Definite Integration — Maths STD 12 Science — Question

$\begin{aligned} & \text { Let } I=\int_0^\pi \frac{x \sin x}{1+\sin x} d x \\ & =\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\sin (\pi-x)} d x\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]\end{aligned}$$\begin{aligned} & =\int_0^\pi \frac{(\pi-x) \sin x}{1+\sin x} d x \\ & =\int_0^\pi \frac{\pi \sin x}{1+\sin x} d x-I \\ & I=\int_0^\pi \frac{\pi \sin x}{1+\sin x} d x-I\end{aligned}$$\begin{aligned} & 2 I=\int_0^\pi \frac{\pi \sin x \cdot(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & 2 I=\int_0^\pi \frac{\pi \sin x \cdot(1-\sin x)}{1-\sin ^2 x} d x \\ & \frac{2 I}{\pi}=\int_0^\pi \frac{\sin x \cdot(1-\sin x)}{\cos ^2 x} d x\end{aligned}$$\begin{aligned} \frac{2 I}{\pi} & =\int_0^\pi \frac{\sin x \cdot-\sin ^2 x}{\cos ^2 x} d x \\ \frac{2 I}{\pi} & =\int_0^\pi \frac{\sin x}{\cos ^2 x} d x-\int_0^\pi \frac{\sin ^2 x}{\cos ^2 x} d x\end{aligned}$
$\begin{aligned} & \frac{2 I}{\pi}=\int_0^\pi \sec x \cdot \tan x d x-\int_0^\pi \tan ^2 x d x \\ & \frac{2 I}{\pi}=[\sec x]_0^\pi-\int_0^\pi\left(\sec ^2 x-1\right) d x\end{aligned}$$\begin{aligned} & \frac{2 I}{\pi}=[\sec \pi-\sec 0]-\int_0^\pi \sec ^2 x \cdot d x+\int_0^\pi 1 d x \\ & \frac{2 I}{\pi}=[-1-1]-[\tan x]_0^\pi-[x]_0^\pi \\ & \frac{2 I}{\pi}=[-2]-[\tan \pi-\tan 0]+\pi \\ & \frac{2 I}{\pi}=[-2]-0+\pi \\ & \therefore I=\frac{(\pi-2) \pi}{2}\end{aligned}$