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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.

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  • $\begingroup$ 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
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    $\begingroup$ $x = \cot t$ 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
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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 $$\int\dfrac1{\sqrt{v-v^2}}\mathrm dv.$$

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