Prove \int_{0}^{\pi}{x.f(\sin x)}dx=\pi \int_{0}^{ \pi /2}{\sin x dx}
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62\int_{0}^{\pi}{x.f(sin x)}+\int_{0}^{\pi}{(\pi-x).f(sin (\pi-x))} \\ =\int_{0}^{\pi}{x.f(sin x)}+\int_{0}^{\pi}{(\pi-x).f(sin (x))} \\ =\int_{0}^{\pi}{(\pi-x+x).f(sin (x))} =\int_{0}^{\pi}{(\pi).f(sin (x))}
It remains to prove
\int_{0}^{\pi}{(\pi).f(sin (x))} = 2\int_{0}^{\pi/2}{(\pi).f(sin (x))}
Which is very obvious?