Let {I_1} = int_a^{pi - a} {xf(sin x)dx} and | vedclass

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(C) Given ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$.Using the property $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }$,we have:${I_1} = \int_a^{\pi -

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$\frac{2}{\pi }{I_1}$

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(C) Given ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$.Using the property $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }$,we have:${I_1} = \int_a^{\pi - a} {(\pi - x)f(\sin(\pi - x))dx}$.Since $\sin(\pi - x) = \sin x$,this becomes:${I_1} = \int_a^{\pi - a} {(\pi - x)f(\sin x)dx}$.${I_1} = \pi \int_a^{\pi - a} {f(\sin x)dx} - \int_a^{\pi - a} {xf(\sin x)dx}$.${I_1} = \pi {I_2} - {I_1}$.$2{I_1} = \pi {I_2}$.Therefore,${I_2} = \frac{2}{\pi }{I_1}$.

Let ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$ and ${I_2} = \int_a^{\pi - a} {f(\sin x)dx}$,then ${I_2}$ is equal to

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