int_0^pi x f(sin x) dx = | vedclass

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(B) Let $I = \int_0^\pi x f(\sin x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (\pi - x) f(\sin(\pi - x)

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$\frac{\pi}{2} \int_0^\pi f(\sin x) dx$

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(B) Let $I = \int_0^\pi x f(\sin x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^\pi (\pi - x) f(\sin(\pi - x)) dx$. Since $\sin(\pi - x) = \sin x$,this becomes: $I = \int_0^\pi (\pi - x) f(\sin x) dx = \pi \int_0^\pi f(\sin x) dx - \int_0^\pi x f(\sin x) dx$. $I = \pi \int_0^\pi f(\sin x) dx - I$. $2I = \pi \int_0^\pi f(\sin x) dx$. $I = \frac{\pi}{2} \int_0^\pi f(\sin x) dx$.

$\int_0^\pi x f(\sin x) dx = $

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