Proof of maximal principle on Laplace Equation involving Poisson integral formula

One approach might be to observe that $$ \varphi(r,\theta) = \frac{a^2-r^2}{2\pi}\int_0^{2\pi} \frac{|M|}{a^2-2ar\cos(\xi - \theta)+r^2 }\mathrm{d}\xi $$ solves the boundary value problem with the boundary condition of constant value $ | M | $. If you have already proven the uniqueness of solutions of the boundary value problem, then since the constant function $ \varphi(r,\theta) = |M| $ is a solution, you can conclude the value of the integral is $ |M| $.

Otherwise, you can just compute the integral. If you know some tools from complex analysis, then it's not a hard contour integral. Otherwise, I think there is some reasonable $ u $-substitution that works. For example, using the double angle formula for $ \cos $ $$ \int \frac{1}{a + b \cos(\theta)} d \theta = \int \frac{1}{a+b \; (2 \cos^2(\theta/2) - 1)} d \theta $$ Then substituting $ u = \tan(\theta/2) $ seems to go somewhere.

Note again that the value of the integral is $ |M| $, which suggests that any approach involving upper-bounding the kernel $ \frac{1}{a^2-2ar\cos(\theta-\xi)+r^2 } $ won't work.