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Homework Help: Trig substitution should be simple but it's driving me nuts

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\int x/sqrt{(x^{2}+4)}[/itex]


    2. Relevant equations
    x=2tanx


    3. The attempt at a solution
    x=2tanx
    [itex]\int[/itex]2tan[itex]\vartheta[/itex]/[itex]\sqrt{tan^2\vartheta}[/itex]+4
    2/2 *[itex]\int[/itex]tan/sec

    [itex]\int[/itex]sin=-cos

    now is the part where i am stuck
    i know from using substitution that the answr should be [itex]\sqrt{x^2+4}[/itex]
    but no matter how i manipulate it, it comes out strange.
    all help is appreciated
     
  2. jcsd
  3. Sep 13, 2011 #2

    vela

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    The integral you're trying to evaluate is
    [tex]\int\frac{x}{\sqrt{x^2+4}}\,dx[/tex]
    Note the presence of the dx. That's what you forgot to account for when you did the trig substitution.
     
  4. Sep 13, 2011 #3
    how do you get rid of the natural log at the end?
     
  5. Sep 13, 2011 #4

    vela

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    What natural log? You apparently made another mistake.
     
  6. Sep 13, 2011 #5
    the new integral is
    int 2tan(x)sec(x)^2/Sqrt(4tan(x)^2+4)
    the sec^2 cancel out making it the integral of 2tan(x)=-2lncos(x)
     
  7. Sep 13, 2011 #6

    vela

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    Slow down and look at it more carefully.
    [tex]\int \frac{2 \tan \theta \ \sec^2\theta}{\sqrt{4\tan^2\theta+4}}\,d\theta = \int \frac{2 \tan \theta \ \sec^2\theta}{2\sqrt{\sec^2\theta}}\,d\theta[/tex]
     
  8. Sep 13, 2011 #7
    I'm not sure that's what you mean, but I hope you noticed that there's a much simpler substitution possible :-)

    Anyway, the others are correct about your mistakes :-)
     
  9. Sep 13, 2011 #8
    thank you all for your timely assistance
     
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