So this is a rather theoretical question as I kinda don't understand how to approach this from the perspective I want or how that approach is flawed (which I suspect might be the case).
Let's consider the uniform distribution of the points $(x,y)$ on the unit circle. Note that it is degenerate as it has zero measure in the 2D space. We are interested in finding out what is the marginal of $x$ under this distribution. By changing to polar coordinates one can easily see that this corresponds to a uniform over the angle $\theta$. Since $x=\cos(\theta)$ by the change of variable formula we have that
$$ p(x) = \frac{1}{2 \pi} \left|\frac{d x}{d \theta}\right|^{-1} = \frac{1}{2 \pi} \frac{1}{\sqrt{1 - x^2}}$$
This, of course, is wrong (a bit) since we have to take into account the fact that the cosine is not bijective so we have to split it to its to branches (where it is the same) and add them so the first factor actually is $\pi^{-1}$ and you can convince yourself this is the correct choice by the fact that $$\int_{-1}^{1} \sqrt{1-x^2} dx = \pi$$
Ok, however now I want to somehow derive that without using polar coordinates. Here is my approach and please anyone corrects me on this or help me finish it.
Using the definition of the unit circle we have that the joint distribution is
$$p(x,y) = \frac{1}{2 \pi} \delta (x^2 + y^2 - 1)$$ where $\delta$ is a dirac measure. Now trying to integrate that over $y$ gives me issues. Speicifically, since outside the dirac measure the function is constant it would seem that $p(x) = \frac{1}{2 \pi}$ which we know is wrong. So why is this wrong and how is the correct way to derive the result (I think I missing some knowledge of how to correctly do this integral).
Let us generalize from the unit circle to a circle with radius $R$ for clarity. Using multiple times the well-known Dirac distribution identity for substitution, we calculate
$$ \begin{align}\langle f \rangle ~&=~\frac{1}{2\pi} \int_{[0,2\pi]}\! d\theta~ f(R\cos\theta,R\sin\theta) \cr ~&=~\frac{1}{2\pi R} \iint_{\mathbb{R}^2}\! r dr~d\theta~ \delta (r-R) ~f(r\cos\theta,r\sin\theta) \cr ~&=~\frac{1}{2\pi R} \iint_{\mathbb{R}^2}\! dx~dy~ \delta (\sqrt{x^2+y^2}-R) ~f(x,y) \cr ~&=~\frac{1}{\pi } \iint_{\mathbb{R}^2}\! dx~dy~ \delta (x^2+y^2-R^2) ~f(x,y) \cr ~&=~\frac{1}{\pi } \int_{[-R,R]}\! dx \int_{\mathbb{R}}\! \frac{dy}{2|y|} \sum_{\pm}\delta (y\mp \sqrt{R^2-x^2}) ~f(x,y) \cr ~&=~\frac{1}{2\pi } \int_{[-R,R]}\! dx \frac{1}{ \sqrt{R^2-x^2}} \sum_{\pm} f(x,\pm \sqrt{R^2-x^2}).\end{align} $$
This last expression displays the correct $x$-distribution. The $\pm$ summation reflects the 2 branches of the square root.