Question

Evaluate: ${\int }_0^{\pi }\frac{x}{1+\sin \alpha \sin x}dx$

Answer 1

Let $I={\int }_0^{\pi }\frac{x}{1+\sin \alpha \sin x}dx$

$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{1+\sin \alpha \sin (\pi -x)}dx$ since ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$\Rightarrow ,I={\int }_0^{\pi }\frac{\pi }{1+\sin \alpha \sin (\pi -x)}dx-{\int }_0^{\pi }\frac{x}{1+\sin \alpha \sin (\pi -x)}dx$

$\Rightarrow ,I={\int }_0^{\pi }\frac{\pi }{1+\sin \alpha \sin (\pi -x)}dx-I$

$\Rightarrow ,2I={\int }_0^{\pi }\frac{\pi }{1+\sin \alpha \sin (\pi -x)}dx$

Substituting $\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}},$ we get

$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}+\sin \alpha \times 2\tan \frac{x}{2}}dx$

Let $\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}=2dt$

Also,

when $x\to 0,t\to \tan 0=0$

when $x\to \pi ,t\to \tan \frac{\pi }{2}=\infty$

$\therefore I=\frac{\pi }{2}{\int }_0^{\infty }\frac{2dt}{t^2+2tsin\alpha +1}$

$\Rightarrow I=\pi {\int }_0^{\infty }\frac{dt}{{(t+\sin \alpha )}^2+{\cos }^2\alpha }$

$\Rightarrow I=\frac{\pi }{\cos \alpha }{\left[{\tan }^{-1}\left(\frac{t+\sin \alpha }{\cos \alpha }\right)\right]}_0^{\infty }$

$\Rightarrow I=\frac{\pi }{\cos \alpha }[{\tan }^{-1}\infty -{\tan }^{-1}\left(\tan \alpha \right)]$

$\Rightarrow I=\frac{\pi }{\cos \alpha }(\frac{\pi }{2}-\alpha )$