# Math Help - This integral is doing my head in!

1. ## This integral is doing my head in!

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!

2. 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

3. 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

$\chi$ $\sigma$

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

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

5. 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?

6. Is...

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

Kind regards

$\chi$ $\sigma$

7. 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.

8. Once again, thanks chisigma.

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

Cheers!