# Hyperbolic Functions

1. Mar 18, 2009

### roam

1. I am asked to calculate $$\int \frac{dx}{cosh(x)}$$

2. Relevant equations

3. The attempt at a solution

I know that $$\frac{1}{cosh(x)}$$ is equivalent to "sech(x)" which by definition is $$\frac{2}{e^x+e^{-x}}$$.

I'm confused & I don't know which one I need to use for this question:

$$\int sech(x) = 2tan{-1}(tanh(\frac{x}{2}))$$

$$\int \frac{2}{e^x + e^{-x}} = 2tan^{-1}(e^x)$$

Any help would be greatly appreciated.

Last edited: Mar 18, 2009
2. Mar 18, 2009

### yyat

The two answers you got only differ by a constant of $$\frac{\pi}{2}$$. Recall that the indefinite integral is only defined up to a constant, so both solutions are correct.