This question resisted attacks at MSE, so I am posting it here.
Here is the graph of $\dfrac{\sin x}{\sin y}=\dfrac{\sin x+\sin y}{\sin(x+y)}$.
Find the area of the region enclosed by the curve and the $x$-axis, from $x=0$ to $x=\pi$.
Context, and why I suspect the answer is $\frac{\pi^2}{8}$
This question arose when I asked myself: "A triangle's vertices are three uniformly random points on a circle. The side lengths are, in random order, $a,b,c$. The triangle inequality tells us that $P(a+b<c)=0$. But what is $P\left(a+b<\left(\frac{a}{b}\right)c\right)$, given that $\frac{a}{b}>1$ ?"
A simulation of $10^7$ random triangles yielded an estimated probability of $0.49998$, suggesting that the probability is $\frac12$, which would imply that the area in this question is $\frac{\pi^2}{8}$. The connection between the probability and the area is explained in the MSE question.
I decided to phrase this question in terms of area, but we can attack it in terms of area or probability.
Progress so far
I used Wolfram (changing $\sin y$ to $y$, and changing $\sin x$ to $a$) to help me change $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ to:
$$y=\arcsin\left(\frac13\left(\sin (2x)-\sin x+f(x)-\frac{g(x)}{f(x)}\right)\right)dx$$
where
$$f(x)=\left(\frac12\left(h(x)+\sqrt{4g(x)^3+h(x)^2}\right)\right)^\frac13$$
$$g(x)=3\sin^2 x-(\sin 2x-\sin x)^2$$
$$h(x)=-16\left(\cos x\right)\sin^5 x+10\left(\cos x\right)\sin^3x+24\sin^5 x+10\sin^3 x$$
In this answer at the MSE question, user @Masd reports that $\int_0^\pi y \, \mathrm dx$ matches $\frac{\pi^2}{8}$ to $11$ decimal places.
${}$
Best Answer
Here is my computation. First of all, I had no idea whatsoever how to integrate implicit trigonometric functions, so I decided to switch to the integration with respect to the sides $x,y,z$ of the triangle $XYZ$. Recall that the conditions for the domain is $xz\le y^2+xy$, $x>y$ plus the triangle inequality $z>x-y$. This triangle inequality is compatible with the first condition if and only if $x< (\sqrt 2+1)y$. So the admissible domain is given by $$ y< x< (\sqrt 2+1)y,\qquad x-y<z<y+\frac {y^2} x\,. $$ Now we have to integrate over the points on the circle, i.e., for the case when the circumradius is $1$ and with respect to the angular variables. We relax the circumradius to $r>0$, place $Z$ on the positive semi-axis and the circle center at the origin (so $Z=r$), and take $Y=re^{2i\varphi}, X=re^{2i\psi}$, so $$ x=2r\sin\varphi,\qquad y=2r\sin\psi,\qquad z=2r\sin(\varphi-\psi) $$ (the last one up to a sign) where $\varphi,\psi$ run independently over $[0,\pi]$. The condition $x>y$ reduces the $d\varphi\,d\psi$ measure to $\frac{\pi^2}2$ and each admissible triple $x,y,z$ is realized twice (up to the flip about the horizontal axes). We can use any measure on $(0,+\infty)$ we want when integrating over $r$, so we will choose $\frac{dr}{r}$.
Now it is time to compute the Jacobian $J$ of the mapping $(r,\varphi,\psi)\mapsto (x,y,z)$. We have $$ J=8\left|\begin{matrix} \sin\varphi & r\cos\varphi & 0 \\ \sin\psi & 0 & r\cos\psi \\ \sin(\varphi-\psi) &r\cos(\varphi-\psi) & -r\cos(\varphi-\psi) \end{matrix}\right| \\ =8r^2[\cos\varphi\cos\psi\sin(\varphi-\psi)-\sin\varphi\cos\psi\cos(\varphi-\psi)+\cos\varphi\sin\psi\cos(\varphi-\psi)] \\ =8r^2[\cos\varphi\cos\psi\sin(\varphi-\psi)- \sin(\varphi-\psi)\cos(\varphi-\psi)] \\ =8r^2\sin(\varphi-\psi)\sin\varphi\sin\psi=\frac{xyz}r\,, $$ i.e. $$ \frac{dr}r\,d\varphi\,d\psi=\frac{dx\,dy\,dz}{xyz}\,, $$ which is pretty neat.
