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Ch. 8 - Techniques of Integration
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. 8 - Techniques of IntegrationProblema 8.4.10
Capitolo 8, Problema 8.4.10

Evaluate the integrals in Exercises 1–14.
∫ 5 dx / √(25x² - 9), where x > 3/5

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Recognize that the integral has the form \( \int \frac{5}{\sqrt{25x^2 - 9}} \, dx \), which resembles an integral involving \( \sqrt{a^2 x^2 - b^2} \). Here, identify \( a^2 = 25 \) and \( b^2 = 9 \), so \( a = 5 \) and \( b = 3 \).
Since the expression under the square root is \( 25x^2 - 9 = (5x)^2 - 3^2 \), consider using a hyperbolic substitution because \( x > \frac{3}{5} \) ensures the expression inside the square root is positive. Use the substitution \( 5x = 3 \cosh(t) \), which implies \( x = \frac{3}{5} \cosh(t) \).
Differentiate \( x = \frac{3}{5} \cosh(t) \) to find \( dx = \frac{3}{5} \sinh(t) \, dt \). Also, rewrite the square root in terms of \( t \): \( \sqrt{25x^2 - 9} = \sqrt{9 \cosh^2(t) - 9} = 3 \sinh(t) \).
Substitute all parts back into the integral: replace \( dx \) and the square root expression, and simplify the integral in terms of \( t \). This should reduce the integral to a simpler form involving hyperbolic functions.
Integrate with respect to \( t \), then substitute back \( t = \cosh^{-1}\left( \frac{5x}{3} \right) \) to express the answer in terms of \( x \).

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Integration of Rational Functions Involving Square Roots

This concept involves integrating functions where the integrand contains a rational expression with a square root in the denominator or numerator. Recognizing the form helps in choosing appropriate substitution or trigonometric identities to simplify the integral.
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Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals involving expressions like √(a²x² - b²). By substituting x with a trigonometric function, the square root simplifies, allowing the integral to be expressed in terms of trigonometric functions that are easier to integrate.
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Introduction to Trigonometric Functions

Domain Restrictions and Absolute Values in Integration

When integrating functions with square roots, domain restrictions (like x > 3/5) ensure the expression under the root is positive. This affects the choice of substitution and the sign of the resulting expressions, especially when dealing with absolute values in antiderivatives.
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Average Value of a Function
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