Answer
See proof
Work Step by Step
Let $I=\int_{0}^\pi{xf(\sin(x))}dx$
If $u=\pi-x$ then for $x=0 \to u=\pi$
if $u=\pi-x$ then for $x=\pi \to u=0$
$$du=-dx \to dx=-du$$
$$\begin{align*}
I&=-\int_{\pi}^0{(\pi-u)f(\sin(\pi-u))}du\\
&=\int_{0}^\pi {(\pi-u)f(\sin(u))}du\\
&=\pi\int_{0}^\pi { f(\sin(u))}du-\int_{0}^\pi { uf(\sin(u))}du
\end{align*}$$
Since $u$ is a variable, replace $u$ by $x$:
$$\int_{0}^\pi{xf(\sin(x))}dx=\pi\int_{0}^\pi { f(\sin(x))}dx-\int_{0}^\pi { xf(\sin(x))}dx$$
$$\int_{0}^\pi{xf(\sin(x))}dx+\int_{0}^\pi { xf(\sin(x))}dx=\pi\int_{0}^\pi { f(\sin(x))}dx$$
$$2\int_{0}^\pi{xf(\sin(x))}dx=\pi\int_{0}^\pi { f(\sin(x))}dx$$
$$\int_{0}^\pi{xf(\sin(x))}dx=\frac{\pi}{2}\int_{0}^\pi { f(\sin(x))}dx$$
