# An integral and a plane

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?

• I have no idea why it would be a plane intersected with a sphere. You did not include any relevant information, such as is $c^2<1$. But I would multiply with $\sin\theta$ both numerator and denominator, then in the denominator I would replace $\sin^2\theta$ by $1-\cos^2\theta$. Finally I would do the substitution $u=\cos\theta$, $du=-\sin\theta$. See if this simplifies your equation. Sep 17, 2022 at 15:30
• I think I get $\int \frac{-c du}{(1- u^2)\sqrt{( 1 -u^2)- c^2}}$ for the integral when I do the substitution . Sep 17, 2022 at 16:40
• if $c^2>1$ then $\sin^2\theta-c^2<1$ so the square root is always imaginary Sep 17, 2022 at 16:44
• ok then it has to be the case that $c^2 < 1$ Sep 17, 2022 at 16:45
• I think a second substitution is needed too Sep 17, 2022 at 16:53

$$\displaystyle\int\frac{c\,d\theta}{\sin(\theta)\sqrt{\sin^2(\theta)-c^2}}\underset{\overbrace{\;t=\cos(\theta)\;}}{=}\int\frac{-c\,dt}{\left(1-t^2\right)\sqrt{1-c^2-t^2}}=$$
$$=-\displaystyle\int\frac{ct\,dt}{t\left(1-t^2\right)\sqrt{1-c^2-t^2}}\underset{\overbrace{\;\;u=\frac{ct}{\sqrt{1-c^2-t^2}}\;\;}}{=}-\int\frac{du}{u^2+1}=$$
$$=-\arctan(u)+\text{constant}=$$
$$=-\arctan\left(\!\dfrac{c\cos(\theta)}{\sqrt{\sin^2(\theta)-c^2}}\!\right)+\text{constant}\,.$$
\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}
• Quanto, your result is correct only in the case that $\,\sin\theta>0\,.$ You should calculate the integral in the case that $\,\sin\theta<0\,$ too. Sep 17, 2022 at 20:49