Class 12: Indefinite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int e^x(\cos x-\sin x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x(\cos x-\sin x)\,dx$
$\displaystyle \text{Let } t=e^x\cos x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\cos x+e^x(-\sin x)$
$\displaystyle \Rightarrow dt=e^x(\cos x-\sin x)\,dx$
$\displaystyle \therefore \int e^x(\cos x-\sin x)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\cos x+C$

$\displaystyle \textbf{Question 2: }~\int e^x\!\left(\frac{1}{x^2}-\frac{2}{x^3}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right)\,dx$
$\displaystyle \text{Also let } t=\frac{e^x}{x^2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{x^2}+e^x\left(-\frac{2}{x^3}\right)$
$\displaystyle \Rightarrow dt=e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right)\,dx$
$\displaystyle \therefore \int e^x\left(\frac{1}{x^2}-\frac{2}{x^3}\right)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =\frac{e^x}{x^2}+C$

$\displaystyle \textbf{Question 3: }~\int e^x\!\left(\frac{1+\sin x}{1+\cos x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)\,dx$
$\displaystyle =\int e^x\left(\frac{1}{1+\cos x}+\frac{\sin x}{1+\cos x}\right)\,dx$
$\displaystyle =\int e^x\left(\frac{1}{2\cos^2\frac{x}{2}}+\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\right)\,dx$
$\displaystyle =\int e^x\left(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2}\right)\,dx$
$\displaystyle \text{Putting } t=e^x\tan\frac{x}{2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\tan\frac{x}{2}+e^x\cdot\frac{1}{2}\sec^2\frac{x}{2}$
$\displaystyle \Rightarrow dt=e^x\left(\tan\frac{x}{2}+\frac{1}{2}\sec^2\frac{x}{2}\right)\,dx$
$\displaystyle \therefore \int e^x\left(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2}\right)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\tan\frac{x}{2}+C$

$\displaystyle \textbf{Question 4: }~\int e^x(\cot x-\mathrm{cosec}^2 x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x(\cot x-\mathrm{cosec}^2 x)\,dx$
$\displaystyle \text{Here } f(x)=\cot x$
$\displaystyle \text{Let } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\cot x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\cot x+e^x(-\mathrm{cosec}^2 x)$
$\displaystyle \Rightarrow dt=e^x(\cot x-\mathrm{cosec}^2 x)\,dx$
$\displaystyle \therefore \int e^x(\cot x-\mathrm{cosec}^2 x)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\cot x+C$

$\displaystyle \textbf{Question 5: }~\int e^x\!\left(\frac{x-1}{2x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\frac{x-1}{2x^2}\right)\,dx$
$\displaystyle =\frac{1}{2}\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)\,dx$
$\displaystyle \text{Here } f(x)=\frac{1}{x}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=\frac{e^x}{x}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{x}+e^x\left(-\frac{1}{x^2}\right)$
$\displaystyle \Rightarrow dt=e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)\,dx$
$\displaystyle \therefore I=\frac{1}{2}\int dt$
$\displaystyle =\frac{t}{2}+C$
$\displaystyle =\frac{e^x}{2x}+C$

$\displaystyle \textbf{Question 6: }~\int e^x\sec x(1+\tan x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\sec x(1+\tan x)\,dx$
$\displaystyle =\int e^x(\sec x+\sec x\tan x)\,dx$
$\displaystyle \text{Here } f(x)=\sec x$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\sec x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\sec x+e^x\sec x\tan x$
$\displaystyle \Rightarrow dt=e^x(\sec x+\sec x\tan x)\,dx$
$\displaystyle \therefore \int e^x(\sec x+\sec x\tan x)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\sec x+C$

$\displaystyle \textbf{Question 7: }~\int e^x(\tan x-\log\cos x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x(\tan x-\log\cos x)\,dx$
$\displaystyle \text{Here } f(x)=-\log\cos x$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=-e^x\log\cos x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=-\left[e^x\log(\cos x)+e^x\frac{1}{\cos x}(-\sin x)\right]$
$\displaystyle \Rightarrow dt=\left[-e^x\log(\cos x)+e^x\tan x\right]dx$
$\displaystyle \Rightarrow dt=e^x(\tan x-\log\cos x)\,dx$
$\displaystyle \therefore \int e^x(\tan x-\log\cos x)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =-e^x\log(\cos x)+C$
$\displaystyle =e^x\log(\sec x)+C$

