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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.37
Chapter 7, Problem 7.3.37

37–56. Integrals Evaluate each integral.
∫ cosh 2x dx

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1
Recall the definition of the hyperbolic cosine function: \(\cosh x = \frac{e^x + e^{-x}}{2}\).
Use the double angle identity for hyperbolic cosine: \(\cosh 2x = 2\cosh^2 x - 1\) or express \(\cosh 2x\) in terms of exponentials as \(\frac{e^{2x} + e^{-2x}}{2}\).
Rewrite the integral using the exponential form: \(\int \cosh 2x \, dx = \int \frac{e^{2x} + e^{-2x}}{2} \, dx\).
Split the integral into two simpler integrals: \(\frac{1}{2} \int e^{2x} \, dx + \frac{1}{2} \int e^{-2x} \, dx\).
Integrate each term using the formula \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\), then combine the results and add the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hyperbolic Functions

Hyperbolic functions, such as cosh(x) and sinh(x), are analogs of trigonometric functions but based on exponential functions. Specifically, cosh(x) = (e^x + e^(-x))/2. Understanding their definitions and properties is essential for integrating expressions involving cosh.
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Integration of Exponential Functions

Since hyperbolic functions are expressed in terms of exponentials, integrating cosh(2x) involves integrating exponential functions like e^(2x). The integral of e^(ax) with respect to x is (1/a)e^(ax) + C, which helps in solving the integral of cosh(2x).
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Integrals of General Exponential Functions

Use of Substitution in Integration

When integrating functions like cosh(2x), substitution is useful to simplify the integral. Letting u = 2x transforms the integral into a simpler form, allowing the use of standard integration rules and then substituting back to the original variable.
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Related Practice
Textbook Question

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