A) \[\frac{\pi }{2}{{I}_{1}}\]
B) \[\pi \,{{I}_{1}}\]
C) \[\frac{2}{\pi }{{I}_{1}}\]
D) \[2{{I}_{1}}\]
Correct Answer: C
Solution :
\[{{I}_{1}}=\int_{a}^{\pi -a}{xf(\sin x)dx}\]\[=\int_{a}^{\pi -a}{(\pi -x)\,f\,(\sin (\pi -x))\,dx}\], \[[\because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)\,dx}}]\] \[=\int_{a}^{\pi -a}{(\pi -x)\,f\,(\sin x)\,dx}\]\[=\int_{a}^{\pi -a}{\pi \,f\,(\sin x)\,dx-{{I}_{1}}}\] \[\Rightarrow 2{{I}_{1}}=\pi \,{{I}_{2}}\,\Rightarrow {{I}_{2}}=\frac{2}{\pi }{{I}_{1}}\].
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