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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.7.71
Chapter 7, Problem 7.7.71

Evaluate the integrals in Exercises 31–78.
71. ∫(from √2/3 to 2/3)dy/(|y|√(9y²-1))

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First, carefully analyze the integral \( \int_{\sqrt{2}/3}^{2/3} \frac{dy}{|y| \sqrt{9y^2 - 1}} \). Notice the absolute value \( |y| \) in the denominator and the square root expression \( \sqrt{9y^2 - 1} \). Since the limits of integration are positive (both \( \sqrt{2}/3 \) and \( 2/3 \) are positive), we can simplify \( |y| = y \) in this interval.
Rewrite the integral as \( \int_{\sqrt{2}/3}^{2/3} \frac{1}{y \sqrt{9y^2 - 1}} \, dy \). To simplify the integrand, consider a substitution that will help handle the \( \sqrt{9y^2 - 1} \) term. A good choice is to let \( 3y = \sec(\theta) \), because \( \sec^2(\theta) - 1 = \tan^2(\theta) \), which will simplify the square root.
Express \( y \) and \( dy \) in terms of \( \theta \): \( y = \frac{\sec(\theta)}{3} \) and \( dy = \frac{\sec(\theta) \tan(\theta)}{3} d\theta \). Also, rewrite the square root: \( \sqrt{9y^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \). Substitute all these into the integral.
After substitution, the integral becomes \( \int \frac{1}{(\frac{\sec(\theta)}{3}) \cdot \tan(\theta)} \cdot \frac{\sec(\theta) \tan(\theta)}{3} d\theta \). Simplify the expression inside the integral by canceling terms carefully.
Finally, determine the new limits of integration by substituting the original \( y \)-values into \( 3y = \sec(\theta) \) to find the corresponding \( \theta \)-values. Then, integrate the simplified function with respect to \( \theta \) over these new limits.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals with Absolute Value

When integrating functions involving absolute values, it is important to consider the sign of the variable within the integration limits. The absolute value function can change the integrand's form depending on whether the variable is positive or negative, which may require splitting the integral at points where the expression inside the absolute value changes sign.
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Definition of the Definite Integral

Integration of Functions Involving Square Roots

Integrals containing square roots, especially of quadratic expressions like √(9y² - 1), often require substitution methods or trigonometric substitutions. Recognizing the form under the root helps in choosing an appropriate substitution to simplify the integral into a standard form.
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Integrals Involving Natural Logs: Substitution

Domain and Limits of Integration

Understanding the domain of the integrand and the given limits is crucial. The expression under the square root must be non-negative for real values, and the limits must be checked to ensure the function is defined and continuous over the interval. This affects the validity and approach to evaluating the definite integral.
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Finding the Domain and Range of a Graph
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