Let X and Y be iid normal distribution with mean $\mu = 0$, let $Z=\frac{|X|}{|Y|}$, what is the distribution of Z?
I'm thinking of using the distribution method but I'm stuck by the integration. I'm also wondering if change of variable technique can be used.
I'm not sure if it has a name; however, the ratio $X/Y$ itself has a well-known, named distribution, so you can just take the absolute value of that random variable.
First off, $X/Y = (X/\sigma)/(Y/\sigma)$, so we may assume without loss of generality that $X,Y$ are standard normal random variables (with mean $0$ and standard deviation $1$)
Then we may write $$\mathbb{P}(X/Y \leq 0) = \mathbb{P}(X\leq 0, Y>0) + \mathbb{P}(X\geq 0, Y<0) = \frac12,$$ so for $r > 0$, $$\mathbb{P}(X/Y \leq r) = \frac12 + \mathbb{P}(0 < X/Y \leq r) = \frac12 + 2\mathbb{P}(X\leq rY, X>0, Y>0).$$ This last expression can be found by a polar substitution: $$\begin{align*}\mathbb{P}(X \leq rY, X>0, Y>0) &= \frac{1}{2\pi}\int_0^\infty\int_0^{ry}e^{-\tfrac12(x^2+y^2)}\,dx\,dy \\ &= \frac1{2\pi}\int_0^\infty\int_0^{\arctan(r)}\rho e^{-\rho^2/2}\,d\theta\,d\rho \\ &= \frac1{2\pi}\arctan(r)\end{align*}$$ giving $$\mathbb{P}(X/Y \leq r) = \frac12 + \frac1\pi\arctan(r).$$
The expression for $r < 0$ can be found by symmetry as $\mathbb{P}(X/Y \leq r) = \mathbb{P}(X/Y \geq -r)$, which gives the same expression as above, just extended to all real $r$.