Two points on a circle

Idea. Fix the first point. The central angle to the second point is uniform, and the chord length is a simple function of that angle, so the answer is one integral.

Step 1. By symmetry, fix the first point. The central angle $\theta$ between the two points is uniform on $[0,2\pi )$.

Step 2. The chord spanning angle $\theta$ is the base of an isosceles triangle with two sides 1, so its length is $2\sin \frac{\theta }{2}$.

Step 3. $$\begin{array}{rl}\mathbb{E}[\text{distance}]& =\frac{1}{2\pi }{\int }_0^{2\pi }2\sin \frac{\theta }{2}d\theta \\ & =\frac{1}{2\pi }[-4\cos \frac{\theta }{2}{]}_0^{2\pi }\\ & =\frac{8}{2\pi }=\frac{4}{\pi }.\end{array}$$

So the expected distance is $\frac{4}{\pi }\approx 1.273$, a bit less than the $\sqrt{2}\approx 1.414$ you get for two points a quarter turn apart.