Let $Z$ be a random variable with cumulative distribution function: \begin{equation} F_Z(x) = \frac{1}{4 \delta^2}\left( \arccos\left( \frac{2-x^2}{2} \right) - (\pi-2\delta)\right)^2, \end{equation} and $V$ be uniformly distributed on $[\mu -\nu, \mu+\nu]$, such that: \begin{equation} f_V(x)=\frac{1}{2\nu}. \end{equation} $Z$ represents distance (of a line in the unit circle with a starting point on the circle boundary uniformly on $[-\delta, \delta]$ and end point uniformly on $[\pi-\delta, \pi+\delta]$, derived by hand and confirmed by simulation) and $V$ represents velocity. Assume $\delta< \frac{1}{2} \pi$ and $\nu<\mu$.
I am looking for the distribution of time $Q:=\frac{Z}{V}$ (time an object spends inside the circle when traveling along a line with distance $Z$ and with velocity $V$). For possible values of $Q$ we have $Q \in [\frac{\sqrt{2-2\cos(\pi-2\delta)}}{\mu+\nu}, \frac{2}{\mu-\nu}]$.
On the wikipedia page https://en.wikipedia.org/wiki/Ratio_distribution, the ratio distribution is derived using (in my notation) $f_Z$ and $f_V$. Finding antiderivatives when $|y|$ is introduced turns out to be quite challenging, so I wanted to try a more intuitive approach.
My reasoning to find the distribution of $Q$ is like this (using only one integral instead of the double integral the wikipedia article uses): \begin{equation} \mathbb{P}(Q \leq x) = \mathbb{P}(\frac{Z}{V} \leq x) \\ = \mathbb{P}(Z \leq xV) \\ = F_Z(xV)\\ = \int_{y=\mu-\nu}^{\mu+\nu}f_Y(y)F_Z(xV|V=y)dy \\ =\int_{y=\mu-\nu}^{\mu+\nu}\frac{1}{2\nu}F_Z(xy)dy. \\ \end{equation} When computing this by hand, I find a long expression that by no means represents a probability distribution. Also, asking Mathematica to compute the integral gives no usable answer.
What am I doing wrong here? Thanks in advance!