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

[FelixHelix]

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

 

[dextercioby]

They use only one substitution: u=tanh(x/2). What's not clear about that ?

 

[FelixHelix]

I suppose I mean the second transformation. I can't see how you get there... What identities are at work...?

 

[dextercioby]

The ones in hyperbolic trigonometry. Check out the wiki page: http://en.wikipedia.org/wiki/Hyperbolic_function

 

[vanhees71]

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}.

 

[FelixHelix]

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