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V 49 – x2 dx. Evaluate x2 Solution Let x = 7 sin(0), where 2 Then dx = 2 d0 and V 49 – x? 49 49 sin?(e) = V 49 cos?(0) = 7|cos(0)| = 7 cos(0). CoS sos".) Thus, the inverse substitution rule gives 2 ーT (Note that cos(0) 2 0 because 2 49- х2 7 cos(0) de dx 3=

V 49 – x2 dx. Evaluate x2 Solution Let x = 7 sin(0), where 2 Then dx = 2 d0 and V 49 – x? 49 49 sin?(e) = V 49 cos?(0) = 7|cos(0)| = 7 cos(0). CoS sos".) Thus, the inverse substitution rule gives 2 ーT (Note that cos(0) 2 0 because 2 49- х2 7 cos(0) de dx 3=

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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Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) or by drawing a diagram, as in the figure below, where 0 is interpreted as an angle of a right triangle. 7 V 49 - x2 Since sin(0) we label the opposite side and the hypotenuse as having lengths x and 7. Then the Pythagorean theorem gives the length of the adjacent side as %3D V 49 – x2, so we can simply read the value of cot(0) from the figure. cot(0) %3! (Although 0 > 0 in the diagram, this expression for cot(0) is valid even when 0 < 0.) Since sin(0) we have 0 = sin and so 49 - x2 dx = + C. 5:51 PM
Transcribed Image Text:Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot(0) in terms of sin(0) or by drawing a diagram, as in the figure below, where 0 is interpreted as an angle of a right triangle. 7 V 49 - x2 Since sin(0) we label the opposite side and the hypotenuse as having lengths x and 7. Then the Pythagorean theorem gives the length of the adjacent side as %3D V 49 – x2, so we can simply read the value of cot(0) from the figure. cot(0) %3! (Although 0 > 0 in the diagram, this expression for cot(0) is valid even when 0 < 0.) Since sin(0) we have 0 = sin and so 49 - x2 dx = + C. 5:51 PM
49 – x2 -x² Evaluate "xp Solution -TT Sos. Then dx = 2 Let x = 7 sin(0), where de and V 49 – x2 = V49 49 sin?(0) = V 49 cos?(0) = 7|cos(0)| = 7 cos(0). %D %D (Note that cos(0) 2 0 because <0 s ".) Thus, the inverse substitution rule gives 2 2 7 cos(0) 49 sin2(0) V 49 – x2 dx = de cos?(0) de cot2(0) de -/(csc"(0) - 1) de + C. muct roturn to the original variable x. This can be done eith
Transcribed Image Text:49 – x2 -x² Evaluate "xp Solution -TT Sos. Then dx = 2 Let x = 7 sin(0), where de and V 49 – x2 = V49 49 sin?(0) = V 49 cos?(0) = 7|cos(0)| = 7 cos(0). %D %D (Note that cos(0) 2 0 because <0 s ".) Thus, the inverse substitution rule gives 2 2 7 cos(0) 49 sin2(0) V 49 – x2 dx = de cos?(0) de cot2(0) de -/(csc"(0) - 1) de + C. muct roturn to the original variable x. This can be done eith
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