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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.5.82

Variations on the substitution method Evaluate the following integrals.                                                                                                        
                                                                                                                                                                    
 ∫ (eˣ ― e⁻ˣ)/ (eˣ + e⁻ˣ) d𝓍

Verified step by step guidance
1
Recognize that the integral is of the form \(\int \frac{e^x - e^{-x}}{e^x + e^{-x}} \, d\!x\), which suggests a substitution involving the denominator or a related function.
Recall the hyperbolic functions: \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\). Notice that the numerator is \(2 \sinh x\) and the denominator is \(2 \cosh x\), so the integrand simplifies to \(\frac{2 \sinh x}{2 \cosh x} = \frac{\sinh x}{\cosh x} = \tanh x\).
Rewrite the integral as \(\int \tanh x \, d\!x\) to simplify the problem.
Recall that the derivative of \(\ln(\cosh x)\) is \(\tanh x\), so the integral of \(\tanh x\) with respect to \(x\) is \(\ln|\cosh x| + C\).
Write the final integral expression as \(\int \frac{e^x - e^{-x}}{e^x + e^{-x}} \, d\!x = \ln|\cosh x| + C\), where \(C\) is 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.

Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform a complicated integral into a basic form. It involves identifying a part of the integrand as a new variable, differentiating it, and rewriting the integral in terms of this variable. This technique is especially useful when the integral contains composite functions.
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Euler's Method

Hyperbolic Functions and Their Properties

Expressions involving eˣ and e⁻ˣ often relate to hyperbolic functions such as sinh(x) and cosh(x). Recognizing these can simplify integration since sinh(x) = (eˣ - e⁻ˣ)/2 and cosh(x) = (eˣ + e⁻ˣ)/2. Using these identities helps rewrite the integral in a more manageable form.
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Properties of Functions

Integration of Rational Functions

Integrals involving ratios of functions, like (eˣ - e⁻ˣ)/(eˣ + e⁻ˣ), require understanding how to manipulate and simplify rational expressions. This often involves algebraic simplification or substitution to reduce the integral to a standard form, enabling straightforward integration.
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Intro to Rational Functions
Related Practice
Textbook Question

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question.


The region bounded by y = 6 cos 𝓍 and the 𝓍-axis between 𝓍 = ―π/2 and 𝓍 = π

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Textbook Question

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

v = [1 / (2t + 1)] (m/s), for 0 ≤ t ≤ 8 ; n = 4

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Textbook Question

Average distance on a parabola What is the average distance between the parabola y = 30𝓍 (20 ― 𝓍 ) and the 𝓍-axis on the interval [0, 20] ?

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Textbook Question

Consider the function

ƒ(t) = { t      if  ―2 ≤ t < 0

t²/2    if    0 ≤ t ≤ 2                                                                                                                                                                       

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(f) Find a constant C such that F(𝓍) = G(𝓍) + C .

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Textbook Question

Evaluating integrals Evaluate the following integrals.


∫₀⁵ |2𝓍―8|d𝓍

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Textbook Question

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.


 ∫₀⁴ ƒ(𝓍) d𝓍, where ƒ(𝓍) = {5      if 𝓍 ≤ 2                                                                                                                                                                                     

                      3𝓍 ― 1  if 𝓍 > 2

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