1}{4}t^{4}+c\)
\(=\frac{1}{4}\sin^{4}{x}+c\) ➜\(\because t=\sin{x}\)
\(Q.2.(ii)\) \(\int{\sin^2{x}\cos{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\sin^3{x}+c\)
[ ঢাঃ২০০২]
উত্তরঃ \(\frac{1}{3}\sin^3{x}+c\)
[ ঢাঃ২০০২]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\sin^2{x}\cos{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\sin^2{x}\cos{x}dx}\)
\(=\int{t^2.dt}\)
\(=\frac{t^{2+1}}{2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{3}}{3}+c\)
\(=\frac{1}{3}t^{3}+c\)
\(=\frac{1}{3}\sin^{3}{x}+c\) ➜\(\because t=\sin{x}\)
\(Q.2.(iii)\) \(\int{\sin^2{\frac{x}{2}}\cos{\frac{x}{2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\sin^3{\frac{x}{2}}+c\)
[ কুঃ২০০৩ ]
উত্তরঃ \(\frac{2}{3}\sin^3{\frac{x}{2}}+c\)
[ কুঃ২০০৩ ]
সমাধানঃ
ধরি,
\(\sin{\frac{x}{2}}=t\)
\(\Rightarrow \frac{d}{dx}\left(\sin{\frac{x}{2}}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{\frac{x}{2}}.\frac{1}{2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{2}\cos{\frac{x}{2}}dx=dt\)
\(\therefore \cos{\frac{x}{2}}dx=2dt\)
\(\int{\sin^2{\frac{x}{2}}\cos{\frac{x}{2}}dx}\)\(\sin{\frac{x}{2}}=t\)
\(\Rightarrow \frac{d}{dx}\left(\sin{\frac{x}{2}}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{\frac{x}{2}}.\frac{1}{2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{2}\cos{\frac{x}{2}}dx=dt\)
\(\therefore \cos{\frac{x}{2}}dx=2dt\)
\(=\int{t^2.2dt}\)
\(=2\int{t^2dt}\)
\(=2\frac{t^{2+1}}{2+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\frac{t^{3}}{3}+c\)
\(=\frac{2}{3}t^{3}+c\)
\(=\frac{2}{3}\sin^{3}{\frac{x}{2}}+c\) ➜\(\because t=\sin{\frac{x}{2}}\)
\(Q.2.(iv)\) \(\int{\sin^3{x}\cos^4{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{7}\cos^7{x}-\frac{1}{5}\cos^5{x}+c\)
[ রাঃ২০০১ ]
উত্তরঃ \(\frac{1}{7}\cos^7{x}-\frac{1}{5}\cos^5{x}+c\)
[ রাঃ২০০১ ]
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\sin^3{x}\cos^4{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\sin^2{x}\cos^4{x}\sin{x}dx}\)
\(=\int{(1-\cos^2{x})\cos^4{x}\sin{x}dx}\)
\(=\int{(1-t^2)t^4(-dt)}\)
\(=-\int{(t^4-t^6)dt}\)
\(=-\int{t^4dt}+\int{t^6dt}\)
\(=\int{t^6dt}-\int{t^4dt}\)
\(=\frac{t^{6+1}}{6+1}-\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{7}}{7}-\frac{t^{5}}{5}+c\)
\(=\frac{1}{7}t^{7}-\frac{1}{5}t^{5}+c\)
\(=\frac{1}{7}\cos^{7}{x}-\frac{1}{5}\cos^{5}{x}+c\) ➜\(\because t=\cos{x}\)
\(Q.2.(v)\) \(\int{\sin^3{x}\cos^3{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\sin^4{x}-\frac{1}{6}\sin^6{x}+c\)
[ যঃ২০০৬ ]
উত্তরঃ \(\frac{1}{4}\sin^4{x}-\frac{1}{6}\sin^6{x}+c\)
[ যঃ২০০৬ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\sin^3{x}\cos^3{x}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\sin^3{x}\cos^2{x}\cos{x}dx}\)
\(=\int{\sin^3{x}(1-\sin^2{x})\cos{x}dx}\)
\(=\int{t^3(1-t^2)dt}\)
\(=\int{(t^3-t^5)dt}\)
\(=\int{t^3dt}-\int{t^5dt}\)
\(=\int{t^3dt}-\int{t^5dt}\)
\(=\frac{t^{3+1}}{3+1}-\frac{t^{5+1}}{5+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{4}}{4}-\frac{t^{6}}{6}+c\)
\(=\frac{1}{4}t^{4}-\frac{1}{6}t^{6}+c\)
\(=\frac{1}{4}\sin^{4}{x}-\frac{1}{6}\sin^{6}{x}+c\) ➜\(\because t=\sin{x}\)
\(Q.2.(vi)\) \(\int{\sin^5{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\cos{x}+\frac{2}{3}\cos^3{x}-\frac{1}{5}\cos^5{x}+c\)
[ রুয়েটঃ২০১০-২০১১ ]
উত্তরঃ \(-\cos{x}+\frac{2}{3}\cos^3{x}-\frac{1}{5}\cos^5{x}+c\)
[ রুয়েটঃ২০১০-২০১১ ]
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\sin^5{x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\sin^4{x}\sin{x}dx}\)
\(=\int{(\sin^2{x})^2\sin{x}dx}\)
\(=\int{(1-\cos^2{x})^2\sin{x}dx}\)
\(=\int{(1-t^2)^2(-dt)}\)
\(=-\int{(1-2t^2+t^4)dt}\)
\(=-\int{dt}+2\int{t^2dt}-\int{t^4dt}\)
\(=-\int{dt}+2\int{t^2dt}-\int{t^4dt}\)
\(=-t+2\frac{t^{2+1}}{2+1}-\frac{t^{4+1}}{4+1}+c\) ➜ \(\because \int{dx}=x, \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-t+2\frac{t^{3}}{3}-\frac{t^{5}}{5}+c\)
\(=-t+2\frac{t^{3}}{3}-\frac{t^{5}}{5}+c\)
\(=-t+2\frac{1}{3}t^{3}-\frac{1}{5}t^{5}+c\)
\(=-\cos{x}+\frac{2}{3}\cos^{3}{x}-\frac{1}{5}\cos^{5}{x}+c\) ➜\(\because t=\cos{x}\)
\(Q.2.(vii)\) \(\int{xe^{x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}e^{x^2}+c\)
[ বঃ২০০৩ ]
উত্তরঃ \(\frac{1}{2}e^{x^2}+c\)
[ বঃ২০০৩ ]
সমাধানঃ
ধরি,
\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
\(\int{xe^{x^2}dx}\)\(x^2=t\)
\(\Rightarrow \frac{d}{dx}(x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\Rightarrow 2xdx=dt\)
\(\therefore xdx=\frac{1}{2}dt\)
\(=\int{e^{x^2}xdx}\)
\(=\int{e^{t}.\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{e^{t}dt}\)
\(=\frac{1}{2}e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}e^{x^2}+c\) ➜\(\because t=x^2\)
\(Q.2.(viii)\) \(\int{x^2a^{x^3}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{a^{x^3}}{3\ln{a}}+c\)
[ মাঃ২০০৯ ]
উত্তরঃ \(\frac{a^{x^3}}{3\ln{a}}+c\)
[ মাঃ২০০৯ ]
সমাধানঃ
ধরি,
\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
\(\int{x^2a^{x^3}dx}\)\(x^3=t\)
\(\Rightarrow \frac{d}{dx}(x^3)=\frac{d}{dx}(t)\)
\(\Rightarrow 3x^2=\frac{dt}{dx}\)
\(\Rightarrow 3x^2dx=dt\)
\(\therefore x^2dx=\frac{1}{3}dt\)
\(=\int{a^{x^3}x^2dx}\)
\(=\int{a^{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{a^{t}dt}\)
\(=\frac{1}{3}\frac{a^{t}}{\ln{a}}+c\) ➜ \(\because \int{a^xdx}=\frac{a^x}{\ln{a}}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{a^{x^3}}{3\ln{a}}+c\) ➜\(\because t=x^3\)
\(Q.2.(ix)\) \(\int{\frac{e^{\sqrt{x}}}{5\sqrt{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{5}e^{\sqrt{x}}+c\)
উত্তরঃ \(\frac{2}{5}e^{\sqrt{x}}+c\)
সমাধানঃ
ধরি,
\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(\int{\frac{e^{\sqrt{x}}}{5\sqrt{x}}dx}\)\(\sqrt{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2\sqrt{x}}dx=dt\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(=\frac{1}{5}\int{e^{\sqrt{x}}\frac{1}{\sqrt{x}}dx}\)
\(=\frac{1}{5}\int{e^{t}2dt}\)
\(=\frac{2}{5}\int{e^{t}dt}\)
\(=\frac{2}{5}e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{2}{5}e^{\sqrt{x}}+c\) ➜\(\because t=\sqrt{x}\)
\(Q.2.(x)\) \(\int{\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{x+\frac{1}{x}}+c\)
[ বিআইটিঃ ২০০১-২০০২]
উত্তরঃ \(e^{x+\frac{1}{x}}+c\)
[ বিআইটিঃ ২০০১-২০০২]
সমাধানঃ
ধরি,
\(x+\frac{1}{x}=t\)
\(\Rightarrow \frac{d}{dx}\left(x+\frac{1}{x}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-\frac{1}{x^2}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\)
\(\therefore \left(1-\frac{1}{x^2}\right)dx=dt\)
\(\int{\left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx}\)\(x+\frac{1}{x}=t\)
\(\Rightarrow \frac{d}{dx}\left(x+\frac{1}{x}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow 1-\frac{1}{x^2}=\frac{dt}{dx}\) ➜\(\because \frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}\)
\(\therefore \left(1-\frac{1}{x^2}\right)dx=dt\)
\(=\int{e^{x+\frac{1}{x}}\left(1-\frac{1}{x^2}\right)dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{x+\frac{1}{x}}+c\) ➜\(\because t=x+\frac{1}{x}\)
\(Q.2.(xi)\) \(\int{\cos{x}e^{\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{\sin{x}}+c\)
[ ঢাঃ ২০১১; রাঃ২০০৮ ]
উত্তরঃ \(e^{\sin{x}}+c\)
[ ঢাঃ ২০১১; রাঃ২০০৮ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\cos{x}e^{\sin{x}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{e^{\sin{x}}\cos{x}dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{\sin{x}}+c\) ➜\(\because t=\sin{x}\)
\(Q.2.(xii)\) \(\int{\sec^2{x}e^{\tan{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{\tan{x}}+c\)
উত্তরঃ \(e^{\tan{x}}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\sec^2{x}e^{\tan{x}}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{e^{\tan{x}}\sec^2{x}dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{\tan{x}}+c\) ➜\(\because t=\tan{x}\)
\(Q.2.(xiii)\) \(\int{\frac{e^{\sin^{-1}{x}}}{\sqrt{1-x^2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{\sin^{-1}{x}}+c\)
[ বুয়েটঃ২০০৬; বুটেক্সঃ২০১০-২০১১; কুয়েটঃ২০০৬-২০০৭; চঃ২০১১ ]
উত্তরঃ \(e^{\sin^{-1}{x}}+c\)
[ বুয়েটঃ২০০৬; বুটেক্সঃ২০১০-২০১১; কুয়েটঃ২০০৬-২০০৭; চঃ২০১১ ]
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\int{\frac{e^{\sin^{-1}{x}}}{\sqrt{1-x^2}}dx}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(=\int{e^{\sin^{-1}{x}}\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{\sin^{-1}{x}}+c\) ➜\(\because t=\sin^{-1}{x}\)
\(Q.2.(xiv)\) \(\int{e^{a\sin^{-1}{x}}.\frac{dx}{\sqrt{1-x^2}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{a}e^{a\sin^{-1}{x}}+c\)
উত্তরঃ \(\frac{1}{a}e^{a\sin^{-1}{x}}+c\)
সমাধানঃ
ধরি,
\(a\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a\frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\Rightarrow a\frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=\frac{1}{a}dt\)
\(\int{e^{a\sin^{-1}{x}}.\frac{dx}{\sqrt{1-x^2}}}\)\(a\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a\frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\Rightarrow a\frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=\frac{1}{a}dt\)
\(=\int{e^{a\sin^{-1}{x}}.\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{e^{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{e^{t}dt}\)
