If ${\int }_0^{\pi }xf(\sin x)dx=A{\int }_0^{\pi /2}f(\sin x)dx,$ then $A$ is

Let $I={\int }_0^{\pi }xf(\sin x)dx.$ Using the property ${\int }_0^a\varphi (x)dx={\int }_0^a\varphi (a-x)dx$:$I={\int }_0^{\pi }(\pi -x)f(\sin (\pi -x))dx$.
$I={\int }_0^{\pi }(\pi -x)f(\sin x)dx$
$I=\pi {\int }_0^{\pi }f(\sin x)dx-{\int }_0^{\pi }xf(\sin x)dx$
$I=\pi {\int }_0^{\pi }f(\sin x)dx-I$
$2I=\pi {\int }_0^{\pi }f(\sin x)dx$
Since $f(\sin (\pi -x))=f(\sin x),$ we can use the property ${\int }_0^{2a}\psi (x)dx=2{\int }_0^a\psi (x)dx$ if $\psi (2a-x)=\psi (x)$:
$2I=\pi \cdot 2{\int }_0^{\pi /2}f(\sin x)dx$.
$I=\pi {\int }_0^{\pi /2}f(\sin x)dx$.
Comparing with the given form $A{\int }_0^{\pi /2}f(\sin x)dx,$ we find $A=\pi$.