Help integrating 1/(cosh(z)+1)

[Jarfi]

Homework Statement

integrate 1/(cosh(z)+1)

Homework Equations

The Attempt at a Solution

integral(1/(cosh(z)+1))=arctan((cosh(z)) but can I also do 1/(cosh(z)+1)=cosh(z)-1/sinh(z) and go from there and get a simpler solution or something?

 

[vela]

Homework Statement

integrate 1/(cosh(z)+1)

Homework Equations

The Attempt at a Solution

integral(1/(cosh(z)+1))=arctan((cosh(z)) but can I also do 1/(cosh(z)+1)=cosh(z)-1/sinh(z) and go from there and get a simpler solution or something?

Do you mean
$$\int \frac{dz}{\cosh z + 1} = \text{arctanh}\cosh z$$ and
$$\frac{1}{\cosh z+1} = \frac{\cosh z-1}{\sinh^2 z}.$$ The first one I figure was a typo, but I don't see how you got your second expression, the one for 1/(cosh z + 1)

 

[SammyS]

...
$$\frac{1}{\cosh z+1} = \frac{\cosh z-1}{\sinh^2 z}.$$
...
, but I don't see how you got your second expression for 1/(cosh z + 1)

$\sinh^2 z = \cosh^2 z - 1$

 

[vela]

But how did the OP get 1/(cosh(z)+1)=cosh(z)-1/sinh(z)?

 

[SammyS]

But how did the OP get 1/(cosh(z)+1)=cosh(z)-1/sinh(z)?

I needed to read that more closely ...

 

[vela]

I didn't word it very well. I changed the wording so it's a bit clearer.

 

[az_lender]

Homework Statement

integrate 1/(cosh(z)+1)

Homework Equations

The Attempt at a Solution

integral(1/(cosh(z)+1))=arctan((cosh(z)) but can I also do 1/(cosh(z)+1)=cosh(z)-1/sinh(z) and go from there and get a simpler solution or something?

The derivatives of both arctan(x) and arctanh(x) have x^2 in their denominators, so I can't agree with your initial result. But note the identity sqrt[cosh(z) + 1] = 2 cosh(z/2), which will help you get a correct answer for the original integral.

A further point: please don't write an integral without a differential, it's meaningless. Thus, one does not EVER integrate 1/[cosh(z)+1], rather one integrates dz/[cosh(z)+1].