1. ## substitution problem

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.

2. From

\displaystyle \begin{aligned} u&=\pi-x\\ du&=-dx \end{aligned}

we may derive

\displaystyle \begin{aligned} x&=\pi-u\\ dx&=-du. \end{aligned}

Hope this helps!