# Calculate integral $\int_{-1}^{1} (1-x^2)^n dx$ [duplicate]

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 :)

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.

• Many thanks to the three of you ! Apr 15, 2017 at 20:35
• @Mélanie De la Cheminée You are welcome. Apr 15, 2017 at 20:41
• I still can't see how to get from the recurrence relation to $\frac{2^{2n+1}{(n!)}^{2}}{(2n+1)!}$. Please could you provide more hints?
– rl4y
Dec 28, 2020 at 19:03
• Please have a look here: en.wikipedia.org/wiki/Wallis'_integrals#Recurrence_relation Dec 29, 2020 at 8:13

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 $\Gamma(n+1)=n!$.

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 : $$I_n=J_{n,n}=\frac{2^{2n+1}(n!)^2}{(2n+1)!}$$

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).