Prove that: _0^ x 1- x d x= ( - )

Question

Prove that: $\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x=\frac{\pi(\pi-\alpha)}{\sin \alpha}$

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Answer

Let $I=\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x$. Then,
$I=\int_0^\pi \frac{(\pi-x)}{1-\cos \alpha \sin (\pi-x)} d x$
By using property of defnite integrals,
$\Rightarrow I=\int_0^\pi \frac{\pi}{1-\cos \alpha \sin x} d x-\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x$
$\Rightarrow I=\pi \int_0^\pi \frac{1}{1-\cos \alpha \sin x} d x-I$
$\Rightarrow 2 I=\pi \int_0^\pi \frac{1}{1-\cos \alpha \sin x} d x$
$\Rightarrow 2 I=\pi \int_0^\pi \frac{1+\tan ^2 x / 2}{\left(1+\tan ^2 x / 2\right)-2 \cos \alpha \tan x / 2} d x$
$\Rightarrow 2 I=\pi \int_0^\pi \frac{\sec ^2 x / 2}{\tan ^2 x / 2-2 \cos \alpha \tan x / 2+1} d x$
$\Rightarrow I=\frac{\pi}{2} \int_0^\pi \frac{\sec ^2 x / 2}{\tan ^2 x / 2-2 \cos \alpha \tan x / 2+1} d x$
Let $ \tan \frac{x}{2}=t$. Then, $d\left(\tan \frac{x}{2}\right)=d t \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t$
Also, $ x=0 \Rightarrow t=\tan 0=0$ and $x=\pi \Rightarrow t=\tan \frac{\pi}{2}=\infty$
$\therefore I=\frac{\pi}{2} \int_0^{\infty} \frac{2 d t}{t^2-2 t \cos \alpha+1}$
$\Rightarrow I=\pi \int_0^{\infty} \frac{1}{(t-\cos \alpha)^2+\left(1-\cos ^2 \alpha\right)} d t$
$\Rightarrow I=\pi \int_0^{\infty} \frac{1}{\sin ^2 \alpha+(t-\cos \alpha)^2} d t$
$\Rightarrow I=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}\left(\frac{t-\cos \alpha}{\sin \alpha}\right)\right]_0^{\infty}$
$\Rightarrow I=\frac{\pi}{\sin \alpha}\left[\tan ^{-1} \alpha-\tan ^{-1}(-\cot \alpha)\right]$
$\Rightarrow I=\frac{\pi}{\sin \alpha}\left[\frac{\pi}{2}+\tan ^{-1}(\cot \alpha)\right]$
$\left.\Rightarrow I=\frac{\pi}{\sin \alpha}\left[\frac{\pi}{2}+\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\alpha\right)\right)\right\}\right]=\frac{\pi}{\sin \alpha}\left[\frac{\pi}{2}+\frac{\pi}{2}-\alpha\right]=\frac{\pi(\pi-\alpha)}{\sin \alpha}$

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