# Hyperbolic integral.

1. Jan 16, 2014

### 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

#### Attached Files:

• ###### DEF 2013 integral solution.jpg
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2. Jan 16, 2014

### dextercioby

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

3. Jan 16, 2014

### FelixHelix

I suppose I mean the second transformation. I cant see how you get there... What identities are at work...?

4. Jan 16, 2014

### dextercioby

5. Jan 17, 2014

### 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
$$\frac{\mathrm{d}}{\mathrm{d} x} \tanh x=\frac{1}{\cosh^2 x}.$$

6. Jan 17, 2014

### 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