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I tried using the substitution of $u=\cos\theta-2$ which gives $\dfrac{d\theta}{du}=\dfrac{-1}{\sin\theta}$. Then,

$\displaystyle\int\dfrac{\sin^3\theta}{\sqrt{\cos\theta-2}}\,d\theta=\displaystyle\int\dfrac{\sin^2\theta\sin\theta}{\sqrt{u}}\left(\dfrac{-1}{\sin\theta}\right)du=-\displaystyle\int\dfrac{\sin^2\theta}{\sqrt{u}}du$.

I am stuck here...perhaps using $\sin^2\theta=1-\cos^2\theta$ might help, but I don't know how.

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    $\begingroup$ If you are working with real numbers, then $\sqrt{\cos \theta -2}$ makes no sense. $\endgroup$ – Beni Bogosel Nov 18 '14 at 11:54
  • $\begingroup$ Yes, I didn't notice that. I probably copied something badly but don't have the original question in front of me. $\endgroup$ – bibo_extreme Nov 18 '14 at 11:57
  • $\begingroup$ You could replace it with $\sqrt{2-\cos \theta}$. Maybe that's what the original question had. $\endgroup$ – Beni Bogosel Nov 18 '14 at 11:58
  • $\begingroup$ Most likely, yes. $\endgroup$ – bibo_extreme Nov 18 '14 at 12:00
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As Beni points out, this only makes sense in the complex setting (and with a choice of branch cut), but you can put the integral in terms of $u$ alone via the Pythagorean Identity: $$\sin^2 \theta = 1 - \cos^2 \theta = 1 - (u + 2)^2 = -(u^2 + 4u + 3).$$

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Let $\sqrt{\cos\theta-2}=u\implies\dfrac{-\sin\theta\ d\theta}{\sqrt{\cos\theta-2}}=du$

and $\cos\theta=u^2+2$

$$\int\frac{\sin^3\theta}{\sqrt{\cos\theta-2}}d\theta=-\int[1-(u^2+2)^2]du$$

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Use $\sin{\theta}\, d\theta = -d(\cos{\theta})$. Then the integral is

$$i\int d(\cos{\theta}) \frac{\sin^2{\theta}}{\sqrt{2-\cos{\theta}}}$$

(Note the $i$ comes from the square root.)

Let $u=\cos{\theta}$; then we have

$$i \int du \frac{1-u^2}{\sqrt{2-u}} = -i \int dv \frac{1-(2-v)^2}{\sqrt{v}} = i \int dv \, v^{-1/2} (3 - 4 v+v^2)$$

I think you can take it from here.

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