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?

Part b) can be obtained just by induction. The result is immediate for $n=1$ and
\begin{align*}
\int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\ 
= & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!}-\frac{1}{2n+2}\int_0^1(1-x^2)^{n+1} dx
\end{align*}
This implies that
$$
\int_0^1 (1-x^2)^{n+1}dx = \frac{2n+2}{2n+3} \frac{2^{2n}(n!)^2}{(2n+1)!}=\frac{(2n+2)^2 2^{2n} (n!)^2}{(2n+1)!(2n+2)(2n+3)}=\dfrac{2^{2n+2}((n+1)!)^2}{(2n+3)!}.
$$

\begin{align*}
\int_0^1 x (-2x) (1-x^2)^n dx = &\left[x \dfrac{(1-x^2)^{n+1}}{n+1} \right]_0^1 - \frac{1}{n+1}\int_0^1 (1-x^2)^{n+1}
\end{align*}
