Suppose that $X$ and $Y$ have a two-dimensional normal distribution with means $0$, variances 1, and correlation coefficient $\rho$, $|\rho| < 1$. Let $(R, \Theta)$ be the polar coordinates.
Determine the distribution of $\Theta$.
I found the joint to be $$\frac{1}{\sqrt{(1-\rho^2)}}r\frac{1}{2\pi}\exp\left\{-\frac{1}{2} \frac{1}{1-\rho^2}r^2(1-\rho \sin(2\theta)) \right\}$$
In order to obtain $f_{\Theta}(\theta)$ I need to take the marginal of the joint, by doing $$\int_{-\infty}^\infty\frac{1}{\sqrt{(1-\rho^2)}}r\frac{1}{2\pi}\exp\left\{-\frac{1}{2} \frac{1}{1-\rho^2}r^2\big(1-\rho \sin(2\theta)\big)\right\}dr$$
I've done so, by making the joint look like the pdf of a gamma distro: also, after setting $u=r^2$, $du=2rdr$ I get
$$
\begin{split}
f_{\Theta}(\theta)&= \frac{1}{\sqrt{(1-\rho^2)}}\frac{1}{4\pi}\int_{0}^{\infty} \exp\left\{-u\left(\frac{1}{2} \frac{1}{1-\rho^2}\big(1-\rho \sin(2\theta)\big) \right)\right\}dr\\
&=\frac{1-\rho^2}{2\pi\sqrt{1-\rho^2}(1-\rho \sin(2\theta))}
\end{split}
$$
But I can't find the range and that is my usual problem. Any tip or trick on how to find the correct range on these type of problems would be more than welcome!
Do I just use my calculus knowledge to find the domain of f?