Evaluate $\int_0^\pi xf(\sin x)dx$ Let $f(\sin x)$ be a given function of $\sin x$. 
How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?

$\int_0^\pi xf(\sin x)dx$
=$\int_0^\pi (\pi-x)f(\sin (\pi - x))dx$
= $\int_0^\pi (\pi-x)f(\sin x)dx$
= $\pi\int_0^\pi f(\sin x)dx - \int_0^\pi xf(\sin x)dx$
$2\int_0^\pi xf(\sin x)d$ =  $\pi\int_0^\pi f(\sin x)dx$
$\int_0^\pi xf(\sin x)d$ =  $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$
I also used:  $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$
