Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7 Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)
Step 3 So far we have the following. * sec ¯ 1 (8) cos(0) sin2(0). de We can now use the substitution u = sin(0), so du = E 0 = น 3' √3 2 (cos(0) 63 Further, when 0 = sec sec¯¹(8), u = 8 cos(0) de. Once more, we must also make a substitution for the limits of integration. When 3√√7 8 Step 4 We have determined that if we let u = sin(0), then du = cos(0) de on the interval √3 63 Applying the substitution gives us the following result. 2 8 L sec-1 (8) Jπ/3 r sec ¯1 (8) cos(0) sin2(0) 63/8 1 de = и du 2 3/2 u² Step 5 We can now evaluate the integral. 63/8 3/2 du = 11 du 42 63/8 68 3/2 2 √3 8 3√7 Evaluate the integral. b 8 dx (x² - 1)3/2 Step 1 Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one. f(x) dx = = [fg f(g(t))g'(t) dt [F(x) We are given the following. 8 dx (x² - 1)3/2 We note that (x² - 1)3/2. = 3 - 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate. Expression Substitution Identity √√√x²-a² x = a sec(0), 0 ≤ 0 < Π 2 oг π ≤ < 3π 2 sec² (0) - 1 = = tan²(0) 3 3 If - 1 = √x² - a then a = 1 1 Therefore, we can let x = sec(0), so dx = sec tan tan (0) sec (0) ᏧᎾ . Further, when 3 πT We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 = 3 x = 8,0 = sec 8 8 Step 2 We have determined that if we let x = sec(0), then dx Π = sec(0) tan(0) do on the interval sec ec¯¹(8)]. Applying the substitution gives us the following result. 8 L dx - 1)/3/2 = √ / 3 r sec¯1 (8) sec(0) tan(0) tan³ (0) ᏧᎾ (x²- To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result. * sec¯¹ (8) sec(0) tan(0) tan³ (0) Jπ/3 cos(0) sec¯¹ (8) de = de Jπ/3 sin² (0) sin² (0)
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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Please help me with the last part of this problem. I am having trouble understanding what to do. Thank you
Transcribed Image Text:Step 3
So far we have the following.
* sec ¯ 1 (8) cos(0)
sin2(0).
de
We can now use the substitution u = sin(0), so du =
E
0 =
น
3'
√3
2
(cos(0)
63
Further, when 0 = sec
sec¯¹(8), u =
8
cos(0)
de. Once more, we must also make a substitution for the limits of integration. When
3√√7
8
Step 4
We have determined that if we let u = sin(0), then du = cos(0) de on the interval
√3
63
Applying the substitution gives us the following result.
2
8
L sec-1 (8)
Jπ/3
r sec ¯1 (8) cos(0)
sin2(0)
63/8
1
de =
и
du
2
3/2
u²
Step 5
We can now evaluate the integral.
63/8
3/2
du =
11 du
42
63/8
68
3/2
2
√3
8
3√7
![Evaluate the integral.
b
8
dx
(x² - 1)3/2
Step 1
Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one.
f(x)Â dx =
= [fg
f(g(t))g'(t) dt
[F(x)
We are given the following.
8
dx
(x² - 1)3/2
We note that (x² - 1)3/2.
=
3
- 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate.
Expression
Substitution
Identity
√√√x²-a²
x = a sec(0), 0 ≤ 0 <
Π
2
oг π ≤ <
3π
2
sec² (0) - 1 =
= tan²(0)
3
3
If
- 1
=
√x²
- a
then a = 1
1
Therefore, we can let x = sec(0), so dx = sec tan
tan (0) sec (0)
ᏧᎾ .
Further, when
3
πT
We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 =
3
x = 8,0 = sec
8
8
Step 2
We have determined that if we let x = sec(0), then dx
Π
=
sec(0) tan(0) do on the interval sec
ec¯¹(8)]. Applying the substitution gives us the following result.
8
L
dx
- 1)/3/2 = √
/ 3
r sec¯1 (8) sec(0) tan(0)
tan³ (0)
ᏧᎾ
(x²-
To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result.
* sec¯¹ (8) sec(0) tan(0)
tan³ (0)
Jπ/3
cos(0)
sec¯¹ (8)
de =
de
Jπ/3
sin² (0)
sin² (0)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc774740b-d90e-4acc-a307-37df5b6fc7ef%2F7ecf0472-5106-415e-afd7-73f8f1407931%2Fqg9hx3l_processed.png&w=3840&q=75)
Transcribed Image Text:Evaluate the integral.
b
8
dx
(x² - 1)3/2
Step 1
Recall the Inverse Substitution Rule, where f and g are differentiable functions and g is one-to-one.
f(x)Â dx =
= [fg
f(g(t))g'(t) dt
[F(x)
We are given the following.
8
dx
(x² - 1)3/2
We note that (x² - 1)3/2.
=
3
- 1 Therefore, the following entry from the Table of Trigonometric Substitutions is appropriate.
Expression
Substitution
Identity
√√√x²-a²
x = a sec(0), 0 ≤ 0 <
Π
2
oг π ≤ <
3π
2
sec² (0) - 1 =
= tan²(0)
3
3
If
- 1
=
√x²
- a
then a = 1
1
Therefore, we can let x = sec(0), so dx = sec tan
tan (0) sec (0)
ᏧᎾ .
Further, when
3
πT
We also must make a substitution for the limits of integration in the definite integral. Since x = sec(0), we note that when x = 2, 0 =
3
x = 8,0 = sec
8
8
Step 2
We have determined that if we let x = sec(0), then dx
Π
=
sec(0) tan(0) do on the interval sec
ec¯¹(8)]. Applying the substitution gives us the following result.
8
L
dx
- 1)/3/2 = √
/ 3
r sec¯1 (8) sec(0) tan(0)
tan³ (0)
ᏧᎾ
(x²-
To evaluate this trigonometric integral, we put everything in terms of sin(0) and cos(0). Doing so gives the following result.
* sec¯¹ (8) sec(0) tan(0)
tan³ (0)
Jπ/3
cos(0)
sec¯¹ (8)
de =
de
Jπ/3
sin² (0)
sin² (0)
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