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

Answer 1

The correct option is A $\pi$

${\int }_0^{\pi }xf\left(sinx\right)dx$

Let $w=\pi -x$ $\therefore$ $\Rightarrow dw=-dx$

$\therefore$ ${\int }_0^{\pi }xf\left(sinx\right)dx=-{\int }_{\pi }^0(\pi -w)f\left(sin(\pi -w)\right)dw$

$={\int }_0^{\pi }(\pi -w)f\left(sinw\right)dw$

$={\int }_0^{\pi }(\pi -x)f\left(sinx\right)dx$

$=\pi {\int }_0^{\pi }f\left(sinx\right)dx-{\int }_0^{\pi }xf\left(sinx\right)dx$

$={\int }_0^{\pi }xf\left(sinx\right)dx=\frac{\pi }{2}{\int }_0^{\pi }f\left(sinx\right)dx$

$=\pi {\int }_0^{\pi /2}f\left(sinx\right)dx$

$\Rightarrow A=\pi$ [A]