Solving Hyperbolic Integral 1/(1+cosh(x)) with Wolfram

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    Hyperbolic Integral
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Discussion Overview

The discussion centers around solving the integral of 1/(1+cosh(x)), with a focus on the transformations used in the solution process, particularly those provided by Wolfram. Participants explore the mathematical identities and substitutions involved in the integration process, specifically in the context of hyperbolic functions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the second transformation used in the integral solution provided by Wolfram.
  • Another participant clarifies that the primary substitution used is u=tanh(x/2), questioning what is unclear about it.
  • A participant refers to the need for understanding hyperbolic identities to grasp the transformations involved.
  • One participant presents a derivation starting from the identity involving cosh, leading to the conclusion of the integral being equal to tanh(x/2), while noting the derivative of tanh.
  • A later reply indicates that the participant has gained clarity on the transformations after further explanation, specifically mentioning the double angle identity and Osbournes Rule.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the transformations, as one participant remains confused while another provides a solution. The discussion reflects differing levels of understanding regarding hyperbolic identities and their application in the integral.

Contextual Notes

Limitations in understanding may stem from varying familiarity with hyperbolic functions and the specific transformations used in the integration process. The discussion does not resolve all uncertainties regarding the identities at play.

FelixHelix
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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
 

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They use only one substitution: u=tanh(x/2). What's not clear about that ?
 
I suppose I mean the second transformation. I can't see how you get there... What identities are at work...?
 
Starting from the identity
\cosh x=\cosh^2 (x/2)+\sinh^2(x/2)=2 \cosh^2(x/2)-1,
we get
\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).
Of course one should now that
\frac{\mathrm{d}}{\mathrm{d} x} \tanh x=\frac{1}{\cosh^2 x}.
 
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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