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### Answers (1)

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

Hints:-  You must know the integral rules of trigonometric functions.

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

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

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

\begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{1+\sin \alpha \sin x} \\ &2 I=\pi \int_{0}^{\pi} \frac{1}{\sin \alpha \sin x+1} d x \end{aligned}

Substituting $\sin x=\frac{2 \tan x / 2}{1+\tan ^{2} x / 2}$

\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{1+\tan ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan ^{2} x / 2} d x \\ &I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} x / 2}{1+\tan ^{2} x / 2+\sin \alpha+2 \tan x / 2} d x \end{aligned}

Let  $\tan \frac{x}{2}=t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { Also when } x \rightarrow 0, t \rightarrow 0$

$\sec ^{2} \frac{x}{2}=2 d t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t \rightarrow \infty$

\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2}+2 t \sin \alpha+1} \\ &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(t+\sin \alpha)^{2}+\cos ^{2} \alpha} d t \\ &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1}\left(\frac{1+\sin \alpha}{\cos \alpha}\right)\right]_{0}^{\infty} \end{aligned}

\begin{aligned} &I=\frac{\pi}{\cos \alpha}\left[\tan ^{-1} \infty-\tan ^{-1}(\tan \alpha)\right] \\ &I=\frac{\pi}{\cos \alpha}\left[\frac{\pi}{2}-\alpha\right] \end{aligned}

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