Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question 14
8.4.79
14
if x>1
secx+C = tan1x -1 +C
Show that
x/>
sec x+C= -tan -1+C if x< -1.
First consider the case where x> 1. Find a trigonometric substitution for x. Choose the correct answer below.
OA. x sec 0
O B. x sin 8
C. x-tan 0
Using the substitution from the previous step, find dx.
dx
Substitute forx and dx in the integral.
dx
= (1)
Rewnite the expression in the radical from the previous step using a trigonometric identity, and then simplify the entire integrand. Note
that because x> 1, the angle e is in the first quadrant.
de (Simplify your answer.)
Integrate the result from the previous step.
dx
+ C
The original indefinite integral is in terms of x, so the final answer must be also in terms of x. Use the trigonometric substitution chosen
for this problem to express 0 in terms of x.
+ C
Use the answer from the previous step and the reference triangle to the right to find an expression for
the integral solution in terms of inverse tangent.
V-
dx
+ C
a
Now consider the case where x< - 1. Note that in this case, 0 must be restricted to the second quadrant. Using the same procedure as
before, find a trigonometric substitution for x, find dx, and make the substitution into the integral.
dx
= (2)
1
Rewrite the expression in the radical from the previous step using a trigonometric identity, and then simplify the entire integrand. Note
that becausex< -1. the angle 8 is in the second quadrant, and tan 0 = - tan 0
de (Simplify your answer.)
Integrate the result from the previous step.
+ C
of 8
2/17/2021, 20:
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Transcribed Image Text:8.4.79 14 if x>1 secx+C = tan1x -1 +C Show that x/> sec x+C= -tan -1+C if x< -1. First consider the case where x> 1. Find a trigonometric substitution for x. Choose the correct answer below. OA. x sec 0 O B. x sin 8 C. x-tan 0 Using the substitution from the previous step, find dx. dx Substitute forx and dx in the integral. dx = (1) Rewnite the expression in the radical from the previous step using a trigonometric identity, and then simplify the entire integrand. Note that because x> 1, the angle e is in the first quadrant. de (Simplify your answer.) Integrate the result from the previous step. dx + C The original indefinite integral is in terms of x, so the final answer must be also in terms of x. Use the trigonometric substitution chosen for this problem to express 0 in terms of x. + C Use the answer from the previous step and the reference triangle to the right to find an expression for the integral solution in terms of inverse tangent. V- dx + C a Now consider the case where x< - 1. Note that in this case, 0 must be restricted to the second quadrant. Using the same procedure as before, find a trigonometric substitution for x, find dx, and make the substitution into the integral. dx = (2) 1 Rewrite the expression in the radical from the previous step using a trigonometric identity, and then simplify the entire integrand. Note that becausex< -1. the angle 8 is in the second quadrant, and tan 0 = - tan 0 de (Simplify your answer.) Integrate the result from the previous step. + C of 8 2/17/2021, 20:
Use the trigonometric substitution chosen for this problem in order to express 0 in terms of x.
https://xlitemprod.pearsoncmg.com/api/v1/print
dx
+C
Finally, use the answer from the previous step and the reference triangle to the right to find an expression for
the integral solution in terms of inverse tangent.
V-a
dx
+C
(1)
sec e
(2)
sec 0 tan 0
de
de
sec 0 tan 0)sec 0-1
(sec 0)/sec 0-1
sec 0 tan e
sec 0
d0
de
(sec 0 tan 0)-/sec 20-1
sec 0 tan 0
(sec 0) sec 0-1
sec 0 tan e
de
(sec 0)/sec e-1
(sec 0)-/sec 20- 1
sec 0 tan 0
sec 0 tan 0
de
(sec 0)/sec 2e-1
(sec 0)-/sec20-1
7 of 8
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Transcribed Image Text:Use the trigonometric substitution chosen for this problem in order to express 0 in terms of x. https://xlitemprod.pearsoncmg.com/api/v1/print dx +C Finally, use the answer from the previous step and the reference triangle to the right to find an expression for the integral solution in terms of inverse tangent. V-a dx +C (1) sec e (2) sec 0 tan 0 de de sec 0 tan 0)sec 0-1 (sec 0)/sec 0-1 sec 0 tan e sec 0 d0 de (sec 0 tan 0)-/sec 20-1 sec 0 tan 0 (sec 0) sec 0-1 sec 0 tan e de (sec 0)/sec e-1 (sec 0)-/sec 20- 1 sec 0 tan 0 sec 0 tan 0 de (sec 0)/sec 2e-1 (sec 0)-/sec20-1 7 of 8
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