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 $$I_{n}=2{{2^{n}(n!)}\over{(2n+1)(2n-1)\cdots 3.1}}=2{{2^{n}(n!)}\over{{{(2n+1)!}\over{2^{n}(n!)}}}}=2{{2^{2n}(n!)^{2}}\over{(2n+1)!}}$$
The idea is to rewrite your problem to $$B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1} dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$ where $Re(p),Re(q)>0$.
Exercise: Rewrite your integral to $$\int_0^1t^{-1/2}(1-t)^ndt$$ For suitable $p$ and $q$ you will obtain your answer.
We want to show that $$\frac{\Gamma(\tfrac{1}{2})\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}={{2^{2n+1}(n!)^{2}}\over{(2n+1)!}}$$ First we know that $\Gamma(\tfrac{1}{2})=\sqrt{\pi}$. Second we know that $\Gamma(n+1)=n\Gamma(n)$. On the other hand we know that $\Gamma(n)=(n-1)!$ so $n\cdot (n-1)!=n!$ hence $$\frac{\Gamma(\tfrac{1}{2})\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}=\frac{n!\sqrt{\pi}}{\Gamma(n+\frac{3}{2})}$$ So we have to deal with $\Gamma(n+\frac{3}{2})$. Now $$\Gamma(n+\tfrac{3}{2})=(n+\tfrac{1}{2})!=(n+\tfrac{1}{2})(n-\tfrac{1}{2})!=(n+\tfrac{1}{2})\Gamma(n+\tfrac{1}{2})$$ An identity from Wiki (https://en.wikipedia.org/wiki/Gamma_function#General) tell us that $$\Gamma(n+\tfrac{1}{2})=\frac{(2n)!}{n!4^n}\sqrt{\pi}$$ Hence $$\frac{n!\sqrt{\pi}}{\Gamma(n+\frac{3}{2})}=\frac{n!\sqrt{\pi}}{(n+\frac{1}{2})\Gamma(n+\tfrac{1}{2})}=\frac{n!\sqrt{\pi}}{(n+\frac{1}{2})\frac{(2n)!}{n!4^n}\sqrt{\pi}}$$ From here it should be easy to see that $$\frac{n!\sqrt{\pi}}{(n+\frac{1}{2})\frac{(2n)!}{n!4^n}\sqrt{\pi}}=\frac{2\cdot 4^{n} (n !)^{2}}{\left(2 n +1\right)!}=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$$ Notice that I used the fact that $n+\frac{1}{2}=\frac{1}{2}(2n+1)$ and $(2n+1)\cdot (2n)!=(2n+1)!$. Hope it helps.