Question

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

Answer 1

$\begin{array}{c}I={\int }_0^{\pi }\frac{x}{1+\sin \alpha \cdot \sin x}dx\left(i\right)\\ ={\int }_0^{\pi }\frac{\left(\pi -x\right)}{1+\sin \alpha \cdot \sin x}dx\left(ii\right)\\ \left(i\right)+\left(ii\right)\\ 2I=\pi {\int }_0^{\pi }\frac{dx}{1+\sin \alpha \cdot \sin x}\\ 2I=\pi {\int }_0^{\pi }\frac{dx}{1+\sin \alpha \cdot \frac{2\tan x/2}{1+{\tan }^{2}x/2}}\\ 2I={\int }_0^{\pi }\frac{{\sec }^{2}x/2dx}{{\tan }^2\frac{x}{2}+2\sin \alpha \cdot \tan \frac{x}{2}+1}\\ \tan \frac{x}{2}=t\Rightarrow 2{\sec }^2\frac{x}{2}dx=dt\\ {\sec }^2\frac{x}{2}dx=\frac{dt}{2}\\ whenx=0,t=0\\ x=\pi ,t=\infty \\ 2I=\frac{\pi }{2}{\int }_0^{\infty }\frac{dt}{t^2+2\sin \alpha \cdot t+{\sin }^2\alpha +1-{\sin }^2\alpha }\\ 2I=\frac{\pi }{2}{\int }_0^{\infty }\frac{dt}{{\left(t+\sin \alpha \right)}^2+{\cos }^2\alpha }=\frac{\pi }{4\cos \alpha }{\left[{\tan }^{-1}\left(\frac{t+\sin \alpha }{\cos \alpha }\right)\right]}_0^{\infty }\end{array}$

or, $2I=\frac{\pi }{4\cos \alpha }(\frac{\pi }{2}-\alpha )$

or, $I=\frac{\pi }{8\cos \alpha }(\frac{\pi }{2}-\alpha )$.