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Evaluate the Integral I 3 X √x² +9 dx

Evaluate the Integral I 3 X √x² +9 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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Hello, I don't understand how I would know to use 3tanx as a sub when doing this, and why does it become squareroot of u-9 squared? 

Evaluate the Integral 1 x³ 3 / √x² +9 2 dx π π Let x = 3 tan (t), where sts 2 2 Then dx = 3 sec² (t) dt. Note that since π π 2 <t< 3 sec² (t) is positive. 2' (3 tan (t))³ √√√(3 tan (t))² +9 Simplify terms. + Tap for more steps... (3 sec² (t)) dt ×
Transcribed Image Text:Evaluate the Integral 1 x³ 3 / √x² +9 2 dx π π Let x = 3 tan (t), where sts 2 2 Then dx = 3 sec² (t) dt. Note that since π π 2 <t< 3 sec² (t) is positive. 2' (3 tan (t))³ √√√(3 tan (t))² +9 Simplify terms. + Tap for more steps... (3 sec² (t)) dt ×
Integrate Using u-Substitution 1 Let u = 1 du √x² +9 2 - / 3 X 2 + Tap for more steps... 2 √u - 9² √u 2 x² + 9. Then du = 2xdx, so xdx. Rewrite using u and du. 12/2 9 2√√/u dx 1 Simplify. + Tap for more steps... du 2 du ×
Transcribed Image Text:Integrate Using u-Substitution 1 Let u = 1 du √x² +9 2 - / 3 X 2 + Tap for more steps... 2 √u - 9² √u 2 x² + 9. Then du = 2xdx, so xdx. Rewrite using u and du. 12/2 9 2√√/u dx 1 Simplify. + Tap for more steps... du 2 du ×
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