$\displaystyle \textbf{Question 8: }~\int e^x[\sec x+\log(\sec x+\tan x)]\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x[\sec x+\log(\sec x+\tan x)]\,dx$
$\displaystyle \text{Here } f(x)=\log(\sec x+\tan x)$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\log(\sec x+\tan x)$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\log(\sec x+\tan x)+e^x\frac{1}{\sec x+\tan x}(\sec x\tan x+\sec^2 x)$
$\displaystyle \Rightarrow \frac{dt}{dx}=e^x[\log(\sec x+\tan x)+\sec x]$
$\displaystyle \Rightarrow dt=e^x[\sec x+\log(\sec x+\tan x)]\,dx$
$\displaystyle \therefore \int e^x[\sec x+\log(\sec x+\tan x)]\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\log(\sec x+\tan x)+C$

$\displaystyle \textbf{Question 9: }~\int e^x(\cot x+\log\sin x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x(\cot x+\log\sin x)\,dx$
$\displaystyle \text{Here } f(x)=\log\sin x$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\log\sin x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\log(\sin x)+e^x\frac{1}{\sin x}\cos x$
$\displaystyle \Rightarrow \frac{dt}{dx}=e^x(\log\sin x+\cot x)$
$\displaystyle \Rightarrow dt=e^x(\cot x+\log\sin x)\,dx$
$\displaystyle \therefore \int e^x(\cot x+\log\sin x)\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\log\sin x+C$

$\displaystyle \textbf{Question 10: }~\int e^x\frac{x-1}{(x+1)^3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left[\frac{x-1}{(x+1)^3}\right]dx$
$\displaystyle =\int e^x\left[\frac{x+1-2}{(x+1)^3}\right]dx$
$\displaystyle =\int e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right]dx$
$\displaystyle \text{Here, } f(x)=\frac{1}{(x+1)^2}$
$\displaystyle \Rightarrow f'(x)=-\frac{2}{(x+1)^3}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=\frac{e^x}{(x+1)^2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{(x+1)^2}+e^x\left(-\frac{2}{(x+1)^3}\right)$
$\displaystyle \Rightarrow dt=e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right]dx$
$\displaystyle \therefore \int e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right]dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =\frac{e^x}{(x+1)^2}+C$

$\displaystyle \textbf{Question 11: }~\int e^x\!\left(\frac{\sin 4x-4}{1-\cos 4x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left[\frac{\sin 4x-4}{1-\cos 4x}\right]dx$
$\displaystyle =\int e^x\left[\frac{2\sin 2x\cos 2x}{2\sin^2 2x}-\frac{4}{2\sin^2 2x}\right]dx$
$\displaystyle =\int e^x\left[\cot 2x-2\,\mathrm{cosec}^2 2x\right]dx$
$\displaystyle \text{Here, } f(x)=\cot 2x$
$\displaystyle \Rightarrow f'(x)=-2\,\mathrm{cosec}^2 2x$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\cot 2x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\cot 2x+e^x\left(-2\,\mathrm{cosec}^2 2x\right)$
$\displaystyle \Rightarrow dt=e^x\left[\cot 2x-2\,\mathrm{cosec}^2 2x\right]dx$
$\displaystyle \therefore \int e^x\left[\cot 2x-2\,\mathrm{cosec}^2 2x\right]dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\cot 2x+C$

$\displaystyle \textbf{Question 12: }~\int \frac{2-x}{(1-x)^2}e^x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left[\frac{(1-x)^2}{(1+x^2)^2}\right]dx$
$\displaystyle =\int e^x\left[\frac{1+x^2-2x}{(1+x^2)^2}\right]dx$
$\displaystyle =\int e^x\left[\frac{1}{1+x^2}-\frac{2x}{(1+x^2)^2}\right]dx$
$\displaystyle \text{Here, } f(x)=\frac{1}{1+x^2}$
$\displaystyle \Rightarrow f'(x)=-\frac{2x}{(1+x^2)^2}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=\frac{e^x}{1+x^2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{1+x^2}+e^x\left(-\frac{2x}{(1+x^2)^2}\right)$
$\displaystyle \Rightarrow dt=e^x\left[\frac{1}{1+x^2}-\frac{2x}{(1+x^2)^2}\right]dx$
$\displaystyle \therefore \int e^x\left[\frac{1}{1+x^2}-\frac{2x}{(1+x^2)^2}\right]dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =\frac{e^x}{1+x^2}+C$

