Let $I=\int_0^\pi \frac{x}{1+\sin x} d x\quad\text{.... (i)}$

and $I=\int_0^\pi \frac{\pi-x}{1+\sin (\pi-x)} d x=\int_0^\pi \frac{\pi-x}{1+\sin x} d x\quad\text{.... (ii)}$

$$\begin{aligned} &\text { On adding Eqs. (i) and (ii), we get }\\ &\begin{aligned} 2 I & =\pi \int_0^\pi \frac{1}{1+\sin x} d x \\ & =\pi \int_0^\pi \frac{(1-\sin x) d x}{(1+\sin x)(1-\sin x)} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\pi \int_0^\pi \frac{(1-\sin x) d x}{\cos ^2 x} \\ & =\pi \int_0^\pi\left(\sec ^2 x-\tan x \cdot \sec x\right) d x \\ & =\pi \int_0^\pi \sec ^2 x d x-\pi \int_0^\pi \sec x x \cdot \tan x d x \\ & =\pi[\tan x]_0^\pi-\pi[\sec x]_0^\pi \\ & =\pi[\tan x-\sec x]_0^\pi \\ & =\pi[\tan \pi-\sec \pi-\tan 0-\sec 0] \end{aligned}$$

$$\begin{aligned} \Rightarrow & & 2 I & =\pi[0+1-0+1] \\ & & 2 I & =2 \pi \\ \therefore & & I & =\pi \end{aligned}$$

Let $I=\int \frac{(2 x-1)}{(x-1)(x+2)(x-3)} d x$

Now, $\quad \frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x-3)}$

$\Rightarrow \quad 2 x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)$

Put $x=3$, then

$$\begin{aligned} 6-1 & =C(3-1)(3+2) \\ \Rightarrow\quad 5 & =10 C \Rightarrow C=\frac{1}{2} \end{aligned}$$

$$\begin{aligned} &\text { Again, put } x=1 \text {, then }\\ &2-1=A(1+2)(1-3) \end{aligned}$$

$\Rightarrow \quad 1=-6 A \Rightarrow A=-\frac{1}{6}$

$$\begin{aligned} &\text { Now, put } x=-2 \text {, then }\\ &\begin{aligned} & -4-1 =B(-2-1)(-2-3) \\ & \Rightarrow -5 =15 B \Rightarrow B=-\frac{1}{3} \end{aligned} \end{aligned}$$

$\therefore \quad I=-\frac{1}{6} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{1}{x+2} d x+\frac{1}{2} \int \frac{1}{x-3} d x$

$$\begin{aligned} & =-\frac{1}{6} \log |(x-1)|-\frac{1}{3} \log |(x+2)|+\frac{1}{2} \log |(x-3)|+C \\ & =-\log |(x-1)|^{1 / 6}-\log |(x+2)|^{1 / 3}+\log |(x-3)|^{1 / 2}+C \\ & =\log \left|\frac{\sqrt{x-3}}{(x-1)^{1 / 6}(x+2)^{1 / 3}}\right|+C \end{aligned}$$

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

$$\begin{aligned} & =\int e^{\tan ^{-1} x}\left(\frac{1+x^2}{1+x^2}+\frac{x}{1+x^2}\right) d x \\ & =\int e^{\tan ^{-1} x} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^2} d x \end{aligned}$$

$I=I_1+I_2\quad\text{.... (i)}$

Now, $I_2=\int \frac{x \mathrm{e}^{\tan ^{-1} x}}{1+x^2} d x$

Put $\tan ^{-1} x=t \Rightarrow x=\tan t$

$\Rightarrow \quad \frac{1}{1+x^2} d x=d t$

$\therefore\quad I=\int_{\mathrm{I}}^{\tan } t \cdot e_{\mathrm{II}}^t d t$

$$\begin{aligned} & =\tan t \cdot e^t-\int \sec ^2 t \cdot e^t d t+C \\ & =\tan t \cdot e^t-\int\left(1+\tan ^2 t\right) e^t d t+C \quad\left[\because \sec ^2 \theta=1+\tan ^2 \theta\right] \end{aligned}$$

