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  • 1 Post By romsek

Math Help - trig sub correct or incorrect

  1. #1
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    trig sub correct or incorrect

    \int \frac{1}{x^2-6x+5}dx I complete the square for \int \frac{1}{(x-3)^2-4}dx let (x-3) = 2 sec theta and dx = 2 sec theta tan theta dtheta

    \int \frac{2 sec\theta tan \theta d\theta}{4(sec^2 \theta -1} \Rightarrow \frac{1}{2} \int \frac {sec \theta}{tan \theta} d\theta

    \Rightarrow \frac{1}{2} \int csc \theta d\theta = \frac{1}{2} ln |csc \theta - cot \theta| + C

     = \frac{1}{2} ln| \frac{x-3}{\sqrt{(x-3)^2 -4}} - \frac{2}{\sqrt{(x-3)^2 -4}}|+C

    is this correct? wolfram did it differently so I just want to check to see if my way is correct as well.
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  2. #2
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    Re: trig sub correct or incorrect

    your way looks correct as well. The plots of your function and wolfram's integral lie on top of each other.

    I have to say though that this is clearly a partial fractions problem and not a trig substitution problem even if you can find an substitution that works.

    $x^2-6x+5=(x-5)(x-1)$ so a bit of work shows

    $\dfrac{1}{x^2-6x+5}=\dfrac{1}{4}\left(\dfrac{1}{x-5}-\dfrac{1}{x-1}\right)$

    and the integral pops right out.
    Last edited by romsek; March 4th 2014 at 12:40 PM.
    Thanks from Jonroberts74
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  3. #3
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    Re: trig sub correct or incorrect

    ah that makes much more sense. its from a improper integral convergence or divergence question. thanks
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