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

Question

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.

Answer 1

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 $n=0$.

Answer 2

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.