Question

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

Answer 1

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

$Onaddingboth\left(1\right)+\left(2\right),weget$

$2I={\int }_0^{\pi }\frac{\pi \sin xdx}{1+\sin x}I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+\sin x}dxI=\frac{\pi }{2}{\int }_0^{\pi }\frac{\left(1-\sin x\right)\sin x}{\left(1+\sin x\right)\left(1-\sin x\right)}dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x-{\sin }^{2}x}{{\cos }^{2}x}dx=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{{\cos }^{2}x}dx-\frac{\pi }{2}{\int }_0^{\pi }\frac{{\sin }^{2}x}{{\cos }^{2}x}dx=\frac{\pi }{2}{\int }_0^{\pi }\tan x\sec xdx-\frac{\pi }{2}{\int }_0^{\pi }\frac{1-{\cos }^{2}x}{{\cos }^{2}x}dx=\frac{\pi }{2}{\left[\sec \right]}_0^{\pi }-\frac{\pi }{2}{\int }_0^{\pi }{\sec }^{2}xdx+\frac{\pi }{2}{\int }_0^{\pi }dx=\frac{\pi }{2}\left[\sec \pi -\sec 0\right]-\frac{\pi }{2}{\left[\tan x\right]}_0^{\pi }+\frac{\pi }{2}{\left[x\right]}_0^{\pi }=\frac{\pi }{2}\left[-1-\right]-\frac{\pi }{2}\left[\tan \pi -\tan 0\right]+\frac{\pi }{2}\left[\pi -0\right]=-\pi -\frac{\pi }{2}\left[0-0\right]+\frac{\pi }{2}\left[\pi \right]=\pi \left[\frac{\pi }{2}-1\right]Thus{\int }_0^{\pi }\frac{x\sin x}{1+\sin x}=\pi \left[\frac{\pi }{2}-1\right]$

Hence, this is the answer.