# Hyperbolic integral

• Jan 10th 2012, 01:09 PM
Mrhappysmile
Hyperbolic integral
Ok so:
Its a 3 part question. I've completed parts 1 and 2.
Part 1)
Draw $tanhx$ and $tanh^-1 x$ on the same graph
Part 2)
Find $INT tanh x dx$ between limits k and 0 (where k > 0) which gives you $ln coshk$

Part 3) though...
I need to show $I tanh^-1 x dx$ between tanhk and 0 is $ktanhk - ln coshk$

If you integrate it by parts you form: $[xtanh^-1 x] tanhk -> 0 = ktanhk$ and I t
hink $- int x/(1-x^2)$ between the same limits. I can't work out how to link lncoshk and that integral. And with the previous parts in the question.

P.S How do you do an integral sign on this forum :p
• Jan 10th 2012, 01:17 PM
pickslides
Re: Hyperbolic integral
What have you tried? Show us some workings.

Quote:

Originally Posted by Mrhappysmile
P.S How do you do an integral sign on this forum :p

use \int command in LaTeX
• Jan 11th 2012, 01:07 AM
tom@ballooncalculus
Re: Hyperbolic integral
Quote:

Originally Posted by Mrhappysmile
I can't work out how to link lncoshk and that integral. And with the previous parts in the question.

http://www.ballooncalculus.org/draw/graph/inverse.png
• Jan 11th 2012, 01:41 AM
Mrhappysmile
Re: Hyperbolic integral
Ah I see so Instead of working it out as an integral. You have a rectangle of total area ktanhk and you're subtracting what you know from the previous parts. (The shaded area which is lncoshk). Hence ktanhk - lncoshk.

Thanks!
• Jan 11th 2012, 01:43 AM
Mrhappysmile
Re: Hyperbolic integral
Quote:

Originally Posted by pickslides

use \int command in LaTeX

Thanks