Solution

$ \begin{aligned} & \sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2} \\ & \Rightarrow \int \sin ^{2}(2 x+5) d x=\int \frac{1-\cos (4 x+10)}{2} d x \\ & =\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x \\ & =\frac{1}{2} x-\frac{1}{2}(\frac{\sin (4 x+10)}{4})+C \\ & =\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C \end{aligned} $

Solution

It is known that, $\sin A \cos B=\frac{1}{2}{\sin (A+B)+\sin (A-B)}$

$\therefore \int \sin 3 x \cos 4 x d x=\frac{1}{2} \int{\sin (3 x+4 x)+\sin (3 x-4 x)} d x$

$ =\frac{1}{2} \int{\sin 7 x+\sin (-x)} d x $

$ \begin{aligned} & =\frac{1}{2} \int{\sin 7 x-\sin x} d x \\ & =\frac{1}{2} \int \sin 7 x d x-\frac{1}{2} \int \sin x d x \\ & =\frac{1}{2}(\frac{-\cos 7 x}{7})-\frac{1}{2}(-\cos x)+C \\ & =\frac{-\cos 7 x}{14}+\frac{\cos x}{2}+C \end{aligned} $

Solution

It is known that, $\cos A \cos B=\frac{1}{2}{\cos (A+B)+\cos (A-B)}$

$ \begin{aligned} \therefore \int \cos 2 x(\cos 4 x \cos 6 x) d x & =\int \cos 2 x[\frac{1}{2}{\cos (4 x+6 x)+\cos (4 x-6 x)}] d x \\ & =\frac{1}{2} \int{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)} d x \\ & =\frac{1}{2} \int{\cos 2 x \cos 10 x+\cos ^{2} 2 x} d x \\ & =\frac{1}{2} \int[{\frac{1}{2} \cos (2 x+10 x)+\cos (2 x-10 x)}+(\frac{1+\cos 4 x}{2})] d x \\ & =\frac{1}{4} \int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x \\ & =\frac{1}{4}[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}]+C \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } I=\int \sin ^{3}(2 x+1) \\ & \Rightarrow \int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ & =\int(1-\cos ^{2}(2 x+1)) \sin (2 x+1) d x \end{aligned} $

Let $\cos (2 x+1)=t$

$\Rightarrow-2 \sin (2 x+1) d x=d t$

$\Rightarrow \sin (2 x+1) d x=\frac{-d t}{2}$

$ \begin{aligned} \Rightarrow I & =\frac{-1}{2} \int(1-t^{2}) d t \\ & =\frac{-1}{2}{t-\frac{t^{3}}{3}} \\ & =\frac{-1}{2}{\cos (2 x+1)-\frac{\cos ^{3}(2 x+1)}{3}} \\ & =\frac{-\cos (2 x+1)}{2}+\frac{\cos ^{3}(2 x+1)}{6}+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \sin ^{3} x \cos ^{3} x \cdot d x \\ & =\int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x \\ & =\int \cos ^{3} x(1-\cos ^{2} x) \sin x \cdot d x \end{aligned} $

Let $\cos x=t$

$ \begin{aligned} \Rightarrow & -\sin x \cdot d x=d t \\ \Rightarrow I & =-\int t^{3}(1-t^{2}) d t \\ & =-\int(t^{3}-t^{5}) d t \\ & =-{\frac{t^{4}}{4}-\frac{t^{6}}{6}}+C \\ & =-{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}}+C \\ & =\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+C \end{aligned} $

Solution

It is known that, $\sin A \sin B=\frac{1}{2}{\cos (A-B)-\cos (A+B)}$

$\therefore \int \sin x \sin 2 x \sin 3 x d x=\int[\sin x \cdot \frac{1}{2}{\cos (2 x-3 x)-\cos (2 x+3 x)}] d x$

$=\frac{1}{2} \int(\sin x \cos (-x)-\sin x \cos 5 x) d x$

$=\frac{1}{2} \int(\sin x \cos x-\sin x \cos 5 x) d x$

$=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$

$=\frac{1}{4}[\frac{-\cos 2 x}{2}]-\frac{1}{2} \int{\frac{1}{2} \sin (x+5 x)+\sin (x-5 x)} d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4}[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{4}]+C$

$=\frac{-\cos 2 x}{8}-\frac{1}{8}[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}]+C$

