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There is numerical evidence that

$$\int_0^\pi\arcsin\left(\frac14\sqrt{8-2\sqrt{10-2\sqrt{17-8\cos x}}}\right)dx=\frac{\pi^2}{5}.$$

How can this be proved?

Context

In another question, three random points on a circle are chosen, and tangents to the circle at those points form a triangle, with side lengths $d,e,f$.

enter image description here

The relevant question here is to prove that $P(de<f^2)=\frac35$.

Assuming the radius is $1$, we have

$d=\tan X-\tan(X+Y)$
$e=\tan Y-\tan(X+Y)$
$f=\tan X+\tan Y$

$P(de<f^2)=P((\tan X-\tan(X+Y))(\tan Y-\tan(X+Y))<(\tan X+\tan Y)^2)$

I rotated the graph $45^\circ$ by letting $X=y-x$ and $Y=y+x$.

Then Wolfram gave me $y$ as a function of $x$. After some minor simplifications, I ended up with the integral in this question.

My attempt

Approachzero does not turn up anything similar. Wolfram does not evaluate the anti-derivative.

Here is the graph of $y=\arcsin\left(\frac14\sqrt{8-2\sqrt{10-2\sqrt{17-8\cos x}}}\right)$.

enter image description here

I tried applying some general advice about related integrals, but I suspect this one may be out of my league.

Dan
  • 35,053

1 Answers1

14

Applying $\arcsin x=\frac12\arccos(1-2x^2)$ followed by $\arccos x=\frac12\arccos(2x^2-1)$ yields: $$ \arcsin\left(\frac14\sqrt{8-2\sqrt{10-2\sqrt{17-8\cos x}}}\right)=\frac14\arccos\left(\frac14-\frac14\sqrt{17-8\cos x}\right)=\frac14f(x)$$

Based on the graph from below, we can rewrite the red part, $\int_0^\pi f(x)dx$, as the total square area $(\pi^2)$ minus the blue part: $\int_{2\pi/3}^\pi f^{-1}(y)dy$, where $f^{-1}(y)=\arccos\left(2\sin^2y+\cos y\right)$.

enter image description here

$$\Rightarrow \frac14\int_0^\pi \arccos\left(\frac14-\frac14\sqrt{17-8\cos x}\right)dx=\frac{\pi^2}{4}-\frac14\int_\frac{2\pi}{3}^\pi\arccos\left(2\sin^2y+\cos y\right)dy$$ $$\overset{y\to \pi-y}=\frac{\pi^2}{4}-\frac14\int_0^\frac{\pi}{3}\arccos\left(2\sin^2y-\cos y\right)dy=\frac{\pi^2}{4}-\frac{\pi^2}{20}=\frac{\pi^2}{5}$$

The last integral can be found here.

Zacky
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