Question

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

• A. $\frac{\pi }{2}{\int }_0^{\pi }f\left(\sin x\right)dx$
• B. $\frac{\pi }{4}{\int }_0^{\pi }f\left(\sin x\right)dx$
• C. $\pi {\int }_0^{\pi /2}f\left(\sin x\right)dx$
• D. $\pi {\int }_0^{\pi /2}f\left(\cos x\right)dx$

Answer 1

Answer: (A), (C), (D)

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

Let $I={\int }_0^{\pi }xf\left(sinx\right)dx$
Using property ${\int }_0^{2a}f\left(x\right)dx={\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f(2a-x)dx$
$I={\int }_0^{\frac{\pi }{2}}xf\left(sinx\right)dx+{\int }_0^{\frac{\pi }{2}}(\pi -x)f\left(sin(\pi -x)\right)dx={\int }_0^{\frac{\pi }{2}}\pi f\left(sinx\right)dx$
Now using property ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$I={\int }_0^{\frac{\pi }{2}}\pi f\left(sin(\frac{\pi }{2}-x)\right)dx=\pi {\int }_0^{\frac{\pi }{2}}f\left(cosx\right)dx$