Question

The value of the integral ${\int }_0^{\pi }\frac{xdx}{1+\cos \alpha \sin x},0<\alpha <\pi$ is

• A. $\frac{\pi \alpha }{\sin \alpha }$
• B. $\frac{\pi \alpha }{1+\sin \alpha }$
• C. $\frac{\pi \alpha }{\cos \alpha }$
• D. $\frac{\pi \alpha }{1+\cos \alpha }$

Answer 1

Answer: (A)

$\frac{\pi \alpha }{\sin \alpha }$

$I={\int }_0^{\pi }\frac{x}{1+\cos \alpha \sin x}dx$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{1+\cos \alpha \sin (\pi -x)}dx$
$\therefore 2I=\pi {\int }_0^{\pi }\frac{1}{1+\cos \alpha \sin x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{(1+{\tan }^2\frac{x}{2})+2\cos \alpha \tan \frac{1}{2}}dx$
Substitute $\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$
$\therefore 2I={\int }_0^{\infty }\frac{2}{1+t^2+2t\cos \alpha }dt=2\pi {\int }_0^{\infty }\frac{2}{{(t+\cos \alpha )}^2+{\sin }^2\alpha }dt$
$\Rightarrow I=\frac{\pi }{\sin \alpha }{\left({\tan }^{-1}\left(\frac{t+\cos \alpha }{\sin \alpha }\right)\right)}_0^{\infty }=\frac{\pi }{\sin \alpha }\left({\tan }^{-1}\infty -{\tan }^{-1}\cot \infty \right)$
$=\frac{\pi }{\sin \alpha }(\frac{\pi }{2}-(\frac{\pi }{2}-\alpha ))$
$\therefore I=\frac{\alpha \pi }{\sin \alpha }$