Determine the recurrence formula

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.

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 $$ I_n = \frac{\sqrt{\pi} \ \Gamma(n + 1)}{\Gamma(n +\frac32)} = \frac{2^{n+1} n!}{(2n+1)!!} $$

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}