A) \[\frac{\pi }{2}{{I}_{1}}\]
B) \[\pi {{I}_{1}}\]
C) \[\frac{2}{\pi }{{I}_{1}}\]
D) \[2{{I}_{1}}\]
Correct Answer: C
Solution :
Since,\[{{I}_{1}}=\int_{a}^{\pi -a}{x\,\,f(\sin x)\,\,dx}\] \[=\int_{a}^{\pi -a}{(\pi -x)f(\sin (\pi -x))}\,\,dx\] \[=\int_{a}^{\pi -a}{(\pi -x)f(\sin x)}\,\,dx\] \[=\int_{a}^{\pi -a}{\pi f(\sin x)}\,\,dx-{{I}_{1}}\] \[\Rightarrow \] \[2{{I}_{1}}={{I}_{2}}\] \[\Rightarrow \] \[{{I}_{2}}=\frac{2}{\pi }{{I}_{1}}\]You need to login to perform this action.
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