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

In summary: I will check up on the elliptic integral and try to derive the area formula using it.In summary, the area of an ellipse can be found using the parametrization
  • #1
cxcxcx0505
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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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  • #2
The trick is to use
[tex]\cos x=1-2\sin^2(x/2) \; \Rightarrow \; 1-\cos x=2 \sin^2(x/2).[/tex]
Then your integral becomes
[tex]I=\int \mathrm{d} x \frac{1}{4\sin^4(x/2)}.[/tex]
From this we get
[tex]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 ].[/tex]
Substituting in the latter integral [itex]u=\cot(x/2)[/itex], this is immediately integrated to
[tex]I=-\frac{1}{2} \cot(x/2)- \frac{1}{6} \cot(x/2)^3].[/tex]
 
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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.
 
  • #4
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
[tex]\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).[/tex]
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,
[tex]A=\pi a b.[/tex]
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
 

1. What is the integration method for ∫ 1/(1-cosx)^2 dx?

The integration method for ∫ 1/(1-cosx)^2 dx is substitution, specifically using the trigonometric identity 1-cosx=2sin²(x/2).

2. How do I substitute in the trigonometric identity for ∫ 1/(1-cosx)^2 dx?

To substitute in the trigonometric identity for ∫ 1/(1-cosx)^2 dx, let u = tan(x/2) and use the identity 1+tan²(x/2) = sec²(x/2) to rewrite the integral in terms of u.

3. Can the integral for ∫ 1/(1-cosx)^2 dx be solved without using substitution?

No, the integral for ∫ 1/(1-cosx)^2 dx cannot be solved without using substitution. Substitution is necessary to simplify the integral and make it solvable.

4. What are the limits of integration for ∫ 1/(1-cosx)^2 dx?

The limits of integration for ∫ 1/(1-cosx)^2 dx are the same as the original integral, which is from 0 to 2π.

5. Is there a shortcut or easier method for integrating ∫ 1/(1-cosx)^2 dx?

No, there is no shortcut or easier method for integrating ∫ 1/(1-cosx)^2 dx. Substitution is the most efficient and accurate method for solving this integral.

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