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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.7.43
Chapter 7, Problem 7.7.43

Evaluate the integrals in Exercises 41–60.
43. ∫6cosh(x/2 - ln3)dx

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Recall the definition of the hyperbolic cosine function: \(\cosh(u) = \frac{e^{u} + e^{-u}}{2}\). This can help in rewriting the integral if needed.
Identify the inner function inside the hyperbolic cosine: here, \(u = \frac{x}{2} - \ln 3\). This will be useful for substitution.
Use substitution by letting \(u = \frac{x}{2} - \ln 3\). Then, compute \(\frac{du}{dx} = \frac{1}{2}\), which implies \(dx = 2 \, du\).
Rewrite the integral in terms of \(u\): \(\int 6 \cosh\left(\frac{x}{2} - \ln 3\right) dx = \int 6 \cosh(u) \cdot 2 \, du = \int 12 \cosh(u) \, du\).
Integrate \(\cosh(u)\) with respect to \(u\): \(\int \cosh(u) \, du = \sinh(u) + C\). So, the integral becomes \(12 \sinh(u) + C\). Finally, substitute back \(u = \frac{x}{2} - \ln 3\) to express the answer in terms of \(x\).

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Key Concepts

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

Hyperbolic Cosine Function (cosh)

The hyperbolic cosine function, cosh(x), is defined as (e^x + e^(-x))/2. It is an even function and appears frequently in calculus problems involving hyperbolic functions. Understanding its properties and derivatives is essential for integrating expressions involving cosh.
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Graph of Sine and Cosine Function

Integration of Composite Functions

When integrating functions like cosh(ax + b), it is important to use substitution or recognize the integral form. The integral of cosh(u) with respect to u is sinh(u), so adjusting for the inner function's derivative is necessary to find the correct antiderivative.
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Evaluate Composite Functions - Special Cases

Logarithmic Constants in Integration

Constants such as ln(3) inside the argument of a function affect the integration as shifts but do not change the integral's form. Recognizing that ln(3) is a constant helps simplify the integral by treating it as a constant shift in the variable.
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Logarithmic Differentiation
Related Practice
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