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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.6.63
Capitolo 7, Problema 7.6.63

Evaluate the integrals in Exercises 53–76.
63. ∫(from -1 to -√2/2)dy/(y√(4y²-1))

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Identify the integral to evaluate: \(\int_{-1}^{-\frac{\sqrt{2}}{2}} \frac{dy}{y \sqrt{4y^{2} - 1}}\).
Recognize that the integrand involves a square root of a quadratic expression, \(\sqrt{4y^{2} - 1}\), which suggests a trigonometric substitution to simplify the square root.
Use the substitution \(y = \frac{1}{2} \sec(\theta)\), because \(4y^{2} - 1 = 4 \left( \frac{1}{2} \sec(\theta) \right)^{2} - 1 = \sec^{2}(\theta) - 1 = \tan^{2}(\theta)\), which simplifies the square root to \(\tan(\theta)\).
Compute \(dy\) in terms of \(d\theta\): \(dy = \frac{1}{2} \sec(\theta) \tan(\theta) d\theta\), and rewrite the integral entirely in terms of \(\theta\).
Change the limits of integration from \(y\) to \(\theta\) by solving \(y = \frac{1}{2} \sec(\theta)\) for the given bounds, then integrate the resulting expression with respect to \(\theta\).

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