Evaluating $\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ in verifying density function Let $(X,Y)$ have a bivariate normal density centered at the origin with $\mathbb{E}(X^2)=\mathbb{E}(Y^2)=1$ and $\mathbb{E}(XY)=\rho$. In polar coordinates, $(X,Y)$ becomes $(R,\Phi)$ where $R^2=X^2+Y^2$. Then it can be shown that $\Phi$ has a density given by $$f_{\Phi}(\phi)=\frac{\sqrt{1-\rho^2}}{2\pi(1-\rho\sin2\phi)},\quad 0<\phi<2\pi$$
But I am having difficulty proving that this is indeed a density, more specifically the fact that $\displaystyle\int_0^{2\pi}f_{\Phi}(\phi)\,\mathrm{d}\phi=1$. Applying the usual substitution $t=\tan\phi$ in the integral $\displaystyle\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ seems to be misleading. Should there be any change in the integration limits for proving that this a density. If so, why?
EDIT.
Another related integral is $I=\displaystyle\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1-\rho\sin{2\theta}}=\frac{\pi-\arccos\rho}{\sqrt{1-\rho^2}}$, which arises while calculating $\mathbb{P}(X>0,Y>0)$ for the above distribution once I transform to polar coordinates. The above probability is then given by $\dfrac{\sqrt{1-\rho^2}}{2\pi}I$. Following the suggestions given in the answers below, I was able to evaluate $I$ correctly and hence calculated the probability by direct integration. 
 A: \begin{align}
t & = \tan\varphi \\[10pt]
dt & = \sec^2\varphi\,d\varphi = (1+\tan^2\varphi)\,d\varphi = (1+t^2)\,d\varphi, \\[10pt]
\text{so } \frac{dt}{1+t^2} & = d\varphi \\[10pt]
\text{and } \sin(2\varphi) & = 2\sin\varphi\cos\varphi = 2\sin(\arctan t)\cos(\arctan t) \\[10pt]
& \phantom{{}= 2\sin\varphi\cos\varphi} = 2\frac{t}{\sqrt{1+t^2}} \cdot\frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2}
\end{align}
Then we have
$$
\int_0^{2\pi} \frac{d\varphi}{1 - \rho\sin(2\varphi)} = \left(\int_{-\infty}^\infty + \int_{-\infty}^\infty 
\right) \frac{\left( 
\dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)}
$$
What do I mean by that?? Simply that as $\varphi$ goes from $0$ to $2\pi,$ then $(\cos(2\varphi),\sin(2\varphi))$ goes around the circle twice, and each revolution causes $t$ to run through the whole real line once. We have a function of period $\pi$ integrated over the interval from $0$ to $2\pi.$ Thus we have
$$
2 \int_{-\infty}^\infty
\frac{\left( \dfrac{dt}{1+t^2} \right)}{1 - \rho\left(\dfrac{2t}{1+t^2}\right)} = 2\int_{-\infty}^\infty \frac{dt}{1+t^2 -2\rho t}.
$$
Now complete the square:
$$
t^2 - 2\rho t + 1 = (t^2 - 2\rho t + \rho^2) + 1 - \rho^2 = (t-\rho)^2 + 1-\rho^2.
$$
We would like a quadratic polynomial whose constant term is $1,$ so this becomes
\begin{align}
& (t-\rho)^2 + 1-\rho^2 = (1-\rho^2)\left( \frac{(t-\rho)^2}{1-\rho^2} + 1 \right) \\[10pt]
= {} & (1-\rho^2) \left( \left( \frac{t-\rho}{\sqrt{1-\rho^2}} \right)^2 + 1 \right) \\[10pt]
= {} & (1-\rho^2) (u^2 + 1) \\[10pt]
\text{so } \sqrt{1-\rho^2}\,\,du & = dt.
\end{align}
Our integral becomes
\begin{align}
2\sqrt{1-\rho^2}\int_{-\infty}^\infty \frac{du}{(1-\rho^2) (u^2+1)} = \frac{2\pi}{\sqrt{1-\rho^2}}.
\end{align}
A: Enforcing the substitution $2\phi \to \phi$ and exploiting the $2\pi$-periodicity of the integrand reveals that 
$$\begin{align}
\int_0^{2\pi}\frac{1}{1-\rho\sin(2\phi)}\,d\phi&=\frac12\int_0^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi\\\\
&=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_{2\pi}^{4\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\
&=\frac12 \left(\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi+\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \right)\\\\
&=\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi \tag 1
\end{align}$$
Next, we exploit the $2\pi$-periodicity and the oddness of the sine function in $(1)$ to obtain
$$\begin{align}
\int_0^{2\pi}\frac{1}{1-\rho\sin(\phi)}\,d\phi& =2\int_0^\pi \frac{1}{1-\rho^2\sin^2(\phi)}\,d\phi\\\\
&=4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi\tag2
\end{align}$$
Applying the tangent half-angle substitution $t=\tan(\phi/2)$ to the integral on the right-hand side of $(2)$ yields
$$\begin{align}
4\int_0^\pi \frac{1}{(2-\rho^2)+\rho^2\cos(\phi)}\,d\phi&=4\int_0^\infty \frac{1}{(2-\rho^2)+\rho^2 \frac{1-t^2}{1+t^2}}\,\frac{2}{1+t^2}\,dt\\\\
&=4\int_0^\infty \frac{1}{1+(1-\rho^2)t^2}\,dt\\\\
&=\frac{2\pi}{\sqrt{1-\rho^2}}
\end{align}$$
A: Let $z=e^{i\phi}$ and then $d\phi=\frac{1}{iz}dz$. So $\sin\phi=\frac{1}{2i}(z-\frac1z)$
\begin{eqnarray}
&&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\
&=&\int_0^{2\pi}\frac{1}{1-\rho \sin{\phi }}\,\mathrm{d}\phi\\
&=&\int_{|z|=1}\frac{1}{1-\rho \frac{1}{2i}(z-\frac1z)}\,\frac{1}{iz}\mathrm{d}z\\
&=&\int_{|z|=1}\frac{2}{2iz-\rho (z^2-1)}\,\mathrm{d}z\\
&=&-\frac{1}{\rho}\int_{|z|=1}\frac{2}{z^2-\frac{2i}\rho z-1}\,\mathrm{d}z\\
&=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\
\end{eqnarray}
where 
$$ z_{1,2}=\frac{i}{\rho}\pm \frac{\sqrt{1-\rho^2}i}{\rho} $$
are the two root of $z^2-\frac{2i}\rho z-1=0$
and only $z_2$ is inside $|z|=1$. So
\begin{eqnarray}
&&\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi\\
&=&-\frac{2}{\rho}\int_{|z|=1}\frac{1}{(z-z_1)(z-z_2)}\,\mathrm{d}z\\
&=&-\frac{2}{\rho}2\pi i\frac{1}{z_2-z_1}\\
&=&\frac{2\pi}{\sqrt{1-\rho^2}}.
\end{eqnarray}
