A) \[2\pi \]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) 0
Correct Answer: B
Solution :
Let \[=\int\limits_{0}^{\pi }{xf}(\sin x)dx=\int\limits_{0}^{\pi }{(\pi -x)f(\sin x)dx}\] \[\therefore \,\,2I=\pi \int\limits_{0}^{\pi }{f(\sin x)dx=\pi .2\int\limits_{0}^{\frac{\pi }{2}}{f(\sin x)\,dx}}\] \[\therefore \,I=\pi \int\limits_{0}^{\frac{\pi }{2}}{f\,(\sin x)dx\Rightarrow A=\pi }\]
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