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