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Hyperbolic Functions

  1. Mar 18, 2009 #1
    1. I am asked to calculate [tex]\int \frac{dx}{cosh(x)}[/tex]



    2. Relevant equations



    3. The attempt at a solution

    I know that [tex]\frac{1}{cosh(x)}[/tex] is equivalent to "sech(x)" which by definition is [tex]\frac{2}{e^x+e^{-x}}[/tex].

    I'm confused & I don't know which one I need to use for this question:

    [tex]\int sech(x) = 2tan{-1}(tanh(\frac{x}{2}))[/tex]

    [tex]\int \frac{2}{e^x + e^{-x}} = 2tan^{-1}(e^x)[/tex]

    Any help would be greatly appreciated.
     
    Last edited: Mar 18, 2009
  2. jcsd
  3. Mar 18, 2009 #2
    The two answers you got only differ by a constant of [tex]\frac{\pi}{2}[/tex]. Recall that the indefinite integral is only defined up to a constant, so both solutions are correct.
     
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