The integral $\int\limits\limits_0^\pi x f(\sin x) d x$ is equal to

$\frac{\pi}{2} \int\limits_0^\pi f(\sin x) d x$

Let $I=\int\limits_0^\pi x f(\sin x) d x$           .....(i)

Then,

$I =\int\limits_0^\pi(\pi-x) f(\sin (\pi-x)) d x$               [Using $\int\limits_0^a f(x)dx = \int\limits_0^a f(a-x) dx$]

$\Rightarrow I =\int\limits_0^\pi(\pi-x) f(\sin x) d x$            .....(ii)

Adding (i) and (ii), we get

$2 I=\pi \int\limits_0^\pi f(\sin x) d x \Rightarrow I=\frac{\pi}{2} \int\limits_0^\pi f(\sin x) d x$