# Thread: Hyperbolic integral

1. ## Hyperbolic integral

Ok so:
Its a 3 part question. I've completed parts 1 and 2.
Part 1)
Draw $\displaystyle tanhx$ and $\displaystyle tanh^-1 x$ on the same graph
Part 2)
Find $\displaystyle INT tanh x dx$ between limits k and 0 (where k > 0) which gives you $\displaystyle ln coshk$

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

If you integrate it by parts you form: $\displaystyle [xtanh^-1 x] tanhk -> 0 = ktanhk$ and I t
hink $\displaystyle - int x/(1-x^2)$ 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

2. ## Re: Hyperbolic integral

What have you tried? Show us some workings.

Originally Posted by Mrhappysmile
P.S How do you do an integral sign on this forum
use \int command in LaTeX

3. ## Re: Hyperbolic integral

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

4. ## 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!

5. ## Re: Hyperbolic integral

Originally Posted by pickslides

use \int command in LaTeX
Thanks