Compute $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$ 
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!
 A: 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$$I_1.I_2=\frac{1}{16}\frac{\Gamma(1/4)\Gamma(1/2)^2}{\Gamma(5/4)}=\frac{\pi}{4}$$
A: Yes, there is another approach – elliptic integrals.
By Byrd and Friedman 214.05 and 318.02, the two integrals evaluate as follows:
$$\left(\frac1{\sqrt2}K(m=1/2)\right)\left(\sqrt2E(1/2)-\frac{\sqrt2}2K(1/2)\right)$$
This can be rewritten as
$$\frac12(K(1-1/2)E(1/2)+K(1/2)E(1-1/2)-K(1/2)K(1-1/2))=\frac\pi4$$
where we have used Legendre's relation.
A: For the antiderivatives, make $x=\sin(t)$
$$I_1=\int \dfrac{dx}{\sqrt{1-x^4}}=\int \dfrac{dt}{\sqrt{1+\sin^2(t)}}=F\left(\left.t\right|-1\right) $$
$$I_2=\int \dfrac{x^2}{\sqrt{1-x^4}}dx=\int \dfrac{1+\sin^2(t)-1}{\sqrt{1+\sin^2(t)}}dt$$ that is to say
$$I_2=\int \sqrt{1+\sin^2(t)}\,dt -I_1=E(t|-1)-F(t|-1)$$
Integrated between $0$ and $\frac \pi 2$
$$I_1=K(-1)\qquad I_2=E(-1)-K(-1)$$ $$I_1\,I_2=K(-1)\Big[E(-1)-K(-1) \Big]=\frac \pi 4$$
