$\int \frac{d x}{e^x+e^{-x}} \quad [e^{-x}=\frac{1}{e^x}]$

Solution:

$= \int \frac{e^x}{e^{2 x}+1} d x \quad [e^x=t] , [e^x d x=d t]$

$= \int \frac{d t}{t^2+1}=\tan ^{-1} t+c$

$= \tan ^{-1} e^x+c$

Solution:

$\quad [\cos 2 x=\cos ^2 x-\sin ^2 x]$

$=\int \frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2} d x$

$=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$

$=\int \frac{d t}{t}=\operatorname{log}|t|+c$

$=\log |\sin x+\cos x|+c$

$I=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$

Solution:

$[A^2-B^2=(A-B).(A+B)]$

Numerator =

$(\sin ^4 x)^2-(\cos ^4 x)^2$

$=\left(\sin ^4 x-\cos ^4 x\right)\left(\sin ^4 x+\cos ^4 x\right)$

$=(1-2 \cos ^2 x)(1+\cos ^4 x-2 \cos ^2 x+\cos ^4 x)$

$=(1-2 \cos ^2 x)(1+2 \cos ^4 x-2 \cos ^2 x)$

$Denominator =1-2 \sin ^2 x \cos ^2 x$

$=1-2\left(1-\cos ^2 x\right) \cos ^2 x$

$=\left(1-2 \cos ^2 x+2 \cos ^4 x\right)$

$I=\int\left(1-2 \cos ^2 x\right) d x$

$[\cos 2 x \ =2 \cos ^2 x-1] [\cos ^2 x=\frac{1+\cos 2 x}{2}]$

= $x-2 [\int \frac{1+\cos 2 x}{2} d x]$

= $x-x-\frac{\sin 2 x}{2}+c$

= $-\sin x \cos x+c$

Solution:

$I_1=\int \frac{\sec ^2 x d x}{a \sec ^2 x+b \tan ^2 x}$

$[1+\tan ^2 x =\sec ^2 x]$

$=\int \frac{\sec ^2 x d x}{a+(a+b) \tan ^2 x}$

$[\tan x=t]$

$[\sec ^2 x d x=d t]$

$=\int \frac{d t}{a+(a+b) t^2}$

$=\frac{1}{(a+b)} \int \frac{d t}{\left(\frac{a}{a+b}\right)+t^2}$

$=\frac{1}{a+b} \int \frac{d t}{\alpha^2+t^2} \quad [\alpha=\sqrt{a / a+b}]$

$=\frac{1}{a+b} \times \frac{1}{\alpha} \cdot \tan ^{-1} \frac{t}{\alpha}+c$

$=\frac{1}{a+b} \times \frac{\sqrt{a+b}}{\sqrt{a}} \tan ^{-1} \frac{t \sqrt{a+b}}{\sqrt{a}}+c$

$=\frac{1}{\sqrt{a(a+b)}} \cdot \tan ^{-1} \frac{t \sqrt{a+b}}{\sqrt{a}}+c$

$I=\int \underset{II}1.\underset{I}\sin ^{-1} \frac{2 x}{1+x^2} d x$

Solution:

$[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^2}}]$

$=\sin ^{-1} \frac{2 x}{1+x^2} \cdot x-\int \frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^2}\right)^2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^2}\right) \times x d x$

$\left[\frac{d}{d x}(\frac{2 x}{1+x^2}) = \frac{\left(1+x^2\right) \times 2-2 {x} \times 2 {x}}{\left(1+x^2\right)^2} =\frac{2\left(1-x^2\right)}{\left(1+x^2\right)^2}\right]$

$\sqrt{1-\left(\frac{2 x}{1+x^2}\right)^2} = \sqrt{1-\frac{4 x^2}{\left(1+x^2\right)^2}}=\sqrt{\frac{\left(1+x^2\right)^2-4 x^2}{\left(1+x^2\right)^2}}$

$=\sqrt{\frac{\left(1-x^2\right)^2}{\left(1+x^2\right)^2}}=\frac{\left(1-x^2\right)}{\left(1+x^2\right)}$

$\Rightarrow I =x \sin ^{-1} \frac{2 x}{1+x^2}-\int \frac{x}{\frac{1-x^2}{(1+x^2)}} \times \frac{2(1-x^2)}{(1+x^2)^2} d x$

$=x \sin ^{-1} \frac{2 x}{1+x^2}-\int \frac{2x}{1+x^2} d x$

$I=\int \cos ^{-1}\left(2 x^2-1\right) d x$

Solution:

$x =\cos \theta \Rightarrow d x=-\sin \theta d \theta$

$I =\int \cos ^{-1}\left(2 \cos ^2 \theta-1\right) (-\sin \theta) d \theta$

$=\int 2 \cdot \theta(-\sin \theta) d \theta$

$=-2 \int \theta \sin \theta d \theta$

$=-2\left[\theta(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right]$

$=+2 \theta \cos \theta-2 \sin \theta+c$

$=2 x \cos ^{-1} x-2 \sqrt{1-x^2}+c$

$I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

Solution:

