Evaluate : $\displaystyle \int_{0}^{\pi} \frac{x}{1+\sin x} \mathrm{~d} x$

\[I = \int_0^\pi \frac{x}{1+\sin x}\,dx = \int_0^\pi \frac{\pi - x}{1+\sin(\pi - x)}\,dx = \pi\int_0^\pi \frac{1}{1+\sin x}\,dx - \int_0^\pi \frac{x}{1+\sin x}\,dx\]\[\therefore\; 2I = \pi\int_0^\pi \frac{dx}{1+\sin x}\]\[\therefore\; I = \frac{\pi}{2}\int_0^\pi \frac{dx}{1+\sin x} = \frac{\pi}{2}\int_0^\pi \frac{1}{1+\cos\left(\dfrac{\pi}{2}-x\right)}\,dx\]\[= \frac{\pi}{2}\int_0^\pi \frac{1}{2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}\,dx\]\[= \frac{\pi}{4}\left[-2\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right]_0^\pi\]\[= \frac{\pi}{4}\left[2-(-2)\right] = \pi\]