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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
Capitolo 8, Problema 8.R.54

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
54. ∫ dx/√(9x² - 25), x > 5/3

Guida verificata passo dopo passo
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Step 1: Recognize the integral's form. The integrand resembles the structure of a trigonometric substitution problem, specifically involving a square root of a quadratic expression. The denominator √(9x² - 25) suggests using a substitution based on the identity for secant: sec²θ - 1 = tan²θ.
Step 2: Rewrite the quadratic expression in the form suitable for substitution. Factor out the constant 9 from the square root: √(9x² - 25) = √(9(x² - 25/9)) = 3√(x² - (5/3)²). This reveals the structure of a difference of squares, which is ideal for trigonometric substitution.
Step 3: Perform the substitution. Let x = (5/3)secθ, which implies dx = (5/3)secθtanθ dθ. Substituting into the integral, the square root √(9x² - 25) becomes 3√((5/3)²sec²θ - (5/3)²) = 3(5/3)tanθ = 5tanθ.
Step 4: Simplify the integral using the substitution. The integral ∫ dx/√(9x² - 25) transforms into ∫ ((5/3)secθtanθ dθ) / (5tanθ). Cancel out the common terms, leaving ∫ (1/3)secθ dθ.
Step 5: Evaluate the simplified integral. The integral of secθ is a standard result: ∫ secθ dθ = ln|secθ + tanθ| + C. Substitute back θ in terms of x using the original substitution x = (5/3)secθ to express the final result in terms of x.

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