[HELP] How to integrate ∫ 1/(1-cosx)^2 dx ?

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Discussion Overview

The discussion revolves around the integration of the function ∫ 1/(1-cosx)^2 dx, exploring methods and related concepts in calculus. Participants also touch on the integration of a similar function involving an elliptical integral.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the identity cos x = 1 - 2sin^2(x/2) to transform the integral into a more manageable form.
  • Another participant inquires about integrating a related function ∫ 1/(1-e*cosx)^2 dx and whether the same method applies.
  • A response indicates that the latter integral is an elliptical integral, which lacks an analytic, closed form solution in terms of elementary functions.
  • Another participant clarifies that the area of an ellipse can be derived using a parametrization method, which is distinct from elliptic integrals.
  • It is noted that while the area of an ellipse can be calculated easily, the arc length cannot be expressed in elementary forms and is related to elliptic functions.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the area of an ellipse and elliptic integrals, with some asserting that they are unrelated while others emphasize the complexity of arc length calculations involving elliptic functions. The discussion remains unresolved regarding the integration of the elliptical integral.

Contextual Notes

Participants reference various mathematical identities and methods without resolving the assumptions or limitations inherent in the transformations and integrals discussed.

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[HELP] How to integrate ∫ 1/(1-cosx)^2 dx ??

Hi all,

A very fundamental question here, but I cannot find solution from calculus books. Anyone know how to integrate ∫ 1/(1-cosx)^2 dx ?? Thanks.
 
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The trick is to use
\cos x=1-2\sin^2(x/2) \; \Rightarrow \; 1-\cos x=2 \sin^2(x/2).
Then your integral becomes
I=\int \mathrm{d} x \frac{1}{4\sin^4(x/2)}.
From this we get
I=\frac{1}{4} \int \mathrm{d} x \left [\frac{1}{\sin^2(x/2)}+\frac{\cos^2(x/2)}{\sin^4(x/2)} \right ].
Substituting in the latter integral u=\cot(x/2), this is immediately integrated to
I=-\frac{1}{2} \cot(x/2)- \frac{1}{6} \cot(x/2)^3].
 
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Hi vanhees71,

Thanks for your solution. I am trying to derive the formula for the area of an ellipse using polar coordinate. And I stuck here. How about if we want to integrate ∫ 1/(1-e*cosx)^2 dx where there is a constant inside, can we still use the same method? Thanks.
 
cxcxcx0505 said:
Hi vanhees71,

Thanks for your solution. I am trying to derive the formula for the area of an ellipse using polar coordinate. And I stuck here. How about if we want to integrate ∫ 1/(1-e*cosx)^2 dx where there is a constant inside, can we still use the same method? Thanks.
No, you can't.
What you have there is what is essentially called an elliptical integral, and has no analytic, closed form solution in terms of elementary functions.

But, the elliptical integral is very well-known, so you can check up on that concept.
 
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The area of an ellipse, however, has nothing to do with elliptic integrals. It's easily found, using the parametrization
\vec{r}(\lambda,\phi)=\lambda \begin{pmatrix} a \cos \phi \\ b \sin \phi \end{pmatrix}, \quad \lambda \in [0,1],\; \phi \in [0,2 \pi).
Just evaluate the Jacobian and do the double integral, which is pretty easy to get the well-known formula for the area of the ellipse,
A=\pi a b.
The arc length of an ellipse cannot be given in terms of elementary forms but only by elliptic functions (except for the special case of a circle of course :-)). That's where the term elliptic functions comes from for the corresponding integrals and related integrals of a similar type.
 
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Thanks all for the guidance
 

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