A) \[\pi \int_{0}^{\pi }{f(\sin x)\,dx}\]
B) \[\frac{\pi }{2}\int_{0}^{\pi }{f(\sin x)\,dx}\]
C) \[\frac{\pi }{2}\int_{0}^{\pi /2}{f(\sin x)\,dx}\]
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{\pi }{x}f(\sin x)dx=\frac{\pi }{2}\int_{0}^{\pi }{f(\sin x)dx}\] Since \[\int_{0}^{a}{xf(x)dx=\frac{1}{2}a\int_{0}^{a}{f(x)dx,}}\]if\[f(a-x)=f(x)\].
You need to login to perform this action.
You will be redirected in
3 sec
