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Arctan(x) in terms of logs

  1. Jul 2, 2009 #1
    my math teacher said that the arctan can be setup in terms of logs
    does any one know how to do this.
  2. jcsd
  3. Jul 2, 2009 #2


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    Solve tan(y)=x. (Prerequisite: you must know how to write tan(y) in terms of exponentials)

    Other methods are possible (e.g. antidifferentiate f(x)=1/(1+x²)), but there are more technical details involved.
  4. Jul 2, 2009 #3
    can i use eulers formula to do it .

    so would it be [(e^(ix)-e^(-ix)]/[(ie^(ix)+ie^(-ix))] = tan(x)
    Last edited: Jul 2, 2009
  5. Jul 3, 2009 #4


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    You'll need to solve a quadratic in the process.
  6. Jul 3, 2009 #5
    what quadratic
  7. Jul 3, 2009 #6


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    You have
    [itex]tan x= \frac{e^x- e^{-x}}{e^x+ e^{-x}}= y[/itex]
    First multiply on both sides of the equation by [itex]e^x+ e^{-x}[/itex].
    Then multiply both sides o the equation by [itex]e^x[/itex]
  8. Jul 4, 2009 #7
    oh i see
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