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Answer:- $\frac{(\pi-2) \pi}{2}$

Hints:-  You must know the integration rules of trigonometric rule.

Given:-   $\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$

Solution : $\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$

$I=\int_{0}^{\pi} \frac{(\pi-x) x \sin (\pi-x)}{1+\sin (\pi-x)} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$

\begin{aligned} &I=\int_{0}^{\pi} \frac{\pi \sin x}{1+\sin x} d x-I \\ &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)}{\left(1-\sin ^{2} x\right)} d x \end{aligned}

\begin{aligned} &\frac{2 I}{\pi}=\int_{0}^{\pi} \frac{\sin x-\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 \\ &\frac{2 I}{\pi}=\int_{0}^{\pi} \sec x \cdot \tan x \cdot d x-\int_{0}^{\pi} \tan ^{2} x \cdot 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 \cdot d x \\ &\frac{2 I}{\pi}=[-1-1]-[\tan ]_{0}^{\pi}-[x]_{0}^{\pi} \end{aligned}

\begin{aligned} &\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}

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