Calculate integral $\int_{-1}^{1} (1-x^2)^n dx $

Hint. By using integration by parts, we have $$ \begin{align} I_n=\int_{-1}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{-1}^{1}+2n\int_{-1}^{1}x^2(1-x^2)^{n-1}dx \\\\&=0+2n\int_{-1}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx \\\\&=2nI_{n-1}-2nI_{n} \end{align} $$ giving $$ I_{n}=\frac{2n}{2n+1}\cdot I_{n-1}, \quad n\ge1, $$ with $I_0=2,\,I_1=\frac43.$

I think you can take it from here.

HINT:

Note that

$$\begin{align} \int_{-1}^1(1-x^2)^n\,dx&=\int_0^1 u^{-1/2}(1-u)^n\,du\\\\ &=B(1/2,n+1)\\\\ &=\frac{\Gamma(1/2)\Gamma(n+1)}{\Gamma(n+3/2)} \end{align}$$

Now, use the functional relationship $\Gamma(x+1)=x\Gamma(x)$ along with $\Gamma(n+1)=n!$.

Let $J_{p,q}=\int_{-1}^1 (1-x)^p(1+x)^q\,dx$. You have, by IPP : $$J_{p,q}=-\frac{1}{p+1}\left[(1-x)^{p+1}(1+x)^q\right]{-1}^1 + \frac{q}{p+1}\int_{-1}^1 (1-x)^{p+1}(1+x)^{q-1}\,dx$$ The first term cancels, so $$J_{p,q}=\frac{q}{p+1}J_{p+1,q-1}$$ By induction, you then prove $$J_{p,q}=\frac{q!p!}{(p+q)!}J_{p+q,0}=\frac{2^{p+q+1}q!p!}{(p+q+1)!}$$ Finally : $$I_n=J_{n,n}=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$$