Is $\int\limits_{0}^{\pi}x\cdot f(\sin x)dx=\pi \cdot \int\limits_{0}^{\pi/2} f(\sin x)dx$?

Question

How can I prove that for all continuous $f$ $$\int\limits_{0}^{\pi}x\cdot f(\sin x)dx$$ $$=\pi \int\limits_{0}^{\pi/2}f(\sin x)dx$$