Find x²√√x² + 16 Solution Let x = 4 tan(0), (1 √x² + 16 = √16 (tan² (0) + 1) = √ 16 sec²(0) = 4|sec(0)| = 4 sec(0). Thus, we have sec(0) tan²(0) dx. TU 2 1 < 0 < TU 2 Then dx = dx 12√2+16/ VX² + 16 16 tan2 (0) 4 sec(0) 1 */( 16 de tan²(0) To evaluate this trigonometric integral we put everything in terms of sin(0) and cos(0). cos²(8) sin²(0) ᏧᎾ . de and
Find x²√√x² + 16 Solution Let x = 4 tan(0), (1 √x² + 16 = √16 (tan² (0) + 1) = √ 16 sec²(0) = 4|sec(0)| = 4 sec(0). Thus, we have sec(0) tan²(0) dx. TU 2 1 < 0 < TU 2 Then dx = dx 12√2+16/ VX² + 16 16 tan2 (0) 4 sec(0) 1 */( 16 de tan²(0) To evaluate this trigonometric integral we put everything in terms of sin(0) and cos(0). cos²(8) sin²(0) ᏧᎾ . de and
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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