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

1. Nov 7, 2013

### cxcxcx0505

[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.

2. Nov 7, 2013

### vanhees71

The trick is to use
$$\cos x=1-2\sin^2(x/2) \; \Rightarrow \; 1-\cos x=2 \sin^2(x/2).$$
$$I=\int \mathrm{d} x \frac{1}{4\sin^4(x/2)}.$$
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
$$I=-\frac{1}{2} \cot(x/2)- \frac{1}{6} \cot(x/2)^3].$$

3. Nov 7, 2013

### cxcxcx0505

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. Nov 8, 2013

### arildno

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.

5. Nov 8, 2013

### vanhees71

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.

6. Nov 9, 2013

### cxcxcx0505

Thanks all for the guidance