Integral with hyperbolic: cosh x

1. Oct 12, 2013

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

2. Oct 12, 2013

arildno

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

3. Oct 12, 2013

FelixHelix

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

4. Oct 12, 2013

arildno

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

5. Oct 12, 2013

arildno

Correct!
So, what must the purported answer be called?

6. Oct 12, 2013

FelixHelix

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

7. Oct 12, 2013

FelixHelix

?? I'm not sure??

8. Oct 12, 2013

arildno

Maybe what is called TOTALLY WRONG!!

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

9. Oct 12, 2013

FelixHelix

yes a typo indeed - apologies. What substitution do u use to get there though - i dont see it!

10. Oct 12, 2013

arildno

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)

11. Oct 12, 2013

arildno

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:
$$1==\cosh^{2}(\frac{x}{2})-\sinh^{2}(\frac{x}{2})$$

12. Oct 12, 2013

FelixHelix

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

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