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

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

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

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

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

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

$=\pi [x{]}_0^{\pi }-2\pi {\int }_0^{\pi /2} \frac{1}{1+\sin x}dx$

$={\pi }^2-2\pi {\int }_0^{\pi /2} \frac{1}{1+\sin \left(\frac{\pi }{2}-x\right)}dx$

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

$={\pi }^2-\pi {\int }_0^{\pi /2} {\sec }^2\frac{x}{2}dx={\pi }^2-\pi {\left[2\tan \frac{x}{2}\right]}_0^{\pi /2}$

$2I={\pi }^2-2\pi \therefore I=\frac{{\pi }^2}{2}-\pi$