Prove that $\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$ Prove that : 
$$\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$$

The trick is induction together with integration by parts. Denote our integral by $I_n$. Then
$$
I_{n+1}=\int_{-1}^1(1-x^2)^n(1-x^2)\text{d}x=I_n-\int_{-1}^1 x^2(1-x^2)^n\text{d}x.
$$
Integrating by parts by $u=x$ and $v'=x(1-x^2)^n$ we arrive at the recurrence $$I_{n+1}=\frac{2n+2}{2n+3}I_n.$$
Therefore $$I_{n+1}=\frac{2(n+1)\cdot 2(n+1)(n!)^2\cdot 2^{2n+1}}{2(2n+3)(n+1)(2n+1)!}=\frac{\bigl((n+1)!\bigr)^2\cdot 2^{2n+3}}{(2n+3)!},$$ which is the induction step. I leave to you the verification of the formula for $n=1$.
