Prove that $\int _0^1x^a\left(1-x\right)^bdx$ = $\int _0^1x^b\left(1-x\right)^adx$, where $a,b\in \mathbb{R}$ Prove that $$\int _0^1x^a\left(1-x\right)^bdx = \int _0^1x^b\left(1-x\right)^adx$$
How can I even get started on this? I evaluate the integral with parts, but it just gets more and more tedious since I'm working with these constants here. 

Substitute $u=(1-x)$. We then have $du=-dx$ and when $x=0$, we have $u=1$, $x=1$ gives $u=0$. Thus
$$\int_0^1x^a(1-x)^bdx=-\int_1^0(1-u)^au^bdu=\int_0^1(1-u)^au^bdu$$          
No integration by parts or anything necessary, just a straight substitution.
