Show that

$\displaystyle \int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$

Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as $\displaystyle I_n$. We have the recurrence relation:

$\displaystyle I_{k + 1} = \left ( \frac{2k + 1}{2k + 3} \right )~ I_k$

-Dan