<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] Here is a way you could do it without induction, using the Beta function and the Gamma function. Let $\sqrt{t}=x$, so that 
$$
I_n=\int_0^1(1-t)^n\frac{t^{1/2}}{2}dt.
$$
This now looks like the Beta function, so we have 
$$
I_n=\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma ((n+1)+1/2)}.
$$
Using the fact that $\Gamma(1/2)=\sqrt{\pi}$, $\Gamma(n+1)=n!$ for positive integer $n$, and $\Gamma((n+1)+1/2)=\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt{\pi}$ for positive integer $n$ (all of which can be found in linked wikipedia page), we have 
$$
I_n=\frac{\sqrt{\pi}n!4^{n+1}(n+1)!}{2(2n+2)!\sqrt{\pi}}=\frac{n!4^{n+1}(n+1)!}{2(2n+2)!}=\frac{n!(n+1)!2^{2n+1}}{(2n+2)!}=\frac{2\cdot4\cdot\dots\cdot 2n}{3\cdot5\cdot\dots 2n+1}
$$
Edit: The last equality is somewhat subtle. Note that $(n+1)!\cdot 2^{n+1}$ cancels the even terms in the denominator. The remaining powers of $2$ distribute over the $n!$.
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