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
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.3: Trigonometric Functions Of Real Numbers
Problem 32E
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Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcdc0c33c-1cee-4754-904a-1b31674d8af7%2F3a266e14-4a79-4256-a77b-830e2d3a4f7b%2Fwixm5zd_processed.png&w=3840&q=75)
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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