Integration and inverse trig functions

[mavsqueen06]

Homework Statement

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!

Homework Equations

possibly an identity of some sort?

The Attempt at a Solution

?

 

[Kurdt]

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.

 

[AlephZero]

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]

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]

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.