<s>[INST] Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution. Find the density of $Y = a/(1 + X^{2})$, where $X$ has the Cauchy distribution.
MY SOLUTION
To begin with, let us remember that the Cauchy probability density function is given by
\begin{align*}
f_{X}(x) = \frac{1}{\pi(1+x^{2})}\quad\text{for}\quad x\in\textbf{R}
\end{align*}
Consequently, the sought distribution is described as
\begin{align*}
F_{Y}(Y\leq y) &= \textbf{P}(Y\leq y) = \textbf{P}\left(\frac{a}{1+X^{2}} \leq y\right) = \textbf{P}\left(1+X^{2}\geq \frac{a}{y}\right)\\\\
& = \textbf{P}\left(|X|\geq \sqrt{\frac{a-y}{y}}\right) = \textbf{P}\left(X\geq \sqrt{\frac{a-y}{y}}\right) + \textbf{P}\left(X \leq -\sqrt{\frac{a-y}{y}}\right)\\\\
& = 1 - F_{X}\left(\sqrt{\frac{a-y}{y}}\right) + F_{X}\left(-\sqrt{\frac{a-y}{y}}\right)
\end{align*}
where $y\in(0,a]$. Finally, we have
\begin{align*}
F_{X}(x) = \int_{-\infty}^{x}f_{X}(u)\mathrm{d}u = \int_{-\infty}^{x}\frac{\mathrm{d}x}{\pi(1+x^{2})} = \frac{\arctan(x)}{\pi} + \frac{1}{2}\end{align*}
I have two questions. Firstly, I would like to know if this approach is correct. Secondly, I would like to know if there is another way to solve this problem. Thanks in advance!
 [/INST] Your approach is fine. 
Perhaps you are aware of this result (very useful to generate a Cauchy variable): if $Z \sim U(-\frac{\pi}{2},\frac{\pi}{2})$ then $X = \tan(Z)$ follows a canonical Cauchy distribution.
Now, we have $Y = \frac{a}{1+X^2}= a \cos^2(Z)$ which is slightly easier. 
Then, assuming $a>0$ $$\begin{align}
f_Y(y) &= 2 \frac{f_Z(z)}{|2 a \cos(z) \sin(z)|} \\
&=\frac{1}{a \pi} \frac{1}{|\cos^2(z)\tan(z)|}\\
&=\frac{1}{a \pi} \frac{1}{|\frac{y}{a}x|}\\
&=\frac{1}{ \pi} \frac{1}{y\sqrt{a/y-1}} \\
&=\frac{1}{ \pi} \frac{1}{\sqrt{y(a-y)}}
\end{align}
$$
for $0<y<a$
</s>