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Math Help - This integral is doing my head in!

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

    I am trying to do the integral

    <br />
\displaystyle\int^\pi_0 F(\theta)\,d\theta<br />

    ...where

    <br />
F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)}<br />

    ...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!
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Astrofiend View Post
    I am trying to do the integral

    <br />
\displaystyle\int^\pi_0 F(\theta)\,d\theta<br />

    ...where

    <br />
F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)}<br />

    ...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
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  3. #3
    MHF Contributor chisigma's Avatar
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    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
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  4. #4
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    Thanks Anonymous1, but chisigma - you are a diamond among gems!

    Thanks very much... I appreciate it a great deal.
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  5. #5
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    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)}<br />


    Excuse my ignorance, but how does that work?
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  6. #6
    MHF Contributor chisigma's Avatar
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    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
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Astrofiend View Post
    I am trying to do the integral

    <br />
\displaystyle\int^\pi_0 F(\theta)\,d\theta<br />

    ...where

    <br />
F(\theta) = \frac{\epsilon(1-cos\theta)}{1+\epsilon(1-cos\theta)}<br />

    ...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.
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  8. #8
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    Once again, thanks chisigma.

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

    Cheers!
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