How to prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$
To prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$ is true, first I started calculating the integral of the left indefinitely
$$ \int {x\,f(\sin x)\,\,dx} $$...