In my textbook we are trying to solve this separation of variables equation:
$$\int {dz\over \sqrt{B^2-z^2}} = \int d\theta$$
Without any explanation the solution is:
$$\cos^{-1}\Big( {z\over B} \Big)=\theta-\theta_0$$
But shouldn't it be $\sin^{-1}?$ All of this is going towards trying to find the equation for gravitational potential, the rest of the work I understand but this part is confusing to me. I know that $\cos^{-1}={\pi\over 2}-\sin^{-1}$, but I don't see how they applied it here. Seeing as this is a physics textbook and not a math one I'm not surprised they don't give any explanation, but even Wolfram gives me a different answer which uses $\tan^{-1}$.
Is this the right answer? If so, how did the authors get it?
Edit:
The book does give a value for $B$:
$$\Big( {mE\over L^2} \Big){\sqrt{1-{{2\beta L^2\over mE^2}}}} $$
So according to the book $B$ is not negative or an absolute value. Using this information makes it hard for me to find a way to make the integral negative so I can use $\cos^{-1}$
You're actually right, that should be $\sin^{-1}$ (or $\ -cos^{-1}$).
Indeed, since $\sin^{-1}$ and $\cos^{-1}$ are linked, as you said, by $\cos^{-1}+\sin^{-1}=\pi/2$, and since a continuous function has an infinite number of antiderivatives differing by a constant, taking $sin^{-1}$ or $cos^{-1}$ makes no real difference, besides from inverting the sign (which is the issue here).
EDIT : a very valid point was made in the comments of Dr. Sonnhard Graubner's answer : The - sign may be coming from the B term.