Question

If ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$, then the value of ${\int }_0^{\pi }xf\left(\sin x\right)dx=$

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

Answer 1

The correct option is B $\frac{\pi }{2}{\int }_0^{\pi }f\left(\sin x\right)dx$

Let $S={\int }_0^{\pi }xf\left(sinx\right)dx$

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

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

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

$S=\pi {\int }_0^{\pi }f\left(sinx\right)dx-S$

$2S=\pi {\int }_0^{\pi }f\left(sinx\right)dx$

$S=\frac{\pi }{2}{\int }_0^{\pi }f\left(sinx\right)dx$

${\int }_0^{\pi }xf\left(sinx\right)dx=\frac{\pi }{2}{\int }_0^{\pi }f\left(sinx\right)dx$