$\displaystyle \textbf{Question 13: }~\int e^x\frac{1+x}{(2+x)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left[\frac{1+x}{(2+x)^2}\right]dx$
$\displaystyle =\int e^x\left[\frac{2+x-1}{(2+x)^2}\right]dx$
$\displaystyle =\int e^x\left[\frac{1}{2+x}-\frac{1}{(2+x)^2}\right]dx$
$\displaystyle \text{Here, } f(x)=\frac{1}{2+x}$
$\displaystyle \Rightarrow f'(x)=-\frac{1}{(2+x)^2}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=\frac{e^x}{2+x}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{2+x}+e^x\left(-\frac{1}{(2+x)^2}\right)$
$\displaystyle \Rightarrow dt=e^x\left[\frac{1}{2+x}-\frac{1}{(2+x)^2}\right]dx$
$\displaystyle \therefore \int e^x\left[\frac{1}{2+x}-\frac{1}{(2+x)^2}\right]dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =\frac{e^x}{2+x}+C$

$\displaystyle \textbf{Question 14: }~\int \frac{\sqrt{1-\sin x}}{1+\cos x}\,e^{-x/2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{\sqrt{1-\sin x}}{1+\cos x}\,e^{-x/2}\,dx$
$\displaystyle =\int \left(\frac{\sqrt{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}}{2\cos^2\frac{x}{2}}\right)e^{-x/2}\,dx$
$\displaystyle =\int \frac{\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}{2\cos^2\frac{x}{2}}\,e^{-x/2}\,dx$
$\displaystyle =\int \frac{\sin\frac{x}{2}-\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\,e^{-x/2}\,dx$
$\displaystyle =\int \left[\frac{1}{2}\sec\frac{x}{2}\tan\frac{x}{2}-\frac{1}{2}\sec\frac{x}{2}\right]e^{-x/2}\,dx$
$\displaystyle =\frac{1}{2}\int \left(\sec\frac{x}{2}\tan\frac{x}{2}-\sec\frac{x}{2}\right)e^{-x/2}\,dx$
$\displaystyle \text{Let } t=e^{-x/2}\sec\frac{x}{2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^{-x/2}\frac{\sec\frac{x}{2}\tan\frac{x}{2}}{2}+\sec\frac{x}{2}\,e^{-x/2}\left(-\frac{1}{2}\right)$
$\displaystyle \Rightarrow dt=\frac{e^{-x/2}}{2}\left[\sec\frac{x}{2}\tan\frac{x}{2}-\sec\frac{x}{2}\right]dx$
$\displaystyle \therefore \frac{1}{2}\int \left(\sec\frac{x}{2}\tan\frac{x}{2}-\sec\frac{x}{2}\right)e^{-x/2}\,dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^{-x/2}\sec\frac{x}{2}+C$

$\displaystyle \textbf{Question 15: }~\int e^x\!\left(\log x+\frac{1}{x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\log x+\frac{1}{x}\right)dx$
$\displaystyle \text{Here, } f(x)=\log x$
$\displaystyle \Rightarrow f'(x)=\frac{1}{x}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\log x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\log x+e^x\frac{1}{x}$
$\displaystyle \Rightarrow dt=e^x\left(\log x+\frac{1}{x}\right)dx$
$\displaystyle \therefore \int e^x\left(\log x+\frac{1}{x}\right)dx=\int dt$
$\displaystyle \Rightarrow I=t+C$
$\displaystyle =e^x\log x+C$

$\displaystyle \textbf{Question 16: }~\int e^x\!\left(\log x+\frac{1}{x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \left(\log x+\frac{1}{x^2}\right)e^x\,dx$
$\displaystyle =\int e^x\left(\log x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^2}\right)dx$
$\displaystyle =\int e^x\left(\log x+\frac{1}{x}\right)dx+\int e^x\left(-\frac{1}{x}+\frac{1}{x^2}\right)dx$
$\displaystyle \text{Let } t=e^x\log x$
$\displaystyle \Rightarrow dt=\left(e^x\log x+\frac{e^x}{x}\right)dx$
$\displaystyle \text{Let } p=-\frac{e^x}{x}$
$\displaystyle \Rightarrow dp=\left(-\frac{e^x}{x}+\frac{e^x}{x^2}\right)dx$
$\displaystyle \therefore I=\int dt+\int dp$
$\displaystyle =t+p+C$
$\displaystyle =e^x\log x-\frac{e^x}{x}+C$
$\displaystyle =e^x\left(\log x-\frac{1}{x}\right)+C$

