Question: 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 $\cos^{-1}$

Answer: 
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.
The first derivative of $\arccos$ is given by $$-\frac{1}{\sqrt{1-x^2}}$$, so the first derivative of $$\arccos\left(\frac{z}{B}\right)$$ is given by $$-\frac{1}{B \sqrt{1-\frac{z^2}{B^2}}}$$