How do we find the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}?$ We would like to find out the closed form for integral $(1)$

$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}\tag1$$

An attempt:
We may write 
$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$ 
Let $x=\sin x$ and $y=\cos x$
$x^6+y^6=(x^4-x^2y^2+y^4)$ 
Simplified down to
$x^6+y^6=\sin^4 x+\cos2x\cos^2x$ 
$x^6+y^6=\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x$ 
$(1)$ becomes 
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x}}\tag2$$
The power has reduced but looked more messier 
How else can we  evaluate $(1)$?
 A: Using  this,$$\sin^6(x)+\cos^6(x)=1-\frac 34 \sin^2(2x)$$  makes $$\int{ dx\over \sqrt{\sin^6(x)+\cos^6(x)}}=\int{ dx\over \sqrt{1-\frac 34\sin^2(2x)}}=\frac{1}{2} F\left(2 x\left|\frac{3}{4}\right.\right)$$  where appears the elliptic integral of the first kind. 
Using the bounds, the result simplifies to $$K\left(\frac{3}{4}\right)\approx 2.15652$$  which is the complete elliptic integral of the first kind.
If I may suggest, have a look here.
A: Using the substitution $x=\arctan t$ the given integral boils down to:
$$ I=\int_{0}^{+\infty}\sqrt{\frac{1+t^2}{1+t^6}}\,dt=\int_{0}^{+\infty}\frac{dt}{\sqrt{t^4-t^2+1}}=\int_{0}^{+\infty}\frac{dt}{\sqrt{(t^2+\omega^2)(t^2+\bar{\omega}^2)}} $$
(with $\omega=\exp\left(\frac{\pi i}{3}\right)$ ) that is a complete elliptic integral of the first kind. Such integral can be computed through the AGM mean $\text{AGM}(a,b)=\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)$:
$$ I = \frac{\pi}{2\,\text{AGM}(\omega,\bar{\omega})} = \color{red}{\frac{\pi}{\text{AGM}(1,2)}}.$$
In particular, the integral is bounded between $\pi(12-8\sqrt{2})$ and $\pi\sqrt[4]{\frac{2}{9}}$.
