Question

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

The correct options are
A $\frac{\pi }{2}{\int }_0^{\pi }f\left(\sin x\right)dx$
C $\pi {\int }_0^{\pi /2}f\left(\sin x\right)dx$
D $\pi {\int }_0^{\pi /2}f\left(\cos x\right)dx$

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

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

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

Hence $=\pi {\int }_0^{\pi }xf\left(\sin x\right)dx=\left(\pi /2\right){\int }_0^{\pi }f\left(\sin x\right)dx$

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

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