If ${\displaystyle \underset{0}{\overset{\pi }{\int }}xf\left(\sin x\right)dx=A{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)}}dx$ then $A=$

$I={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)f\left(\sin x\right)}dx$

$=\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}f\left(\sin x\right)dx-{\displaystyle \underset{0}{\overset{\pi }{\int }}xf\left(\sin x\right)}}dx$

$2I=\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}f\left(\sin x\right)}dx$

$2I=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)}dx$

${\displaystyle \underset{0}{\overset{\pi }{\int }}xf\left(\sin x\right)dx=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)}}dx$

$A=\pi$