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(D) 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 get: $I = \int_0^{\pi} (\pi - x) f(\

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

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(D) 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 get: $I = \int_0^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx$. Since $\sin(\pi - x) = \sin x$,this becomes: $I = \int_0^{\pi} (\pi - x) f(\sin x) \, dx = \pi \int_0^{\pi} f(\sin x) \, dx - \int_0^{\pi} x f(\sin x) \, dx$. $I = \pi \int_0^{\pi} f(\sin x) \, dx - I$. $2I = \pi \int_0^{\pi} f(\sin x) \, dx$. Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$,here $2a = \pi$,so $a = \frac{\pi}{2}$. Since $f(\sin(\pi - x)) = f(\sin x)$,we have: $2I = \pi \cdot 2 \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$. $I = \pi \int_0^{\frac{\pi}{2}} f(\sin x) \, dx$.

$\int_0^{\pi} x f(\sin x) \, dx$ is equal to

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