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Integration Using the arctan rule

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Our question is \int (from 0 to 4) dt/ ((4t^2) + 9)
    (For a better representation, because I fail at LaTeX: http://www.wolframalpha.com/input/?...&f5=4&f=DefiniteIntegralCalculator.rangeend_4 )


    2. Relevant equations

    So we have the rule:

    \int du/(a^2 + u^2) = (1/a) arctan (u/a) + C

    3. The attempt at a solution

    So, using a = 3, and u = 2t (because the rule says that at the bottom, there is a^2 + u^2, so we take the square root of 9, and 4t^2)
    I get 1/3 arctan( 2t/3) from 4 to 0
    1/3 arctan( 8/3) - 1/3 arctan (0)
    =.404..
    However, Wolfram Alpha, and my graphing calculator seem to both give me .202, and wolfram alpha says the coefficient for the final solution is 1/6 arctan..., whereas I got 1/3 arctan..., and I cannot see how it becomes 1/6..

    Thanks for reading!
     
  2. jcsd
  3. Nov 29, 2012 #2

    Dick

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    When you substitute u=2t, you also have to substitute du=2*dt. So also replace dt with du/2. That's the factor of two you are missing.
     
  4. Nov 29, 2012 #3
    Thanks, me and a couple of friends were stumped on that one.
     
  5. Nov 29, 2012 #4

    haruspex

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    When you put u = 2t, what do you put for dt?
     
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