Prove $\int_0^{\sqrt2/4}\frac1{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx=\frac{\pi^2}{8}$ (from a probability question)

Question

Let $$I=\int_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx$$

Prove that $I=\dfrac{\pi^2}{8}$.

Wolfram suggests that it's true but does not find the antiderivative.

Here is the graph of the function being integrated.

Equivalent forms

Using $\space \arcsin \theta=\arccos\sqrt{1-\theta^2}$, we have $$I=\int_0^{\sqrt2/4}\frac1{\sqrt{x-x^2}}\arccos\sqrt{\frac{x^2+(x^2-x)\sqrt{9-16x}}{2x-1}}dx$$

Using $\space \arcsin \theta=\frac12\arccos\left(1-2\theta^2\right)$ (from @Srini's comment), we have $$I=\int_0^{\sqrt2/4}\frac1{2\sqrt{x-x^2}}\arccos\left(\frac{1+(2x^2-2x)\left(1+\sqrt{9-16x}\right)}{2x-1}\right)dx$$

Context

The integral comes from the following probability question: "The vertices of a triangle are uniformly random points on a circle. The side lengths in random order are $a,b,c$. Simulations suggest that $P(ab^3+a^3b<c^4)=\dfrac12$. Can this be proved?"

By letting $a=\sin X,\space b=\sin Y,\space c=|\sin(X+Y)|$ where $X$ and $Y$ are uniformly random in $(0,\pi)$, then letting $X=x-y$ and $Y=x+y$, the probability question is equivalent to proving $I=\dfrac{\pi^2}{8}$.

(The probability question was inspired by "A typo leads to a discovery", in particular a comment by @Nilotpal Sinha.)

Related integrals

Many probability questions about a random triangle inscribed in a circle are equivalent to some integral question.

• $P(ab<c^2)=\dfrac35$ is equivalent to $\displaystyle\int_0^{\pi/3}\arccos\left(2 \sin^{2}\left(x\right)-\cos\left(x\right)\right)dx=\frac{\pi^{2}}{5}$, which is proved here.
• $P(ab<cR)=\dfrac12$, where $R$ is the circle's cradius, is equivalent to $\displaystyle\int_{\pi/4}^{\pi/2}\arccos\left(\cos x\sqrt{1+\tan x}\right)dx=\frac{\pi^2}{16}$, which is proved here.
• $P(a^2+b^2<c(a+b))=\dfrac12$ is equivalent to $\displaystyle\int_0^{\pi/6}\arccos\left(\frac{\sin x}{\cos 2x}\right)dx=\frac{\pi^2}{16}$, which is proved here.
• $P\left(\dfrac{1}{a}+\dfrac{1}{b}<\dfrac{1}{c}\right)=\dfrac15$ is equivalent to $\displaystyle\int_{\arccos(1/4)}^{\pi/2}\arccos\left(\cos x\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right)dx=\frac{\pi^2}{40}$, which is proved here.
• $P(\text{triangle's centroid lies inside triangle's incircle})=4-\dfrac{24}{\pi}\arctan\dfrac12$ is equivalent to $\displaystyle\int_0^1\frac{1}{\sqrt{1-x^2}}\arccos\left(\frac{3x^3-3x+4x^2\sqrt{2-x^2}}{5x^2-1}\right)dx=\frac{\pi}{2}\arctan\frac{17}{31}$, which is proved here.

These examples suggest that proving $I=\dfrac{\pi^2}{8}$ will involve a lot of substitutions, but I don't know how.

Answer 1

Starting with Srini's tip, then applying the identity $\arccos x=2\arctan\sqrt{\dfrac{1-x}{1+x}}$ and following Zacky's method in the fifth link, we wind up with

$$I = \int_{2\sqrt2-1}^3 \int_0^{f(y)} \frac{2x}{h(x) + g(x)\,y^2} \, dy \, dx$$

where $f(x)=\sqrt{\dfrac{x^2+2x-7}{x^2+2x+9}}$, $g(x)=7+x^2$, and $h(x)=9-x^2$. Replacing $y\to f(y)$ and applying Fubini's theorem then yields a log integral that can be evaluated as follows.

$$\begin{align*} I &= \int_{2\sqrt2-1}^3 \frac{y+1}{\sqrt{y^2+2y-7}\sqrt{y^2+2y+9}} \log \frac{2(y+1)}{y^2+2y-7} \, dy \\ &= \int_\tfrac1{\sqrt2}^1 \frac1{y\sqrt{1-y^4}} \log\frac{y/\sqrt2}{1-y^2} \, dy & y\to\frac{2\sqrt2}y-1 \\ &= \frac14 \int_\tfrac1{\sqrt2}^1 \frac{1+y^2}{y\left(1-y^2\right)} \log\frac{1+\sqrt{1-y^4}}{1-\sqrt{1-y^4}} \, dy & \text{by parts} \\ &= \frac1{16} \int_1^{\left(2+\sqrt3\right)^2} \frac{\left(\sqrt z+1\right)^3}{z(z+1)\left(\sqrt z-1\right)} \log z \, dz & z=\frac{1+\sqrt{1-y^4}}{1-\sqrt{1-y^4}} \\ &= \frac14 \int_1^{2+\sqrt3} \left(\frac4{z-1} - \frac1z - \frac{2z}{1+z^2}\right) \log z \, dz & z\to z^2 \end{align*}$$

The remaining antiderivatives are straightforward if equipped with the dilogarithm,

$$\begin{align*} \int\frac{\log z}{1-z}\,dz &= \operatorname{Li}_2(1-z) + C\\ \int\frac{\log z}z\,dz &= \frac12(\log z)^2 + C \\ \int\frac{2z}{1+z^2}\log z \, dz &\stackrel{\rm IBP}= \log z\log\left(1+z^2\right) + \frac14 \operatorname{Li}_2\left(-z^2\right) + C \end{align*}$$

Pooling these results together produces the closed form,

$$I = -\frac{\pi^2}{96} - \frac18 \log^2\left(2+\sqrt3\right) - \frac14 \log\left(2+\sqrt3\right) \log\left(8+4\sqrt3\right) \\ - \operatorname{Li}_2\left(-1-\sqrt3\right) - \frac18 \operatorname{Li}_2\left(-\left(2+\sqrt3\right)^2\right)$$

Now, for $\xi=2+\sqrt3$ (see the link in Jacob's comment below),

$$8\operatorname{Li}_2(1-\xi)+\operatorname{Li}_2\left(-\xi^2\right) = -\frac{13\pi^2}{12} - \left(\log\xi\right)^2 - 2\log\xi\log\left(1+\xi^2\right)$$

$$\implies I = \boxed{\frac{\pi^2}8} + \frac12 \log \xi \log\underbrace{\frac{\sqrt{1+\xi^2}}{2\sqrt \xi}}_{\equiv1}$$