\(=\frac{1}{a}e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{a\sin^{-1}{x}}+c\) ➜\(\because t=a\sin^{-1}{x}\)
\(Q.2.(xv)\) \(\int{e^{\tan^{-1}{x}}.\frac{dx}{1+x^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(e^{\tan^{-1}{x}}+c\)
[ ঢাঃ২০০৯; মাঃ২০১২,২০১৪ ]
উত্তরঃ \(e^{\tan^{-1}{x}}+c\)
[ ঢাঃ২০০৯; মাঃ২০১২,২০১৪ ]
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(\int{e^{\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(=\int{e^{\tan^{-1}{x}}.\frac{1}{1+x^2}dx}\)
\(=\int{e^{t}dt}\)
\(=e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=e^{\tan^{-1}{x}}+c\) ➜\(\because t=\tan^{-1}{x}\)
\(Q.2.(xvi)\) \(\int{e^{a\tan^{-1}{x}}.\frac{dx}{1+x^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{a}e^{a\tan^{-1}{x}}+c\)
উত্তরঃ \(\frac{1}{a}e^{a\tan^{-1}{x}}+c\)
সমাধানঃ
ধরি,
\(a\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow a\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=\frac{1}{a}dt\)
\(\int{e^{a\tan^{-1}{x}}.\frac{dx}{1+x^2}}\)\(a\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow a\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=\frac{1}{a}dt\)
\(=\int{e^{a\tan^{-1}{x}}.\frac{1}{1+x^2}dx}\)
\(=\int{e^{t}.\frac{1}{a}dt}\)
\(=\frac{1}{a}\int{e^{t}dt}\)
\(=\frac{1}{a}e^{t}+c\) ➜ \(\because \int{e^xdx}=e^x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a}e^{a\tan^{-1}{x}}+c\) ➜\(\because t=a\tan^{-1}{x}\)
\(Q.2.(xvii)\) \(\int{\sin^5{\theta}\cos{\theta}d\theta}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{6}\sin^6{\theta}+c\)
উত্তরঃ \(\frac{1}{6}\sin^6{\theta}+c\)
সমাধানঃ
ধরি,
\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
\(\int{\sin^5{\theta}\cos{\theta}d\theta}\)\(\sin{\theta}=t\)
\(\Rightarrow \frac{d}{d\theta}(\sin{\theta})=\frac{d}{d\theta}(t)\)
\(\Rightarrow \cos{\theta}=\frac{dt}{d\theta}\)
\(\therefore \cos{\theta}d\theta=dt\)
\(=\int{t^5dt}\)
\(=\frac{t^{5+1}}{5+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{6}}{6}+c\)
\(=\frac{1}{6}t^{6}+c\)
\(=\frac{1}{6}\sin^6{\theta}+c\) ➜\(\because t=\sin{\theta}\)
\(Q.2.(xviii)\) \(\int{\cos^3{x}\sqrt{\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{2}{3}\sin^{\frac{3}{2}}{x}-\frac{2}{7}\sin^{\frac{7}{2}}{x}+c\)
উত্তরঃ \(\frac{2}{3}\sin^{\frac{3}{2}}{x}-\frac{2}{7}\sin^{\frac{7}{2}}{x}+c\)
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\cos^3{x}\sqrt{\sin{x}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\cos^2{x}\sqrt{\sin{x}}\cos{x}dx}\)
\(=\int{(1-\sin^2{x})\sqrt{\sin{x}}\cos{x}dx}\) ➜\(\because \cos^2{x}=1-\sin^2{x}\)
\(=\int{(1-t^2)\sqrt{t}dt}\)
\(=\int{(1-t^2)t^{\frac{1}{2}}dt}\)
\(=\int{\left(t^{\frac{1}{2}}-t^{2+\frac{1}{2}}\right)dt}\)
\(=\int{\left(t^{\frac{1}{2}}-t^{\frac{4+1}{2}}\right)dt}\)
\(=\int{\left(t^{\frac{1}{2}}-t^{\frac{5}{2}}\right)dt}\)
\(=\int{t^{\frac{1}{2}}dt}-\int{t^{\frac{5}{2}}dt}\)
\(=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}-\frac{t^{\frac{5+2}{2}}}{\frac{5+2}{2}}+c\)
\(=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+c\)
\(=\frac{2}{3}t^{\frac{3}{2}}-\frac{2}{7}t^{\frac{7}{2}}+c\)
\(=\frac{2}{3}\sin^{\frac{3}{2}}{x}-\frac{2}{7}\sin^{\frac{7}{2}}{x}+c\) ➜\(\because t=\sin{\theta}\)
\(Q.2.(xix)\) \(\int{\sin^2{x}\cos^2{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{8}x-\frac{1}{8}\sin{4x}+c\)
উত্তরঃ \(\frac{1}{8}x-\frac{1}{8}\sin{4x}+c\)
সমাধানঃ
ধরি,
\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(\int{\sin^2{x}\cos^2{x}dx}\)\(4x=t\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t)\)
\(\Rightarrow 4.1=\frac{dt}{dx}\)
\(\Rightarrow 4=\frac{dt}{dx}\)
\(\Rightarrow 4dx=dt\)
\(\therefore dx=\frac{1}{4}dt\)
\(=\frac{1}{4}\int{4\sin^2{x}\cos^2{x}dx}\)
\(=\frac{1}{4}\int{(2\sin{x}\cos{x})^2dx}\)
\(=\frac{1}{4}\int{(\sin{2x})^2dx}\) ➜\(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(=\frac{1}{4}\int{\sin^2{2x}dx}\)
\(=\frac{1}{8}\int{2\sin^2{2x}dx}\)
\(=\frac{1}{8}\int{(1-\cos{4x})dx}\) ➜\(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{8}\int{\cos{4x}dx}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{8}\int{\cos{t}.\frac{1}{4}dt}\)
\(=\frac{1}{8}\int{dx}-\frac{1}{32}\int{\cos{t}dt}\)
\(=\frac{1}{8}x-\frac{1}{8}\sin{t}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{8}x-\frac{1}{8}\sin{4x}+c\) ➜\(\because t=4x\)
\(Q.2.(xx)\) \(\int{\sin^2{x}\cos{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
উত্তরঃ \(\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
সমাধানঃ
ধরি,
\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার
ধরি,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(\int{\sin^2{x}\cos{2x}dx}\)\(2x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 2.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2=\frac{dt_{1}}{dx}\)
\(\Rightarrow 2dx=dt_{1}\)
\(\therefore dx=\frac{1}{2}dt_{1}\)
আবার
ধরি,
\(4x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 4.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4=\frac{dt_{2}}{dx}\)
\(\Rightarrow 4dx=dt_{2}\)
\(\therefore dx=\frac{1}{4}dt_{2}\)
\(=\frac{1}{2}\int{2\sin^2{x}\cos{2x}dx}\)
\(=\frac{1}{2}\int{(1-\cos{2x})\cos{2x}dx}\) ➜\(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{2}\int{(\cos{2x}-\cos^2{2x})dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{2}\int{\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{2\cos^2{2x}dx}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{(1+\cos{4x})dx}\) ➜\(\because 2\cos^2{A}=1+\cos{2A}\)
\(=\frac{1}{2}\int{\cos{2x}dx}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{4x}dx}\)
\(=\frac{1}{2}\int{\cos{t_{1}}\frac{1}{2}dt_{1}}-\frac{1}{4}\int{dx}-\frac{1}{4}\int{\cos{t_{2}}\frac{1}{4}dt_{2}}\)
\(=\frac{1}{4}\int{\cos{t_{1}}dt_{1}}-\frac{1}{4}\int{dx}-\frac{1}{16}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{1}{4}\sin{t_{1}}-\frac{1}{4}x-\frac{1}{16}\sin{t_{2}}+c\) ➜ \(\because \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\sin{2x}-\frac{1}{4}x-\frac{1}{16}\sin{4x}+c\) ➜\(\because t_{1}=2x, t_{2}=4x\)
\(=\frac{1}{16}(4\sin{2x}-4x-\sin{4x})+c\)
\(Q.2.(xxi)\) \(\int{\sin^4{x}\cos^4{x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{1024}(24x-8\sin{4x}+\sin{8x})+c\)
উত্তরঃ \(\frac{1}{1024}(24x-8\sin{4x}+\sin{8x})+c\)
সমাধানঃ
ধরি,
\(4x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার
ধরি,
\(8x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(\int{\sin^4{x}\cos^4{x}dx}\)\(4x=t_{1}\)
\(\Rightarrow \frac{d}{dx}(4x)=\frac{d}{dx}(t_{1})\)
\(\Rightarrow 4.1=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4=\frac{dt_{1}}{dx}\)
\(\Rightarrow 4dx=dt_{1}\)
\(\therefore dx=\frac{1}{4}dt_{1}\)
আবার
ধরি,
\(8x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(8x)=\frac{d}{dx}(t_{2})\)
\(\Rightarrow 8.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8=\frac{dt_{2}}{dx}\)
\(\Rightarrow 8dx=dt_{2}\)
\(\therefore dx=\frac{1}{8}dt_{2}\)
\(=\frac{1}{16}\int{16\sin^4{x}\cos^4{x}dx}\)
\(=\frac{1}{16}\int{(2\sin{x}\cos{x})^4dx}\)
\(=\frac{1}{16}\int{(\sin{2x})^4dx}\) ➜\(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(=\frac{1}{16}\int{(\sin^2{2x})^2dx}\)
\(=\frac{1}{64}\int{(2\sin^2{2x})^2dx}\)
\(=\frac{1}{64}\int{(1-\cos{4x})^2dx}\) ➜\(\because 2\sin^2{A}=1-\cos{2A}\)
\(=\frac{1}{64}\int{(1-2\cos{4x}+\cos^2{4x})dx}\)
\(=\frac{1}{64}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{64}\int{\cos^2{4x}dx}\)
\(=\frac{1}{64}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{2\cos^2{4x}dx}\)
\(=\frac{1}{64}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{(1+\cos{8x})dx}\)
\(=\frac{1}{64}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{dx}+\frac{1}{128}\int{\cos{8x}dx}\)
\(=\frac{1}{64}\int{dx}+\frac{1}{128}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{\cos{8x}dx}\)
\(=\frac{1+2}{128}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{\cos{8x}dx}\)
\(=\frac{3}{128}\int{dx}-\frac{1}{32}\int{\cos{4x}dx}+\frac{1}{128}\int{\cos{8x}dx}\)
\(=\frac{3}{128}\int{dx}-\frac{1}{32}\int{\cos{t_{1}}\frac{1}{4}dt_{1}}+\frac{1}{128}\int{\cos{t_{2}}\frac{1}{8}dt_{2}}\)
\(=\frac{3}{128}\int{dx}-\frac{1}{128}\int{\cos{t_{1}}dt_{1}}+\frac{1}{1024}\int{\cos{t_{2}}dt_{2}}\)
\(=\frac{3}{128}x-\frac{1}{128}\sin{t_{1}}+\frac{1}{1024}\sin{t_{2}}+c\) ➜ \(\because \int{dx}=x, \int{\cos{x}dx}=\sin{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{3}{128}x-\frac{1}{128}\sin{4x}+\frac{1}{1024}\sin{8x}+c\) ➜\(\because t_{1}=4x, t_{2}=8x\)
\(=\frac{1}{1024}(24x-8\sin{4x}+\sin{8x})+c\)
\(Q.2.(xxii)\) \(\int{\frac{\sin{4x}}{\sin^4{x}+\cos^4{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-ln{|\sin^4{x}+\cos^4{x}|}+c\)
উত্তরঃ \(-ln{|\sin^4{x}+\cos^4{x}|}+c\)
সমাধানঃ
ধরি,
\(\sin^4{x}+\cos^4{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^4{x}+\cos^4{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 4\sin^3{x}\cos{x}+4\cos^3{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow 4\sin^3{x}\cos{x}-4\sin{x}\cos^3{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}\cos{x}(\cos^2{x}-\sin^2{x})=\frac{dt}{dx}\)
