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Apply the trigonometric substitution x = √3 sec(0). When determining the lower limit, you will have two possible choices for 0. Pick the value of that has its terminal side in the quadrant that ensures √x² - 3 will be positive. -√3 6x²-3 dx = X 6√3 sec²(0) - 3 √3 sec(0) X X ) de

Apply the trigonometric substitution x = √3 sec(0). When determining the lower limit, you will have two possible choices for 0. Pick the value of that has its terminal side in the quadrant that ensures √x² - 3 will be positive. -√3 6x²-3 dx = X 6√3 sec²(0) - 3 √3 sec(0) X X ) de

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Apply the trigonometric substitution x = √3 sec(0). When determining the lower limit, you will have two possible choices for 0. Pick the value of that has its terminal side in the quadrant that ensures x² - 3 will be positive. [-√³6√x²-3 dx = 1² X -2 X X 6√3 sec²(0) - 3 √3 sec(0) de X ) de
Transcribed Image Text:Apply the trigonometric substitution x = √3 sec(0). When determining the lower limit, you will have two possible choices for 0. Pick the value of that has its terminal side in the quadrant that ensures x² - 3 will be positive. [-√³6√x²-3 dx = 1² X -2 X X 6√3 sec²(0) - 3 √3 sec(0) de X ) de
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