I wish to find the expected area of a triangle with vertices independently and uniformly at random chosen from a unit circle. My first attempt here had a mistake which was found. I have devised a new method using signed area. Unfortunately, now I'm stuck.
Attempt: As before, let $O$ be the center of the circle and $P_1, P_2, P_3$ be the vertices. This time, we assign $[OP_1 P_2]$ a sign: negative if $P_1 P_2 P_3$ does not contain $O$ and $P_3, O$ lie on opposite sides of the line $P_1 P_2,$ else positive. Now the equation $[P_1 P_2 P_3] = [OP_1P_2] + [OP2 P_3] + [OP_3 P_1]$ really holds and we have $\mathbb{E}([P_1 P_2 P_3]) = 3\mathbb{E}([OP_1 P_2])$ as before. Let $T$ be the plain unsigned area of $[OP_1 P_2],$ let $\theta = \angle P_1 O P_2,$ let $a=OP_1, b = OP_2, c = P_1 P_2,$ and let $S', S$ be the ", area of" the region of the circle cut by $P_1 P_2$ that does not include $O$ respectively. The sign is negative iff $P_3 \in S',$ so we have $\mathbb{E}([OP_1P_2]) = \frac{1}{\pi}\mathbb{E}(T(-1 \cdot S + 1 \cdot (\pi - S)) = \mathbb{E}[T(\pi - 2S)].$
Clearly, $T = 0.5ab\sin \theta.$ Some computation reveals that $\pi - 2S = 2\sin^{-1}(h) + h\sqrt{1-h^2}$ where $h$ is the distance from $O$ to the chord through $P_1 P_2.$ Unfortunately, $h = \frac{2T}{c} = \frac{ab\sin \theta}{\sqrt{a^2+b^2-2ab\cos \theta}}.$ The probability density function of $a, b, \theta$ is $2x, 2y, \theta/\pi$ over $[0,1], [0,1], [0,\pi]$ respectively and $a, b, \theta$ are pairwise independent, so the joint pdf is $\frac{4}{\pi} xy\theta$ and we need to compute the integrals
$$\frac{2}{\pi}\int_0^1 \int_0^1 \int_0^{\pi} x^2y^2 \theta \sin\theta \sin^{-1}\left(\frac{xy\sin \theta}{\sqrt{x^2+y^2-2xy\cos\theta}}\right) \, d\theta \, dx \, dy$$ and $$\frac{2}{\pi}\int_0^1 \int_0^1 \int_0^{\pi} x^3 y^3 \theta \sin^2 \theta \frac{\sqrt{x^2+y^2-2xy\cos\theta - x^2 y^2 \sin^2\theta}}{x^2+y^2-2xy\cos\theta} \, d\theta \, dx \, dy.$$
I have low hopes of computing these integrals directly, so I'm wondering if there's a substitution or secret tactic that will greatly help. Integration by parts will allow you to integrate the first one with respect to $\theta,$ but that makes the resulting double integral too much of a mess.
