An integral and a plane

Question

I have an integral

$\displaystyle\int \frac{c\,d \theta}{\sin( \theta) \sqrt{\sin^2 (\theta) - c^2}}$

I don't know how to solve this but also I am led to believe that it will describe a plane passing through a sphere after transforming the relation between $\theta$ and $\phi$ back to cartesian coordinates i.e from

$ z = r\cos (\theta)$

$ x = r\sin(\theta)\cos(\phi)$

$ y = r\sin(\theta)\sin(\phi)$

How shall one do this?

Answer 1

$\displaystyle\int\frac{c\,d\theta}{\sin(\theta)\sqrt{\sin^2(\theta)-c^2}}\underset{\overbrace{\;t=\cos(\theta)\;}}{=}\int\frac{-c\,dt}{\left(1-t^2\right)\sqrt{1-c^2-t^2}}=$

$=-\displaystyle\int\frac{ct\,dt}{t\left(1-t^2\right)\sqrt{1-c^2-t^2}}\underset{\overbrace{\;\;u=\frac{ct}{\sqrt{1-c^2-t^2}}\;\;}}{=}-\int\frac{du}{u^2+1}=$

$=-\arctan(u)+\text{constant}=$

$=-\arctan\left(\!\dfrac{c\cos(\theta)}{\sqrt{\sin^2(\theta)-c^2}}\!\right)+\text{constant}\,.$

Answer 2

\begin{align} \int \frac{c}{\sin \theta \sqrt{\sin^2 \theta - c^2}}d\theta = &\int \frac{c}{\sin^2 \theta \sqrt{1-c^2 \csc^2 \theta}}d\theta\\ =&- \int \frac{c}{\sqrt{1-c^2 - c^2\cot^2 \theta}} d(\cot \theta)\\ =& - \sin^{-1} \frac {c\cot \theta}{\sqrt{1-c^2}} +C \end{align}