Let $f(x)$ be continuous in $[0,1]$ $$\int_{0}^\pi xf(\sin x) dx = \frac{\pi}{2} \int_{0}^\pi f(\sin x) dx $$
I tried integration by parts and substitution, in vain. How else could I try to prove this?
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Use the idea of https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx/439856#439856 – lab bhattacharjee Mar 01 '18 at 13:19
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1@YuriyS Note that $\sin([0,\pi])=[0,1]$ – Robert Z Mar 01 '18 at 13:19
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We must prove that $$\int_{0}^{\pi}(x-\frac{\pi}{2})f(\sin x)dx=0$$by substituting $u=x-\dfrac{\pi}{2}$ we have$$\int_{0}^{\pi}(x-\frac{\pi}{2})f(\sin x)dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}uf(\cos u)du=0$$since $uf(\cos u)$ is an odd function and its integral over a symmetric interval is zero
Mostafa Ayaz
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Hint. By letting $t=\pi-x$ we get $$\int_{0}^\pi xf(\sin x) dx=\int_{\pi}^0 (\pi-t)f(\sin (\pi-t)) (-dt).$$ Can you take it from here?
Robert Z
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