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.

  • $\begingroup$ Ignore my previous comment, which was incorrect. This integral can be done, but it is tricky. $\endgroup$ – David H Oct 29 '13 at 16:39
  • $\begingroup$ $x = \sin t$ may help. I get a rather ugly but perhaps tractable integral. $\endgroup$ – rogerl Oct 29 '13 at 16:53
  • 1
    $\begingroup$ $x = \cot t$ will transform the integral to something familiar. $\endgroup$ – achille hui Oct 29 '13 at 17:01
  • $\begingroup$ I would recommend $u=\cos t$, or even $u=\cos t/\sqrt{\sin^2t-c^2}$. $\endgroup$ – mickep Jan 3 '16 at 21:05

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 $$\int\dfrac1{\sqrt{v-v^2}}\mathrm dv.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.