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

Let $I={\int }_0^{\pi } \frac{x}{1+\sin x}dx.......\left(1\right)$

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

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

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

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

$=\pi {\int }_0^{\pi } \frac{1-\sin x}{{\cos }^{2}x}dx$

$2I=\pi {\int }_0^{\pi } \left[\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}\right]dx$

$=\pi {\int }_0^{\pi } \left({\sec }^{2}x-\sec x\tan x\right)dx=\pi (\tan x-\sec x{)}_0^{\pi }$

$=\pi \left[\right(\tan \pi -\sec \pi )-(\tan \left(0\right)-\sec \left(0\right)]$

$=\pi [0-(-1)-(0-1\left)\right]=\pi [1+1]$

$\therefore 2I=2\pi \Rightarrow I=\pi$