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Help integrating 1/(cosh(z)+1)

  1. Apr 6, 2014 #1
    1. The problem statement, all variables and given/known data

    integrate 1/(cosh(z)+1)

    2. Relevant equations




    3. The attempt at a solution

    integral(1/(cosh(z)+1))=arctan((cosh(z)) but can I also do 1/(cosh(z)+1)=cosh(z)-1/sinh(z) and go from there and get a simpler solution or something?
     
  2. jcsd
  3. Apr 6, 2014 #2

    vela

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    Do you mean
    $$\int \frac{dz}{\cosh z + 1} = \text{arctanh}\cosh z$$ and
    $$\frac{1}{\cosh z+1} = \frac{\cosh z-1}{\sinh^2 z}.$$ The first one I figure was a typo, but I don't see how you got your second expression, the one for 1/(cosh z + 1)
     
    Last edited: Apr 7, 2014
  4. Apr 6, 2014 #3

    SammyS

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    ## \sinh^2 z = \cosh^2 z - 1 ##
     
  5. Apr 6, 2014 #4

    vela

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    But how did the OP get 1/(cosh(z)+1)=cosh(z)-1/sinh(z)?
     
  6. Apr 6, 2014 #5

    SammyS

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    I needed to read that more closely ...
     
  7. Apr 7, 2014 #6

    vela

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    I didn't word it very well. I changed the wording so it's a bit clearer.
     
  8. Apr 7, 2014 #7
    The derivatives of both arctan(x) and arctanh(x) have x^2 in their denominators, so I can't agree with your initial result. But note the identity sqrt[cosh(z) + 1] = 2 cosh(z/2), which will help you get a correct answer for the original integral.

    A further point: please don't write an integral without a differential, it's meaningless. Thus, one does not EVER integrate 1/[cosh(z)+1], rather one integrates dz/[cosh(z)+1].
     
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