Evaluate $\int_0^\pi xf(\sin x)dx$

If you make the substitution $w = \pi-x$, so that $dw = -dx$, you get $$\int_0^\pi xf(\sin x)dx=-\int_\pi^0 (\pi-w)f(\sin(\pi-w))dw$$ $$ = \int_0^\pi (\pi-x)f(\sin(x))dx$$ $$ = \pi\int_0^\pi f(\sin(x))dx - \int_0^\pi xf(\sin(x))dx$$ which gives the result you want.

$\int_0^\pi xf(\sin x)dx$

=$\int_0^\pi (\pi-x)f(\sin (\pi - x))dx$

= $\int_0^\pi (\pi-x)f(\sin x)dx$

= $\pi\int_0^\pi f(\sin x)dx - \int_0^\pi xf(\sin x)dx$

$2\int_0^\pi xf(\sin x)d$ = $\pi\int_0^\pi f(\sin x)dx$

$\int_0^\pi xf(\sin x)d$ = $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$

I also used: $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$