∫limits_0^πf(sin x) dx is equal to

$I=\int\limits_{0}^{\pi}f(sin x) dx$

$=\int\limits_{0}^{\pi}(\pi-x)f\,sin(\pi-x)dx=\pi\int\limits_{0}^{\pi}f(sinx)dx-I$

$⇒I=\frac{\pi}{2}f(sinx)dx$.

Again, $I =\frac{\pi}{2}\int\limits_{0}^{\pi}f(sinx) dx = 2\frac{\pi}{2}\int\limits_{0}^{\pi/2}f(sinx) dx = \pi\int\limits_{0}^{\pi/2}f(sinx) dx$.

Hence (B) is the correct answer.