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#23) y'(x) = 4 tan 2x-3 los x Saj (x) dx = S 4 tom 2x - 3 los x dx 3(x) = 4 S sin 2x dx - S3 losx. dx los 2x = 4 S sin 2x μ ک 2 = 4.1 Six M S = - 2 sin 2x du -3 Seosx dx - sinxte I du du Sink = -2 ln u + C =-2 ln los 2x + C - 3. Sin x + C = -2 ln los 2x-3 sin x + C u = los 2x du = -sin 2x. 2 dx =dx du -2 Sin 2x = 2 In sec 2 x - 3 Sin x + C → The book has this ansurer, how did he get this

#23) y'(x) = 4 tan 2x-3 los x Saj (x) dx = S 4 tom 2x - 3 los x dx 3(x) = 4 S sin 2x dx - S3 losx. dx los 2x = 4 S sin 2x μ ک 2 = 4.1 Six M S = - 2 sin 2x du -3 Seosx dx - sinxte I du du Sink = -2 ln u + C =-2 ln los 2x + C - 3. Sin x + C = -2 ln los 2x-3 sin x + C u = los 2x du = -sin 2x. 2 dx =dx du -2 Sin 2x = 2 In sec 2 x - 3 Sin x + C → The book has this ansurer, how did he get this

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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4 of 4 #23) y'(x) = 4 tan 2x-3 los x Sag'(x) dx = √ 4 tam 2x - 3 los x de y ( x) = 4 √ sin 2x dx - √3.10sx dx S losx los 2x = 4 S 2 sin 2x u = 4.1 J $/1 S sim M S du -2 sin 2x I du -|3 du Sinza -3 Seosx dx J - 3 Sin x +C : -2 ln u + C =-2 ln cos2x+c = 3 Sin x +C = -2 ln sos 2x - 3 Sin x + C м: лов 2х du = -Sin 2x. 2 dx du -2 Sin 2x = 2 ln suc 2x - 3 Sinx + C → The book has this ansuler, how did he get this : dx
Transcribed Image Text:4 of 4 #23) y'(x) = 4 tan 2x-3 los x Sag'(x) dx = √ 4 tam 2x - 3 los x de y ( x) = 4 √ sin 2x dx - √3.10sx dx S losx los 2x = 4 S 2 sin 2x u = 4.1 J $/1 S sim M S du -2 sin 2x I du -|3 du Sinza -3 Seosx dx J - 3 Sin x +C : -2 ln u + C =-2 ln cos2x+c = 3 Sin x +C = -2 ln sos 2x - 3 Sin x + C м: лов 2х du = -Sin 2x. 2 dx du -2 Sin 2x = 2 ln suc 2x - 3 Sinx + C → The book has this ansuler, how did he get this : dx
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