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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.7.59

Evaluate the integrals in Exercises 41–60.
59. ∫(from -ln2 to 0)cosh²(x/2) dx

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Recall the definition of the hyperbolic cosine function: \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). However, for integration, it is often easier to use the identity involving \(\cosh^2\).
Use the identity \(\cosh^2(u) = \frac{1 + \cosh(2u)}{2}\) to rewrite the integrand. Here, let \(u = \frac{x}{2}\), so the integral becomes \(\int_{-\ln 2}^0 \cosh^2\left(\frac{x}{2}\right) dx = \int_{-\ln 2}^0 \frac{1 + \cosh(x)}{2} dx\).
Split the integral into two simpler integrals: \(\int_{-\ln 2}^0 \frac{1}{2} dx + \int_{-\ln 2}^0 \frac{\cosh(x)}{2} dx\).
Integrate each part separately: the integral of \(\frac{1}{2}\) with respect to \(x\) is \(\frac{x}{2}\), and the integral of \(\frac{\cosh(x)}{2}\) with respect to \(x\) is \(\frac{\sinh(x)}{2}\).
Evaluate the resulting expression \(\left[ \frac{x}{2} + \frac{\sinh(x)}{2} \right]\) at the limits \(x = -\ln 2\) and \(x = 0\), and subtract to find the value of the definite integral.

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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 shares properties similar to the cosine function but relates to hyperbolic geometry. Understanding its definition and behavior is essential for integrating expressions involving cosh.
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Integration of Hyperbolic Functions

Integrating hyperbolic functions often involves using identities or rewriting the function in exponential form. For example, cosh²(x) can be expressed using the identity cosh²(x) = (1 + cosh(2x))/2, which simplifies the integral into more manageable terms.
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Definite Integrals and Limits of Integration

A definite integral calculates the net area under a curve between two specific points. Evaluating from -ln(2) to 0 requires substituting these limits into the antiderivative after integration, ensuring careful handling of logarithmic and exponential expressions.
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