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

$\pi \int\limits_0^{\pi / 2} f(\cos x) d x$

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

Then,

$I =\int\limits_0^\pi(\pi-x)(\sin (\pi-x)) d x$

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

Adding (i) and (ii), we get

$2 I=\pi \int\limits_0^\pi f(\sin x) d x$

$\Rightarrow 2 I=2 \pi \int\limits_0^{\pi / 2} f(\sin x) d x$            $\left[\begin{array}{l}\text { Using: } \int\limits_0^{2 a} f(x) d x=2 \int\limits_0^a f(x) d x \\ \text { when } f(2 a-x)=f(x)\end{array}\right]$

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

$\Rightarrow I=\pi \int\limits_0^{\pi r r 2} f(\sin (\pi / 2-x)) d x=\pi \int\limits_0^{\pi / 2} f(\cos x) d x$