arctan of ratio of two normal variables is uniform

Let $R=\sqrt{X^2+Y^2}$. We want to calculate the distribution of $(R,\theta)$, i.e. the polar cordinates of $(X,Y)$. For $r\ge 0$ and $\alpha\in [-\pi,\pi]$ we have $\Bbb P[R\le r,\, \theta\le\alpha]=\Bbb P[(X,Y)\in\{(x,y)\in \Bbb R^2|\; x$ and $y$ have polar cordinates $(\rho , \beta)$ with $\rho \le r$ and $\beta\le\alpha\}=\Bbb P[(X,Y)\in \Omega]$.

So, since $X$ and $Y$ are indipendents, density of $(X,Y)$ is the product of the densities of $X$ and $Y$. So, passing in polar cordinates:$$\Bbb P[R\le r,\, \theta\le\alpha]=\int\int_{\Omega} \frac 1{2\pi} e^{-\frac {x^2+y^2}2}dxdy=\int_{-\pi}^{\alpha}\int_0^r \frac 1{2\pi} \rho e^{-\frac{\rho}2}d\rho d\theta=\frac {\alpha}{2\pi}(1-e^{-\frac{r^2}2})$$ Now we have that $(R,\theta)$ has density $f(r,\alpha)=\frac r{2\pi} e^{-\frac{r^2}2}\chi _{[0,+\infty)\times [-\pi,\pi]} (r,\alpha)$, from this, finally we obtain the density of $\theta$: $\int_0^{+\infty} \frac r{2\pi} e^{-\frac{r^2}2} \chi _{[0,+\infty)\times [-\pi,\pi]} (r,\alpha) dr=\frac 1{2\pi}\chi _{[-\pi,\pi]} (\alpha)$.