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Hyperbolic integral.

  1. Jan 16, 2014 #1
    Hi there. I've been trying to solve the integral of 1/(1+cosh(x)). I use Wolfram to give me a detailed solution but I still don't understand second transformations they use.

    I've attached a a screen grab of the workings and hoped someone could run through it with me.

    I've used the tan x = t but never for hyperbolic...

    Thanks F
     

    Attached Files:

  2. jcsd
  3. Jan 16, 2014 #2

    dextercioby

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    They use only one substitution: u=tanh(x/2). What's not clear about that ?
     
  4. Jan 16, 2014 #3
    I suppose I mean the second transformation. I cant see how you get there... What identities are at work...?
     
  5. Jan 16, 2014 #4

    dextercioby

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  6. Jan 17, 2014 #5

    vanhees71

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    Starting from the identity
    [tex]\cosh x=\cosh^2 (x/2)+\sinh^2(x/2)=2 \cosh^2(x/2)-1,[/tex]
    we get
    [tex]\int \mathrm{d} x \frac{1}{1+\cosh x}=\frac{1}{2} \int \mathrm{d} x \frac{1}{\cosh^2(x/2)}=\tanh(x/2).[/tex]
    Of course one should now that
    [tex]\frac{\mathrm{d}}{\mathrm{d} x} \tanh x=\frac{1}{\cosh^2 x}.[/tex]
     
  7. Jan 17, 2014 #6
    Ahhh, I see now. Double angle identity and then using Osbournes Rule... easy when you see it!

    Thanks vanhees71 for taking the time to show the working - It's much appreciated....

    FH
     
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