Determine the recurrence formula $\int_{-1}^{1}(1-x^2)^ndx$
I have trouble with finding recurrence formula for this integral. $n$ is natural parameter. I've tried to split up $(1-x^2)^n = (1+x)^n(1-x)^n$ and then to integrate partially, but it only makes things more complicate.
Maybe substitution $x=sint$ can lead to solution?
When I apply it I get: 
$\int_{-\pi/2}^{\pi/2}(cost)^{n+1}dt$
What to do next then?
 A: You can integrate by parts directly to obtain
\begin{align}
\int \limits_{-1}^1 (1-x^2)^n \, \mathrm{d} x &= \left[x (1-x^2)^n \right]_{x=-1}^{x=1} + 2n \int \limits_{-1}^1 x^2 (1-x^2)^{n-1} \, \mathrm{d} x \\
&= 2 n \left[\int \limits_{-1}^1 (1-x^2)^{n-1} \, \mathrm{d} x - \int \limits_{-1}^1 (1-x^2)^n \, \mathrm{d} x\right]
\end{align}
for $n \in \mathbb{N}$, which is the recurrence relation you are after.
A: Not a recurrence, but it seems that
$$
I_n = \int_{-1}^{1}(1-x^2)^ndx = \frac{2^{a_n}}{b_n}
$$
where $a_n$ is OEIS/A030303 and $b_n$ is OEIS/A001803.
Also,
$$
\frac{1}{(1 - x)^{3/2}} = \sum_{n=0}^{\infty} \frac{2}{I_n} x^n
$$
WA gives
$$
I_n = \frac{\sqrt{\pi} \ \Gamma(n + 1)}{\Gamma(n +\frac32)}
= \frac{2^{n+1} n!}{(2n+1)!!}
$$
A: On the path of Nikola Mijušković...
\begin{align}I_n&=\int_{-1}^1 (1-x^2)^n\, dx\\&=2\int_{0}^1 (1-x^2)^n\, dx\end{align}
Perform the change of variable $x=\sin t$,
\begin{align}I_n&=2\int_{0}^{\frac{\pi}{2}} \cos^{2n+1}t\, dt\\
&=2W_{2n+1}
\end{align}
$W_n$ is the n-th Wallis number (see: https://en.wikipedia.org/wiki/Wallis%27_integrals )
It is known that, for $n\geq 0$,
\begin{align}W_1&=1\\
W_{n+2}&=\frac{n+1}{n+2}W_n\end{align}
Therefore,
\begin{align}I_{n+1}&=2W_{2n+3}\\
&=\frac{2(2n+2)}{2n+3}W_{2n+1}\\
&=\frac{2(n+1)}{2n+3}I_{n}
\end{align}
