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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 8 - Integration TechniquesProblema 8.4.71
Capitolo 8, Problema 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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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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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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Graph of Sine and Cosine Function
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108. Draining a tank Water is drained from a 3000-gal tank at a rate that starts at 100 gal/hr and decreases continuously by 5%/hr. If the drain is left open indefinitely, how much water drains from the tank? Can a full tank be emptied at this rate? 

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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.

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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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87-92. An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution u = tan(x/2) or, equivalently, x = 2 tan⁻¹u. The following relations are used in making this change of variables.

A: dx = 2/(1 + u²) du

B: sin x = 2u/(1 + u²)

C: cos x = (1 - u²)/(1 + u²)

88. Evaluate ∫ dx/(2 + cos x).

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7–84. Evaluate the following integrals.

62. ∫ from 0 to π/2 √(1 + cosθ) dθ

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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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