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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 $(1)$?

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  • 2
    $\begingroup$ $$\sin^6x+\cos^6x=1-3\sin^2x\cos^2x$$ $$\dfrac1{\sqrt{1-3\sin^2x\cos^2x}}=\dfrac{\sec^2x}{\sqrt{\sec^4x-3\tan^2x}}=\dfrac{\sec^2x}{\sqrt{1-\tan^2x+\tan^4x}}$$ $\endgroup$ – lab bhattacharjee Mar 13 '17 at 7:28
  • $\begingroup$ wolfram alpha says that $$\sin^6 x+\cos^6 x=\frac38\cos(4x)+\frac58$$ The identity surely comes from rearrange the expression using the exponential definition of sine and cosine function. $\endgroup$ – Masacroso Mar 13 '17 at 7:30
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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.

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  • $\begingroup$ can $K(3/4)$ can it be represent in terms of $\Gamma$ and $\pi$, I am sure I have see this number 2.156... has a closed form on one of wolfram page. $\endgroup$ – gymbvghjkgkjkhgfkl Mar 13 '17 at 8:09
  • $\begingroup$ @Bui. Inverse symbolic calculators do not find anything. $\endgroup$ – Claude Leibovici Mar 13 '17 at 8:19
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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 $\pi\sqrt[4]{\frac{2}{9}}$.

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