For $r\in \mathbb R$ and $\theta \in [0,2\pi]$ and a compactly supported function $f$ on $\mathbb R^2$ one defines the radon transform of it as $$\mathcal Rf(r,\theta):= \int_{-\infty}^{\infty} f(r \cos \theta- s \sin \theta, r \sin \theta + s \cos \theta) \, ds.$$
It is very common to express it with the dirac notation:
$$\mathcal Rf(r,\theta):= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(x,y)\delta(r-x\cos\theta-y\sin\theta) \, dx\,dy.$$
How does one derive the second expression from the first one? I normally see an intuitive geometrical explanation of the first expression but then in the literature it is often just stated "... or equivalently" for the second expression.
Let me start from $$ \mathcal Rf(r,\theta) := \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(x,y) \,\delta(r-x\cos\theta-y\sin\theta) \, dx\,dy. $$ Making the coordinate transformation/substitution $$ \begin{cases} x &= t\cos\theta - s\sin\theta \\ y &= t\sin\theta + s\cos\theta \\ \end{cases} $$ gives $$\begin{align} \mathcal Rf(r,\theta) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t\cos\theta - s\sin\theta, t\sin\theta + s\cos\theta) \,\delta(r-t) \,dt \,ds. \end{align}$$ We can now easily evaluate the $t$ integral giving: $$\begin{align} \mathcal Rf(r,\theta) &= \int_{-\infty}^{\infty} f(r\cos\theta - s\sin\theta, r\sin\theta + s\cos\theta) \,ds. \end{align}$$