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.
 A: The same working with $n$ replaced by $n-1$ and then $n-2$ gives
$$(2n-1)\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\alpha\,d\alpha
  =(2n-2)\int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha$$
and
$$(2n-3)\int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha
  =(2n-4)\int_{-\pi/2}^{\pi/2}\cos^{2n-5}\alpha\,d\alpha\ .$$
Therefore
$$\eqalign{
  \int_{-\pi/2}^{\pi/2}\cos^{2n+1}\alpha\,d\alpha
  &=\frac{2n}{2n+1}\int_{-\pi/2}^{\pi/2}\cos^{2n-1}\alpha\,d\alpha\cr
  &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}
    \int_{-\pi/2}^{\pi/2}\cos^{2n-3}\alpha\,d\alpha\cr
  &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}
    \int_{-\pi/2}^{\pi/2}\cos^{2n-5}\alpha\,d\alpha\ .\cr}$$
Continuing in the same way gives eventually
$$\eqalign{\int_{-\pi/2}^{\pi/2}\cos^{2n+1}\alpha\,d\alpha
  &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}\cdots\frac45\frac23
    \int_{-\pi/2}^{\pi/2}\cos\alpha\,d\alpha\cr
  &=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\frac{2n-4}{2n-3}\cdots\frac45\frac232
    \ .\cr}$$
This is the value of the integral: we now need to simplify.  First multiply top and bottom by $(2n)(2n-2)(2n-4)\cdots2$, then take out as many factors of $2$ as possible: the integral is
$$I=\frac{2^{2n+1}n^2(n-1)^2(n-2)^2\cdots1^2}{(2n+1)(2n)(2n-1)\cdots(3)(2)}
  =\frac{2^{2n+1}(n!)^2}{(2n+1)!}\ .$$
Comments


*

*This is the method of integration by reduction formula (look it up).

*You don't actually need the trig substitution as you can integrate
$$\int_{-1}^1(1-x^2)^n\,dx
  =\bigl[x(1-x^2)^n\bigr]_{-1}^1+2n\int_{-1}^1 x^2(1-x^2)^{n-1}\,dx\ ,$$
which leads to the alternative reduction formula
$$\int_{-1}^1(1-x^2)^n\,dx
  =\frac{2n}{2n+1}\int_{-1}^1 (1-x^2)^{n-1}\,dx\ .$$
Then use this in the same way as above.

A: Consider the integral $\int_{0}^{+1} (1-x^2)^n dx$
Let $I(n)$ be the above integral do the substitution $x=\sin t$. Then $dx=\cos t dt$ and 
$1-x^2=cos^2 t$.  
Also $t=0$ for $x=0$ and $t=\pi/2$ for $x=1$ so
$I(n) = \int_0^{\pi/2}  \cos^{2n+1} t \ dt = \int_0^{\pi/2} \cos^{2n} t \ \cos t \ dt$.
To integrate by parts do the substitution  
$$u = \cos^{2n} t$$ and $$dv=\cos t dt$$ From where we get
$$du = 2n \cos^{2n-1} t (-\sin t) dt  $$     and
$$v=\sin t$$
Now, $uv$ is zero at both limits so  
$$I(n) = 2n \int \cos^{2n-1} t \sin^2 t dt = 2n \int \cos^{2n-1} t (1-\cos^2 t) dt = 2n [ I(n-1) - I(n) ]$$ from where we get
$$I(n) = \frac{2n}{2n+1}I(n-1)$$
It is clear that $I(0)=1$ so 
$$I(n) = \frac{(2n) 2(n-1) 2(n-2) ... 2}{ [ (2n+1) (2n-1) ... 3]}$$ 
Multiplying both the numerator and the denominator by $(2n) 2(n-1) 2(n-2) ... 2$ the denominator becomes $(2n+1)!$ where as the numerator is the square of $2^n n!$ So, finally 
$I(n)= \frac{2^{2n} (n!^2)} { (2n+1)!}$
The asked integral is just the double of the computed integral.
A: If I try to prove it for $n=1$ I get:
$$\int_{-1}^1(1-x^2)dx=(x-x^3/3)|_{-1}^{1}=2-2/3=4/3\ne\frac{3!1!^2}{2^3}=\frac{3}{4}$$
