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Math Help - Trig substitution integral

  1. #1
    DBA
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    Trig substitution integral

    <br />
\int \frac{\sqrt{x^{2}-4}}{x^2}dx<br />

    I started with setting

    x=2sec \theta and dx=2sec \theta tan \theta
    and x^{2} = 4sec^{2} \theta

     \tan^{2} \theta = \sec^{2} \theta -1

    Then I rewrite it...

    <br />
\int \frac{2tan\theta}{4sec^{2}\theta} * 2 sec \theta tan\theta d\theta = \int \frac{tan^{2}\theta}{sec\theta}d\theta = \int \frac{sin^{2}\theta}{cos\theta}d\theta <br />

    When I go on I do not get anything what helps me solving it.
    Thanks for any help...
    Last edited by mr fantastic; February 27th 2010 at 03:39 AM. Reason: Fixed a latex tag
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  2. #2
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    How many anti-derivatives are you meant to know - because there is an anti-derivative here that you could use.
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  3. #3
    DBA
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    I can not think of any anti-derivative I could use. I am not sure if I understand your question.
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  4. #4
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    If you make the substitution, 1-sin^{2}(\theta) into your equation, it will reduce down to sec(\theta)-cos(\theta). Secant isn't a "top of my head" integral, but it is good to know.
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  5. #5
    DBA
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    Sorry but I do not get it.. Where can I make the substitution of
    1-sin^{2}(\theta) = cos^{2} \theta

    I have
    \int \frac{sin^{2}\theta}{cos\theta}d\theta

    I do not knwo how you get to ...
    sec(\theta)-cos(\theta)

    Even if I use
    1-cos^{2}(\theta) = sin^{2} \theta
    I don't get a term that helps me...
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  6. #6
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    \frac{\sin^2x}{\cos x}=\frac{1-\cos^2x}{\cos x}=\frac{1}{\cos x}-\frac{\cos^2x}{\cos x}=\sec x-\cos x
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