Q:

Find probability density of $\arctan(Y/X)$ if $(X, Y)$ is multivariate normal distributed, where $EX=EY=0$ and $EX^2=EY^2=2$ and $EXY=1$

I calculated density function for (X, Y):
$\frac{1}{2\sqrt{3}\pi}e^{-\frac{x^2-xy+y^2}{3}}$
But if i try to write probability function via integral, it becomes to hard to integrate:
$$\int_{v > \frac{u}{\operatorname{tg}(t)}} e^{-\frac{u^2}{3}}e^{\frac{vu}{3}}e^{-\frac{v^2}{3}}$$
And after integration I'll need to differentiate it to find density... 

A:

It is tempting to utilize the polar transformation:
\begin{equation}
\begin{cases}
X = R\cos\Theta, \\
Y = R\sin\Theta. 
\end{cases}
\end{equation}
Then $\arctan(Y/X) = \Theta$. To find the distribution of $\Theta$, it is standard to start with the joint distribution of $(R, \Theta)$. The Jacobian is
\begin{equation}
\frac{\partial(x, y)}{\partial(r, \theta)} = r
\end{equation}
is standard.  Hence the joint density of $(R, \Theta)$ is 
\begin{align}
f(r, \theta) = \frac{r}{2\sqrt{3}\pi}\exp\left(-\frac{1}{3}(r^2 - r^2\sin\theta\cos\theta)\right), \quad r > 0, \theta \in (0, 2\pi]. 
\end{align}
Can you take it from here?