$=\frac{1}{8}[\frac{\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x]+C$

Solution

It is known that, $\sin A \sin B=\frac{1}{2} \cos (A-B)-\cos (A+B)$ $\therefore \int \sin 4 x \sin 8 x d x=\int{\frac{1}{2} \cos (4 x-8 x)-\cos (4 x+8 x)} d x$

$ \begin{aligned} & =\frac{1}{2} \int(\cos (-4 x)-\cos 12 x) d x \\ & =\frac{1}{2} \int(\cos 4 x-\cos 12 x) d x \\ & =\frac{1}{2}[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}] \end{aligned} $

Solution

$ \begin{aligned} & \frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} x} \\ & {[2 \sin ^{2} \frac{x}{2}=1-\cos x \text{ and } 2 \cos ^{2} \frac{x}{2}=1+\cos x]} \\ & =\tan ^{2} \frac{x}{2} \\ & =(\sec ^{2} \frac{x}{2}-1) \\ & \therefore \int \frac{1-\cos x}{1+\cos x} d x=\int(\sec ^{2} \frac{x}{2}-1) d x \\ & =[\frac{\tan \frac{x}{2}}{1}-x]+C \\ & =2 \tan \frac{x}{2}-x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos x}{1+\cos x}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \quad[\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2} \text{ and } \cos x=2 \cos ^{2} \frac{x}{2}-1] \\ & =\frac{1}{2}[1-\tan ^{2} \frac{x}{2}] \\ & \begin{aligned} \therefore \int \frac{\cos x}{1+\cos x} d x & =\frac{1}{2} \int(1-\tan ^{2} \frac{x}{2}) d x \\ & =\frac{1}{2} \int(1-\sec ^{2} \frac{x}{2}+1) d x \\ & =\frac{1}{2} \int(2-\sec ^{2} \frac{x}{2}) d x \\ & =\frac{1}{2}[2 x-\frac{\tan ^{2}}{2}]+C \\ & =x-\tan \frac{x}{2}+C \end{aligned} \end{aligned} $

Solution

$ \begin{aligned} & \sin ^{4} x=\sin ^{2} x \sin ^{2} x \\ & =(\frac{1-\cos 2 x}{2})(\frac{1-\cos 2 x}{2}) \\ & =\frac{1}{4}(1-\cos 2 x)^{2} \\ & =\frac{1}{4}[1+\cos ^{2} 2 x-2 \cos 2 x] \\ & =\frac{1}{4}[1+(\frac{1+\cos 4 x}{2})-2 \cos 2 x] \\ & =\frac{1}{4}[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] \\ & =\frac{1}{4}[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] \\ & \therefore \int \sin ^{4} x d x=\frac{1}{4} \int[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x] d x \\ & =\frac{1}{4}[\frac{3}{2} x+\frac{1}{2}(\frac{\sin 4 x}{4})-\frac{2 \sin 2 x}{2}]+C \\ & =\frac{1}{8}[3 x+\frac{\sin 4 x}{4}-2 \sin 2 x]+C \\ & =\frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C \end{aligned} $

Solution

$ \begin{aligned} \cos ^{4} 2 x & =(\cos ^{2} 2 x)^{2} \\ & =(\frac{1+\cos 4 x}{2})^{2} \\ & =\frac{1}{4}[1+\cos ^{2} 4 x+2 \cos 4 x] \\ & =\frac{1}{4}[1+(\frac{1+\cos 8 x}{2})+2 \cos 4 x] \\ & =\frac{1}{4}[1+\frac{1}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x] \\ & =\frac{1}{4}[\frac{3}{2}+\frac{\cos 8 x}{2}+2 \cos 4 x] \end{aligned} $

$ \begin{aligned} \therefore \int \cos ^{4} 2 x d x & =\int(\frac{3}{8}+\frac{\cos 8 x}{8}+\frac{\cos 4 x}{2}) d x \\ & =\frac{3}{8} x+\frac{\sin 8 x}{64}+\frac{\sin 4 x}{8}+C \end{aligned} $

