Question

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

Answer 1

Let $I={\int }_0^{\pi }\frac{x}{1+\sin x}dx$ ...(1)
Substituting $x\to \pi -x$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{1+\sin (\pi -x)}dx$
$[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right]$
$\Rightarrow I={\int }_0^{\pi }\frac{\pi -x}{1+\sin x}dx$ ....(2)
Adding equations (1) and (2), we get
$\Rightarrow 2I={\int }_0^{\pi }\frac{\pi }{1+\sin x}dx$
$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\sin x}dx$
$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}dx$
$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }\frac{1-\sin x}{1-{\sin }^{2}x}dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx$
$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }[\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}]dx$
$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }[{\sec }^{2}x-\tan x\sec x]dx$
$\Rightarrow I=\frac{\pi }{2}[\tan x-\sec x{]}_0^{\pi }$
$\Rightarrow I=\frac{\pi }{2}\left[\right(\tan \pi -\sec \pi )-(\tan 0-\sec 0\left)\right]$
$\Rightarrow I=\frac{\pi }{2}\left[\right(0+1)-(0-1\left)\right]=\pi$
Hence, the value of required integration is $\pi$