Question

If ${\int }_0^{\pi }xf\left(sinx\right)dx=A{\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx,$ then A is equals to

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

Answer 1

The correct option is B $\pi$

$LetI={\int }_0^{\pi }xf\left(sinx\right)dx------\left(1\right)$
$I={\int }_0^{\pi }(\pi -x)f\left[sin\right(\pi -x\left)\right]dx$
$\Rightarrow I={\int }_0^{\pi }(\pi -x)f\left[sin\right(x\left)\right]dx-----\left(2\right)$
$Adding\left(1\right)and\left(2\right)$
$\Rightarrow 2I=\pi {\int }_0^{\pi }f\left(sinx\right)dx$
$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
$\Rightarrow A{\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx=\pi {\int }_0^{\frac{\pi }{2}}f\left(sinx\right)dx$
$\Rightarrow A=\pi$