$I=\int \sin (a x+b) \cos (a x+b) d x$

solution:

case1

$\int \sin x \cos x d x$

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

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

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

$I=-\frac{1}{4} \frac{\cos 2(a x+b)}{a}+c$

case2

$a x+b=t \Rightarrow a d x=d t$

$I=\int \sin t \cos t \frac{d t}{a}$

$=\frac{1}{a} \int \sin t \cos t d t$

$\sin t=u \Rightarrow \cos t d t=d u$

$=\frac{1}{a} \int u d u=\frac{u^2}{2 a}+c$

$=\frac{\sin ^2(a x+b)}{2 a}+c$

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

$I =\int \frac{\tan ^4 \sqrt{x} {\sec ^2 \sqrt{x}}}{\sqrt{x}} d x$

solution:

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

$=\int {\tan ^4 t. \sec ^2 t. 2 d t}$

$[Put \tan t=u,$ $\quad$$\sec ^2 t d t=d u]$

$=2 \int u^4 d u$ $\quad$

$=2 \frac{u^5}{5}+c=\frac{2}{5} \tan ^5 \sqrt{x}+c .$

Alter.

$\tan \sqrt{x}=t \Rightarrow \sec ^2 \sqrt{x} \times \frac{1}{2 \sqrt{x}} d x=d t$

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

$I=2 \int t^4 d t=\frac{2}{5} t^{5}+c$

$=\frac{2}{5} \tan ^5 \sqrt{x}+c$

1. $I =\int \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{x^8+1} d x$

$I=\int 10^{5^x} \cdot 5^x \cdot d x$

$[Put 5^x=t$$5^x \log _e 5 d x=d t]$

$=\int 10^t \frac{d t}{\log _e 5}$

$=\frac{1}{\log _e 5} \cdot \int 10^t d t=\frac{1}{\log _e{5}} \frac{10^t}{\log _e{10}}+C$

$=\frac{1}{\log _e{5}} \frac{10^{5^x}}{\log _e{10}}+C$

i)

$I=\int \tan x d x=\int \frac{\sin x}{\cos x} d x$

$\quad \cos x=t \Rightarrow-\sin x d x=d t$

$I=\int-\frac{d t}{t}=-\log |t|+C$

$=-\log |\cos x|+c=\log |\sec x|+C$

$\int \tan x d x=\log |\sec x|+C$

ii) $\quad$ $\int \cot x d x=\log |\sin x|+C$

$\text { iii) } \quad$$I =\int \sec x d x=\int \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} d x$

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

[Put , $\sec x+\tan x=t,$ $\Rightarrow (\sec^2 x+\sec x.\tan x) d x=dt]$

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

$=\log |t|+C$ $\quad$

${\int \sec x d x=\log |\sec x+\tan x|+C}$

iv) $\int \operatorname{cosec} x d x$

$=\int \frac{\operatorname{cosec} x(\operatorname{cosec} x+\cot x)}{\operatorname{cosec} x+\cot x} d x$

Choose

$\operatorname{cosec} x+\cot x=t$

$(-\operatorname{cosec} x \cot x-\operatorname{cosec}^2 x) d x=d t$

$=-\int \frac{d t}{t}=-\log |t|+C$

$=\log \left|\frac{1}{\operatorname{cosec} x+\cot x}\right|+C$ $\quad$

$[\because \operatorname{cosec}^2 x- {\cot^2 x=1}]$

$\int \frac{\sin x}{\sin (x+a)} d x$

$x+a=t \Rightarrow d x=d t$

$\int \frac{\sin (t-a)}{\sin t} d t=\int\left(\frac{\sin t \cos a-\cos t \sin a}{\sin t}\right) d t$

$=\cos a \int 1 \cdot d t-\sin a \int \cot t d t \quad$

$[\because \int \cot t d t$,$=\log |\sin t|+c]$

$= \cos a\left(t+c_1\right)-\sin a\left(\log |\sin t|+c_2\right)$

$= \left(x+a+c_1\right) \cos a-\sin a\left(\log |\sin (x+a)|+c_2\right)$

$= x \cos a-\sin a \log |\sin (x+a)|+c$

Example:

$I=\int \sin ^3 x \cos ^3 x d x$

$=\int \sin ^3 x \cdot \cos ^2 x \cdot \cos x d x,$ $\quad$ ${[ \because \sin ^2 x+ \cos ^2 x=1]}$

$=\int \sin ^3 x\left(1-\sin ^2 x\right) \cos x d x$

$=\int t^3\left(1-t^2\right) d t$ $\quad$ $[Put \sin x=t, \cos x d x=d t]$

Example: $I=\int \sin 4 x \sin 8 x d x$

$=\frac{1}{2} \int 2 \sin 4 x \sin 8 x d x$

$I =\frac{1}{2} \int[\cos (4-8) x-\cos (4+8) x] d x$

$=\frac{1}{2} \int(\cos 4 x-\cos 12 x) d x$

$={\frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]+C}$