\(\Rightarrow -2.2\sin{x}\cos{x}(\cos^2{x}-\sin^2{x})=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{2x}\cos{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}, \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow -\sin{4x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow -\sin{4x}dx=dt\)
\(\therefore \sin{4x}dx=-dt\)
\(\int{\frac{\sin{4x}}{\sin^4{x}+\cos^4{x}}dx}\)\(\sin^4{x}+\cos^4{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^4{x}+\cos^4{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 4\sin^3{x}\cos{x}+4\cos^3{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow 4\sin^3{x}\cos{x}-4\sin{x}\cos^3{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}\cos{x}(\cos^2{x}-\sin^2{x})=\frac{dt}{dx}\)
\(\Rightarrow -2.2\sin{x}\cos{x}(\cos^2{x}-\sin^2{x})=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{2x}\cos{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}, \cos^2{A}-\sin^2{A}=\cos{2A}\)
\(\Rightarrow -\sin{4x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow -\sin{4x}dx=dt\)
\(\therefore \sin{4x}dx=-dt\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-ln{|\sin^4{x}+\cos^4{x}|}+c\) ➜\(\because t=\sin^4{x}+\cos^4{x}\)
\(Q.2.(xxiii)\) \(\int{\frac{\tan{x}dx}{1+\cos^2{x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}ln{|2+\tan^2{x}|}+c\)
উত্তরঃ \(\frac{1}{2}ln{|2+\tan^2{x}|}+c\)
সমাধানঃ
ধরি,
\(2+\tan^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(2+\tan^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+2\tan{x}\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow 2\tan{x}\sec^2{x}dx=dt\)
\(\therefore \tan{x}\sec^2{x}dx=\frac{1}{2}dt\)
\(\int{\frac{\tan{x}dx}{1+\cos^2{x}}}\)\(2+\tan^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(2+\tan^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+2\tan{x}\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow 2\tan{x}\sec^2{x}dx=dt\)
\(\therefore \tan{x}\sec^2{x}dx=\frac{1}{2}dt\)
\(=\int{\frac{\tan{x}}{\cos^2{x}\left(\frac{1}{\cos^2{x}}+1\right)}dx}\)
\(=\int{\frac{\tan{x}\frac{1}{\cos^2{x}}}{\frac{1}{\cos^2{x}}+1}dx}\)
\(=\int{\frac{\tan{x}\sec^2{x}}{\sec^2{x}+1}dx}\)
\(=\int{\frac{\tan{x}\sec^2{x}}{1+\tan^2{x}+1}dx}\)
\(=\int{\frac{1}{2+\tan^2{x}}\tan{x}\sec^2{x}dx}\)
\(=\int{\frac{1}{t}\frac{1}{2}dt}\)
\(=\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\frac{1}{2}ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}ln{|2+\tan^2{x}|}+c\) ➜\(\because t=2+\tan^2{x}\)
\(Q.2.(xxiv)\) \(\int{\frac{\sin{x}}{\sin{(x+a)}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(x\cos{a}-\sin{a}\ln{|\sin{(x+a)}|}+c\)
উত্তরঃ \(x\cos{a}-\sin{a}\ln{|\sin{(x+a)}|}+c\)
সমাধানঃ
ধরি,
\(\sin{x}\cos{a}+\cos{x}\sin{a}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}\cos{a}+\cos{x}\sin{a})=\frac{d}{dx}t\)
\(\Rightarrow \cos{x}\cos{a}-\sin{x}\sin{a}=\frac{dt}{dx}\)
\(\therefore (\cos{x}\cos{a}-\sin{x}\sin{a})dx=dt\)
\(\int{\frac{\sin{x}}{\sin{(x+a)}}dx}\)\(\sin{x}\cos{a}+\cos{x}\sin{a}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}\cos{a}+\cos{x}\sin{a})=\frac{d}{dx}t\)
\(\Rightarrow \cos{x}\cos{a}-\sin{x}\sin{a}=\frac{dt}{dx}\)
\(\therefore (\cos{x}\cos{a}-\sin{x}\sin{a})dx=dt\)
\(=\int{\frac{\sin{x}}{\sin{x}\cos{a}+\cos{x}\sin{a}}dx}\)
ধরি,
\(\sin{x}=p(\sin{x}\cos{a}+\cos{x}\sin{a})+q(\cos{x}\cos{a}-\sin{x}\sin{a})+r .....(1)\)
\(\Rightarrow \sin{x}=p\sin{x}\cos{a}+p\cos{x}\sin{a}+q\cos{x}\cos{a}-q\sin{x}\sin{a}+r\)
\(\Rightarrow \sin{x}=\sin{x}(p\cos{a}-q\sin{a})+\cos{x}(p\sin{a}+q\cos{a})+r\)
\(\Rightarrow p\cos{a}-q\sin{a}=1, p\sin{a}+q\cos{a}=0, r=0\) ➜ উভয় পার্শ হতে \(\sin{x}, \cos{x}\)এর সহগ ও ধ্রুবক রাশির সমতা নিয়ে।
\(\Rightarrow p\cos{a}-q\sin{a}=1 ....(2), p\sin{a}+q\cos{a}=0 ....(3)\)
\(\Rightarrow p\cos^2{a}-q\sin{a}\cos{a}+p\sin^2{a}-q\sin{a}\cos{a}=\cos{a}\) ➜ \((2)\times{\cos{a}}+(3)\times{\sin{a}}\) এর সাহায্যে।
\(\Rightarrow p\cos^2{a}+p\sin^2{a}=\cos{a}\)
\(\Rightarrow p(\sin^2{a}+\cos^2{a})=\cos{a}\)
\(\Rightarrow p.1=\cos{a}\)
\(\therefore p=\cos{a}\)
\((3)\) হতে
\(\cos{a}\sin{a}+q\cos{a}=0\)
\(\Rightarrow \cos{a}(\sin{a}+q)=0\)
\(\Rightarrow \cos{a}\ne{0}, \sin{a}+q=0\)
\(\therefore q=-\sin{a}\)
\(\therefore p=\cos{a}, q=-\sin{a}, r=0\)
\((1)\) হতে
\(\sin{x}=\cos{a}(\sin{x}\cos{a}+\cos{x}\sin{a})-\sin{a}(\cos{x}\cos{a}-\sin{x}\sin{a})\)
এখন,
\(=\int{\frac{\cos{a}(\sin{x}\cos{a}+\cos{x}\sin{a})-\sin{a}(\cos{x}\cos{a}-\sin{x}\sin{a})}{\sin{x}\cos{a}+\cos{x}\sin{a}}dx}\)
\(=\int{\left(\cos{a}-\frac{\sin{a}(\cos{x}\cos{a}-\sin{x}\sin{a})}{\sin{x}\cos{a}+\cos{x}\sin{a}}\right)dx}\)
\(=\cos{a}\int{dx}-\sin{a}\int{\frac{\cos{x}\cos{a}-\sin{x}\sin{a}}{\sin{x}\cos{a}+\cos{x}\sin{a}}dx}\)
\(=\cos{a}\int{dx}-\sin{a}\int{\frac{1}{\sin{x}\cos{a}+\cos{x}\sin{a}}(\cos{x}\cos{a}-\sin{x}\sin{a})dx}\)
\(=\cos{a}\int{dx}-\sin{a}\int{\frac{1}{t}dt}\)
\(=\cos{a}.x-\sin{a}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\cos{a}-\sin{a}\ln{|\sin{x}\cos{a}+\cos{x}\sin{a}|}+c\) ➜\(\because t=\sin{x}\cos{a}+\cos{x}\sin{a}\)
\(=x\cos{a}-\sin{a}\ln{|\sin{(x+a)}|}+c\)
\(Q.2.(xxv)\) \(\int{(\sqrt{\tan{x}}+\sqrt{\cot{x}})dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\sqrt{2}\sin^{-1}{\sin{x}-\cos{x}}+c\)
উত্তরঃ \(\sqrt{2}\sin^{-1}{\sin{x}-\cos{x}}+c\)
সমাধানঃ
ধরি,
\(\sin{x}-\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}-\cos{x})=\frac{d}{dx}t\)
\(\Rightarrow \sin{x}+\cos{x}=\frac{dt}{dx}\)
\(\therefore (\sin{x}+\cos{x})dx=dt\)
\(\int{(\sqrt{\tan{x}}+\sqrt{\cot{x}})dx}\)\(\sin{x}-\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}-\cos{x})=\frac{d}{dx}t\)
\(\Rightarrow \sin{x}+\cos{x}=\frac{dt}{dx}\)
\(\therefore (\sin{x}+\cos{x})dx=dt\)
\(=\int{\left(\frac{\sqrt{\sin{x}}}{\sqrt{\cos{x}}}+\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}}\right)dx}\)
\(=\int{\left(\frac{\sin{x}+\cos{x}}{\sqrt{\sin{x}\cos{x}}}\right)dx}\)
\(=\int{\left(\frac{\sqrt{2}(\sin{x}+\cos{x})}{\sqrt{2\sin{x}\cos{x}}}\right)dx}\)
\(=\sqrt{2}\int{\left(\frac{\sin{x}+\cos{x}}{\sqrt{1-1+2\sin{x}\cos{x}}}\right)dx}\)
\(=\sqrt{2}\int{\left(\frac{\sin{x}+\cos{x}}{\sqrt{1-(1-2\sin{x}\cos{x})}}\right)dx}\)
\(=\sqrt{2}\int{\left(\frac{\sin{x}+\cos{x}}{\sqrt{1-(\sin^2{x}+\cos^2{x}-2\sin{x}\cos{x})}}\right)dx}\)
\(=\sqrt{2}\int{\left(\frac{\sin{x}+\cos{x}}{\sqrt{1-(\sin{x}-\cos{x})^2}}\right)dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-(\sin{x}-\cos{x})^2}}(\sin{x}+\cos{x})dx}\)
\(=\sqrt{2}\int{\frac{1}{\sqrt{1-t^2}}dt}\)
\(=\sqrt{2}\sin^{-1}{t}+c\) ➜ \(\because \int{\frac{1}{\sqrt{1-x^2}}dx}=\sin^{-1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\sqrt{2}\sin^{-1}{\sin{x}-\cos{x}}+c\) ➜\(\because t=\sin{x}-\cos{x}\)
\(Q.2.(xxvi)\) \(\int{\frac{dx}{\sin{x}\cos^3{x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}\tan^2{x}+\ln{|\tan{x}|}+c\)
উত্তরঃ \(\frac{1}{2}\tan^2{x}+\ln{|\tan{x}|}+c\)
সমাধানঃ
ধরি,
\(\tan{x}=t_{1}\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}t_{1}\)
\(\Rightarrow \sec^2{x}=\frac{dt_{1}}{dx}\)
\(\therefore \sec^2dx=dt_{1}\)
আবার
ধরি,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}t_{2}\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\therefore 2dx=dt_{2}\)
\(\int{\frac{dx}{\sin{x}\cos^3{x}}}\)\(\tan{x}=t_{1}\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}t_{1}\)
\(\Rightarrow \sec^2{x}=\frac{dt_{1}}{dx}\)
\(\therefore \sec^2dx=dt_{1}\)
আবার
ধরি,
\(2x=t_{2}\)
\(\Rightarrow \frac{d}{dx}(2x)=\frac{d}{dx}t_{2}\)
\(\Rightarrow 2.1=\frac{dt_{2}}{dx}\)
\(\Rightarrow 2=\frac{dt_{2}}{dx}\)
\(\therefore 2dx=dt_{2}\)
\(=\int{\frac{1}{\sin{x}\cos^3{x}}dx}\)
\(=\int{\frac{\sin^2{x}+\cos^2{x}}{\sin{x}\cos^3{x}}dx}\)
\(=\int{\left(\frac{\sin^2{x}}{\sin{x}\cos^3{x}}+\frac{\cos^2{x}}{\sin{x}\cos^3{x}}\right)dx}\)
\(=\int{\left(\frac{\sin{x}}{\cos{x}}.\frac{1}{\cos^2{x}}+\frac{1}{\sin{x}\cos{x}}\right)dx}\)
\(=\int{\left(\tan{x}\sec^2{x}+\frac{2}{2\sin{x}\cos{x}}\right)dx}\)
\(=\int{\left(\tan{x}\sec^2{x}+\frac{2}{\sin{2x}}\right)dx}\)
\(=\int{(\tan{x}\sec^2{x}+2cosec \ {2x})dx}\)
\(=\int{\tan{x}\sec^2{x}dx}+\int{cosec \ {2x}2dx}\)
\(=\int{t_{1}dt_{1}}+\int{cosec \ {t_{2}}dt_{2}}\)
\(=\frac{t^2_{1}}{2}+\ln{|\tan{\frac{t_{2}}{2}}|}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{cosec \ {x}dx}=\ln{|\tan{\frac{x}{2}}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}t^2_{1}+\ln{|\tan{\frac{t_{2}}{2}}|}+c\)
\(=\frac{1}{2}\tan^2{x}+\ln{|\tan{\frac{2x}{2}}|}+c\) ➜\(\because t_{1}=\tan{x}, t_{2}=2x\)
\(=\frac{1}{2}\tan^2{x}+\ln{|\tan{x}|}+c\)
\(Q.2.(xxvii)\) \(\int{\cos^2{x}\sin{2x}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{2}\cos^{4}{x}+c\)
উত্তরঃ \(-\frac{1}{2}\cos^{4}{x}+c\)
সমাধানঃ
ধরি,
\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}t\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\cos^2{x}\sin{2x}dx}\)\(\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x})=\frac{d}{dx}t\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\cos^2{x}.2\sin{x}\cos{x}dx}\)
\(=2\int{\cos^3{x}\sin{x}dx}\)
\(=2\int{t^3(-dt)}\)
\(=-2\int{t^3dt}\)
\(=-2\frac{t^{3+1}}{3+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-2\frac{t^{4}}{4}+c\)
\(=-2\frac{1}{4}t^{4}+c\)
