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3. Using trigonometric substitution, what is 1 sec-²¹ ( 137 ) + √x²-9 2x² + C A. 1 300² (2) √²-90 (3)- sec-1 + C 6x² B. 1/² 50c-1( 37) - √x ²2 - 9 sec + C 2x² sec-¹ (33)₁ + √x²-9 6x² C. D. √√X²=9₁ x3 + C dx?

3. Using trigonometric substitution, what is 1 sec-²¹ ( 137 ) + √x²-9 2x² + C A. 1 300² (2) √²-90 (3)- sec-1 + C 6x² B. 1/² 50c-1( 37) - √x ²2 - 9 sec + C 2x² sec-¹ (33)₁ + √x²-9 6x² C. D. √√X²=9₁ x3 + C dx?

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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3. Using trigonometric substitution, what is — sec-1 ( 5 ) + √x ²2-9 3 + C 2x² 1 sec-¹ (3-√² (3) -√√x ²-9 6x² 1 —- sec-¹ (3) - √x ²2 - 9 6 + C 2x² 1 — sec-¹ (53) + √²-200 + + C 6x² A. B. C. D. + C x²-9 x3 Sv dx?
Transcribed Image Text:3. Using trigonometric substitution, what is — sec-1 ( 5 ) + √x ²2-9 3 + C 2x² 1 sec-¹ (3-√² (3) -√√x ²-9 6x² 1 —- sec-¹ (3) - √x ²2 - 9 6 + C 2x² 1 — sec-¹ (53) + √²-200 + + C 6x² A. B. C. D. + C x²-9 x3 Sv dx?
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