Now let $\Omega$ be the set of all good triangles and $\omega$ be the set of good triangles with $r=1$. Then the back of the envelope computation is the following: $$ \left[\int_0^\infty\frac{dr}{r}\right]\int_\omega d\varphi\,d\psi= 2\int_{\Omega}\frac{dr}{r}\,d\varphi\,d\psi =2\int_{\Omega}\frac{dx\,dy\,dz}{xyz} \\ = 2\int_0^\infty\frac{dy}{y}\left[\int_{y}^{(\sqrt 2+1)y}\frac{dx}x\left(\int_{x-y}^{y+\frac{y^2}x}\frac{dz}z\right)\right] $$ Substituting $x=ty$, $t\in(1,\sqrt 2+1)$, we evaluate the inner integrals to $$ \int_1^{\sqrt 2+1}\log\frac{t+1}{t(t-1)}\frac{dt}t\,, $$ which is independent of $y$, so we finally get $$2\left[\int_0^\infty\frac{dy}y\right] \int_1^{\sqrt 2+1}\log\frac{t+1}{t(t-1)}\frac{dt}t\,, $$ Now it remains to cancel $\int_0^\infty\frac{ds}s$ to get the claimed identity.
The little (but somewhat irritating) problem is that the integral $\int_0^\infty \frac{ds}s$ diverges. However, as Landau (the physicist) used to say, "A chicken is not a bird and a logarithm is not infinity". So we will fix that with a little bit more elaborate argument. ${}{}$
We shall impose one extra condition on the admissible triangles, namely, we will demand that $y$ is comparable to $r$. We always trivially have $r\ge \frac y2$, so we'll fix a big $A>0$ and demand that $r<Ay$. Notice that it is a condition on the shape only, not on the size. Denoting by $\omega_A$ the set of admissible triangles with circumradius $1$ and by $\Omega_A$ the set of all admissible triangles and choosing a huge finite $R>0$, we can write, as before, $$ \left[\int_1^R\frac{dr}r\right]\int_{\omega_A}d\varphi\,d\psi=2\int_{\Omega_A\cap\{1<r<R\}}\frac{dx\,dy\,dz}{xyz}\,. $$ Now notice that $$ \Omega_A\cap\{2<y<\frac RA\}\subset \Omega_A\cap\{1<r<R\}\subset\Omega_A\cap\{\frac 1A<y<2R\}\,, $$ so instead of one identity, we write two inequalities and conclude that $\log R\int_{\omega_A}d\varphi\,d\psi$ is squeezed between $\log\frac{R}{2A}$ times $2\int_1^{\sqrt 2+1}\left(\int_{E(t,A)}\frac{ds}{s}\right)\frac{dt}t$ and $\log(2AR)$ times the same quantity, where we put $x=ty, z=sy$ and defined $E(t,A)$ to be the set of $s$ for which the triangle with the sides $1,t,s$ is admissible.
Now, dividing by $\log R$, letting $R\to+\infty$ and using the squeeze theorem, we conclude that $$ \int_{\omega_A}d\varphi\,d\psi= 2\int_1^{\sqrt 2+1}\left(\int_{E(t,A)}\frac{ds}{s}\right)\frac{dt}t $$ for every fixed $A$. But when $A\to+\infty$, the set $\omega_A$ expands to $\omega$ and for each $t$, the set $E(t,A)$ expands to $[t-1,1+\frac 1t]$, so the monotone convergence theorem finishes the story.
That's it. I tried to find a simple change of variables that would avoid using non-trivial identities for $\mathrm{Li}_2$ to get the final $\frac{\pi^2}8$ answer, but failed so far. Still, we have the result, so, I hope, somebody will eventually provide a simpler proof. The value $\frac 12$ of the probability is too nice to be there without a clear reason. :-)