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$\int_0^\pi x f(\sin x) d x=$
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$\frac{\pi}{2} \int_0^\pi f(\sin x) d x$
$\begin{aligned} & \int_0^\pi x f(\sin x) d x=\frac{\pi}{2} \int_0^\pi f(\sin x) d x \\ & \text {Since } \int_0^a x f(x) d x=\frac{1}{2} a \int_0^{a} f(x) d x \text {, if } f(a-x)=f(x) .\end{aligned}$
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