Solution

$ \begin{aligned} \frac{\sin ^{2} x}{1+\cos x} & =\frac{(2 \sin \frac{x}{2} \cos \frac{x}{2})^{2}}{2 \cos ^{2} \frac{x}{2}}[\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} ; \cos x=2 \cos ^{2} \frac{x}{2}-1] \\ & =\frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} \\ & =2 \sin ^{2} \frac{x}{2} \\ & =1-\cos ^{2} x \\ \therefore \int \frac{\sin ^{2} x}{1+\cos x} d x & =\int(1-\cos x) d x \\ & =x-\sin x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}=\frac{-2 \sin \frac{2 x+2 \alpha}{2} \sin \frac{2 x-2 \alpha}{2}}{-2 \sin \frac{x+\alpha}{2} \sin \frac{x-\alpha}{2}} \quad[\cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}] \\ &=\frac{\sin (x+\alpha) \sin (x-\alpha)}{\sin (\frac{x+\alpha}{2}) \sin (\frac{x-\alpha}{2})} \\ &=\frac{[2 \sin (\frac{x+\alpha}{2}) \cos (\frac{x+\alpha}{2})][2 \sin (\frac{x-\alpha}{2}) \cos (\frac{x-\alpha}{2})]}{\sin (\frac{x+\alpha}{2}) \sin (\frac{x-\alpha}{2})} \\ &=4 \cos (\frac{x+\alpha}{2}) \cos (\frac{x-\alpha}{2}) \\ &=2[\cos (\frac{x+\alpha}{2}+\frac{x-\alpha}{2})+\cos \frac{x+\alpha}{2}-\frac{x-\alpha}{2}] \\ &=2[\cos (x)+\cos \alpha] \\ &=2 \cos x+2 \cos \alpha \\ & \therefore \int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int 2 \cos x+2 \cos \alpha \\ &=2[\sin x+x \cos \alpha]+C \end{aligned} $

Solution

$ \begin{aligned} \frac{\cos x-\sin x}{1+\sin 2 x} & =\frac{\cos x-\sin x}{(\sin ^{2} x+\cos ^{2} x)+2 \sin x \cos x} \\ & =\frac{\cos x-\sin x}{.(\sin x+\cos x)^{2} x+\cos ^{2} x=1 ; \sin 2 x=2 \sin x \cos x]} \end{aligned} $

Let $\sin x+\cos x=t$

$\therefore(\cos x-\sin x) d x=d t$

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

$ =\int \frac{d t}{t^{2}} $

$ \begin{aligned} & =\int t^{-2} d t \\ & =-t^{-1}+C \\ & =-\frac{1}{t}+C \\ & =\frac{-1}{\sin x+\cos x}+C \end{aligned} $

Solution

$ \begin{aligned} & \tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 x \sec 2 x \\ & =(\sec ^{2} 2 x-1) \tan 2 x \sec 2 x \\ & =\sec ^{2} 2 x \cdot \tan 2 x \sec 2 x-\tan 2 x \sec 2 x \\ & \therefore \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x \\ & =\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+C \end{aligned} $

Let $\sec 2 x=t$

$\therefore 2 \sec 2 x \tan 2 x d x=d t$

$ \begin{aligned} \therefore \int \tan ^{3} 2 x \sec 2 x d x & =\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+C \\ & =\frac{t^{3}}{6}-\frac{\sec 2 x}{2}+C \\ & =\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+C \end{aligned} $

Solution

$\tan ^{4} x$

$=\tan ^{2} x \cdot \tan ^{2} x$

$=(\sec ^{2} x-1) \tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-\tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-(\sec ^{2} x-1)$

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

$\therefore \int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 \cdot d x$

$=\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C$

Consider $\int \sec ^{2} x \tan ^{2} x d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}$

From equation (1), we obtain

$\int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C$

Solution

$ \begin{aligned} & \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}=\frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}+\frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} \\ & =\frac{\sin x}{\cos ^{2} x}+\frac{\cos x}{\sin ^{2} x} \\ & =\tan x \sec x+\cot x cosec x \\ & \therefore \quad \int \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x=\int(\tan x \sec x+\cot x cosec x) d x \\ & =\sec x-cosec x+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} \\ & =\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x} \quad[\cos 2 x=1-2 \sin ^{2} x] \\ & =\frac{1}{\cos ^{2} x} \\ & =\sec ^{2} x \\ & \therefore \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x=\tan x+C \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\sin x \cos ^{3} x} & =\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos ^{3} x} \\ & =\frac{\sin x}{\cos ^{3} x}+\frac{1}{\sin x \cos x} \\ & =\tan x \sec ^{2} x+\frac{1 \cos ^{2} x}{\frac{\sin x \cos x}{\cos ^{2} x}} \\ & =\tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x} \end{aligned} $

$\therefore \int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x & =\int t d t+\int_t^{1} d t \\ & =\frac{t^{2}}{2}+\log |t|+C \\ & =\frac{1}{2} \tan ^{2} x+\log |\tan x|+C \end{aligned} $

