Tricky looking integration (after separation of variables)? I've come across something in my notes that jumps from:
$${d\rho \over dz} = \sqrt{\left({\rho \over C}\right)^2 - 1}$$
to:
$$\rho(z) = C \cosh\frac{z-z_0}{C}$$
I know that separation of variables is used so what I'm really asking is how to integrate this part:
$$\int{1 \over \left(\left(\dfrac{\rho}{C}\right)^2 - 1\right)^{1/2}} \, d\rho ,$$
where $\rho$ is a function of $z$ and $C$ is a constant.
Help with this one would be greatly appreciated, thanks! 
 A: \begin{align}
\frac \rho C & = \sec\theta \\[10pt]
\sqrt{\left(\frac\rho C\right)^2 -1} & = \sqrt{\sec^2\theta-1}= \tan\theta \\[10pt]
\frac{d\rho}{C} & = \sec^2\theta\,d\theta
\end{align}
\begin{align}
\int{1 \over \sqrt{\left(\dfrac{\rho}{C}\right)^2 - 1}} \, d\rho & = \int\frac{1}{\tan\theta} \Big( C\sec^2\theta\,d\theta\Big) = C\int \frac{d\theta}{\sin\theta\cos\theta} \\[15pt]
& = C\int\frac{2\,d\theta}{2\sin\theta\cos\theta} = C\int\frac{2\,d\theta}{\sin(2\theta)} = C\int\frac{d\eta}{\sin\eta} \\[15pt]
& = -C\log(\csc\theta+\cot\theta)+\text{constant}.
\end{align}
\begin{align}
\csc\operatorname{arcsec}\frac\rho C & = \frac{\rho}{\sqrt{\rho^2-C^2}} \\[15pt]
\cot\operatorname{arcsec}\frac\rho C & =  \frac{C}{\sqrt{\rho^2-C^2}}
\end{align}
So you get
$$
-C\log\frac{\rho+C}{\sqrt{\rho^2-C^2}} = z + \text{constant} = z - z_0.
$$
Consequently
$$
\frac{\rho+C}{\sqrt{\rho^2-C^2}} = \exp\left(- \frac{z-z_0}{C}  \right) = A
$$
$$
\rho+C = A \sqrt{\rho^2-C^2}
$$
Now if you square both sides you get an equation that is quadratic in $\rho$.
$$
\rho^2+2\rho C+C^2 = A^2\rho^2 - A^2 C^2
$$
$$
\rho^2(1-A^2)+2\rho C + C^2(1+A^2) = 0
$$
Now apply the "quadratic formula":
$$
\rho = \frac{-2C \pm \sqrt{4C^2 - 4(1-A^4)}}{2(1-A^2)} = \frac{-C\pm\sqrt{C^2+A^4-1}}{1-A^2}.
$$
Notice that
\begin{align}
1-A^2  & = 1 - \exp\left( -2\frac{z-z_0}{C} \right) \\[10pt]
& = 2\exp\left( -\frac{z-z_0}{C} \right)\left( \frac{\exp\left( +\frac{z-z_0}{C}\right) - \exp\left( -\frac{z-z_0}{C} \right)}{2} \right) \\[10pt]
& = 2\exp\left( -\frac{z-z_0}{C} \right) \cosh\left(\frac{z-z_0}{C}\right).
\end{align}
(Maybe I'll be back later; I may want to check details . . . )
[ . . . ]
