This integral is doing my head in!

**Astrofiend** · April 6th 2010, 05:40 PM

I am trying to do the integral

$ \displaystyle\int^\pi_0 F(\theta)\,d\theta $

...where

$ F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)} $

...and epsilon is a constant.

I have absolutely no idea how to even start on this one! Perhaps making a substitution $u = 1-cos\theta$ ? Then integration by parts?

I've tried working through some of this, but get nowhere quickly. Any help or pointers in the right direction would be most appreciated!

**Anonymous1** · April 6th 2010, 07:01 PM

Originally Posted by Astrofiend

I am trying to do the integral

$ \displaystyle\int^\pi_0 F(\theta)\,d\theta $

...where

$ F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)} $

...and epsilon is a constant.

I have absolutely no idea how to even start on this one! Perhaps making a substitution $u = 1-cos\theta$ ? Then integration by parts?

I've tried working through some of this, but get nowhere quickly. Any help or pointers in the right direction would be most appreciated!

Looks like a trig substitution problem...

Wolfram Mathematica Online Integrator

**chisigma** · April 6th 2010, 07:39 PM

The integral can be written as...

$I = \int_{0}^{\pi} d\theta - \int_{0}^{\pi} \frac{d\theta}{1+ \varepsilon \cdot (1-\cos \theta)}$ (1)

Clearly we have to concentrate the efforts to the solution of the second integral and in such a case the most efficient approach is the variable change $t = \tan \frac{\theta}{2}$ so that we have...

$\cos \theta = \frac{1-t^{2}}{1+t^{2}}$

$d \theta = \frac{2\cdot dt}{1+t^{2}}$ (2)

... and the integral becomes...

$\int_{0}^{\pi} \frac{d\theta}{1+ \varepsilon \cdot (1-\cos \theta)} = \int_{0}^{\infty} \frac{2\cdot dt}{1+(1+2\cdot \varepsilon) \cdot t^{2}} =$

$= 2\cdot | \frac{\tan^{-1} (\sqrt{1+2\cdot \varepsilon}\cdot t)}{\sqrt{1 + 2\cdot \varepsilon}}|_{0}^{\infty} = \frac{\pi}{\sqrt{1 + 2\cdot \varepsilon}}$ (3)

... so that, taking into account (1), is...

$\int_{0}^{\pi} \frac{\varepsilon \cdot (1-\cos \theta)}{1+ \varepsilon \cdot (1-\cos \theta)}\cdot d\theta = \pi \cdot \frac{\sqrt{1 + 2\cdot \varepsilon}-1}{\sqrt{1 + 2\cdot \varepsilon}}$ (4)

Kind regards

$\chi$ $\sigma$

**Astrofiend** · April 6th 2010, 07:52 PM

Thanks Anonymous1, but chisigma - you are a diamond among gems!

Thanks very much... I appreciate it a great deal.

**Astrofiend** · April 6th 2010, 09:11 PM

Actually, I don't understand the first step that you did when transforming the integral to the form

$I = \int_{0}^{\pi} d\theta - \int_{0}^{\pi} \frac{d\theta}{1+ \varepsilon \cdot (1-\cos \theta)} $

Excuse my ignorance, but how does that work?

**chisigma** · April 6th 2010, 10:21 PM

Is...

$\frac{\varepsilon\cdot (1 - \cos \theta)}{1+\varepsilon\cdot (1 - \cos \theta)}= 1 - \frac{1}{1+\varepsilon\cdot (1 - \cos \theta)}$

Kind regards

$\chi$ $\sigma$

**Drexel28** · April 6th 2010, 10:31 PM

Originally Posted by Astrofiend

I am trying to do the integral

$ \displaystyle\int^\pi_0 F(\theta)\,d\theta $

...where

$ F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)} $

...and epsilon is a constant.

I have absolutely no idea how to even start on this one! Perhaps making a substitution $u = 1-cos\theta$ ? Then integration by parts?

I've tried working through some of this, but get nowhere quickly. Any help or pointers in the right direction would be most appreciated!

Define $J(\varepsilon)=\int_0^{\pi}\frac{\varepsilon(1-\cos(\theta))}{1+\varepsilon(1-\cos(\theta))}d\theta$ . I'm not sure (haven't tried it) but this looks like a prime candidate for differentiation under the integral sign.

**Astrofiend** · April 6th 2010, 10:40 PM

Once again, thanks chisigma.

And thanks Drexel28 - I'll take a look into that too.

Cheers!

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