definite-integral

Question: Let ${{I}_{1}}=\int_{a}^{\pi -a}{xf(\sin x)dx,\,{{I}_{2}}=\int_{a}^{\pi -a}{\,\,f(\sin x)dx}}$, then ${{I}_{2}}$ is equal to



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

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