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