definite-integral

Question: If $\int_{0}^{\pi }{xf(\sin x)dx=A}\int_{0}^{\pi /2}{f(\sin x)dx}$, then A is



1) $2\pi $
2) $\pi $
3) $\frac{\pi }{4}$
4) 0
Solution: Explanation: No Explanation
Properties of definite integration

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