Calculating \int \frac{dx}{cosh(x)}

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SUMMARY

The integral \(\int \frac{dx}{\cosh(x)}\) can be computed using the hyperbolic secant function, represented as \(\text{sech}(x)\). The integral evaluates to \(\int \text{sech}(x) \, dx = 2 \tan^{-1}(\tanh(\frac{x}{2}))\) or alternatively \(\int \frac{2}{e^x + e^{-x}} \, dx = 2 \tan^{-1}(e^x)\). Both forms are valid as they differ only by a constant of \(\frac{\pi}{2}\), highlighting that indefinite integrals are defined up to an arbitrary constant.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically \(\cosh(x)\) and \(\text{sech}(x)\).
  • Familiarity with integration techniques involving trigonometric and hyperbolic identities.
  • Knowledge of the properties of indefinite integrals and their constants.
  • Basic algebraic manipulation of exponential functions.
NEXT STEPS
  • Study the derivation and properties of hyperbolic functions, particularly \(\text{sech}(x)\).
  • Learn about the relationship between hyperbolic functions and their inverse functions.
  • Explore the concept of indefinite integrals and the significance of integration constants.
  • Practice solving integrals involving exponential functions and hyperbolic identities.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for clear explanations of hyperbolic integrals.

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1. I am asked to calculate [tex]\int \frac{dx}{cosh(x)}[/tex]



2. Homework Equations



3. The Attempt at a Solution

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

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

[tex]\int sech(x) = 2tan{-1}(tanh(\frac{x}{2}))[/tex]

[tex]\int \frac{2}{e^x + e^{-x}} = 2tan^{-1}(e^x)[/tex]

Any help would be greatly appreciated.
 
Last edited:
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The two answers you got only differ by a constant of [tex]\frac{\pi}{2}[/tex]. Recall that the indefinite integral is only defined up to a constant, so both solutions are correct.
 

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