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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 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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