$\displaystyle \textbf{Question 17: }~\int \frac{e^x}{x}\{x(\log x)^2+2\log x\}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{e^x}{x}\left[x(\log x)^2+2\log x\right]dx$
$\displaystyle =\int e^x\left[(\log x)^2+\frac{2\log x}{x}\right]dx$
$\displaystyle \text{Here, } f(x)=(\log x)^2$
$\displaystyle \Rightarrow f'(x)=\frac{2\log x}{x}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x(\log x)^2$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x(\log x)^2+e^x\frac{2\log x}{x}$
$\displaystyle \Rightarrow dt=e^x\left[(\log x)^2+\frac{2\log x}{x}\right]dx$
$\displaystyle \therefore I=\int dt$
$\displaystyle =t+C$
$\displaystyle =e^x(\log x)^2+C$

$\displaystyle \textbf{Question 18: }~\int e^x\frac{\sqrt{1-x^2}\sin^{-1}x+1}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left[\frac{\sqrt{1-x^2}\sin^{-1}x+1}{\sqrt{1-x^2}}\right]dx$
$\displaystyle =\int e^x\left[\sin^{-1}x+\frac{1}{\sqrt{1-x^2}}\right]dx$
$\displaystyle \text{Here, } f(x)=\sin^{-1}x$
$\displaystyle \Rightarrow f'(x)=\frac{1}{\sqrt{1-x^2}}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\sin^{-1}x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\sin^{-1}x+e^x\frac{1}{\sqrt{1-x^2}}$
$\displaystyle \Rightarrow dt=e^x\left[\sin^{-1}x+\frac{1}{\sqrt{1-x^2}}\right]dx$
$\displaystyle \therefore I=\int dt$
$\displaystyle =t+C$
$\displaystyle =e^x\sin^{-1}x+C$

$\displaystyle \textbf{Question 19: }~\int e^{2x}(-\sin x+2\cos x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{2x}(-\sin x+2\cos x)\,dx$
$\displaystyle \text{Put } t=e^{2x}\cos x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=2e^{2x}\cos x+e^{2x}(-\sin x)$
$\displaystyle \Rightarrow dt=e^{2x}(-\sin x+2\cos x)\,dx$
$\displaystyle \therefore I=\int dt$
$\displaystyle =t+C$
$\displaystyle =e^{2x}\cos x+C$

$\displaystyle \textbf{Question 20: }~\int e^x\!\left(\tan^{-1}x+\frac{1}{1+x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }F(x)=e^{x}\tan^{-1}x.$
$\displaystyle \frac{d}{dx}\big(F(x)\big)=\frac{d}{dx}\left(e^{x}\tan^{-1}x\right).$
$\displaystyle \frac{d}{dx}\left(e^{x}\tan^{-1}x\right)=\left(\frac{d}{dx}e^{x}\right)\tan^{-1}x+e^{x}\left(\frac{d}{dx}\tan^{-1}x\right).$
$\displaystyle =e^{x}\tan^{-1}x+e^{x}\cdot\frac{1}{1+x^{2}}.$
$\displaystyle =e^{x}\left(\tan^{-1}x+\frac{1}{1+x^{2}}\right).$
$\displaystyle \text{Hence, }\int e^{x}\left(\tan^{-1}x+\frac{1}{1+x^{2}}\right)\,dx=e^{x}\tan^{-1}x+C.$

$\displaystyle \textbf{Question 21: }~\int e^x\!\left(\frac{\sin x\cos x-1}{\sin^2 x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\frac{\sin x\cos x-1}{\sin^2 x}\right)dx$
$\displaystyle =\int e^x\left(\cot x-\mathrm{cosec}^2 x\right)dx$
$\displaystyle \text{Here, } f(x)=\cot x$
$\displaystyle \Rightarrow f'(x)=-\mathrm{cosec}^2 x$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=e^x\cot x$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle dt=e^x(\cot x-\mathrm{cosec}^2 x)\,dx$
$\displaystyle \therefore I=\int dt$
$\displaystyle =t+C$
$\displaystyle =e^x\cot x+C$

