PDF or CDF of $D = \sqrt{1 - X^2 \sin^2{\theta}} - X \cos{\theta}$ How to calculate the PDF or the CDF of $D$ where:
$$D = \sqrt{1 - X^2 \sin^2{\theta}} - X \cos{\theta}$$
If $X$ is uniform distributed on $[-1, 1]$, $\theta$ is uniformly distributed on $[0, 2 \pi]$ and they are independent.
I know that:
$$F_D(d) = \iint\limits_D \, f_\theta(\theta) f_X(x) \mathrm{d}x\,\mathrm{d}\theta $$
But I don't know how to find the ranges I should integrate on!
 A: For $F_D(d)$ you need to integrate over the region of the $(x,\theta)$ plane where $-1 \le x \le 1$, $0 \le \theta \le 2 \pi$, and $\sqrt{1-x^2 \sin^2 \theta} - x \cos \theta \le d$.  Here, for example (in red) is the region for $d = 1/2$:

And here is $d = 1$:

A: The change of variables $z=\sqrt{1-x^2\sin^2\theta}-x\cos\theta$ and $u=\sqrt{1-x^2\sin^2\theta}$ indicates that the probability density function of $D$ is
$$
f_D(z)=\int \frac{u\mathrm{d}u}{\sqrt{1-u^2}\sqrt{1-2zu+z^2}}.
$$
This integral might be transformed into an elliptic integral of the first kind but I did not check carefully enough the bounds on $z$ this change of variables involves to conclude. 
To the OP: what did you try? And what makes you think the result can be expressed using only usual functions?
A: If you are not interested in a particularly elegant expression, then you can very easily and simply calculate the CDF of $D$ as follows:
$$
{\rm P}(D \le s) = \frac{1}{{4\pi }}\int_{[ - 1,1] \times [0,2\pi ]} {\mathbf{1}\big(\sqrt {1 - x^2 \sin ^2 \theta }  - x\cos \theta  \le s\big)\,dx d\theta } ,\;\; 0 \leq s \leq 2,
$$
where $\mathbf{1}$ denotes the indicator function. Note that $D$ is supported on the set $[0,2]$ (hence the restriction  $0 \leq s \leq 2$ above). Indeed, on the one hand, $D \geq 0$ since
$$
\sqrt {1 - x^2 \sin ^2 \theta }  \ge x\cos \theta 
$$
for any $x \in [-1,1]$ and $\theta \in [0,2\pi]$ (this is trivial if $x\cos\theta \leq 0$; otherwise take squares on both sides), and, on the other hand, $D \leq 2$ since
$$
\sqrt {1 - x^2 \sin ^2 \theta }  - x\cos \theta  \le 1 + 1 = 2
$$
(for $x$ and $\theta$ as above). Further note that the choices $(x,\theta)=(1,0)$ and $(x,\theta)=(-1,0)$ correspond to $D=0$ and $D=2$, respectively.
Finally, it should be noted that the double integral above can be calculated very quickly and accurately. For example,
it took about a second to obtain the approximation
$$
{\rm P}(D \leq 1) \approx 0.5813759999978363,
$$
indicating that ${\rm P}(D \leq 1) = 0.581376$, whereas Monte Carlo simulations ($10^7$ repetitions) yielded a much less accurate approximation 
$$
{\rm P}(D \leq 1) \approx 0.5813805,
$$
in about $40$ seconds.
