A) \[0\]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) \[2\pi \]
E) \[3\pi \]
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi }{x}f(\sin x)dx\] ...(i) \[=\int_{0}^{\pi }{(\pi -x)}f[\sin (\pi -x)]dx\] \[\Rightarrow \]\[=\int_{0}^{\pi }{(\pi -x)}f(\sin x)dx\] ...(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi }{\pi }f(\sin x)dx\] \[\Rightarrow \]\[I=\frac{\pi }{2}\int_{0}^{\pi }{f(\sin x)dx=\pi }\int_{0}^{\pi /2}{f(\sin x)}dx\] \[\Rightarrow \]\[A\int_{0}^{\pi /2}{f(\sin x)dx=\pi }\int_{0}^{\pi /2}{f(\sin x)}dx\] \[\Rightarrow \] \[A=\pi \]You need to login to perform this action.
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