int_{0}^{pi} x f(sin x) dx = | vedclass

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(D) Let $I = \int_{0}^{\pi} x f(\sin x) dx$  $(1)$Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:$I = \int_{0}^{\pi} (\pi -

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$\pi \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$

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(D) Let $I = \int_{0}^{\pi} x f(\sin x) dx$  $(1)$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 $\sin(\pi - x) = \sin x$,we have:$I = \int_{0}^{\pi} (\pi - x) f(\sin x) dx$$I = \pi \int_{0}^{\pi} f(\sin x) dx - \int_{0}^{\pi} x f(\sin x) dx$$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$Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$:Here $f(\sin(\pi - x)) = f(\sin x)$,so $\int_{0}^{\pi} f(\sin x) dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$$I = \frac{\pi}{2} 	imes 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get $\int_{0}^{\frac{\pi}{2}} f(\sin x) dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$Therefore,$I = \pi \int_{0}^{\frac{\pi}{2}} f(\cos x) dx$.

$\int_{0}^{\pi} x f(\sin x) dx = $

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