Show that:
$$\int_{-1}^1(1-x^2)^{1/2}x^{2n}\ dx = \pi\frac{(2n-1)!!}{(2n+2)!!}$$
for $n=1,2,3,\cdots$, and $\pi/2$ for $n=0$
Well, we know that:
$$$B(m+1,n+1)$ = \frac{m!n!}{(m+n+1)!} = \int_0^1 t^m(1-t)^n\ dt$$
Let's look at that integral again:
$$\int_{-1}^1(1-x^2)^{1/2}x^{2n}\ dx = 2 \int_{0}^1(1-x^2)^{1/2}x^{2n}\ dx$$
now do $x^2 = t \implies x = t^{1/2}\implies dx = \frac{1}{2}t^{-1/2}$ so the integral becomes: $$\frac{2}{2}\int_{0}^1(1-t)^{1/2}t^{n-1/2}\ dt$$
By comparing with our integral for the Beta function, we have:
$$\int_{0}^1(1-t)^{1/2}t^{n-1/2}\ dt$$ $$\int_0^1 (1-t)^n t^m\ dt$$
a little confusion with $m$ and $n$ can be undone because $B(m,n) = B(n,m)$, so we should have:
$$\int_{0}^1(1-t)^{1/2}t^{n-1/2}\ dt = \frac{\left(n-\frac{1}{2}\right)!\frac{1}{2}!}{(n-\frac{1}{2}+\frac{1}{2}+1)!} = \frac{\left(n-\frac{1}{2}\right)!\frac{1}{2}!}{(n-1)!}$$
It's known that
$$ \left(n-\frac{1}{2}\right)! = \Gamma\left(n+\frac{1}{2}\right) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}$$
And also:
$$\frac{1}{2}! = \Gamma\left(\frac{3}{2}\right) = \frac{1}{2}\Gamma\left(\frac{1}{2}\right) = \frac{1}{2}\sqrt{\pi}$$
but what about $(n-1)!$?