Solution

$\begin{aligned} & \frac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2}=\frac{\cos 2 \mathrm{x}}{\cos ^2 \mathrm{x}+\sin ^2 \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}}=\frac{\cos 2 \mathrm{x}}{1+\sin 2 \mathrm{x}} \\ & \therefore \int \frac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2} \mathrm{dx}=\int \frac{\cos 2 \mathrm{x}}{(1+\sin 2 \mathrm{x})} \mathrm{dx} \end{aligned}$

Let $1+\sin 2 x=t \Rightarrow 2 \cos 2 x d x=d t$

$\begin{aligned} & \therefore \int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x=\frac{1}{2} \int \frac{1}{t} d t \\ & =\frac{1}{2} \log |t|+C \\ & =\frac{1}{2} \log |1+\sin 2 x|+C \\ & =\frac{1}{2} \log \left|(\sin x+\cos x)^2\right|+C \\ & =\log |\sin x+\cos x|+C \\ \end{aligned}$

Solution

$\sin ^{-1}(\cos x)$

Let $\cos x=t$

Then, $\sin x=\sqrt{1-t^{2}}$ $\Rightarrow(-\sin x) d x=d t$

$d x=\frac{-d t}{\sin x}$

$d x=\frac{-d t}{\sqrt{1-t^{2}}}$

$\therefore \int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} t(\frac{-d t}{\sqrt{1-t^{2}}})$

$=-\int \frac{\sin ^{-1} t}{\sqrt{1-t^{2}}} d t$

Let $\sin ^{-1} t=u$

$\Rightarrow \frac{1}{\sqrt{1-t^{2}}} d t=d u$

$\therefore \int \sin ^{-1}(\cos x) d x=\int 4 d u$

$$ \begin{align*} & =-\frac{u^{2}}{2}+C \\ & =\frac{-(\sin ^{1} t)^{2}}{2}+C \\ & =\frac{-[\sin ^{-1}(\cos x)]^{2}}{2}+C \tag{1} \end{align*} $$

It is known that,

$ \begin{aligned} & \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ & \therefore \sin ^{-1}(\cos x)=\frac{\pi}{2}-\cos ^{-1}(\cos x)=(\frac{\pi}{2}-x) \end{aligned} $

Substituting in equation (1), we obtain

$ \begin{aligned} \int \sin ^{-1}(\cos x) d x & =\frac{-[\frac{\pi}{2}-x]^{2}}{2}+C \\ & =-\frac{1}{2}(\frac{\pi^{2}}{2}+x^{2}-\pi x)+C \\ & =-\frac{\pi^{2}}{8}-\frac{x^{2}}{2}+\frac{1}{2} \pi x+C \\ & =\frac{\pi x}{2}-\frac{x^{2}}{2}+(C-\frac{\pi^{2}}{8}) \\ & =\frac{\pi x}{2}-\frac{x^{2}}{2}+C_1 \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\cos (x-a) \cos (x-b)} & =\frac{1}{\sin (a-b)}[\frac{\sin (a-b)}{\cos (x-a) \cos (x-b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin [(x-b)-(x-a)]}{\cos (x-a) \cos (x-b)}] \\ & =\frac{1}{\sin (a-b)} \frac{[\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)]}{\cos (x-a) \cos (x-b)} \\ & =\frac{1}{\sin (a-b)}[\tan (x-b)-\tan (x-a)] \\ \Rightarrow \int \frac{1}{\cos (x-a) \cos (x-b)} d x & =\frac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x \\ & =\frac{1}{\sin (a-b)}[-\log |\cos (x-b)|+\log |\cos (x-a)|] \\ & =\frac{1}{\sin (a-b)}[\log |\frac{\cos (x-a)}{\cos (x-b)}|]+C \end{aligned} $

Solution

$ \begin{aligned} \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x & =\int(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}) d x \\ & =\int(\sec ^{2} x-cosec^{2} x) d x \\ & =\tan x+\cot x+C \end{aligned} $

Hence, the correct Answer is A.

Solution

$\int \frac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x$

Let $e x^{x}=t$

$ \begin{aligned} & \Rightarrow(e^{x} \cdot x+e^{x} \cdot 1) d x=d t \\ & e^{x}(x+1) d x=d t \\ & \begin{aligned} \therefore \int \frac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x & =\int \frac{d t}{\cos ^{2} t} \\ & =\int \sec ^{2} t d t \\ & =\tan t+C \\ & =\tan (e^{x} \cdot x)+C \end{aligned} \end{aligned} $

Hence, the correct Answer is B.