This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skippe the skipped part. Tutorial Exercise Evaluate the integral. Step 1 To evaluate the integral, we will first attempt to rewrite the integrand so that involves powers of sine and cosine in a form where we have only c do so, we recall the following double angle formula. cos(2x) = 2 cos²(x) - 1 Applying this gives the following result. -π/6 Step 3 1 0. rπ/6 Step 2 We can now simplify the result as follows. π/6 √ ² / 0 √₁+( F1 √1 + cos(2x) dx √ 1 + cos(2x) dx = Q A √ 1 + (2 cos² (x) - 1) dx 2 π/6 [²²6 √ 1 + ( 2 cos² (x) − 1 - F2 W S = # 3 π/6 *°v@2 2 20 F3 E D $ 4 J Q F4 R FL TC 2 cos² (x) - 1 rπ/6 2√ √ cos²³(x) dx % 5 2 cos² (x) dx ů F5 T MacBook Air < C 6 G F6 Y & f 7 H F7 U * 00 8 DII F8 J 9 F9 K C Applying this gives the following result. π/6 rπ/6 = √²²0 √ ₁ + ( Step 3 Step 2 We can now simplify the result as follows. π/6 1 √1 + cos(2x) dx = :0 F1 √1+ (2 cos²(x) - 1) dx = Q A π/6 √² √2/6 √ cos²(x) dx = √² 1/2 √₂ fő™ π Simplifying the integrand and noting that cos(x) > 0 for 0 ≤ x ≤ 6 2 cos(x) dx = √2 F2 W S 1 +(2 cos² (x) - 1 = = || π/6 2 6₁² °√(² 3 2 π/6 π/6 -√₂/²0 (1 cos(x) Step 4 We have successfully rewritten the integrand in a format that we can easily evaluate. Doing so gives the following result. π/6 1π/6 √₂ √2²². 80 F3 cos(x) E D 10 $ 4 F4 J R 2 F LL 2 cos² (x) - 1 с оро 2 rπ/6 cos(x) % 5 cos(x) allows us to continue as follows. ✓ cos² (x) dx cos² (x) dx O F5 T dx dx G MacBook Air ^ 6 F6 Y & 7 H F7 U * 0 8 J DII F8 ( | 9 T F9 K 0 0
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skippe the skipped part. Tutorial Exercise Evaluate the integral. Step 1 To evaluate the integral, we will first attempt to rewrite the integrand so that involves powers of sine and cosine in a form where we have only c do so, we recall the following double angle formula. cos(2x) = 2 cos²(x) - 1 Applying this gives the following result. -π/6 Step 3 1 0. rπ/6 Step 2 We can now simplify the result as follows. π/6 √ ² / 0 √₁+( F1 √1 + cos(2x) dx √ 1 + cos(2x) dx = Q A √ 1 + (2 cos² (x) - 1) dx 2 π/6 [²²6 √ 1 + ( 2 cos² (x) − 1 - F2 W S = # 3 π/6 *°v@2 2 20 F3 E D $ 4 J Q F4 R FL TC 2 cos² (x) - 1 rπ/6 2√ √ cos²³(x) dx % 5 2 cos² (x) dx ů F5 T MacBook Air < C 6 G F6 Y & f 7 H F7 U * 00 8 DII F8 J 9 F9 K C Applying this gives the following result. π/6 rπ/6 = √²²0 √ ₁ + ( Step 3 Step 2 We can now simplify the result as follows. π/6 1 √1 + cos(2x) dx = :0 F1 √1+ (2 cos²(x) - 1) dx = Q A π/6 √² √2/6 √ cos²(x) dx = √² 1/2 √₂ fő™ π Simplifying the integrand and noting that cos(x) > 0 for 0 ≤ x ≤ 6 2 cos(x) dx = √2 F2 W S 1 +(2 cos² (x) - 1 = = || π/6 2 6₁² °√(² 3 2 π/6 π/6 -√₂/²0 (1 cos(x) Step 4 We have successfully rewritten the integrand in a format that we can easily evaluate. Doing so gives the following result. π/6 1π/6 √₂ √2²². 80 F3 cos(x) E D 10 $ 4 F4 J R 2 F LL 2 cos² (x) - 1 с оро 2 rπ/6 cos(x) % 5 cos(x) allows us to continue as follows. ✓ cos² (x) dx cos² (x) dx O F5 T dx dx G MacBook Air ^ 6 F6 Y & 7 H F7 U * 0 8 J DII F8 ( | 9 T F9 K 0 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...
Related questions
Question
Solve step 4

Transcribed Image Text:This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skippe
the skipped part.
Tutorial Exercise
Evaluate the integral.
Step 1
To evaluate the integral, we will first attempt to rewrite the integrand so that involves powers of sine and cosine in a form where we have only c
do so, we recall the following double angle formula.
cos(2x) = 2 cos²(x) - 1
Applying this gives the following result.
-π/6
Step 3
1
0.
rπ/6
Step 2
We can now simplify the result as follows.
π/6
√ ² / 0 √₁+(
F1
√1 + cos(2x) dx
√ 1 + cos(2x) dx =
Q
A
√ 1 + (2 cos² (x) - 1) dx
2
π/6
[²²6 √ 1 + ( 2 cos² (x) − 1
-
F2
W
S
=
#
3
π/6
*°v@2
2
20
F3
E
D
$
4
J
Q
F4
R
FL
TC
2 cos² (x) - 1
rπ/6
2√ √ cos²³(x) dx
%
5
2
cos² (x) dx
ů
F5
T
MacBook Air
< C
6
G
F6
Y
&
f
7
H
F7
U
* 00
8
DII
F8
J
9
F9
K
C

Transcribed Image Text:Applying this gives the following result.
π/6
rπ/6
= √²²0 √ ₁ + (
Step 3
Step 2
We can now simplify the result as follows.
π/6
1
√1 + cos(2x) dx =
:0
F1
√1+ (2 cos²(x) - 1) dx =
Q
A
π/6
√² √2/6 √ cos²(x) dx = √²
1/2
√₂ fő™
π
Simplifying the integrand and noting that cos(x) > 0 for 0 ≤ x ≤
6
2
cos(x) dx = √2
F2
W
S
1 +(2 cos² (x) - 1
=
=
||
π/6
2
6₁² °√(²
3
2
π/6
π/6
-√₂/²0 (1 cos(x)
Step 4
We have successfully rewritten the integrand in a format that we can easily evaluate. Doing so gives the following result.
π/6
1π/6
√₂ √2²².
80
F3
cos(x)
E
D
10
$
4
F4
J
R
2
F
LL
2 cos² (x) - 1
с оро
2
rπ/6
cos(x)
%
5
cos(x)
allows us to continue as follows.
✓ cos² (x) dx
cos² (x) dx
O
F5
T
dx
dx
G
MacBook Air
^
6
F6
Y
&
7
H
F7
U
* 0
8
J
DII
F8
(
|
9
T
F9
K
0
0
Expert Solution
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