# Prove that $\int\limits_0^{\pi}xf(\sin x)dx =\frac{\pi}{2}\int\limits_0^{\pi}f(\sin x)dx$ [duplicate]

I tried solving it. By substituting $\sin x$ with $t$ then $dt=\cos xdx$ but I am stuck after that. Where to go next?

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You know $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ (Hint: Substitute $y=a+b-x$, then $dy=-dx$)
Now, $$I=\int\limits_0^{\pi}xf(\sin x)dx =\int\limits_0^{\pi}(\pi-x)f(\sin (\pi-x))dx=\int\limits_0^{\pi}(\pi-x)f(\sin x)dx\\=\pi\int\limits_0^{\pi}f(\sin x)dx-\int\limits_0^{\pi}xf(\sin x)dx...(a)$$ $$2I=\int\limits_0^{\pi}xf( \sin x)dx +\pi\int\limits_0^{\pi}f(sin x)dx -\int\limits_0^{\pi}xf(\sin x)dx\\=\pi\int\limits_0^{\pi}f(\sin x)dx\\I=\frac{\pi}{2}\int\limits_0^{\pi}f(\sin x)dx$$