[SOLVED] Area of a hyperbolic circle

Prove that a hyperbolic circle with hyperbolic radius $\displaystyle R$ in the hyperbolic upper half plane is $\displaystyle 2\pi (\cosh R - 1)$

The problem suggests to place a coordinate system at the Euclidean of the circle and work out the double integral in polar coordinates. However, and I believe I must be messing up somewhere, if we let the Euclidean center of the circle be $\displaystyle (h,k)$ and have Euclidean radius $\displaystyle \rho<k$, then we have in polar coordinates centered at the Euclidean center $\displaystyle x= \rho\cos\theta + h$ and $\displaystyle y=\rho\sin\theta + k$, and so we have the hyperbolic area of the circle being:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\rho}\, \frac{rdr d\theta}{(r\sin\theta + k)^{2}}$

However, this integral seems to be undefined because along the line somewhere, I get a $\displaystyle \cot\theta$, which is undefined at $\displaystyle 0$ and $\displaystyle 2\pi$.

Are my parameterizations wrong? Or are they correct and this integral does exist? Should I not deal with the Euclidean radius and approach it some other way? Thanks.