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

• A. $0$
• B. $\pi$
• C. $\pi /4$
• D. $2\pi$

Answer 1

The correct option is B $\pi$

$I={\int }_0^{\pi }xf\left(\sin x\right)dx$

Using $I={\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

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

$\therefore 2I=\pi {\int }_0^{\pi }f\left(\sin x\right)dx$

Now using, $I={\int }_0^{2a}f\left(x\right)dx={\int }_0^a(f\left(x\right)+f(2a-x))dx$

$2I=\pi {\int }_0^{\frac{\pi }{2}}(f\left(\sin x\right)+f\left(\sin (\pi -x)\right))dx=\pi {\int }_0^{\frac{\pi }{2}}(f\left(\sin x\right)dx+f\left(\sin (\pi -x)\right))$

$\therefore I=\pi {\int }_0^{\frac{\pi }{2}}f\left(\sin x\right)dx$

Hence, $A=\pi$