Show that $$\int_0^\pi{d\theta\over(a-b\cos\theta)^2}={a\pi\over (a^2-b^2)^{3\over 2}}$$ where $a>b>0$. I'm not sure how to simplify this integral or evaluate it. Any solutions or hints are greatly appreciated.


closed as off-topic by Travis Willse, C. Dubussy, imranfat, heropup, Martin R Apr 21 '16 at 21:25

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    $\begingroup$ With all due respect I think you ask lots of questions but you don't seem to be willing to show what did you attempt. This is important because that way people can focus on your particular problems and also because many people here expect you to show some effort. $\endgroup$ – DonAntonio Apr 21 '16 at 20:44
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    $\begingroup$ You should also start accepting (at least some of) the answers that the MSE community kindly provided you. $\endgroup$ – Jack D'Aurizio Apr 21 '16 at 20:45
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    $\begingroup$ @JackD'Aurizio Jack, thank you for having the courage to post this comment. If I had a dollar for every answer posted for which neither others who posted solutions nor I received any up votes, well ... you know the rest of this... ;-)) -Mark $\endgroup$ – Mark Viola Apr 21 '16 at 20:50
  • $\begingroup$ I'm sorry everyone. I will heed your advice. $\endgroup$ – Happy Apr 21 '16 at 20:55
  • $\begingroup$ This integral is Kepler's third law! It's the situation for which the eccentric anomaly $$\sin E=\frac{\sqrt{1-e^2}\sin\theta}{1+e\cos\theta}$$ was invented. In terms of $E$ with $e=\frac ba$ the integral transforms to $$\frac1{a^2(1-e^2)^{\frac32}}\int_0^{\pi}(1-e\cos E)dE=\left.\frac{a}{(a^2-b^2)^{\frac32}}(E-e\sin E)\right|_0^{\pi}=\frac{\pi a}{(a^2-b^2)^{\frac32}}$$ $\endgroup$ – user5713492 Apr 22 '16 at 3:53


Note that $$\int_0^\pi \frac{1}{(a-b\cos(\theta))^2}\,d\theta=-\frac{\partial}{\partial a}\int_0^\pi \frac{1}{a-b\cos(\theta)}\,d\theta$$

Then, evaluate the integral on the right-hand side using either the classical Tangent Half-Angle Substitution or use contour integration.

  • $\begingroup$ @derpmagoo Please let me know how I can improve my answer. I really want to give you the best answr I can. -Mark $\endgroup$ – Mark Viola Apr 21 '16 at 20:51
  • $\begingroup$ My knowledge of integration is pretty elementary. I have never seen this before. I'll have to read that link. Thank you. $\endgroup$ – Happy Apr 21 '16 at 20:57
  • $\begingroup$ Derp, may I ask respectfully, "If your knowledge of integration is pretty elementary, then why are you tackling integrals such as the one here?" While this can be handled by elementary methodologies, this might not be the best place to start. ;-)) -Mark $\endgroup$ – Mark Viola Apr 21 '16 at 21:10
  • $\begingroup$ Dr.MV: +1 for giving a good hint and not the answer. $\endgroup$ – RRL Apr 21 '16 at 22:40
  • $\begingroup$ @RRL Thank you! Much appreciative. $\endgroup$ – Mark Viola Apr 21 '16 at 22:51

Under the assumption $a>b>0$, let us start with: $$ I(a,b) = \int_{0}^{\pi}\frac{dt}{a-b\cos t} = 2\int_{0}^{\pi/2}\frac{dt}{(a+b)-2b\cos^2 t} $$ that through the substitution $t=\arctan u$ becomes: $$ I(a,b) = 2\int_{0}^{+\infty}\frac{du}{(a+b)(1+u^2)-2b}=\frac{\pi}{\sqrt{a^2-b^2}}. $$ Our integral is just $-\frac{\partial}{\partial a} I(a,b)$, hence it equals $\frac{\pi a}{(a^2-b^2)^{3/2}}$.

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    $\begingroup$ Now this looks a bit familiar ... ;-)) $\endgroup$ – Mark Viola Apr 21 '16 at 20:52

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