\(=-\frac{1}{2}\cos^{4}{x}+c\) ➜\(\because t=\cos{x}\)
অনুশীলনী \(10.C / Q.3\)-এর বর্ণনামূলক প্রশ্নসমুহ
যোজিত ফল নির্ণয় করঃ
\(\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}\) ও \(\int{\frac{f^{\prime}(x)}{f(x)}dx}\) আকার।
\(\int{\frac{f^{\prime}(x)}{\sqrt{f(x)}}dx}\) ও \(\int{\frac{f^{\prime}(x)}{f(x)}dx}\) আকার।
\(Q.3.(i)\) \(\int{\frac{x}{\sqrt{1-x^2}}dx}\)
উত্তরঃ \(-\sqrt{1-x^2}+c\)
[ ঢাঃ২০১৪; দিঃ২০১১ ]
\(Q.3.(ii)\) \(\int{\frac{x^3dx}{\sqrt{1-2x^4}}}\)
উত্তরঃ \(-\frac{1}{4}\sqrt{1-2x^4}+c\)
[ চঃ২০১০ ]
\(Q.3.(iii)\) \(\int{\frac{dx}{x\sqrt{1+\ln{|x|}}}}\)
উত্তরঃ \(2\sqrt{1+\ln{|x|}}+c\)
[ কুঃ২০০৩ ]
\(Q.3.(iv)\) \(\int{\frac{\cos{x}}{\sqrt{\sin{x}}}dx}\)
উত্তরঃ \(2\sqrt{\sin{x}}+c\)
[ রাঃ২০১০; কুঃ২০০৫ ]
\(Q.3.(v)\) \(\int{\frac{dx}{\cos^2{x}\sqrt{\tan{x}-1}}}\)
উত্তরঃ \(2\sqrt{\tan{x}-1}+c\)
[ সিঃ২০০২ ]
\(Q.3.(vi)\) \(\int{\frac{\sqrt{\tan{x}}}{\sin{x}\cos{x}}dx}\)
উত্তরঃ \(2\sqrt{\tan{x}}+c\)
[ দিঃ২০১৪; রাঃ,কুঃ২০০৯; সিঃ২০০৭ ]
\(Q.3.(vii)\) \(\int{\frac{dx}{\sqrt{\sin^{-1}{x}}\sqrt{1-x^2}}}\)
উত্তরঃ \(2\sqrt{\sin^{-1}{x}}+c\)
\(Q.3.(viii)\) \(\int{\frac{dx}{(1+x^2)\sqrt{\tan^{-1}{x}+3}}}\)
উত্তরঃ \(2\sqrt{\tan^{-1}{x}+3}+c\)
\(Q.3.(ix)\) \(\int{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(\ln{(1+x^2)}+c\)
\(Q.3.(x)\) \(\int{\frac{e^{3x}}{e^{3x}-1}dx}\)
উত্তরঃ \(\frac{1}{3}\ln{|e^{3x}-1|}+c\)
\(Q.3.(xi)\) \(\int{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx}\)
উত্তরঃ \(\ln{(e^{x}+e^{-x})}+c\)
[ দিঃ২০১০ ]
\(Q.3.(xii)\) \(\int{\frac{1}{e^{x}+1}dx}\)
উত্তরঃ \(-\ln{(1+e^{-x})}+c\)
[ বঃ২০১৭; যঃ২০১০ ]
\(Q.3.(xiii)\) \(\int{\frac{1}{x\ln{|x|}}dx}\)
উত্তরঃ \(\ln{\ln{|x|}}+c\)
\(Q.3.(xiv)\) \(\int{\frac{1}{x(1+\ln{|x|})}dx}\)
উত্তরঃ \(\ln{(|1+\ln{|x|}|)}+c\)
[ বঃ২০০৯ ]
\(Q.3.(xv)\) \(\int{\frac{\tan{x}}{\ln{|\cos{x}|}}dx}\)
উত্তরঃ \(-\ln{(\ln{|\cos{x}|)}}+c\)
[ রাঃ২০০৯ ; কুঃ২০০৯,২০০১; সিঃ২০০৯,২০০৭; দিঃ২০০৯; যঃ২০০৮,২০০৪; চঃ,বঃ২০০৬ ]
\(Q.3.(xvi)\) \(\int{\frac{\sin{x}}{1+\cos{x}}dx}\)
উত্তরঃ \(-\ln{|1+\cos{x}|}+c\)
\(Q.3.(xvii)\) \(\int{\frac{\sin{x}}{4-\cos{x}}dx}\)
উত্তরঃ \(\ln{|4-\cos{x}|}+c\)
\(Q.3.(xviii)\) \(\int{\frac{\sin{x}}{3+4\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\ln{|3+4\cos{x}|}+c\)
[ ঢাঃ২০১৫,২০০৭; বঃ২০১৩ ]
\(Q.3.(xix)\) \(\int{\frac{\sin{x}}{1+2\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{2}\ln{|1+2\cos{x}|}+c\)
[ রাঃ২০০৩ ]
উত্তরঃ \(-\sqrt{1-x^2}+c\)
[ ঢাঃ২০১৪; দিঃ২০১১ ]
\(Q.3.(ii)\) \(\int{\frac{x^3dx}{\sqrt{1-2x^4}}}\)
উত্তরঃ \(-\frac{1}{4}\sqrt{1-2x^4}+c\)
[ চঃ২০১০ ]
\(Q.3.(iii)\) \(\int{\frac{dx}{x\sqrt{1+\ln{|x|}}}}\)
উত্তরঃ \(2\sqrt{1+\ln{|x|}}+c\)
[ কুঃ২০০৩ ]
\(Q.3.(iv)\) \(\int{\frac{\cos{x}}{\sqrt{\sin{x}}}dx}\)
উত্তরঃ \(2\sqrt{\sin{x}}+c\)
[ রাঃ২০১০; কুঃ২০০৫ ]
\(Q.3.(v)\) \(\int{\frac{dx}{\cos^2{x}\sqrt{\tan{x}-1}}}\)
উত্তরঃ \(2\sqrt{\tan{x}-1}+c\)
[ সিঃ২০০২ ]
\(Q.3.(vi)\) \(\int{\frac{\sqrt{\tan{x}}}{\sin{x}\cos{x}}dx}\)
উত্তরঃ \(2\sqrt{\tan{x}}+c\)
[ দিঃ২০১৪; রাঃ,কুঃ২০০৯; সিঃ২০০৭ ]
\(Q.3.(vii)\) \(\int{\frac{dx}{\sqrt{\sin^{-1}{x}}\sqrt{1-x^2}}}\)
উত্তরঃ \(2\sqrt{\sin^{-1}{x}}+c\)
\(Q.3.(viii)\) \(\int{\frac{dx}{(1+x^2)\sqrt{\tan^{-1}{x}+3}}}\)
উত্তরঃ \(2\sqrt{\tan^{-1}{x}+3}+c\)
\(Q.3.(ix)\) \(\int{\frac{2x}{1+x^2}dx}\)
উত্তরঃ \(\ln{(1+x^2)}+c\)
\(Q.3.(x)\) \(\int{\frac{e^{3x}}{e^{3x}-1}dx}\)
উত্তরঃ \(\frac{1}{3}\ln{|e^{3x}-1|}+c\)
\(Q.3.(xi)\) \(\int{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx}\)
উত্তরঃ \(\ln{(e^{x}+e^{-x})}+c\)
[ দিঃ২০১০ ]
\(Q.3.(xii)\) \(\int{\frac{1}{e^{x}+1}dx}\)
উত্তরঃ \(-\ln{(1+e^{-x})}+c\)
[ বঃ২০১৭; যঃ২০১০ ]
\(Q.3.(xiii)\) \(\int{\frac{1}{x\ln{|x|}}dx}\)
উত্তরঃ \(\ln{\ln{|x|}}+c\)
\(Q.3.(xiv)\) \(\int{\frac{1}{x(1+\ln{|x|})}dx}\)
উত্তরঃ \(\ln{(|1+\ln{|x|}|)}+c\)
[ বঃ২০০৯ ]
\(Q.3.(xv)\) \(\int{\frac{\tan{x}}{\ln{|\cos{x}|}}dx}\)
উত্তরঃ \(-\ln{(\ln{|\cos{x}|)}}+c\)
[ রাঃ২০০৯ ; কুঃ২০০৯,২০০১; সিঃ২০০৯,২০০৭; দিঃ২০০৯; যঃ২০০৮,২০০৪; চঃ,বঃ২০০৬ ]
\(Q.3.(xvi)\) \(\int{\frac{\sin{x}}{1+\cos{x}}dx}\)
উত্তরঃ \(-\ln{|1+\cos{x}|}+c\)
\(Q.3.(xvii)\) \(\int{\frac{\sin{x}}{4-\cos{x}}dx}\)
উত্তরঃ \(\ln{|4-\cos{x}|}+c\)
\(Q.3.(xviii)\) \(\int{\frac{\sin{x}}{3+4\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\ln{|3+4\cos{x}|}+c\)
[ ঢাঃ২০১৫,২০০৭; বঃ২০১৩ ]
\(Q.3.(xix)\) \(\int{\frac{\sin{x}}{1+2\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{2}\ln{|1+2\cos{x}|}+c\)
[ রাঃ২০০৩ ]
\(Q.3.(xx)\) \(\int{\frac{1+\cos{x}}{x+\sin{x}}dx}\)
উত্তরঃ \(\ln{|x+\sin{x}|}+c\)
[যঃ২০০৪ ]
\(Q.3.(xxi)\) \(\int{\frac{1-\tan{x}}{1+\tan{x}}dx}\)
উত্তরঃ \(\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxii)\) \(\int{\frac{\sec^2{x}}{3-4\tan{x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\ln{|3-4\tan{x}|}+c\)
[ দিঃ২০১১ ]
\(Q.3.(xxiii)\) \(\int{\frac{\sec{x}dx}{\ln{(|\sec{x}+\tan{x}|)}}}\)
উত্তরঃ \(\ln{\{\ln{(|\sec{x}+\tan{x}|)}\}}+c\)
\(Q.3.(xxiv)\) \(\int{\frac{dx}{(1+x^2)\tan^{-1}{x}}}\)
উত্তরঃ \(\ln{|\tan^{-1}{x}|}+c\)
[ সিঃ২০১১; চঃ২০১০; বঃ২০০৪ ]
\(Q.3.(xxv)\) \(\int{\frac{dx}{x(x^4-1)}}\)
উত্তরঃ \(\frac{1}{4}\ln{\left|1-\frac{1}{x^4}\right|}+c\)
[ চুয়েটঃ২০০৭-২০০৮ ]
\(Q.3.(xxvi)\) \(\int{cosec \ {\frac{x}{2}}dx}\)
উত্তরঃ \(2\ln{\left|cosec \ {\frac{x}{2}}-\cot{\frac{x}{2}}\right|}+c\)
\(Q.3.(xxvii)\) \(\int{\sec{\frac{x}{2}}dx}\)
উত্তরঃ \(2\ln{\left|\sec{\frac{x}{2}}+\tan{\frac{x}{2}}\right|}+c\)
\(Q.3.(xxviii)\) \(\int{\frac{\sin{x}}{a+b\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{b}\ln{|a+b\cos{x}|}+c\)
\(Q.3.(xxix)\) \(\int{\frac{\sin{2x}}{a\sin^2{x}+b\cos^2{x}}dx}\)
উত্তরঃ \(\frac{1}{a-b}\ln{|a\sin^2{x}+b\cos^2{x}|}+c\)
\(Q.3.(xxx)\) \(\int{\frac{\sin{2x}}{3+5\cos^2{x}}dx}\)
উত্তরঃ \(-\frac{1}{5}\ln{|3+5\cos^2{x}|}+c\)
\(Q.3.(xxxi)\) \(\int{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{1}{2}x-\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxxii)\) \(\int{\frac{\cos{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxxiii)\) \(\int{\frac{dx}{1+\tan{x}}}\)
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
[ ঢাঃ২০১০; কুঃ২০১০; সিঃ২০১৪,২০১০; বঃ২০১৪; যঃ২০১২; দিঃ২০১১ ]
\(Q.3.(xxxiv)\) \(\int{\frac{\cos{x}dx}{(5-2\sin{x})^2}}\)
উত্তরঃ \(\frac{1}{2(5-2\sin{x})}+c\)
\(Q.3.(xxxv)\) \(\int{\frac{\sec^2{(\cot^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(-\frac{1}{x}+c\)
[ বুয়েটঃ২০১০-২০১১ ]
\(Q.3.(xxxvi)\) \(\int{\frac{dx}{(e^x-1)^2}}\)
উত্তরঃ \(-\frac{1}{1-e^{-x}}-\ln{|1-e^{-x}|}+c\)
\(Q.3.(xxxvii)\) \(\int{\frac{dx}{1+e^{-x}}}\)
উত্তরঃ \(\ln{(e^x+1)}+c\)
\(Q.3.(xxxviii)\) \(\int{\frac{dx}{\sqrt{x}+x}}\)
উত্তরঃ \(2\ln{(1+\sqrt{x})}+c\)
\(Q.3.(xxxix)\) \(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\)
উত্তরঃ \(2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)
[ চঃ২০১০,২০১৫; রাঃ২০০৭; যঃ২০০০ ]
উত্তরঃ \(\ln{|x+\sin{x}|}+c\)
[যঃ২০০৪ ]
\(Q.3.(xxi)\) \(\int{\frac{1-\tan{x}}{1+\tan{x}}dx}\)
উত্তরঃ \(\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxii)\) \(\int{\frac{\sec^2{x}}{3-4\tan{x}}dx}\)
উত্তরঃ \(-\frac{1}{4}\ln{|3-4\tan{x}|}+c\)
[ দিঃ২০১১ ]
\(Q.3.(xxiii)\) \(\int{\frac{\sec{x}dx}{\ln{(|\sec{x}+\tan{x}|)}}}\)
উত্তরঃ \(\ln{\{\ln{(|\sec{x}+\tan{x}|)}\}}+c\)
\(Q.3.(xxiv)\) \(\int{\frac{dx}{(1+x^2)\tan^{-1}{x}}}\)
উত্তরঃ \(\ln{|\tan^{-1}{x}|}+c\)
[ সিঃ২০১১; চঃ২০১০; বঃ২০০৪ ]
\(Q.3.(xxv)\) \(\int{\frac{dx}{x(x^4-1)}}\)
উত্তরঃ \(\frac{1}{4}\ln{\left|1-\frac{1}{x^4}\right|}+c\)
[ চুয়েটঃ২০০৭-২০০৮ ]
\(Q.3.(xxvi)\) \(\int{cosec \ {\frac{x}{2}}dx}\)
উত্তরঃ \(2\ln{\left|cosec \ {\frac{x}{2}}-\cot{\frac{x}{2}}\right|}+c\)
\(Q.3.(xxvii)\) \(\int{\sec{\frac{x}{2}}dx}\)
উত্তরঃ \(2\ln{\left|\sec{\frac{x}{2}}+\tan{\frac{x}{2}}\right|}+c\)
\(Q.3.(xxviii)\) \(\int{\frac{\sin{x}}{a+b\cos{x}}dx}\)
উত্তরঃ \(-\frac{1}{b}\ln{|a+b\cos{x}|}+c\)
\(Q.3.(xxix)\) \(\int{\frac{\sin{2x}}{a\sin^2{x}+b\cos^2{x}}dx}\)
উত্তরঃ \(\frac{1}{a-b}\ln{|a\sin^2{x}+b\cos^2{x}|}+c\)
\(Q.3.(xxx)\) \(\int{\frac{\sin{2x}}{3+5\cos^2{x}}dx}\)
উত্তরঃ \(-\frac{1}{5}\ln{|3+5\cos^2{x}|}+c\)
\(Q.3.(xxxi)\) \(\int{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{1}{2}x-\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxxii)\) \(\int{\frac{\cos{x}}{\sin{x}+\cos{x}}dx}\)
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
\(Q.3.(xxxiii)\) \(\int{\frac{dx}{1+\tan{x}}}\)
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
[ ঢাঃ২০১০; কুঃ২০১০; সিঃ২০১৪,২০১০; বঃ২০১৪; যঃ২০১২; দিঃ২০১১ ]
\(Q.3.(xxxiv)\) \(\int{\frac{\cos{x}dx}{(5-2\sin{x})^2}}\)
উত্তরঃ \(\frac{1}{2(5-2\sin{x})}+c\)
\(Q.3.(xxxv)\) \(\int{\frac{\sec^2{(\cot^{-1}{x})}}{1+x^2}dx}\)
উত্তরঃ \(-\frac{1}{x}+c\)
[ বুয়েটঃ২০১০-২০১১ ]
\(Q.3.(xxxvi)\) \(\int{\frac{dx}{(e^x-1)^2}}\)
উত্তরঃ \(-\frac{1}{1-e^{-x}}-\ln{|1-e^{-x}|}+c\)
\(Q.3.(xxxvii)\) \(\int{\frac{dx}{1+e^{-x}}}\)
উত্তরঃ \(\ln{(e^x+1)}+c\)
\(Q.3.(xxxviii)\) \(\int{\frac{dx}{\sqrt{x}+x}}\)
উত্তরঃ \(2\ln{(1+\sqrt{x})}+c\)
\(Q.3.(xxxix)\) \(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\)
উত্তরঃ \(2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)
[ চঃ২০১০,২০১৫; রাঃ২০০৭; যঃ২০০০ ]
\(Q.3.(i)\) \(\int{\frac{x}{\sqrt{1-x^2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\sqrt{1-x^2}+c\)
[ ঢাঃ২০১৪; দিঃ২০১১ ]
উত্তরঃ \(-\sqrt{1-x^2}+c\)
