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Integration and inverse trig functions

  1. Dec 7, 2006 #1
    1. The problem statement, all variables and given/known data
    getting confused with integration of trig functions.
    I am finding the integral of sinhx/1+coshx and I'm not sure how to start. should i use an identity?

    help is appreciated!


    2. Relevant equations
    possibly an identity of some sort?

    3. The attempt at a solution
    ?
     
  2. jcsd
  3. Dec 7, 2006 #2

    Kurdt

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    Well the obvious chamge would be;

    [tex] \frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x) [/tex]

    Which makes it a little easier.

    EDIT: The functions are not inverse trig functions they are hyperbolic functions incidentally.
     
    Last edited: Dec 7, 2006
  4. Dec 7, 2006 #3

    AlephZero

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    Or, change variables: y = cosh(x), dy = sinh(x)dx.
     
  5. Dec 7, 2006 #4
    That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.
     
  6. Dec 8, 2006 #5

    dextercioby

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    Do you see why

    [tex] \int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C} [/tex]

    The derivative of the denominator is the numerator.

    Daniel.
     
  7. Dec 8, 2006 #6

    Kurdt

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    Thats a rather embarrassing schoolboy error that I wish I could blame on how late I was up last night, but its far too simple for that. :blushing:
     
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