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

Solution:

Let $I={\int }_0^{\pi } xf(\sin x)dx\text{. }$

$\begin{array}{l}I={\int }_0^{\pi } (\pi -x)f(\sin (\pi -x\left)\right)dx \left[\because {\int }_0^a f\left(x\right)dx={\int }_0^a f(a-x)dx\right]\\ \Rightarrow I=\pi {\int }_0^{\pi } f(\sin x)dx-I\\ \Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi } f(\sin x)dx\end{array}$

$\begin{array}{l}\Rightarrow I=\frac{\pi }{2}\times 2{\int }_0^{\pi /2} f(\sin x)dx \left[\because {\int }_0^{2a} f\left(x\right)dx=2{\int }_0^a f\left(x\right)dx\text{ if, }f(2a-x)=f\left(x\right)\right]\\ \Rightarrow I=\pi {\int }_0^{\pi /2} f(\sin x)\\ \therefore {\int }_0^{\pi } xf(\sin x)dx=A{\int }_0^{\pi /2} f(\sin x)dx\\ \Rightarrow \pi {\int }_0^{\pi /2} f(\sin x)dx=A{\int }_0^{\pi /2} f(\sin x)dx\\ =A=\pi \end{array}$