Ch. 8 - Integration Techniques

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo

Briggs 3rd Edition

Ch. 8 - Integration Techniques

Problema 8.9.28

Capitolo 8, Problema 8.9.28

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
28. ∫ (from 1 to ∞) tan⁻¹(s)/(s² + 1) ds

Guida verificata passo dopo passo

Identify the integral as an improper integral because the upper limit of integration is infinity: \(\int_{1}^{\infty} \frac{\tan^{-1}(s)}{s^{2} + 1} \, ds\).

Rewrite the integral as a limit to handle the infinite upper bound: \(\lim_{t \to \infty} \int_{1}^{t} \frac{\tan^{-1}(s)}{s^{2} + 1} \, ds\).

Consider using substitution to simplify the integral. Notice that the derivative of \(\tan^{-1}(s)\) is \(\frac{1}{s^{2} + 1}\), which appears in the denominator. Let \(u = \tan^{-1}(s)\), then \(du = \frac{1}{s^{2} + 1} ds\).

Rewrite the integral in terms of \(u\): \(\int \frac{\tan^{-1}(s)}{s^{2} + 1} ds = \int u \, du\).

Integrate \(\int u \, du\) to get \(\frac{u^{2}}{2} + C\), then substitute back \(u = \tan^{-1}(s)\), and finally evaluate the limit \(\lim_{t \to \infty} \left[ \frac{(\tan^{-1}(t))^{2}}{2} - \frac{(\tan^{-1}(1))^{2}}{2} \right]\) to determine convergence or divergence.

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Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, we use limits to define the integral as a limit of definite integrals over finite intervals. Determining convergence or divergence is essential before finding the value.

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Improper Integrals: Infinite Intervals

Inverse Trigonometric Functions

Inverse trigonometric functions, like arctangent (tan⁻¹), are the inverses of trigonometric functions and often appear in integrals. Understanding their properties and derivatives helps in simplifying integrals and applying substitution or integration techniques.

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Derivatives of Other Inverse Trigonometric Functions

Comparison Test for Convergence

The comparison test helps determine if an improper integral converges by comparing it to a known convergent or divergent integral. If the integrand is smaller than a convergent integral's integrand for large values, the integral converges; if larger than a divergent one, it diverges.

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Percorso guidato

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Direct Comparison Test

Pratica correlata

Domanda del libro di testo

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

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Domanda del libro di testo

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

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7–84. Evaluate the following integrals.

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68. Different methods

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