$[ \cos 2 \theta=1-2 \sin ^2 \theta =2 \cos ^2 \theta-1 ]$

[$x=\cos 2 \theta$

$d x=-2 \sin 2 \theta d \theta$]

$\frac{1-x}{1+x}=\frac{1-\left(1-2 \sin ^2 \theta\right)}{1+\left(2 \cos ^2 \theta-1\right)}=\frac{\sin ^2 \theta}{\cos ^2 \theta}$

$I=\int \tan ^{-1}(\tan \theta) \times(-2 \sin 2 \theta) d \theta$

$=-2 \int \underset{I}{\theta} \underset{II}{\sin 2 \theta} d \theta$

$\theta \rightarrow x$

$\quad I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$

Solution:

[$\sqrt{x}=\cos 2 \theta$

$\therefore \left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\frac{\sin ^2 \theta}{\cos ^2 \theta},$ ] [$\frac{1}{2 \sqrt{x}} d x$

$=2 \sin 2 \theta d \theta$

$d x=-4 \cos 2 \theta \sin 2 \theta d \theta$

$=-2 \sin 4 \theta d \theta$ ]

$I =\int \tan \theta \times(-2 \sin 4 \theta) d \theta$

$=-2 \int \frac{\sin \theta}{\cos \theta} \times 2 \sin 2 \theta \cos 2 \theta d \theta$

$=-4 \int \frac{\sin \theta}{\cos \theta} \times 2 \sin \theta \cos \theta \cdot \cos 2 \theta d \theta$

$I=\int e^x({f(x)}+f^{\prime}(x)) d x$

Solution:

$=\int \underset{I_1}{e^x f(x)} d x+\int \underset{I_2}{e^x f^{\prime}}(x) d x$

$I_1 =\int \underset{II}{e^x} \underset{I}{f(x)} d x$

$=f(x) e^x- \underset{I_2}{\int f^{\prime}(x) e^x} d x+c$

$I_1 =f(x) \cdot e^x-I_2 + c$

$\Rightarrow I =f(x) \cdot e^x-I_2+I_2+c$

$\Rightarrow I=e^x f(x)+c$

$I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$

Solution

[Here, $\frac{1}{x} = f(x) ;-\frac{1}{x^2} = f^{\prime} (x)]$

$I =\int e^x \cdot \frac{1}{x} d x-\int e^x \frac{1}{x^2} d x$

$=\frac{1}{x} e^x+\int \frac{1}{x^2} e^x d x-\int \frac{e^x}{x^2} d x+c$

$={\frac{e^x}{x}+c}$

$I=\int\left[\log (\log x) +\frac{1}{(\log x)^2}\right] d x$

Solution:

$\log x=t \Rightarrow x=e^t$

$\Rightarrow \frac{1}{x} d x=d t$

$d x=x dt=e^t d t$

$I =\int\left(\log t+\frac{1}{t^2}\right) e^t d t$

$=\int e^t \log t d t+\int \frac{e^t}{t^2} d t$

$=\log t . e^t-\int \frac{1}{t}. e^t d t+\int \frac{e^t}{t^2} d t$

$=e^t \log t-\left(\int e^t\left(\frac{1}{t}-\frac{1}{t^2}\right) d t\right)$

$=e^t \log t-e^t \cdot \frac{1}{t}+c$

$=x \log (\log x)-x .\frac{1}{\log x}+c$

$=e^t \log t-e^t \cdot \frac{1}{t}+c$

$=x \log (\log x)-x \cdot \frac{1}{\log x}+c$

$I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x$

Solution:

$\left(\frac{1+\sin x}{1+\cos x}\right)=\frac{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{1+2 \cos ^2 \frac{x}{2} -1}$

$=\frac{1}{2} \frac{(\cos x / 2 +\sin x / 2)^2}{\cos ^2 x / 2}$

$=\frac{1}{2}(1+\tan x / 2)^2$

$=\frac{1}{2}\left(1+\tan ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right)$

$=\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}$

[Here, $\tan \frac{x}{2} = f(x)$ ; $\frac{1}{2} \sec ^2 \frac{x}{2} = f^{\prime} (x)$]

$\Rightarrow I=\int e^x\left(\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x$

$I=e^x \tan \frac{x}{2}+c$

$I=\int(1+x-\frac{1}{x}) e^{x+\frac{1}{x}} d x$

Solution:

$=\int e^{x+\frac{1}{x}} d x$

$+\int \underset{I}{x}.\underset{II}{\frac{1}{x}(x-\frac{1}{x}) e^{x+\frac{1}{x}}} dx$

[$Put, x+\frac{1}{x}=t ,\Rightarrow (1-\frac{1}{x^2}) d x=d t ,\Rightarrow \frac{x^2-1}{x^2} d x=d t$

$\Rightarrow \int \frac{1}{x}(x-\frac{1}{x}) e^{x+\frac{1}{x}} d x$

$= \int e^t dt = e^t$

$= e^{x+\frac{1}{x}}$

$\Rightarrow I =\int e^{x+\frac{1}{x}} d x+x e^{x+\frac{1}{x}}-\int 1.e^{x+1 / x} d x + c$

$=x e^{x+\frac{1}{x}}+c$