# Integral with hyperbolic: cosh x

• FelixHelix
In summary, the conversation is about trying to solve an integral related to discounting investments and the use of substitutions and tricks to find the answer. The correct answer is tanh(x/2)+C, which can be obtained by using hyperbolic identities for half-arguments and the fundamental identity.

#### FelixHelix

I cannot reach the answer for this integral which is part of a bigger question related to discounting investments. I know what the answer to the integral is and have tried all the substitutions and tricks I know. Any pointer would be great!

∫(1/(1+cosh(x))) = tanh(x) + C

Thanks, Felix

Well, does:
$$\frac{d}{dx}tanh(x)=\frac{1}{\cosh^{2}(x)}$$ equal your integrand?

No, the integrand is 1/(1+cosh(x))

The integral is readily evaluated by using:
$$\int\frac{dx}{1+\cosh(x)}dx=\int\frac{dx}{1+\cosh(x)}\frac{1-\cosh(x)}{1-\cosh(x)}dx=\int\frac{1-\cosh(x)}{-\sinh^{2}(x)}dx$$

FelixHelix said:
No, the integrand is 1/(1+cosh(x))
Correct!
So, what must the purported answer be called?

I see,thats a good trick i hadnt thought of using. But how would you manipulate the integral you end up with?

arildno said:
Correct!
So, what must the purported answer be called?
?? I'm not sure??

FelixHelix said:
?? I'm not sure??

Maybe what is called TOTALLY WRONG!

To give you a hint:
The correct answer is tanh(x/2)+C, not tanh(x)+C

yes a typo indeed - apologies. What substitution do u use to get there though - i don't see it!

First off:
You can easily do the integral the way I indicated.
THEN, find the appropriate hyperbolic identities for half-arguments (very analogous to the more well-known half-angle formulas for the trig functions). The result will simplify to tanh(x/2)

We have:
$$\sinh(x)=2\cosh(\frac{x}{2})\sinh(\frac{x}{2}),\cosh(x)=\cosh^{2}(\frac{x}{2})+\sinh^{2}(\frac{x}{2})$$
Note the similarity to the trig identities!
You'll probably also need the fundamental identity:
$$1==\cosh^{2}(\frac{x}{2})-\sinh^{2}(\frac{x}{2})$$

Ill take a look in the morning - thanks for your help!