# Integration and inverse trig functions

#### mavsqueen06

1. The problem statement, all variables and given/known data
getting confused with integration of trig functions.
I am finding the integral of sinhx/1+coshx and I'm not sure how to start. should i use an identity?

help is appreciated!

2. Relevant equations
possibly an identity of some sort?

3. The attempt at a solution
?

#### Kurdt

Staff Emeritus
Gold Member
Well the obvious chamge would be;

$$\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)$$

Which makes it a little easier.

EDIT: The functions are not inverse trig functions they are hyperbolic functions incidentally.

Last edited:

#### AlephZero

Homework Helper
Or, change variables: y = cosh(x), dy = sinh(x)dx.

#### d_leet

Well the obvious chamge would be;

$$\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)$$

Which makes it a little easier.
That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.

#### dextercioby

Homework Helper
Do you see why

$$\int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C}$$

The derivative of the denominator is the numerator.

Daniel.

#### Kurdt

Staff Emeritus
Gold Member
That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.
Thats a rather embarrassing schoolboy error that I wish I could blame on how late I was up last night, but its far too simple for that. ### The Physics Forums Way

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