I tried the following:
$$ P[Y \leq y] = P[X \leq \frac{1}{\pi}\arctan(\frac{y}{\alpha})]$$
By looking plugging $x = \frac{1}{\pi}\arctan(\frac{y}{\alpha}) $ in the CDF of $X$, I "guessed" that the edges $-1$ and $1$ mapped to $-\infty$ and $+\infty$, and also obtained:
$$ F_{Y} (y) = \frac{1}{2\pi}\arctan(\frac{y}{\alpha}) + \frac{1}{2}$$
Since $-1 \leq x \leq 1$ and we're evaluating $\tan(\pi x)$, the function actually maps onto the real numbers twice, so I multiply by two:
$$ F_{Y} = \frac{1}{\pi}\arctan(\frac{y}{\alpha}) + 1$$
And differentiating I get the pdf:
$$f_{Y}(y) = \frac{\alpha}{\pi}\frac{1}{\alpha^2 + y^2}$$
I believe the result is right, but I'd like to know if there's a less clumsy way of obtaining it. I'm not sure that the cumulative function method was well suited for this problem.
Thanks.