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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 7.7.49
Chapter 7, Problem 7.7.49

Evaluate the integrals in Exercises 41–60.
49. ∫(sech(√t)tanh(√t)dt)/√t

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Start by examining the integral: \(\int \frac{\text{sech}(\sqrt{t}) \tanh(\sqrt{t})}{\sqrt{t}} \, dt\). Notice that the integrand involves \(\sqrt{t}\) inside the hyperbolic functions and also in the denominator.
Use the substitution method to simplify the integral. Let \(u = \sqrt{t}\), which means \(u = t^{1/2}\). Then, differentiate both sides with respect to \(t\) to find \(du\) in terms of \(dt\): \(du = \frac{1}{2\sqrt{t}} dt\) or equivalently, \(dt = 2u \, du\).
Rewrite the integral in terms of \(u\). Substitute \(\sqrt{t} = u\), \(dt = 2u \, du\), and replace the integrand accordingly: the denominator \(\sqrt{t}\) becomes \(u\), and the numerator remains \(\text{sech}(u) \tanh(u)\). So the integral becomes \(\int \frac{\text{sech}(u) \tanh(u)}{u} \cdot 2u \, du\).
Simplify the expression inside the integral. The \(u\) in the denominator and numerator cancel out, leaving \(\int 2 \text{sech}(u) \tanh(u) \, du\).
Recognize that the derivative of \(\text{sech}(u)\) is \(-\text{sech}(u) \tanh(u)\). Use this fact to rewrite the integral in terms of \(\text{sech}(u)\) and then integrate accordingly.

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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 like sech(x) and tanh(x) are analogs of trigonometric functions but based on hyperbolas. Understanding their definitions, properties, and derivatives is essential for manipulating and integrating expressions involving these functions.
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Asymptotes of Hyperbolas

Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. Recognizing an inner function and its derivative within the integrand allows for effective substitution, especially when dealing with composite functions.
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Euler's Method

Chain Rule and Its Role in Integration

The chain rule relates the derivative of a composite function to the derivatives of its inner and outer functions. In integration, recognizing the chain rule in reverse helps identify suitable substitutions and simplifies the integral of composite functions.
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Intro to the Chain Rule
Related Practice
Textbook Question

Evaluate the integrals in Exercises 87–96.

95. ∫₂⁴ x^(2x) (1 + ln x) dx

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Textbook Question

80. Volume The region enclosed by the curve y=sech(x), the x-axis, and the lines x=±ln√3 is revolved about the x-axis to generate a solid. Find the volume of the solid.

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Textbook Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

15. y = 2√t tanh(√t)

Textbook Question

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.

120. y = x^(sin x)

Textbook Question

13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x→∞? Give reasons for your answer.

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Textbook Question

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞

cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1

tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1

sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1

csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1

coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1

Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.


65. sech⁻¹(3/5)