$\displaystyle \textbf{Question 22: }~\int \{\tan(\log x)+\sec^2(\log x)\}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int\left[\tan(\log x)+\sec^2(\log x)\right]dx$
$\displaystyle \text{Put } \log x=t$
$\displaystyle \Rightarrow x=e^t$
$\displaystyle \Rightarrow dx=e^t\,dt$
$\displaystyle \therefore I=\int (\tan t+\sec^2 t)e^t\,dt$
$\displaystyle \text{Here, } f(t)=\tan t$
$\displaystyle \Rightarrow f'(t)=\sec^2 t$
$\displaystyle \text{Let } p=e^t f(t)$
$\displaystyle \Rightarrow p=e^t\tan t$
$\displaystyle \text{Differentiating both sides w.r.t. } t$
$\displaystyle \frac{dp}{dt}=e^t(\tan t+\sec^2 t)$
$\displaystyle \Rightarrow e^t(\tan t+\sec^2 t)\,dt=dp$
$\displaystyle \therefore I=\int dp$
$\displaystyle =p+C$
$\displaystyle =e^t\tan t+C$
$\displaystyle =x\tan(\log x)+C$

$\displaystyle \textbf{Question 23: }~\int e^x\frac{x-4}{(x-2)^3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^x\left(\frac{x-4}{(x-2)^3}\right)dx$
$\displaystyle =\int e^x\left(\frac{x-2-2}{(x-2)^3}\right)dx$
$\displaystyle =\int e^x\left(\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right)dx$
$\displaystyle \text{Here, } f(x)=\frac{1}{(x-2)^2}$
$\displaystyle \Rightarrow f'(x)=-\frac{2}{(x-2)^3}$
$\displaystyle \text{Put } t=e^x f(x)$
$\displaystyle \Rightarrow t=\frac{e^x}{(x-2)^2}$
$\displaystyle \text{Differentiating both sides w.r.t. } x$
$\displaystyle \frac{dt}{dx}=e^x\frac{1}{(x-2)^2}+e^x\left(-\frac{2}{(x-2)^3}\right)$
$\displaystyle \Rightarrow dt=e^x\left(\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right)dx$
$\displaystyle \therefore I=\int dt$
$\displaystyle =t+C$
$\displaystyle =\frac{e^x}{(x-2)^2}+C$

$\displaystyle \textbf{Question 24: }~\int e^{2x}\!\left(\frac{1-\sin 2x}{1-\cos 2x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle I=\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx$
$\displaystyle =\int e^{2x}\left(\frac{1-2\sin x\cos x}{2\sin^2 x}\right)dx$
$\displaystyle \text{Put } t=2x$
$\displaystyle \Rightarrow dt=2\,dx$
$\displaystyle \therefore I=\frac{1}{2}\int e^t\left(\frac{1-2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}\right)dt$
$\displaystyle =\frac{1}{4}\int e^t\left(\frac{1-2\sin\frac{t}{2}\cos\frac{t}{2}}{\sin^2\frac{t}{2}}\right)dt$
$\displaystyle =\frac{1}{4}\int e^t\left(\frac{1}{\sin^2\frac{t}{2}}-\frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{\sin^2\frac{t}{2}}\right)dt$
$\displaystyle =\frac{1}{4}\int e^t\left(\mathrm{cosec}^2\frac{t}{2}-2\cot\frac{t}{2}\right)dt$
$\displaystyle =-\frac{1}{4}\int e^t\left(2\cot\frac{t}{2}-\mathrm{cosec}^2\frac{t}{2}\right)dt$
$\displaystyle \text{Consider } f(x)=2\cot\frac{t}{2}$
$\displaystyle \Rightarrow f'(x)=-\mathrm{cosec}^2\frac{t}{2}$
$\displaystyle \text{Thus, the given integral is of the form } e^x[f(x)+f'(x)]$
$\displaystyle \therefore I=-\frac{1}{4}\left(2\cot\frac{t}{2}\right)e^t+C$
$\displaystyle =-\frac{1}{4}\left(2\cot x\right)e^{2x}+C$
$\displaystyle =-\frac{1}{2}(\cot x)e^{2x}+C$