<s>[INST] Find the value of a given integral sequence: $I_n=\int_{0}^{1}(1-x^2)^ndx$ Let $(I_n)_{n \geq 1}$ be a sequence such that:
$$I_n = \int_0^1 (1-x^2)^n dx$$
Find the value of $I_n$ (The solution is $\frac{2}{3} \cdot \frac{4}{5} \cdot ... \cdot \frac{2n}{2n+1}$).
I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.
Thank you!
 [/INST] Here is an approach that relies on the Beta function $\operatorname{B}(m,n)$, namely
$$\operatorname{B} (m,n) = \int_0^1 x^{m - 1} (1 - x)^{m - 1} \, dx.$$
In the integral for $I_n$ we begin by enforcing a substitution of $x \mapsto \sqrt{x}$. Thus
\begin{align}
I_n &= \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx\\
&= \frac{1}{2} \int_0^1 x^{\frac{1}{2} - 1} (1 - x)^{(n + 1) - 1} \, dx\\
&= \frac{1}{2} \operatorname{B} \left (\frac{1}{2}, n + 1 \right )\\
&= \frac{1}{2} \frac{\Gamma (\frac{1}{2}) \Gamma (n + 1)}{\Gamma (n + \frac{3}{2} )}\\
&= \frac{\sqrt{\pi}}{2} \frac{\Gamma (n + 1)}{(n + \frac{1}{2}) \Gamma (n + \frac{1}{2})}. 
\end{align}
Since $n$ is a positive integer, we have
$$\Gamma (n + 1) = n! \quad \text{and} \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$$
then
$$I_n = \frac{2^{2n}}{2n + 1} \frac{(n!)^2}{(2n)!} = \frac{2^{2n}}{(2n + 1) \binom{2n}{n}}.$$
Here $\binom{2n}{n}$ denotes the central binomial coefficient.
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