<s>[INST] Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$ 
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$

I tried to integrate that function for $n=1,2,3$ to see whether there's a pattern (recurrence relation), but I just can't figure out how to write the pattern and how to prove that statement. Must I use induction?
 [/INST] As Byron Schmuland has pointed out a recursive pattern was developed here. The process in this solution will be a connection to the Beta function.
Consider the integral
\begin{align}
I_{n} = \int_{0}^{1} \left(1-x^{2}\right)^{n} \, dx.
\end{align}
Let $t = x^{2}$ to obtain
\begin{align}
I_{n} &= \frac{1}{2} \, \int_{0}^{1} t^{-\frac{1}{2}} \, (1-t)^{n} \, dt \\
&= \frac{1}{2} \, B\left(\frac{1}{2}, n + 1 \right) \\
&= \frac{\Gamma\left(\frac{3}{2}\right) \, \Gamma\left(n + 1\right)}{\Gamma\left(n + \frac{3}{2}\right)} = \frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.
\end{align}
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