[ ঢাঃ২০১৪; দিঃ২০১১ ]
সমাধানঃ
ধরি,
\(1-x^2=t\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2x=\frac{dt}{dx}\)
\(\Rightarrow -2xdx=dt\)
\(\therefore xdx=-\frac{1}{2}dt\)
\(\int{\frac{x}{\sqrt{1-x^2}}dx}\)\(1-x^2=t\)
\(\Rightarrow \frac{d}{dx}(1-x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2x=\frac{dt}{dx}\)
\(\Rightarrow -2xdx=dt\)
\(\therefore xdx=-\frac{1}{2}dt\)
\(=\int{\frac{1}{\sqrt{1-x^2}}xdx}\)
\(=\int{\frac{1}{\sqrt{t}}\left(-\frac{1}{2}dt\right)}\)
\(=-\frac{1}{2}\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=-\frac{1}{2}\int{t^{-\frac{1}{2}}dt}\)
\(=-\frac{1}{2}\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=-\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=-t^{\frac{1}{2}}+c\)
\(=-\sqrt{t}+c\)
\(=-\sqrt{1-x^2}+c\) ➜ \(\because t=1-x^2\)
\(Q.3.(ii)\) \(\int{\frac{x^3dx}{\sqrt{1-2x^4}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4}\sqrt{1-2x^4}+c\)
[ চঃ২০১০ ]
উত্তরঃ \(-\frac{1}{4}\sqrt{1-2x^4}+c\)
[ চঃ২০১০ ]
সমাধানঃ
ধরি,
\(1-2x^4=t\)
\(\Rightarrow \frac{d}{dx}(1-2x^4)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-8x^3=\frac{dt}{dx}\)
\(\Rightarrow -8x^3dx=dt\)
\(\therefore x^3dx=-\frac{1}{8}dt\)
\(\int{\frac{x^3dx}{\sqrt{1-2x^4}}}\)\(1-2x^4=t\)
\(\Rightarrow \frac{d}{dx}(1-2x^4)=\frac{d}{dx}(t)\)
\(\Rightarrow 0-8x^3=\frac{dt}{dx}\)
\(\Rightarrow -8x^3dx=dt\)
\(\therefore x^3dx=-\frac{1}{8}dt\)
\(=\int{\frac{1}{\sqrt{1-2x^4}}x^3dx}\)
\(=\int{\frac{1}{\sqrt{t}}\left(-\frac{1}{8}dt\right)}\)
\(=-\frac{1}{8}\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=-\frac{1}{8}\int{t^{-\frac{1}{2}}dt}\)
\(=-\frac{1}{8}\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{8}\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=-\frac{1}{8}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=-\frac{1}{8}.2t^{\frac{1}{2}}+c\)
\(=-\frac{1}{4}\sqrt{t}+c\)
\(=-\frac{1}{4}\sqrt{1-2x^4}+c\) ➜ \(\because t=1-2x^4\)
\(Q.3.(iii)\) \(\int{\frac{dx}{x\sqrt{1+\ln{|x|}}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{1+\ln{|x|}}+c\)
[ কুঃ২০০৩ ]
উত্তরঃ \(2\sqrt{1+\ln{|x|}}+c\)
[ কুঃ২০০৩ ]
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{dx}{x\sqrt{1+\ln{|x|}}}}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\frac{1}{\sqrt{1+\ln{|x|}}}.\frac{1}{x}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{1+\ln{|x|}}+c\) ➜ \(\because t=1+\ln{|x|}\)
\(Q.3.(iv)\) \(\int{\frac{\cos{x}}{\sqrt{\sin{x}}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{\sin{x}}+c\)
[ রাঃ২০১০; কুঃ২০০৫ ]
উত্তরঃ \(2\sqrt{\sin{x}}+c\)
[ রাঃ২০১০; কুঃ২০০৫ ]
সমাধানঃ
ধরি,
\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(\int{\frac{\cos{x}}{\sqrt{\sin{x}}}dx}\)\(\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}=\frac{dt}{dx}\)
\(\therefore \cos{x}dx=dt\)
\(=\int{\frac{1}{\sqrt{\sin{x}}}\cos{x}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{\sin{x}}+c\) ➜ \(\because t=\sin{x}\)
\(Q.3.(v)\) \(\int{\frac{dx}{\cos^2{x}\sqrt{\tan{x}-1}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{\tan{x}-1}+c\)
[ সিঃ২০০২ ]
উত্তরঃ \(2\sqrt{\tan{x}-1}+c\)
[ সিঃ২০০২ ]
সমাধানঃ
ধরি,
\(\tan{x}-1=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x}-1)=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}-0=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\frac{dx}{\cos^2{x}\sqrt{\tan{x}-1}}}\)\(\tan{x}-1=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x}-1)=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}-0=\frac{dt}{dx}\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{\frac{1}{\sqrt{\tan{x}-1}}.\frac{1}{\cos^2{x}}dx}\)
\(=\int{\frac{1}{\sqrt{\tan{x}-1}}.\sec^2{x}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{\tan{x}-1}+c\) ➜ \(\because t=\tan{x}-1\)
\(Q.3.(vi)\) \(\int{\frac{\sqrt{\tan{x}}}{\sin{x}\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{\tan{x}}+c\)
[ দিঃ২০১৪; রাঃ,কুঃ২০০৯; সিঃ২০০৭ ]
উত্তরঃ \(2\sqrt{\tan{x}}+c\)
[ দিঃ২০১৪; রাঃ,কুঃ২০০৯; সিঃ২০০৭ ]
সমাধানঃ
ধরি,
\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(\int{\frac{\sqrt{\tan{x}}}{\sin{x}\cos{x}}dx}\)\(\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \sec^2{x}=\frac{dt}{dx}\)
\(\therefore \sec^2{x}dx=dt\)
\(=\int{\frac{\sqrt{\tan{x}}}{\frac{\sin{x}}{\cos{x}}\cos^2{x}}dx}\)
\(=\int{\frac{\sqrt{\tan{x}}}{\tan{x}\cos^2{x}}dx}\)
\(=\int{\frac{\sqrt{\tan{x}}}{\sqrt{\tan{x}}\sqrt{\tan{x}}\cos^2{x}}dx}\)
\(=\int{\frac{1}{\sqrt{\tan{x}}\cos^2{x}}dx}\)
\(=\int{\frac{1}{\sqrt{\tan{x}}}.\frac{1}{\cos^2{x}}dx}\)
\(=\int{\frac{1}{\sqrt{\tan{x}}}.\sec^2{x}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{\tan{x}}+c\) ➜ \(\because t=\tan{x}\)
\(Q.3.(vii)\) \(\int{\frac{dx}{\sqrt{\sin^{-1}{x}}\sqrt{1-x^2}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{\sin^{-1}{x}}+c\)
উত্তরঃ \(2\sqrt{\sin^{-1}{x}}+c\)
সমাধানঃ
ধরি,
\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(\int{\frac{dx}{\sqrt{\sin^{-1}{x}}\sqrt{1-x^2}}}\)\(\sin^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sqrt{1-x^2}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\)
\(=\int{\frac{1}{\sqrt{\sin^{-1}{x}}\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{\sin^{-1}{x}}}.\frac{1}{\sqrt{1-x^2}}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{\sin^{-1}{x}}+c\) ➜ \(\because t=\sin^{-1}{x}\)
\(Q.3.(viii)\) \(\int{\frac{dx}{(1+x^2)\sqrt{\tan^{-1}{x}+3}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{\tan^{-1}{x}+3}+c\)
উত্তরঃ \(2\sqrt{\tan^{-1}{x}+3}+c\)
সমাধানঃ
ধরি,
\(\tan^{-1}{x}+3=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x}+3)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{(1+x^2)}=\frac{dt}{dx}\)
\(\therefore \frac{1}{(1+x^2)}dx=dt\)
\(\int{\frac{dx}{(1+x^2)\sqrt{\tan^{-1}{x}+3}}}\)\(\tan^{-1}{x}+3=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x}+3)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{(1+x^2)}=\frac{dt}{dx}\)
\(\therefore \frac{1}{(1+x^2)}dx=dt\)
\(=\int{\frac{1}{(1+x^2)\sqrt{\tan^{-1}{x}+3}}dx}\)
\(=\int{\frac{1}{\sqrt{\tan^{-1}{x}+3}}.\frac{1}{(1+x^2)}dx}\)
\(=\int{\frac{1}{\sqrt{t}}dt}\)
\(=\int{\frac{1}{t^{\frac{1}{2}}}dt}\)
\(=\int{t^{-\frac{1}{2}}dt}\)
\(=\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{t^{\frac{-1+2}{2}}}{\frac{-1+2}{2}}+c\)
\(=\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c\)
\(=2t^{\frac{1}{2}}+c\)
\(=2\sqrt{t}+c\)
\(=2\sqrt{\tan^{-1}{x}+3}+c\) ➜ \(\because t=\tan^{-1}{x}+3\)
\(Q.3.(ix)\) \(\int{\frac{2x}{1+x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(1+x^2)}+c\)
উত্তরঃ \(\ln{(1+x^2)}+c\)
সমাধানঃ
ধরি,
\(1+x^2=t\)
\(\Rightarrow \frac{d}{dx}(1+x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 0+2x=\frac{dt}{dx}\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\therefore 2xdx=dt\)
\(\int{\frac{2x}{1+x^2}dx}\)\(1+x^2=t\)
\(\Rightarrow \frac{d}{dx}(1+x^2)=\frac{d}{dx}(t)\)
\(\Rightarrow 0+2x=\frac{dt}{dx}\)
\(\Rightarrow 2x=\frac{dt}{dx}\)
\(\therefore 2xdx=dt\)
\(=\int{\frac{1}{1+x^2}.2xdx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{1+x^2}+c\) ➜ \(\because t=1+x^2\)
\(Q.3.(x)\) \(\int{\frac{e^{3x}}{e^{3x}-1}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{3}\ln{|e^{3x}-1|}+c\)
উত্তরঃ \(\frac{1}{3}\ln{|e^{3x}-1|}+c\)
সমাধানঃ
ধরি,
\(e^{3x}-1=t\)
\(\Rightarrow \frac{d}{dx}(e^{3x}-1)=\frac{d}{dx}(t)\)
\(\Rightarrow e^{3x}.3-0=\frac{dt}{dx}\)
\(\Rightarrow 3e^{3x}=\frac{dt}{dx}\)
\(\Rightarrow 3e^{3x}dx=dt\)
\(\therefore e^{3x}dx=\frac{1}{3}dt\)
\(\int{\frac{e^{3x}}{e^{3x}-1}dx}\)\(e^{3x}-1=t\)
\(\Rightarrow \frac{d}{dx}(e^{3x}-1)=\frac{d}{dx}(t)\)
\(\Rightarrow e^{3x}.3-0=\frac{dt}{dx}\)
\(\Rightarrow 3e^{3x}=\frac{dt}{dx}\)
\(\Rightarrow 3e^{3x}dx=dt\)
\(\therefore e^{3x}dx=\frac{1}{3}dt\)
\(=\int{\frac{1}{e^{3x}-1}e^{3x}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{3}dt}\)
\(=\frac{1}{3}\int{\frac{1}{t}dt}\)
\(=\frac{1}{3}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{3}\ln{|e^{3x}-1|}+c\) ➜ \(\because t=e^{3x}-1\)
\(Q.3.(xi)\) \(\int{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(e^{x}+e^{-x})}+c\)
[ দিঃ২০১০ ]
উত্তরঃ \(\ln{(e^{x}+e^{-x})}+c\)
[ দিঃ২০১০ ]
সমাধানঃ
ধরি,
\(e^{x}+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(e^{x}+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{x}+e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow e^{x}-e^{-x}=\frac{dt}{dx}\)
\(\therefore (e^{x}-e^{-x})dx=dt\)
\(\int{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx}\)\(e^{x}+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(e^{x}+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow e^{x}+e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow e^{x}-e^{-x}=\frac{dt}{dx}\)
\(\therefore (e^{x}-e^{-x})dx=dt\)
\(\int{\frac{1}{e^{x}+e^{-x}}(e^{x}-e^{-x})dx}\)
\(\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{e^{x}+e^{-x}}+c\) ➜ \(\because t=e^{x}+e^{-x}\)
\(Q.3.(xii)\) \(\int{\frac{1}{e^{x}+1}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\ln{(1+e^{-x})}+c\)
[ বঃ২০১৭; যঃ২০১০ ]
উত্তরঃ \(-\ln{(1+e^{-x})}+c\)
[ বঃ২০১৭; যঃ২০১০ ]
সমাধানঃ
ধরি,
\(1+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(1+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}dx=dt\)