$$\begin{aligned} I_2 & =\tan t \cdot e^t-\int\left(1+x^2\right) \frac{e^{\tan ^{-1} x}}{1+x^2} d x+C \\ I_2 & =\tan t \cdot e^t-\int e^{\tan ^{-1} x} d x+C \\ \therefore\quad I & =\int e^{\tan ^{-1} x} d x+\tan t \cdot e^t-\int e^{\tan ^{-1} x} d x+C \\ & =\tan t \cdot e^t+C \\ & =x e^{\tan ^{-1} x}+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x \\ \text{Put}\quad x & =a \tan ^2 \theta \\ \Rightarrow\quad d x & =2 a \tan \theta \sec ^2 \theta d \theta \\ \therefore\quad I & =\int \sin ^{-1} \sqrt{\frac{a \tan ^2 \theta}{a+a \tan ^2 \theta}}\left(2 a \tan \theta \cdot \sec ^2 \theta\right) d \theta \\ & =2 a \int \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) \tan \theta \cdot \sec ^2 \theta d \theta \\ & =2 a \int \sin ^{-1}(\sin \theta) \tan \theta \cdot \sec ^2 \theta d \theta \end{aligned}$$

$$\begin{aligned} & =2 a \int_{\text {I }}^\theta \theta \cdot \tan \theta \sec ^2 \theta d \theta \\ & =2 a\left[\theta \cdot \int \tan \theta \cdot \sec ^2 \theta d \theta-\int\left(\frac{d}{d \theta} \theta \cdot \int \tan \theta \cdot \sec ^2 \theta d \theta\right) d \theta\right] \end{aligned}$$

$\left[\begin{array}{lrl}\text { Put } & \tan \theta & =t \\ \Rightarrow & \sec \theta \cdot \tan \theta \cdot d \theta & =d t \\ \Rightarrow & \int \tan \theta \sec ^2 \theta d \theta & =\int t d t\end{array}\right]$

$$\begin{aligned} & =2 a\left[\theta \cdot \frac{\tan ^2 \theta}{2}-\int \frac{\tan ^2 \theta}{2} d \theta\right] \\ & =a \theta \tan ^2 \theta-a \int\left(\sec ^2 \theta-1\right) d \theta \\ & =a \theta \cdot \tan ^2 \theta-a \tan \theta+a \theta+C \\ & =a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}\right]+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x \\ & =\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^2 \sqrt{1+\cos x}} d x \\ & =\int_{\pi 3}^{\pi 2} \frac{1}{\left(1-\cos ^2 x\right)} d x=\int_{\pi 3}^{\pi 2} \frac{1}{\sin ^2 x} d x \\ & =\int_{\pi 3}^{\pi 2} \operatorname{cosec}^2 x d x=[-\cot x]_{\pi / 3}^{\pi / 2} \\ & =-\left[\cot \frac{\pi}{2}-\cot \frac{\pi}{3}\right]=-\left[0-\frac{1}{\sqrt{3}}\right]=+\frac{1}{\sqrt{3}} \end{aligned}$$

Alternate Method

Let $I=\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5 / 2}} d x=\int_{\pi / 3}^{\pi / 2} \frac{\left(2 \cos ^2 \frac{x}{2}\right)^{1 / 2}}{\left(2 \sin ^2 \frac{x}{2}\right)^{5 / 2}} d x$

$=\frac{\sqrt{2}}{4 \sqrt{2}} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x=\frac{1}{4} \int_{\pi / 3}^{\pi / 2} \frac{\cos \left(\frac{x}{2}\right)}{\sin ^5\left(\frac{x}{2}\right)} d x$

$$\begin{array}{lr} \text { Put } & \sin \frac{x}{2}=t \\ \Rightarrow & \cos \frac{x}{2} \cdot \frac{1}{2} d x=d t \\ \Rightarrow & \cos \frac{x}{2} d x=2 d t \end{array}$$

As $\quad x \rightarrow \frac{\pi}{3}$, then $t \rightarrow \frac{1}{2}$

and $x \rightarrow \frac{\pi}{2}$, then $t \rightarrow \frac{1}{\sqrt{2}}$

$$\begin{aligned} \therefore\quad I & =\frac{2}{4} \int_{1 / 2}^{1 \sqrt{2}} \frac{d t}{t^5}=\frac{1}{2}\left[\frac{t^{-5+1}}{-5+1}\right]_{1 / 2}^{1 / \sqrt{2}} \\ & =-\frac{1}{8}\left[\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4}-\frac{1}{\left(\frac{1}{2}\right)^4}\right]^8 \\ & =-\frac{1}{8}(4-16)=\frac{12}{8}=\frac{3}{2} \end{aligned}$$