The integral ${\int }_0^{\pi } xf(\sin x)dx$ is equal to 

Let $I={\int }_0^{\pi } xf(\sin x)dx$

Then,

$\begin{array}{r}I={\int }_0^{\pi } (\pi -x)f(\sin (\pi -x\left)\right)dx\\ \Rightarrow I={\int }_0^{\pi } (\pi -x)f(\sin x)dx\end{array}$

Adding (i) and (ii), we get

$2I=\pi {\int }_0^{\pi } f(\sin x)dx\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi } f(\sin x)dx$