Question

If ${\int }_0^{\pi }xf\left(\sin x\right)dx=A{\int }_0^{\pi /2}f\left(\sin x\right)dx,$ then $A$ is equal to:

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

Answer 1

The correct option is B $\pi$

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