TITLE: Determine the recurrence formula
QUESTION [5 upvotes]: $\int_{-1}^{1}(1-x^2)^ndx$
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?

REPLY [2 votes]: 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}