Need help to calculate the integral $\int\limits_{-1}^{1}{(1-x^{2})^{n}dx}$ I want to calculate this integral
$$\int\limits_{-1}^{1}{(1-x^{2})^{n}dx}$$
but I don't know how to start. Should I use the method of changing variables? I somehow know that its value should be $${{2^{2n+1}(n!)^{2}}\over{(2n+1)!}}$$
but I don't know how to prove it
Edit: Using the link that Martin introduced, I managed to calculate the integral. integrating by parts we get
$$I_{n}=\int\limits_{-1}^{1}{(1-x^{2})^{n}dx}=\left[{x(1-x^{2})^{n}}\right]\matrix{
{x=1}\cr
{x=-1}\cr
}-\int\limits_{-1}^{1}{-2nx^{2}}(1-x^{2})^{n-1}dx=2n\int\limits_{-1}^{1}{x^{2}}(1-x^{2})^{n-1}dx$$
Now we transform the expression inside the integral in this way
$$I_{n}=2n\int\limits_{-1}^{1}{x^{2}}(1-x^{2})^{n-1}dx=2n\int\limits_{-1}^{1}{(1+x^{2}-1)(1-x^{2})^{n-1}}dx$$
$$=2n\left[{\int\limits_{-1}^{1}{(1-x^{2})^{n-1}dx-\int\limits_{-1}^{1}{(1-x^{2})^{n}dx}}}\right]=2n(I_{n-1}-I_{n})$$
$$\rightarrow I_{n}=2n(I_{n-1}-I_{n})\rightarrow I_{n}+2nI_{n}=2nI_{n-1}\rightarrow I_{n}={{2n}\over{2n+1}}I_{n-1}$$
Using this recursive relation, We can write
$$I_{n}={{2n}\over{2n+1}}{{2(n-1)}\over{2n-1}}I_{n-2}=\cdots={{2n}\over{2n+1}}{{2(n-1)}\over{2n-1}}\cdots{{2(2)}\over{5}}{{2(1)}\over{3}}I_{0}$$
where $I_{0}=2$. Thus
$$I_{n}=2{{2^{n}(n!)}\over{(2n+1)(2n-1)\cdots 3.1}}=2{{2^{n}(n!)}\over{{{(2n+1)!}\over{2^{n}(n!)}}}}=2{{2^{2n}(n!)^{2}}\over{(2n+1)!}}$$

Use that
$$
\begin{gathered}
\int_{-1}^{1}\left[\left(\begin{array}{l}
n \\
0
\end{array}\right) x^0+\left(\begin{array}{l}
n \\
1
\end{array}\right) x^1+\left(\begin{array}{l}
n \\
2
\end{array}\right) x^2+\cdots+\left(\begin{array}{l}
n \\
n
\end{array}\right) x^n\right] d x=\int_{-1}^{1}(1+x)^n d x \\
{\left[x+\frac{1}{2}\left(\begin{array}{l}
n \\
1
\end{array}\right) x^2+\frac{1}{3}\left(\begin{array}{c}
n \\
2
\end{array}\right) x^3+\cdots+\frac{1}{n+1}\left(\begin{array}{l}
n \\
n
\end{array}\right) x^{n+1}\right]_{-1}^{1}=\left[\frac{1}{n+1}(1+x)^{n+1}\right]_{-1}^{1}}
\end{gathered}
$$
