<s>[INST] Evaluating $\int_0^1 (1-x^2)^n dx$ 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?
 [/INST] My idea here is to use the Beta function:
$$ \int_0^1 t^{m}(1 - t)^n \;dt = \frac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}. $$
Where the Gamma function satisfies $\Gamma(n + 1) = n!$ if $n$ is an integer, and also $\Gamma$ satisfies the recurrence $\Gamma(n + 1) = n\Gamma(n)$.
So if we substitute $t = x^2$, $dt = 2x\; dx \iff \frac{1}{2\sqrt{t}}\;dt=dx$ then
\begin{align}
\int_0^1 (1 - x^2)^n \; dx &= \frac{1}{2} \int_0^1 t^{-1/2} (1 - t)^n \;dt \\
&= \frac{1}{2} \frac{\Gamma(\frac12)n!}{\Gamma(n+\frac32)}. \tag{1}
\end{align}
And now using the recurrence $\Gamma(n + 1) = n\Gamma(n)$, we find that
\begin{align}
\Gamma(n+\tfrac32) &= (n+\tfrac12)\Gamma(n + \tfrac12) \\
&= (n+\tfrac12)(n - \tfrac12) \Gamma(n - \tfrac12) \\
&\hspace{2.5mm}\vdots \\
&= (n+\tfrac12)(n - \tfrac12) \cdots \tfrac{5}{2} \cdot \tfrac{3}{2} \cdot \tfrac12 \cdot \Gamma(\tfrac12) \\
&= \frac{1}{2^n} (2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 \cdot \Gamma(\tfrac12) \tag{2}
\end{align}
The product $(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1$ is also known as the double factorial $(2n + 1)!!$. It is well-known that we can rewrite it in terms of $(2n + 1)!$ as
\begin{align}
(2n + 1)(2n - 1) \cdots 5 \cdot 3 \cdot 1 &= \frac{(2n + 1)(2n)(2n - 1)(2n - 2) \cdots 3 \cdot 2 \cdot 1}{(2n)(2n-2) \cdots 2} \\
&= \frac{(2n+1)!}{2^n \cdot n!}. \tag{3}
\end{align}
So now finally, if you combine $(1), (2), (3)$, you get the result you want.
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