If int_0^pi {xf(sin x)dx = A} int_0^{pi /2} { | vedclass

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(B) Let $I = \int_0^\pi {xf(\sin x)dx}$.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}

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$\pi $

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(B) Let $I = \int_0^\pi {xf(\sin x)dx}$.Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we have:$I = \int_0^\pi {(\pi - x)f(\sin(\pi - x))dx} = \int_0^\pi {(\pi - x)f(\sin x)dx}$.Adding the two expressions for $I$:$2I = \int_0^\pi {xf(\sin x)dx} + \int_0^\pi {(\pi - x)f(\sin x)dx} = \int_0^\pi {\pi f(\sin x)dx} = \pi \int_0^\pi {f(\sin x)dx}$.Using the property $\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(2a-x) = f(x)$,we have $\int_0^\pi {f(\sin x)dx} = 2\int_0^{\pi /2} {f(\sin x)dx}$.Thus,$2I = \pi 	imes 2 \int_0^{\pi /2} {f(\sin x)dx} = 2\pi \int_0^{\pi /2} {f(\sin x)dx}$.Therefore,$I = \pi \int_0^{\pi /2} {f(\sin x)dx}$.Comparing this with $A \int_0^{\pi /2} {f(\sin x)dx}$,we get $A = \pi$.

If $\int_0^\pi {xf(\sin x)dx = A} \int_0^{\pi /2} {f(\sin x)dx} $,then $A$ is

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