Compute $\frac{4}{5}\pi \int_0^{2\pi}\int_0^1\sqrt{v^2-2 v\cos\theta+1}\>v{\rm d}v\, {\rm d}\theta$ The integral
$$I=\frac{4}{5}\pi \int_0^{2\pi}\int_0^1\sqrt{v^2-2 v\cos\theta+1}\>v{\rm d}v\, {\rm d}\theta$$
arises from the calculation of the expected distance between two random points inside a unit circle. Can't figure out the change in variables to transform it to
$$I=\frac{4}{5}\pi \int_{-\pi/2}^{\pi/2}\int_0^{\sqrt{2(1+\cos2\theta')}} v'^2\,{\rm d}v'\,{\rm d}\theta'$$
Change of  variable
${}$


Referring to the graph above, the variable changes used are
$$v = \sqrt{{v'}^2-2{v'}\cos\theta'+1},\>\>\>\>\>\>\>\frac{\sin\theta}{\sin\theta'}=\frac{v'}{v}
$$
Then, according to the Jacobian determinant
$$dv d\theta = \frac{v'}{v}dv' d\theta'$$
with the ranges $\theta'\in[-\frac\pi2, \frac\pi2]$ and $v'\in [0, \sqrt{2(1+\cos2\theta')}]$. Thus,
the integral is transformed to
$$I=\frac{4}{5}\pi \int_{-\pi/2}^{\pi/2}\int_0^{\sqrt{2(1+\cos2\theta')}} v'^2\,{\rm d}v'\,{\rm d}\theta'=\frac{128\pi}{45}$$
