TITLE: Evaluating $\int_0^1 (1-x^2)^n dx$
QUESTION [7 upvotes]: In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.
b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:
$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$
and so in part (b) i got:
$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$
I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

REPLY [4 votes]: Beta Integral
$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\frac12\int_0^1\left(1-t\right)^nt^{-1/2}\mathrm{d}t\tag{1a}\\
&=\frac{\Gamma(n+1)\Gamma(1/2)}{2\Gamma(n+3/2)}\tag{1b}\\
&=2^{2n}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\tag{1c}\\
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: substitute $t\mapsto t^{1/2}$
$\text{(1b)}$: Beta Integral
$\text{(1c)}$: Gauss' Multiplication Formula
$\text{(1d)}$: convert to Binomial Coefficients via Factorials

Integration by Parts
$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\int_0^12nt^2\left(1-t^2\right)^{n-1}\,\mathrm{d}t\tag{2a}\\
&=\frac{2n}{2n+1}\int_0^1\left(1-t^2\right)^{n-1}\mathrm{d}t\tag{2b}\\
&=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{2c}\\
&=\color{#090}{\frac1{2n+1}\frac{2^nn!}{(2n)!}}\color{#C00}{2^nn!}\tag{2d}\\[3pt]
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{2e}
\end{align}
$$
Explanation:
$\text{(2a)}$: integrate by parts
$\text{(2b)}$: add $\frac{2n}{2n+1}$ of the left side of $\text{(2a)}$ to $\frac1{2n+1}$ of the right side
$\text{(2c)}$: induction
$\text{(2d)}$: evaluate the products
$\text{(2e)}$: simplify