Question

${\int }_0^{\pi }xf\left(\sin x\right)dx$ is equal to

• A. $\pi {\int }_0^{x}f\left(\cos x\right)dx$
• B. $\pi {\int }_0^{x}f\left(\sin x\right)dx$
• C. $\frac{\pi }{2}{\int }_0^{x/2}f\left(\sin x\right)dx$
• D. $\pi {\int }_0^{\pi /2}f\left(\cos x\right)dx$

Answer 1

The correct option is D $\pi {\int }_0^{\pi /2}f\left(\cos x\right)dx$

Let $I={\int }_0^{\pi }xf\left(\sin x\right)dx$ (i)
$\Rightarrow$ $I={\int }_0^{\pi }(\pi -x)f\left(\sin (\pi -x)\right)dx$
$\Rightarrow$ $I={\int }_0^{\pi }(\pi -x)f\left(\sin x\right)dx$ (ii)
Adding (i) and (ii), we get
$2I={\int }_0^{\pi }\pi f\left(\sin x\right)dx$ $\Rightarrow$

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