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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.6.102
Capitolo 7, Problema 7.6.102

Evaluate the integrals in Exercises 91–102.
102. ∫(from -1/3 to 1/√3)(cos(arctan 3x))/(1+9x²) dx

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Recognize that the integral involves the expression \( \cos(\arctan(3x)) \) and the denominator \( 1 + 9x^2 \). This suggests a trigonometric substitution related to the angle \( \theta = \arctan(3x) \).
Set \( \theta = \arctan(3x) \), which implies \( 3x = \tan(\theta) \). From this, express \( x \) in terms of \( \theta \): \( x = \frac{\tan(\theta)}{3} \).
Calculate the differential \( dx \) in terms of \( d\theta \). Since \( x = \frac{\tan(\theta)}{3} \), then \( dx = \frac{1}{3} \sec^2(\theta) d\theta \).
Rewrite the integral limits in terms of \( \theta \) by substituting the original \( x \) limits into \( \theta = \arctan(3x) \). For \( x = -\frac{1}{3} \), \( \theta = \arctan(-1) = -\frac{\pi}{4} \). For \( x = \frac{1}{\sqrt{3}} \), \( \theta = \arctan\left(3 \cdot \frac{1}{\sqrt{3}}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3} \).
Substitute all parts into the integral: replace \( \cos(\arctan(3x)) \) with \( \cos(\theta) \), replace \( 1 + 9x^2 \) with \( 1 + \tan^2(\theta) = \sec^2(\theta) \), and replace \( dx \) with \( \frac{1}{3} \sec^2(\theta) d\theta \). Simplify the integrand and integral accordingly before integrating with respect to \( \theta \) from \( -\frac{\pi}{4} \) to \( \frac{\pi}{3} \).

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