Question

If $\underset{0}{\overset{\pi }{\int }}xf(\sin x)dx=A\underset{0}{\overset{\pi /2}{\int }}f(\sin x)dx,$ then $A$ is

• A. 0
• B. $2\pi$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

Answer: (D)

Let $I={\int }_0^{\pi }xf(\sin x)dx\dots (1)$
$I={\int }_0^{\pi }(\pi -x)f(\sin (\pi -x))dx={\int }_0^{\pi }(\pi -x)f(\sin x)dx$...(2)
Adding (1) and (2)
$2I={\int }_0^{\pi }(x+\pi -x)f(\sin x)dx=\pi {\int }_0^{\pi }f(\sin x)dx$
$\Rightarrow 2I=2\pi {\int }_0^{\pi /2}f(\sin x)dx\Rightarrow I=\pi {\int }_0^{\pi /2}f(\sin x)dx$
$\Rightarrow A=\pi$