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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.4.71
Chapter 8, Problem 8.4.71

{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,
∫sec³u du = 1/2 (sec u tan u + ln |sec u + tan u|) + C


Graph the following functions and find the area under the curve on the given interval.
f(x) = (9 - x²) ⁻², [0, 3/2]

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1
Identify the integral you need to evaluate: the area under the curve of the function \(f(x) = (9 - x^{2})^{-2}\) on the interval \([0, \frac{3}{2}]\). This means you want to compute the definite integral \(\int_{0}^{\frac{3}{2}} (9 - x^{2})^{-2} \, dx\).
Consider a substitution to simplify the integral. Since the integrand involves \(9 - x^{2}\), use the substitution \(x = 3 \sin u\), which implies \(dx = 3 \cos u \, du\). This substitution transforms the integral into terms of \(u\).
Rewrite the integral in terms of \(u\): replace \(x\) and \(dx\) accordingly, and express the limits of integration in terms of \(u\). When \(x=0\), \(u=\arcsin(0)=0\); when \(x=\frac{3}{2}\), \(u=\arcsin(\frac{1}{2})=\frac{\pi}{6}\).
Simplify the integrand after substitution. Note that \(9 - x^{2} = 9 - 9 \sin^{2} u = 9 \cos^{2} u\), so \((9 - x^{2})^{-2} = (9 \cos^{2} u)^{-2} = \frac{1}{81 \cos^{4} u}\). Incorporate \(dx = 3 \cos u \, du\) to rewrite the integral as \(\int_{0}^{\frac{\pi}{6}} \frac{3 \cos u}{81 \cos^{4} u} \, du = \int_{0}^{\frac{\pi}{6}} \frac{3}{81 \cos^{3} u} \, du\).
Recognize that \(\frac{1}{\cos^{3} u} = \sec^{3} u\), so the integral becomes \(\frac{1}{27} \int_{0}^{\frac{\pi}{6}} \sec^{3} u \, du\). Use the given reduction formula for \(\int \sec^{3} u \, du = \frac{1}{2} (\sec u \tan u + \ln |\sec u + \tan u|) + C\) to evaluate this definite integral between \(0\) and \(\frac{\pi}{6}\).

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Key Concepts

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

Integration of Trigonometric Functions Using Reduction Formulas

Reduction formulas simplify integrals of powers of trigonometric functions by expressing them in terms of lower powers. For example, the integral of sec³u can be evaluated using a specific reduction formula, breaking it down into simpler parts involving sec u tan u and a logarithmic term. This technique is essential for handling complex trigonometric integrals.
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Introduction to Trigonometric Functions

Definite Integrals and Area Under a Curve

A definite integral calculates the net area between a function's graph and the x-axis over a specified interval. It involves evaluating the antiderivative at the interval's endpoints and subtracting. Understanding this concept is crucial for finding the exact area under curves like f(x) = (9 - x²)⁻² on [0, 3/2].
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Definition of the Definite Integral

Graphing Functions to Understand Behavior

Graphing functions helps visualize their shape, continuity, and behavior over intervals. For f(x) = (9 - x²)⁻², graphing reveals how the function behaves near critical points and boundaries, aiding in understanding the integral's geometric interpretation and ensuring correct evaluation of the area.
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Related Practice
Textbook Question

76–83. Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals.

76. ∫ [cosθ / (sin³θ - 4sinθ)] dθ

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Textbook Question

Visual proof Let F(x)=∫₀ˣ √(a²−t²) dt. The figure shows that F(x)= area of sector OAB+ area of triangle OBC.

a. Use the figure to prove that

F(x) = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2

b. Conclude that ∫ √(a²−x²) dx = (a² sin ⁻¹(x/a))/2 + x√(a²−x²)/2 + C.

Textbook Question

102–106. Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by F(s) = ∫[0 to ∞] e^(-st) f(t) dt, where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^(-t), the following improper integral is evaluated using integration by parts:

F(s) = ∫[0 to ∞] e^(-st) e^(-t) dt = ∫[0 to ∞] e^(-(s+1)t) dt = 1/(s+1).

Verify the following Laplace transforms, where a is a real number.

106. f(t) = cos(at) → F(s) = s/(s² + a²)

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Textbook Question

7–84. Evaluate the following integrals.

62. ∫ from 0 to π/2 √(1 + cosθ) dθ

Textbook Question

70. Different methods Let I=∫(x+2)/(x+4)dx.

b. Evaluate I without performing long division on the integrand.

Textbook Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.

42. ∫ (from 3 to 4) 1/(x-3)³ᐟ² dx

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