\(\therefore e^{-x}dx=-dt\)
\(\int{\frac{1}{e^{x}+1}dx}\)\(1+e^{-x}=t\)
\(\Rightarrow \frac{d}{dx}(1+e^{-x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+e^{-x}(-1)=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}dx=dt\)
\(\therefore e^{-x}dx=-dt\)
\(=\int{\frac{e^{-x}}{e^{x}.e^{-x}+e^{-x}}dx}\) ➜ লব ও হরের সহিত \(e^{-x}\) গুণ করে।
\(=\int{\frac{e^{-x}}{e^{x-x}+e^{-x}}dx}\)
\(=\int{\frac{e^{-x}}{e^{0}+e^{-x}}dx}\)
\(=\int{\frac{e^{-x}}{1+e^{-x}}dx}\) ➜\(\because e^{0}=1\)
\(=\int{\frac{1}{1+e^{-x}}e^{-x}dx}\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{(1+e^{-x})}+c\) ➜ \(\because t=1+e^{-x}\)
\(Q.3.(xiii)\) \(\int{\frac{1}{x\ln{|x|}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{\ln{|x|}}+c\)
উত্তরঃ \(\ln{\ln{|x|}}+c\)
সমাধানঃ
ধরি,
\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{1}{x\ln{|x|}}dx}\)\(\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\frac{1}{\ln{|x|}}.\frac{1}{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\ln{|x|}}+c\) ➜ \(\because t=\ln{|x|}\)
\(Q.3.(xiv)\) \(\int{\frac{1}{x(1+\ln{|x|})}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(|1+\ln{|x|}|)}+c\)
[ বঃ২০০৯ ]
উত্তরঃ \(\ln{(|1+\ln{|x|}|)}+c\)
[ বঃ২০০৯ ]
সমাধানঃ
ধরি,
\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(\int{\frac{1}{x(1+\ln{|x|})}dx}\)\(1+\ln{|x|}=t\)
\(\Rightarrow \frac{d}{dx}(1+\ln{|x|})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\frac{1}{x}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{x}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x}dx=dt\)
\(=\int{\frac{1}{1+\ln{|x|}}.\frac{1}{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{1+\ln{|x|}}+c\) ➜ \(\because t=1+\ln{|x|}\)
\(Q.3.(xv)\) \(\int{\frac{\tan{x}}{\ln{|\cos{x}|}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\ln{(\ln{|\cos{x}|)}}+c\)
[ রাঃ২০০৯ ; কুঃ২০০৯,২০০১; সিঃ২০০৯,২০০৭; দিঃ২০০৯; যঃ২০০৮,২০০৪; চঃ,বঃ২০০৬ ]
উত্তরঃ \(-\ln{(\ln{|\cos{x}|)}}+c\)
[ রাঃ২০০৯ ; কুঃ২০০৯,২০০১; সিঃ২০০৯,২০০৭; দিঃ২০০৯; যঃ২০০৮,২০০৪; চঃ,বঃ২০০৬ ]
সমাধানঃ
ধরি,
\(\ln{|\cos{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|\cos{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\cos{x}}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow -\frac{\sin{x}}{\cos{x}}=\frac{dt}{dx}\)
\(\Rightarrow -\tan{x}=\frac{dt}{dx}\)
\(\Rightarrow -\tan{x}dx=dt\)
\(\therefore \tan{x}dx=-dt\)
\(\int{\frac{\tan{x}}{\ln{|\cos{x}|}}dx}\)\(\ln{|\cos{x}|}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{|\cos{x}|})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\cos{x}}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow -\frac{\sin{x}}{\cos{x}}=\frac{dt}{dx}\)
\(\Rightarrow -\tan{x}=\frac{dt}{dx}\)
\(\Rightarrow -\tan{x}dx=dt\)
\(\therefore \tan{x}dx=-dt\)
\(=\int{\frac{1}{\ln{|\cos{x}|}}\tan{x}dx}\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{(\ln{|\cos{x}|})}+c\) ➜ \(\because t=\ln{|\cos{x}|}\)
\(Q.3.(xvi)\) \(\int{\frac{\sin{x}}{1+\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\ln{|1+\cos{x}|}+c\)
উত্তরঃ \(-\ln{|1+\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(\int{\frac{\sin{x}}{1+\cos{x}}dx}\)\(1+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-dt\)
\(=\int{\frac{1}{1+\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}(-dt)}\)
\(=-\int{\frac{1}{t}dt}\)
\(=-\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\ln{|1+\cos{x}|}+c\) ➜ \(\because t=1+\cos{x}\)
\(Q.3.(xvii)\) \(\int{\frac{\sin{x}}{4-\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{|4-\cos{x}|}+c\)
উত্তরঃ \(\ln{|4-\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(4-\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(4-\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=dt\)
\(\int{\frac{\sin{x}}{4-\cos{x}}dx}\)\(4-\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(4-\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow \sin{x}=\frac{dt}{dx}\)
\(\therefore \sin{x}dx=dt\)
\(=\int{\frac{1}{4-\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|4-\cos{x}|}+c\) ➜ \(\because t=4-\cos{x}\)
\(Q.3.(xviii)\) \(\int{\frac{\sin{x}}{3+4\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4}\ln{|3+4\cos{x}|}+c\)
[ ঢাঃ২০১৫,২০০৭; বঃ২০১৩ ]
উত্তরঃ \(-\frac{1}{4}\ln{|3+4\cos{x}|}+c\)
[ ঢাঃ২০১৫,২০০৭; বঃ২০১৩ ]
সমাধানঃ
ধরি,
\(3+4\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(3+4\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{4}dt\)
\(\int{\frac{\sin{x}}{3+4\cos{x}}dx}\)\(3+4\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(3+4\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{4}dt\)
\(=\int{\frac{1}{3+4\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}\left(-\frac{1}{4}dt\right)}\)
\(=-\frac{1}{4}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{4}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\ln{|3+4\cos{x}|}+c\) ➜ \(\because t=3+4\cos{x}\)
\(Q.3.(xix)\) \(\int{\frac{\sin{x}}{1+2\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{2}\ln{|1+2\cos{x}|}+c\)
[ রাঃ২০০৩ ]
উত্তরঃ \(-\frac{1}{2}\ln{|1+2\cos{x}|}+c\)
[ রাঃ২০০৩ ]
সমাধানঃ
ধরি,
\(1+2\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+2\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{2}dt\)
\(\int{\frac{\sin{x}}{1+2\cos{x}}dx}\)\(1+2\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(1+2\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{2}dt\)
\(=\int{\frac{1}{1+2\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}\left(-\frac{1}{2}dt\right)}\)
\(=-\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{2}\ln{|1+2\cos{x}|}+c\) ➜ \(\because t=1+2\cos{x}\)
\(Q.3.(xx)\) \(\int{\frac{1+\cos{x}}{x+\sin{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{|x+\sin{x}|}+c\)
[যঃ২০০৪ ]
উত্তরঃ \(\ln{|x+\sin{x}|}+c\)
[যঃ২০০৪ ]
সমাধানঃ
ধরি,
\(x+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(x+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 1+\cos{x}=\frac{dt}{dx}\)
\(\therefore (1+\cos{x})dx=dt\)
\(\int{\frac{1+\cos{x}}{x+\sin{x}}dx}\)\(x+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(x+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 1+\cos{x}=\frac{dt}{dx}\)
\(\therefore (1+\cos{x})dx=dt\)
\(=\int{\frac{1}{x+\sin{x}}(1+\cos{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|x+\sin{x}|}+c\) ➜ \(\because t=x+\sin{x}\)
\(Q.3.(xxi)\) \(\int{\frac{1-\tan{x}}{1+\tan{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{|\sin{x}+\cos{x}|}+c\)
উত্তরঃ \(\ln{|\sin{x}+\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(\cos{x}+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x}+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}+\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\therefore (\cos{x}-\sin{x})dx=dt\)
\(\int{\frac{1-\tan{x}}{1+\tan{x}}dx}\)\(\cos{x}+\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cos{x}+\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\sin{x}+\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\therefore (\cos{x}-\sin{x})dx=dt\)
\(=\int{\frac{1-\frac{\sin{x}}{\cos{x}}}{1+\frac{\sin{x}}{\cos{x}}}dx}\)
\(=\int{\frac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}}dx}\) ➜ লব ও হরের সহিত \(\cos{x}\) গুণ করে।
\(=\int{\frac{1}{\cos{x}+\sin{x}}(\cos{x}-\sin{x})dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\cos{x}+\sin{x}|}+c\) ➜ \(\because t=\cos{x}+\sin{x}\)
\(Q.3.(xxii)\) \(\int{\frac{\sec^2{x}}{3-4\tan{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{4}\ln{|3-4\tan{x}|}+c\)
[ দিঃ২০১১ ]
উত্তরঃ \(-\frac{1}{4}\ln{|3-4\tan{x}|}+c\)
[ দিঃ২০১১ ]
সমাধানঃ
ধরি,
\(3-4\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(3-4\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sec^2{x}dx=dt\)
\(\therefore \sec^2{x}dx=-\frac{1}{4}dt\)
\(\int{\frac{\sec^2{x}}{3-4\tan{x}}dx}\)\(3-4\tan{x}=t\)
\(\Rightarrow \frac{d}{dx}(3-4\tan{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-4\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sec^2{x}=\frac{dt}{dx}\)
\(\Rightarrow -4\sec^2{x}dx=dt\)
\(\therefore \sec^2{x}dx=-\frac{1}{4}dt\)
\(=\int{\frac{1}{3-4\tan{x}}\sec^2{x}dx}\)
\(=\int{\frac{1}{t}\left(-\frac{1}{4}dt\right)}\)
\(=-\frac{1}{4}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{4}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{4}\ln{|3-4\tan{x}|}+c\) ➜ \(\because t=3-4\tan{x}\)
\(Q.3.(xxiii)\) \(\int{\frac{\sec{x}dx}{\ln{(|\sec{x}+\tan{x}|)}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{\{\ln{(|\sec{x}+\tan{x}|)}\}}+c\)
উত্তরঃ \(\ln{\{\ln{(|\sec{x}+\tan{x}|)}\}}+c\)
সমাধানঃ
ধরি,
\(\ln{(|\sec{x}+\tan{x}|)}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{(|\sec{x}+\tan{x}|)})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}(\sec{x}\tan{x}+\sec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}\{\sec{x}(\sec{x}+\tan{x})\}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}dx=dt\)
\(\int{\frac{\sec{x}dx}{\ln{(|\sec{x}+\tan{x}|)}}}\)\(\ln{(|\sec{x}+\tan{x}|)}=t\)
