<s>[INST] Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$ Prove that: $\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}=\sum_{0}^{+\infty}\frac{C_{2n}^{n}}{4^n(4n+1)}$
Could you give me some hint? Thank for helping.
 [/INST] Using 
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{2n}}{4^{n}}$$
then
\begin{align}
\int \frac{dx}{\sqrt{1-x^4}} &= \int \left( \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n}}{4^{n}} \right) \, dx \\
&= \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{x^{4n+1}}{4^{n} \, (4n+1)}
\end{align}
With limits the integral is:
\begin{align}
\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}} &= \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{1}{4^{n} \, (4n+1)}
\end{align} 
This integral is related to the Beta function and can be seen to have the compact form
$$\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}} = \frac{\sqrt{\pi} \, \Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$
and leads to
$$ \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{1}{4^{n} \, (4n+1)} = \frac{\sqrt{\pi} \, \Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}. $$
</s>