If $\int_0^\pi x f(\sin x) d x=A \int_0^{\pi / 2} f(\sin x) d x$, then $A$ is
If $\int_0^\pi x f(\sin x) d x=A \int_0^{\pi / 2} f(\sin x) d x$, then $A$ is
0
$\pi$
$\frac{\pi}{4}$
$2 \pi$
Solution
Let $I=\int_0^\pi x f(\sin x) d x=\int_0^\pi(\pi-x) f(\sin x) d x=\pi \int_0^\pi f(\sin x) d x-I \quad(\sin c e f(2 a-x)=f(x))$
$\Rightarrow I=\pi \int_0^{\pi / 2} f(\sin x) d x \Rightarrow A=\pi$