\(\Rightarrow \frac{d}{dx}(\ln{(|\sec{x}+\tan{x}|)})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}(\sec{x}\tan{x}+\sec^2{x})=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{\sec{x}+\tan{x}}\{\sec{x}(\sec{x}+\tan{x})\}=\frac{dt}{dx}\)
\(\Rightarrow \sec{x}=\frac{dt}{dx}\)
\(\therefore \sec{x}dx=dt\)
\(=\int{\frac{1}{\ln{(|\sec{x}+\tan{x}|)}}\sec{x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{\{\ln{(|\sec{x}+\tan{x}|)}\}}+c\) ➜ \(\because t=\ln{(|\sec{x}+\tan{x}|)}\)
\(Q.3.(xxiv)\) \(\int{\frac{dx}{(1+x^2)\tan^{-1}{x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{|\tan^{-1}{x}|}+c\)
[ সিঃ২০১১; চঃ২০১০; বঃ২০০৪ ]
উত্তরঃ \(\ln{|\tan^{-1}{x}|}+c\)
[ সিঃ২০১১; চঃ২০১০; বঃ২০০৪ ]
সমাধানঃ
ধরি,
\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(\int{\frac{dx}{(1+x^2)\tan^{-1}{x}}}\)\(\tan^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\tan^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\therefore \frac{1}{1+x^2}dx=dt\)
\(=\int{\frac{1}{\tan^{-1}{x}}.\frac{1}{(1+x^2)}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{|\tan^{-1}{x}|}+c\) ➜ \(\because t=\tan^{-1}{x}\)
\(Q.3.(xxv)\) \(\int{\frac{dx}{x(x^4-1)}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{4}\ln{\left|1-\frac{1}{x^4}\right|}+c\)
[ চুয়েটঃ২০০৭-২০০৮ ]
উত্তরঃ \(\frac{1}{4}\ln{\left|1-\frac{1}{x^4}\right|}+c\)
[ চুয়েটঃ২০০৭-২০০৮ ]
সমাধানঃ
ধরি,
\(1-\frac{1}{x^4}=t\)
\(\Rightarrow \frac{d}{dx}\left(1-\frac{1}{x^4}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{d}{dx}(1-x^{-4})=\frac{dt}{dx}\)
\(\Rightarrow 0+4x^{-4-1})=\frac{dt}{dx}\)
\(\Rightarrow 4x^{-5})=\frac{dt}{dx}\)
\(\Rightarrow 4\frac{1}{x^5}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x^5}dx=\frac{1}{4}dt\)
\(\int{\frac{dx}{x(x^4-1)}}\)\(1-\frac{1}{x^4}=t\)
\(\Rightarrow \frac{d}{dx}\left(1-\frac{1}{x^4}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{d}{dx}(1-x^{-4})=\frac{dt}{dx}\)
\(\Rightarrow 0+4x^{-4-1})=\frac{dt}{dx}\)
\(\Rightarrow 4x^{-5})=\frac{dt}{dx}\)
\(\Rightarrow 4\frac{1}{x^5}=\frac{dt}{dx}\)
\(\therefore \frac{1}{x^5}dx=\frac{1}{4}dt\)
\(=\int{\frac{dx}{x.x^4\left(1-\frac{1}{x^4}\right)}}\)
\(=\int{\frac{1}{x^5\left(1-\frac{1}{x^4}\right)}dx}\)
\(=\int{\frac{1}{1-\frac{1}{x^4}}.\frac{1}{x^5}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{4}dt}\)
\(=\frac{1}{4}\int{\frac{1}{t}dt}\)
\(=\frac{1}{4}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{4}\ln{\left|1-\frac{1}{x^4}\right|}+c\) ➜ \(\because t=1-\frac{1}{x^4}\)
\(Q.3.(xxvi)\) \(\int{cosec \ {\frac{x}{2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\ln{\left|cosec \ {\frac{x}{2}}-\cot{\frac{x}{2}}\right|}+c\)
উত্তরঃ \(2\ln{\left|cosec \ {\frac{x}{2}}-\cot{\frac{x}{2}}\right|}+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}dx=dt\)
\(\therefore dx=2dt\)
\(\int{cosec \ {\frac{x}{2}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}dx=dt\)
\(\therefore dx=2dt\)
\(=\int{cosec \ {t}.2dt}\)
\(=2\int{cosec \ {t}dt}\)
\(=2\ln{\left|cosec \ {t}-\cot{t}\right|}+c\) ➜ \(\because \int{cosec \ {x}dx}=\ln{\left|cosec \ {x}-\cot{x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{\left|cosec \ {\frac{x}{2}}-\cot{\frac{x}{2}}\right|}+c\) ➜ \(\because t=\frac{x}{2}\)
\(Q.3.(xxvii)\) \(\int{\sec{\frac{x}{2}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\ln{\left|\sec{\frac{x}{2}}+\tan{\frac{x}{2}}\right|}+c\)
উত্তরঃ \(2\ln{\left|\sec{\frac{x}{2}}+\tan{\frac{x}{2}}\right|}+c\)
সমাধানঃ
ধরি,
\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}dx=dt\)
\(\therefore dx=2dt\)
\(\int{\sec{\frac{x}{2}}dx}\)\(\frac{x}{2}=t\)
\(\Rightarrow \frac{d}{dx}\left(\frac{x}{2}\right)=\frac{d}{dx}(t)\)
\(\Rightarrow \frac{1}{2}=\frac{dt}{dx}\)
\(\Rightarrow \frac{1}{2}dx=dt\)
\(\therefore dx=2dt\)
\(=\int{\sec{\frac{x}{2}}dx}\)
\(=\int{\sec{t}.2dt}\)
\(=2\int{\sec{t}dt}\)
\(=2\ln{\left|\sec{t}+\tan{t}\right|}+c\) ➜ \(\because \int{\sec{x}dx}=\ln{\left|\sec{x}+\tan{x}\right|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{\left|\sec{\frac{x}{2}}+\tan{\frac{x}{2}}\right|}+c\) ➜ \(\because t=\frac{x}{2}\)
\(Q.3.(xxviii)\) \(\int{\frac{\sin{x}}{a+b\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{b}\ln{|a+b\cos{x}|}+c\)
উত্তরঃ \(-\frac{1}{b}\ln{|a+b\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(a+b\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(a+b\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-b\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -b\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -b\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{b}dt\)
\(\int{\frac{\sin{x}}{a+b\cos{x}}dx}\)\(a+b\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(a+b\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-b\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -b\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -b\sin{x}dx=dt\)
\(\therefore \sin{x}dx=-\frac{1}{b}dt\)
\(=\int{\frac{1}{a+b\cos{x}}\sin{x}dx}\)
\(=\int{\frac{1}{t}\left(-\frac{1}{b}dt\right)}\)
\(=-\frac{1}{b}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{b}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{b}\ln{|a+b\cos{x}|}+c\) ➜ \(\because t=a+b\cos{x}\)
\(Q.3.(xxix)\) \(\int{\frac{\sin{2x}}{a\sin^2{x}+b\cos^2{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{a-b}\ln{|a\sin^2{x}+b\cos^2{x}|}+c\)
উত্তরঃ \(\frac{1}{a-b}\ln{|a\sin^2{x}+b\cos^2{x}|}+c\)
সমাধানঃ
ধরি,
\(a\sin^2{x}+b\cos^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\sin^2{x}+b\cos^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a.2\sin{x}\cos{x}+b.2\cos{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow a.2\sin{x}\cos{x}-b.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow a\sin{2x}-b\sin{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow (a-b)\sin{2x}=\frac{dt}{dx}\)
\(\Rightarrow (a-b)\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=\frac{1}{a-b}dt\)
\(\int{\frac{\sin{2x}}{a\sin^2{x}+b\cos^2{x}}dx}\)\(a\sin^2{x}+b\cos^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(a\sin^2{x}+b\cos^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow a.2\sin{x}\cos{x}+b.2\cos{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow a.2\sin{x}\cos{x}-b.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow a\sin{2x}-b\sin{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow (a-b)\sin{2x}=\frac{dt}{dx}\)
\(\Rightarrow (a-b)\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=\frac{1}{a-b}dt\)
\(=\int{\frac{1}{a\sin^2{x}+b\cos^2{x}}\sin{2x}dx}\)
\(=\int{\frac{1}{t}.\frac{1}{a-b}dt}\)
\(=\frac{1}{a-b}\int{\frac{1}{t}dt}\)
\(=\frac{1}{a-b}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{a-b}\ln{|a\sin^2{x}+b\cos^2{x}|}+c\) ➜ \(\because t=a\sin^2{x}+b\cos^2{x}\)
\(Q.3.(xxx)\) \(\int{\frac{\sin{2x}}{3+5\cos^2{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{5}\ln{|3+5\cos^2{x}|}+c\)
উত্তরঃ \(-\frac{1}{5}\ln{|3+5\cos^2{x}|}+c\)
সমাধানঃ
ধরি,
\(3+5\cos^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(3+5\cos^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+5.2\cos{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow -5.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -5\sin{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow -5\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=-\frac{1}{5}dt\)
\(\int{\frac{\sin{2x}}{3+5\cos^2{x}}dx}\)\(3+5\cos^2{x}=t\)
\(\Rightarrow \frac{d}{dx}(3+5\cos^2{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0+5.2\cos{x}(-\sin{x})=\frac{dt}{dx}\)
\(\Rightarrow -5.2\sin{x}\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -5\sin{2x}=\frac{dt}{dx}\) ➜ \(\because 2\sin{A}\cos{A}=\sin{2A}\)
\(\Rightarrow -5\sin{2x}dx=dt\)
\(\therefore \sin{2x}dx=-\frac{1}{5}dt\)
\(=\int{\frac{1}{3+5\cos^2{x}}\sin{2x}dx}\)
\(=\int{\frac{1}{t}(-\frac{1}{5}dt)}\)
\(=-\frac{1}{5}\int{\frac{1}{t}dt}\)
\(=-\frac{1}{5}\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{5}\ln{|3+5\cos^2{x}|}+c\) ➜ \(\because t=3+5\cos^2{x}\)
\(Q.3.(xxxi)\) \(\int{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}x-\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
উত্তরঃ \(\frac{1}{2}x-\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(\int{\frac{\sin{x}}{\sin{x}+\cos{x}}dx}\)\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(=\int{\frac{2\sin{x}}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\frac{\sin{x}+\cos{x}+\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\left\{\frac{\sin{x}+\cos{x}}{2(\sin{x}+\cos{x})}+\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\int{\left\{\frac{1}{2}+\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\frac{1}{2}\int{\left(1+\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{\sin{x}+\cos{x}}(\sin{x}-\cos{x})dx}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{t}(-dt)}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\frac{1}{2}x-\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x-\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\) ➜ \(\because t=\sin{x}+\cos{x}\)
\(Q.3.(xxxii)\) \(\int{\frac{\cos{x}}{\sin{x}+\cos{x}}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
সমাধানঃ
ধরি,
