# Integral with hyperbolic: cosh x

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

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

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

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

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

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

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

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?? 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

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

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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)

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