Proving $\int_{-1}^{1}(1-x^2)^n\,\mathrm dx=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$ for $n=0,1,2,3...$ 
Prove that
  $$\int_{-1}^{1}(1-x^2)^ndx=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$$ for $n=0,1,2,3...$

I tried to substitute $x=\sin\theta$:
$$\int_{-1}^{1}(1-x^2)^n\,\mathrm dx=\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\theta\,\mathrm d\theta=2\int_{0}^{\pi/2}\cos^{2n+1}\theta\,\mathrm d\theta$$
but I could not continue further.
 A: This answer is inspired by the usual calculation one does with the Wallis' product.
Let
$$
I_n=\int_{-1}^1(1-x^2)^n\,dx.
$$
Then, integrating by parts,
$$
\begin{aligned}
I_{n+1}&=\int_{-1}^1 1\cdot (1-x^2)^{n+1}\,dx\\
&=\bigl[x(1-x^2)^{n+1}\bigr]_{-1}^1-\int_{-1}^1 x(n+1)(1-x^2)^n(-2x)\,dx\\
&=-2(n+1)\int_{-1}^{1}(1-x^2-1)(1-x^2)^n\,dx\\
&=-2(n+1)I_{n+1}+2(n+1)I_n.
\end{aligned}
$$
Hence,
$$
I_{n+1}=\frac{2n+2}{2n+3}I_n.
$$
Now, check that the right-hand side satisfies the same recursion equation, and that $I_0$ agrees with the right-hand side you have when $n=0$.
A: Using Euler's Beta function:
$$ \int_{-1}^{1}(1-x^2)^n\,dx = 2\int_{0}^{1}(1-x^2)^n\,dx = \int_{0}^{1}z^{-1/2}(1-z)^{n}\,dz = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{3}{2}\right)}$$
then, since $\Gamma(z+1)=z\,\Gamma(z)$:
$$\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n+1\right)}{\Gamma\left(n+\frac{3}{2}\right)}=\frac{\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}\right)}\prod_{k=1}^{n}\frac{n}{n+\frac{1}{2}}=2\cdot\frac{(2n)!!}{(2n+1)!!}=\frac{2^{2n+1}n!^2}{(2n+1)!} $$
as wanted.