\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(\int{\frac{\cos{x}}{\sin{x}+\cos{x}}dx}\)\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(=\int{\frac{2\cos{x}}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\frac{(\sin{x}+\cos{x})-(\sin{x}-\cos{x})}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\left\{\frac{\sin{x}+\cos{x}}{2(\sin{x}+\cos{x})}-\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\int{\left\{\frac{1}{2}-\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\frac{1}{2}\int{\left(1-\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{1}{\sin{x}+\cos{x}}(\sin{x}-\cos{x})dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{1}{t}(-dt)}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\frac{1}{2}x+\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\) ➜ \(\because t=\sin{x}+\cos{x}\)
\(Q.3.(xxxiii)\) \(\int{\frac{dx}{1+\tan{x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
[ ঢাঃ২০১০; কুঃ২০১০; সিঃ২০১৪,২০১০; বঃ২০১৪; যঃ২০১২; দিঃ২০১১
উত্তরঃ \(\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\)
[ ঢাঃ২০১০; কুঃ২০১০; সিঃ২০১৪,২০১০; বঃ২০১৪; যঃ২০১২; দিঃ২০১১
সমাধানঃ
ধরি,
\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(\int{\frac{dx}{1+\tan{x}}}\)\(\sin{x}+\cos{x}=t\)
\(\Rightarrow \frac{d}{dx}(\sin{x}+\cos{x})=\frac{d}{dx}(t)\)
\(\Rightarrow \cos{x}-\sin{x}=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})=\frac{dt}{dx}\)
\(\Rightarrow -(\sin{x}-\cos{x})dx=dt\)
\(\therefore (\sin{x}-\cos{x})dx=-dt\)
\(=\int{\frac{dx}{1+\frac{\sin{x}}{\cos{x}}}}\)
\(=\int{\frac{dx}{\frac{\cos{x}+\sin{x}}{\cos{x}}}}\)
\(=\int{\frac{\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\int{\frac{2\cos{x}}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\frac{(\sin{x}+\cos{x})-(\sin{x}-\cos{x})}{2(\sin{x}+\cos{x})}dx}\)
\(=\int{\left\{\frac{\sin{x}+\cos{x}}{2(\sin{x}+\cos{x})}-\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\int{\left\{\frac{1}{2}-\frac{\sin{x}-\cos{x}}{2(\sin{x}+\cos{x})}\right\}dx}\)
\(=\frac{1}{2}\int{\left(1-\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}\right)dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{\sin{x}-\cos{x}}{\sin{x}+\cos{x}}dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{1}{\sin{x}+\cos{x}}(\sin{x}-\cos{x})dx}\)
\(=\frac{1}{2}\int{dx}-\frac{1}{2}\int{\frac{1}{t}(-dt)}\)
\(=\frac{1}{2}\int{dx}+\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=\frac{1}{2}x+\frac{1}{2}\ln{|t|}+c\) ➜ \(\because \int{dx}=x, \int{\frac{1}{x}dx}=\ln{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2}x+\frac{1}{2}\ln{|\sin{x}+\cos{x}|}+c\) ➜ \(\because t=\sin{x}+\cos{x}\)
\(Q.3.(xxxiv)\) \(\int{\frac{\cos{x}dx}{(5-2\sin{x})^2}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\frac{1}{2(5-2\sin{x})}+c\)
উত্তরঃ \(\frac{1}{2(5-2\sin{x})}+c\)
সমাধানঃ
ধরি,
\(5-2\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(5-2\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\cos{x}dx=dt\)
\(\therefore \cos{x}dx=-\frac{1}{2}dt\)
\(\int{\frac{\cos{x}dx}{(5-2\sin{x})^2}}\)\(5-2\sin{x}=t\)
\(\Rightarrow \frac{d}{dx}(5-2\sin{x})=\frac{d}{dx}(t)\)
\(\Rightarrow 0-2\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\cos{x}=\frac{dt}{dx}\)
\(\Rightarrow -2\cos{x}dx=dt\)
\(\therefore \cos{x}dx=-\frac{1}{2}dt\)
\(=\int{\frac{1}{(5-2\sin{x})^2}\cos{x}dx}\)
\(=\int{\frac{1}{t^2}\left(-\frac{1}{2}dt\right)}\)
\(=-\frac{1}{2}\int{\frac{1}{t^2}dt}\)
\(=-\frac{1}{2}\left(-\frac{1}{t}\right)+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\frac{1}{2t}+c\)
\(=\frac{1}{2(5-2\sin{x})}+c\) ➜ \(\because t=5-2\sin{x}\)
\(Q.3.(xxxv)\) \(\int{\frac{\sec^2{(\cot^{-1}{x})}}{1+x^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{x}+c\)
[ বুয়েটঃ২০১০-২০১১ ]
উত্তরঃ \(-\frac{1}{x}+c\)
[ বুয়েটঃ২০১০-২০১১ ]
সমাধানঃ
ধরি,
\(\cot^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=-dt\)
\(\int{\frac{\sec^2{(\cot^{-1}{x})}}{1+x^2}dx}\)\(\cot^{-1}{x}=t\)
\(\Rightarrow \frac{d}{dx}(\cot^{-1}{x})=\frac{d}{dx}(t)\)
\(\Rightarrow -\frac{1}{1+x^2}=\frac{dt}{dx}\)
\(\Rightarrow -\frac{1}{1+x^2}dx=dt\)
\(\therefore \frac{1}{1+x^2}dx=-dt\)
\(=\int{\sec^2{(\cot^{-1}{x})}\frac{1}{1+x^2}dx}\)
\(=\int{\sec^2{t}(-dt)}\)
\(=-\int{\sec^2{t}dt}\)
\(=-\tan{t}+c\) ➜ \(\because \int{\sec^2{x}dx}=\tan{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\tan{\left(\cot^{-1}{x}\right)}+c\) ➜ \(\because t=\cot^{-1}{x}\)
\(=-\tan{\left(\tan^{-1}{\frac{1}{x}}\right)}+c\) ➜ \(\because \cot^{-1}{x}=\tan^{-1}{\frac{1}{x}}\)
\(=-\frac{1}{x}+c\)
\(Q.3.(xxxvi)\) \(\int{\frac{dx}{(e^x-1)^2}dx}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(-\frac{1}{1-e^{-x}}-\ln{|1-e^{-x}|}+c\)
উত্তরঃ \(-\frac{1}{1-e^{-x}}-\ln{|1-e^{-x}|}+c\)
সমাধানঃ
ধরি,
\(1-e^{-x}=t\)
\(\Rightarrow -e^{-x}=t-1\)
\(\Rightarrow e^{-x}=1-t\)
\(\Rightarrow \frac{d}{dx}(e^{-x})=\frac{d}{dx}(1-t)\)
\(\Rightarrow e^{-x}(-1)=0-\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=-\frac{dt}{dx}\)
\(\Rightarrow e^{-x}=\frac{dt}{dx}\)
\(\therefore e^{-x}dx=dt\)
\(\int{\frac{dx}{(e^x-1)^2}}\)\(1-e^{-x}=t\)
\(\Rightarrow -e^{-x}=t-1\)
\(\Rightarrow e^{-x}=1-t\)
\(\Rightarrow \frac{d}{dx}(e^{-x})=\frac{d}{dx}(1-t)\)
\(\Rightarrow e^{-x}(-1)=0-\frac{dt}{dx}\)
\(\Rightarrow -e^{-x}=-\frac{dt}{dx}\)
\(\Rightarrow e^{-x}=\frac{dt}{dx}\)
\(\therefore e^{-x}dx=dt\)
\(=\int{\frac{dx}{e^{2x}\left(1-\frac{1}{e^x}\right)^2}}\)
\(=\int{\frac{e^{-2x}dx}{(1-e^{-x})^2}}\)
\(=\int{\frac{e^{-x}.e^{-x}dx}{(1-e^{-x})^2}}\)
\(=\int{\frac{(1-t).dt}{t^2}}\)
\(=\int{\left(\frac{1}{t^2}-\frac{t}{t^2}\right)dt}\)
\(=\int{\left(\frac{1}{t^2}-\frac{1}{t}\right)dt}\)
\(=\int{\frac{1}{t^2}dt}-\int{\frac{1}{t}dt}\)
\(=-\frac{1}{t}-\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x^2}dx}=-\frac{1}{x}, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=-\frac{1}{1-e^{-x}}-\ln{|1-e^{-x}|}+c\) ➜ \(\because t=1-e^{-x}\)
\(Q.3.(xxxvii)\) \(\int{\frac{dx}{1+e^{-x}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(\ln{(e^x+1)}+c\)
উত্তরঃ \(\ln{(e^x+1)}+c\)
সমাধানঃ
ধরি,
\(e^x+1=t\)
\(\Rightarrow \frac{d}{dx}(e^x+1)=\frac{d}{dx}(1-t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(\int{\frac{dx}{1+e^{-x}}}\)\(e^x+1=t\)
\(\Rightarrow \frac{d}{dx}(e^x+1)=\frac{d}{dx}(1-t)\)
\(\Rightarrow e^x=\frac{dt}{dx}\)
\(\therefore e^xdx=dt\)
\(=\int{\frac{e^xdx}{e^x(1+e^{-x})}}\) ➜ লব ও হরের সহিত \(e^x\) গুণ করে।
\(=\int{\frac{e^xdx}{e^x+e^{-x+x}}}\)
\(=\int{\frac{e^xdx}{e^x+e^{0}}}\)
\(=\int{\frac{e^xdx}{e^x+1}}\)
\(=\int{\frac{1}{e^x+1}\times{e^x}dx}\)
\(=\int{\frac{1}{t}dt}\)
\(=\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\ln{(e^x+1)}+c\) ➜ \(\because t=e^x+1\)
\(Q.3.(xxxviii)\) \(\int{\frac{dx}{\sqrt{x}+x}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\ln{(1+\sqrt{x})}+c\)
উত্তরঃ \(2\ln{(1+\sqrt{x})}+c\)
সমাধানঃ
ধরি,
\(1+\sqrt{x}=t\)
\(\Rightarrow \sqrt{x}=t-1\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t-1)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}-0\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(\int{\frac{dx}{\sqrt{x}+x}}\)\(1+\sqrt{x}=t\)
\(\Rightarrow \sqrt{x}=t-1\)
\(\Rightarrow \frac{d}{dx}(\sqrt{x})=\frac{d}{dx}(t-1)\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}-0\)
\(\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}\)
\(\therefore \frac{1}{\sqrt{x}}dx=2dt\)
\(=\int{\frac{dx}{\sqrt{x}+\sqrt{x}.\sqrt{x}}}\)
\(=\int{\frac{dx}{\sqrt{x}(1+\sqrt{x})}}\)
\(=\int{\frac{1}{1+\sqrt{x}}.\frac{dx}{\sqrt{x}}}\)
\(=\int{\frac{1}{t}.2dt}\)
\(=2\int{\frac{1}{t}dt}\)
\(=2\ln{|t|}+c\) ➜ \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2\ln{(1+\sqrt{x})}+c\) ➜ \(\because t=1+\sqrt{x}\)
\(Q.3.(xxxix)\) \(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\) এর যোজিত ফল নির্ণয় কর।
উত্তরঃ \(2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)
উত্তরঃ \(2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)
সমাধানঃ
ধরি,
\(x^{\frac{1}{4}}=t\)
\(\Rightarrow x=t^4\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^4)\)
\(\Rightarrow 1=4t^3\frac{dt}{dx}\)
\(\therefore dx=4t^3dt\)
\(\int{\frac{dx}{x^{\frac{1}{2}}-x^{\frac{1}{4}}}}\)\(x^{\frac{1}{4}}=t\)
\(\Rightarrow x=t^4\)
\(\Rightarrow \frac{d}{dx}(x)=\frac{d}{dx}(t^4)\)
\(\Rightarrow 1=4t^3\frac{dt}{dx}\)
\(\therefore dx=4t^3dt\)
\(=\int{\frac{dx}{(x^{\frac{1}{4}})^2-x^{\frac{1}{4}}}}\)
\(=\int{\frac{4t^3dt}{t^2-t}}\)
\(=4\int{\frac{t^3dt}{t(t-1)}}\)
\(=4\int{\frac{t^2}{t-1}dt}\)
\(=4\int{\frac{t^2-1+1}{t-1}dt}\)
\(=4\int{\frac{(t+1)(t-1)+1}{t-1}dt}\)
\(=4\int{\left\{\frac{(t+1)(t-1)}{t-1}+\frac{1}{t-1}\right\}dt}\)
\(=4\int{\left\{t+1+\frac{1}{t-1}\right\}dt}\)
\(=4\int{tdt}+4\int{dt}+4\int{\frac{1}{t-1}dt}\)
\(=4\frac{t^2}{2}+4t+4\ln{|t-1|}+c\) ➜ \(\because \int{x^ndx}=\frac{x^{n+1}}{n+1}, \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=2t^2+4t+4\ln{|t-1|}+c\)
\(=2(x^{\frac{1}{4}})^2+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\) ➜ \(\because t=x^{\frac{1}{4}}\)
\(=2x^{\frac{1}{2}}+4x^{\frac{1}{4}}+4\ln{|x^{\frac{1}{4}}-1|}+c\)
\(=2\sqrt{x}+4\sqrt[4]{x}+4\ln{|\sqrt[4]{x}-1|}+c\)
