Find dx. 49 SOLUTION. Let x = 7 tan 0, so dx - do and Vx² + 49 = V 49 tan² @ + 49 = 49 sec² @ - 7 sec 0. tan 8 = Thus, de de. 49 tan? e: 7 sec e tan e To evaluate this trigonometric integral we put everything in terms of sin e and cos 8: cos? e tan? o cos 0 sin? e sin? o° sec e cos e Therefore, making the substitution u= we have du sin cos 0 de - 49 49 49 + C. 49 sin e --- • († -) We use the figure to determine that sin e- and so dx + C. 49

Calculus: Early Transcendentals
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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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Find
dx.
+ 49
SOLUTION. Let x = 7 tan 0, so dx =
de and
/x² + 49 = V 49 tan² @ + 49 = V 49 sec? e - 7 sec 0.
tan 8 =
Thus,
do
do.
49
49 tan? e. 7 sec 8
tan? e
To evaluate this trigonometric integral we put everything in terms of sin e
and cos 6:
cos² e
cos 0 sin? e sin? ®'
cos e
sec e
tan? 0
Therefore, making the substitution u =
we have
du
dx
cos 0
de=
+ 49
sin2 @
49
49
+ C.
49 sin e
-• (무-)우
We use the figure to determine that
sin e =
and so
dx
+ C.
49
Transcribed Image Text:Find dx. + 49 SOLUTION. Let x = 7 tan 0, so dx = de and /x² + 49 = V 49 tan² @ + 49 = V 49 sec? e - 7 sec 0. tan 8 = Thus, do do. 49 49 tan? e. 7 sec 8 tan? e To evaluate this trigonometric integral we put everything in terms of sin e and cos 6: cos² e cos 0 sin? e sin? ®' cos e sec e tan? 0 Therefore, making the substitution u = we have du dx cos 0 de= + 49 sin2 @ 49 49 + C. 49 sin e -• (무-)우 We use the figure to determine that sin e = and so dx + C. 49
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