If f is continuous on [0,$\displaystyle \pi$], use the substitution $\displaystyle u = \pi - x$ to prove that:

$\displaystyle \int_0^{\pi}xf(sinx)dx = \frac{\pi}{2}\int_0^{\pi}f(sinx)dx$

Completely lost here. $\displaystyle du = -1dx $ and I don't know how to start from there.