I need help again, after pondering this for a while

If a and b are positive numbers, prove that

$\displaystyle \int^1_0 x^a(1-x)^b\ dx = \int^1_0 x^b(1-x)^a\ dx $

I think I should make $\displaystyle u = 1 - x$ and $\displaystyle x = 1 - u$, then I get:

$\displaystyle -\int^1_0 u^b(1-u)^a\ du $

This is almost correct, but after the substitution $\displaystyle u=1-x$ , the integral limits get reversed: $\displaystyle x=0\Longrightarrow u=1\,,\,\,x=1\Longrightarrow u=0$ , so all is fine but the last integral's lower limit is 1 and the upper one is 0. Tonio
However, I have been stuck here for a while.

I don't know how to get rid of the negative nor how to substitute in back x without messing it back up.