Calculate integral $\int_{-1}^{1} (1-x^2)^n dx $ How to show that :
$I_n = \int_{-1}^{1} (1-x^2)^n dx $ is equal to $\dfrac{2^{2n+1}(n!)^2}{(2n+1)!}$ ?
With integration by parts ? I don't know how to prove this equality. Someone could help me ? Thank you in advance :)
 A: Hint. By using integration by parts, we have
$$
\begin{align}
I_n=\int_{-1}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{-1}^{1}+2n\int_{-1}^{1}x^2(1-x^2)^{n-1}dx
\\\\&=0+2n\int_{-1}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx
\\\\&=2nI_{n-1}-2nI_{n}
\end{align}
$$ giving
$$
I_{n}=\frac{2n}{2n+1}\cdot I_{n-1}, \quad n\ge1,
$$ with $I_0=2,\,I_1=\frac43.$
I think you can take it from here.
A: HINT:
Note that 
$$\begin{align}
\int_{-1}^1(1-x^2)^n\,dx&=\int_0^1 u^{-1/2}(1-u)^n\,du\\\\
&=B(1/2,n+1)\\\\
&=\frac{\Gamma(1/2)\Gamma(n+1)}{\Gamma(n+3/2)}
\end{align}$$
Now, use the functional relationship $\Gamma(x+1)=x\Gamma(x)$ along with $\Gamma(n+1)=n!$.
A: Let $J_{p,q}=\int_{-1}^1 (1-x)^p(1+x)^q\,dx$. You have, by IPP :
$$J_{p,q}=-\frac{1}{p+1}\left[(1-x)^{p+1}(1+x)^q\right]{-1}^1 + \frac{q}{p+1}\int_{-1}^1 (1-x)^{p+1}(1+x)^{q-1}\,dx$$
The first term cancels, so 
$$J_{p,q}=\frac{q}{p+1}J_{p+1,q-1}$$
By induction, you then prove 
$$J_{p,q}=\frac{q!p!}{(p+q)!}J_{p+q,0}=\frac{2^{p+q+1}q!p!}{(p+q+1)!}$$
Finally :
$$I_n=J_{n,n}=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
A: I would be tempted to use the substitution $x=\sin t$ to get it as
$$\int_{-\pi/2}^{\pi/2}\cos^{2n+1}t\,dt.$$
There are many ways to do this, but integration by parts leads to a
"reduction formula" relating $I_n$ (this integral) to $I_{n-1}$
(which I'm sure will be the same as Olivier Oloa's).
