Tricky Integral Involve Sine

Question

How would I integrate the following: $$ \int \frac{c}{\sin(t)\sqrt{\sin^2(t) - c^2}} \, dt $$

Here $c$ is a constant. I have tried numerous substitutions, but I just can't seem to get the right one. Integration by parts does not seem to be of any help either.

Ignore my previous comment, which was incorrect. This integral can be done, but it is tricky. — David H, Oct 29 '13 at 16:39
$x = \sin t$ may help. I get a rather ugly but perhaps tractable integral. — rogerl, Oct 29 '13 at 16:53
$x = \cot t$ will transform the integral to something familiar. — achille hui, Oct 29 '13 at 17:01
I would recommend $u=\cos t$, or even $u=\cos t/\sqrt{\sin^2t-c^2}$. — mickep, Jan 03 '16 at 21:05

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

Hint:
As suggested by @achille hui, one may use the substitution $u\leadsto\cot t$, to get $\mathrm{d}t=-\sin^2t\,\mathrm{d}u$, and using the fact that $$\sin t=\dfrac1{\sqrt{1+\cot^2t}}=\dfrac1{\sqrt{1+u^2}},\tag{$0\lt t\lt\pi$}$$ we have $$\int \dfrac{c}{\sin t\sqrt{\sin^2 t-c^2}}\,\mathrm{d}t=-c\int\dfrac{\tfrac{1}{\sqrt{1+u^2}}}{\sqrt{\left(\tfrac{1}{\sqrt{1+u^2}}\right)^2-c^2}}\,\mathrm{d}u=-c\int\dfrac1{\sqrt{1-c^2(1+u^2)}}\,\mathrm{d}u.$$ Making use of the substitution $1+u^2\leadsto v$ shows that the evaluation of this integral boils down to computing the familiar $$\int\dfrac1{\sqrt{v-v^2}}\mathrm dv.$$

Tricky Integral Involve Sine

1 Answers1