I have homework problem very similar to Probability Theory relating distance and the Cauchy Distribution:
A radioactive source emits particles in a random direction (with all directions being equally likely). It is held at a distance a from a vertical infinite plane photographic plate. Show that, given the particle hits the plate, the horizontal coordinate of its point of impact (with the point nearest the source as origin) has the Cauchy density function $a/(π(a^2 + x^2))$.
When I attempted the problem, I considered the plane as a 2D plane. Noting that
$f_{\Theta, \Phi}(\theta, \phi) = \frac{1}{\pi^2} I_{[0,\frac{\pi}{2})}(\theta) I_{[0,2\pi)}(\phi)$,
where $f$ denotes a probability density function, and $I$ denotes an indicator function, I used the following parametrisation:
$X = a\tan{\Theta}\cos{\Phi}, \: Y = a\tan{\Theta}\sin{\Phi}$.
Then we have
$f_{X, Y}(x, y)\begin{Vmatrix}\frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi}\end{Vmatrix} = f_{\Theta, \Phi}(\theta, \phi) = \frac{1}{\pi^2} I_{[0,\frac{\pi}{2})}(\theta) I_{[0,2\pi)}(\phi)$.
Now
$\begin{Vmatrix}\frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi}\end{Vmatrix} = a^2 \sec^2{\theta} \tan{\theta} =\frac{(x^2+y^2+a^2)\sqrt{x^2+y^2}}{a}$
and the region $\{(\theta, \phi) : \theta \in [0, \frac{\pi}{2}), \phi \in [0, 2\pi)\}$ maps to the entire $\mathbb{R}^2$, so
$f_{X, Y}(x, y) = \frac{a}{\pi^2} \frac{1}{x^2+y^2+a^2} \frac{1}{\sqrt{x^2+y^2}}$.
Then I attempted to recover $f_X$ by integrating through $y$, but instead of getting the required result, I got
$f_X(x) = \frac{1}{\pi^2\sqrt{x^2+a^2}} \ln{\left(\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+a^2}-a}\right)}$.
At which stage did I make an error? Any help will be appreciated. Thanks.