integral of $\frac{c}{\sin(x)\sqrt{\sin^2(x)-c^2}}$

Question

In a script I found that the indefinite integral of $\int{\frac{c}{\sin(x)\sqrt{\sin^2(x)-c^2}}dx}$ is $\sin^{-1}(\frac{\cot(x)}{c})$. I know that $\frac{1}{\sin^2(x)} = \cot^2(x)+1$. I wanted to use this to substitute $y=\frac{\cot(x)}{c}$ and hoped to then get something of the form $\int{\frac{1}{\sqrt{1-y^2}}dy}$ but it did not work. Am I completely on the wrong track or did I miss something? Thank you!

Answer 1

Okay, I think I got it: First, rewrite the integrand as $$\dfrac1{\sin x\sqrt{\sin^2x-c^2}}=\dfrac{\sin x}{\sin^2x\sqrt{\sin^2x-c^2}}=\dfrac{-\cos'x}{(1-\cos^2x)\sqrt{1-\cos^2x-c^2}}~,$$ then, after doing the obvious trigonometric substitution $t=\cos x,~$ let $a^2=1-c^2,~$ and

$t=au.~$ Then another obvious trigonometric substitution, followed by Weierstrass and

some partial fraction decomposition, then you're done ! :-$)$