I was doing an exercise in Pressley's Elementary Differential Geometry that asks us the derive the Mercator Projection (i.e. come up with a conformal parametrization of the sphere) and I came accross this nonlinear ODE: $\cos \psi = \frac{\text d \psi}{\text d u}$. In fact I don't really have to solve the ODE, I have to get $\cos \psi$ and $\sin \psi$ as functions of $u$.
In the answers at the back of the book they seem to take the inverse of this expression ($\sec \psi = \frac{\text d u}{\text d \psi}$) and then integrate both sides to get $u=\ln (\tan \psi + \sec \psi)$ (notice the missing absolute values). Using this they get $2\cosh u = e^u + e^{-u}=2\sec \psi$ which is enough because then you can easily get $\cos \psi = \text{sech}u$ and $\sin \psi = \tanh u$.
This is all good and dandy but it relies on the integral of secant without an absolute value! If you add the absolute value nothing works anymore... Does this all work out by luck? Is there a better way to solve this? Am I just missing a small step that allows me to remove the absolute value in this particular case?
Nevermind, as soon as I posted this I realized I completely overlooked the fact that $\psi \in (-\frac{\pi}{2}, \frac{\pi}{2})$, where $\tan \psi + \sec \psi$, is actually always positive, back to first year calculus I guess.