<s>[INST] How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ I'am trying to prove that
$$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
Here what I have done so far. We know that:
\begin{equation} 
\sin^{2}x + \cos^{2}x = 1
\end{equation}
Let $x = \sin\alpha$ so $\mathrm{d}x = \cos\alpha\:\mathrm{d}\alpha$ 
\begin{equation}
 x^{2} = \sin^{2}\alpha
\end{equation}
\begin{equation} 
\cos^{2}x = 1 -\sin^{2}x
\end{equation}
We have $x = 1 \Rightarrow \alpha = \frac{\pi}{2}$ and $x = -1 \Rightarrow \alpha = - \frac{\pi}{2}$, so we replace the original  integral by:
$$\int_{-1}^{1} (x-1^2)^n\:\mathrm{d}x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha$$
Now integrating by parts,
$$u = \cos^{2n}\alpha \quad \Rightarrow\:\mathrm{d}v = \cos\alpha\:\mathrm{d}\alpha \\
\mathrm{d}u = -2n\cos^{2n-1}\alpha\sin\alpha\:\mathrm{d}\alpha \quad \Rightarrow v = \sin\alpha \\
\int udv = uv - \int vdu$$
Then,
\begin{align}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha &= \underbrace{\cos^{2n}\alpha \sin\alpha \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}_{\text{= 0}} + 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha \sin^{2} \alpha\:\mathrm{d}\alpha \\[10pt]
&= 2n \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha (1-\cos^2\alpha)\:\mathrm{d}\alpha \\
&= 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha\:\mathrm{d}\alpha - 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha
\end{align}
Now we see that
$$(2n+1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha = 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos^{2n-1}\alpha\:\mathrm{d}\alpha$$
I am stuck here, it's almost the same integral as the original one. Any suggestions or corrections will be very welcome. Thanks.
 [/INST] If I try to prove it for $n=1$ I get:
$$\int_{-1}^1(1-x^2)dx=(x-x^3/3)|_{-1}^{1}=2-2/3=4/3\ne\frac{3!1!^2}{2^3}=\frac{3}{4}$$
</s>