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 [AMU 2000]

Let \[{{I}_{1}}=\int_{a}^{\pi -a}{xf(\sin x)dx,\,{{I}_{2}}=\int_{a}^{\

1.	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 [AMU 2000]
A.	\[\frac{\pi }{2}{{I}_{1}}\]
B.	\[\pi \,{{I}_{1}}\]
C.	\[\frac{2}{\pi }{{I}_{1}}\]
D.	\[2{{I}_{1}}\]
Answer» D. \[2{{I}_{1}}\]

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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 [AMU 2000]

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