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