TITLE: Compute $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$
QUESTION [10 upvotes]: Compute$$\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$$.

The primitive function of each one seems not elementary function.
I tried like this
$$
\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} + \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\int_0^1\dfrac{\mathrm{d}(x-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-x^2}},
$$
but it still does not work.
Actually, I can compute it by Gamma and Beta function. But I wonder is there other approach? Appreciate any help!

REPLY [6 votes]: Use $\beta$ function: https://en.wikipedia.org/wiki/Beta_function
Let $x=\sin^{1/2} t$, then
$$I_1=\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}}=\frac{1}{2} \int_{0}^{\pi/2} \sin^{-1/2}t ~ dt=\frac{1}{4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}$$
Next, $$I_2=\int_{0}^{1} \frac{x^2dx}{\sqrt{1-x^4}}=\frac{1}{2} \int_{0}^{\pi/2} \sin^{1/2}t~ dt=\frac{1}{4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}=\frac{1}{4}\frac{\Gamma(3/4)\Gamma(1/2)}{\Gamma(5/4)}$$
Then$$I_1.I_2=\frac{1}{16}\frac{\Gamma(1/4)\Gamma(1/2)^2}{\Gamma(5/4)}=\frac{\pi}{4}$$