1- cos 2x 2 1+ cos 2x 1 (1- cos? 2x)(1+cos 2x) dx = 8. 8.3.25 sin2 z cos a dx = 1+ 8 dx = cos 2x - cos 2x - cos 2x da = 1+ cos 2x - 8. sin 2x 1+ cos 4.x cos 2x dx = 8. + cos 2x - Cos 4x dx - 1 (1- sin 2r) cos 2x dr = 8 sin 4x 1 (1-sin2 2x) cos 2x d.x. To compute this last integral, 8. 16 64 we let u = sin 2x, so that du = 2 cos 2x dx. Then 16 sin 2x fa- 1 (1 - u?) du = u3 +C = 1 sin 2x - sin 2x) cos 2x dx = +C. 3 Thus, our original given integral is equal to sin 2x + C. sin 2r sin 4.x sin° 2x sin 4x sin 2x – 3 +C = 16 16 16 64 8. 64 48
1- cos 2x 2 1+ cos 2x 1 (1- cos? 2x)(1+cos 2x) dx = 8. 8.3.25 sin2 z cos a dx = 1+ 8 dx = cos 2x - cos 2x - cos 2x da = 1+ cos 2x - 8. sin 2x 1+ cos 4.x cos 2x dx = 8. + cos 2x - Cos 4x dx - 1 (1- sin 2r) cos 2x dr = 8 sin 4x 1 (1-sin2 2x) cos 2x d.x. To compute this last integral, 8. 16 64 we let u = sin 2x, so that du = 2 cos 2x dx. Then 16 sin 2x fa- 1 (1 - u?) du = u3 +C = 1 sin 2x - sin 2x) cos 2x dx = +C. 3 Thus, our original given integral is equal to sin 2x + C. sin 2r sin 4.x sin° 2x sin 4x sin 2x – 3 +C = 16 16 16 64 8. 64 48
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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No need to show any other work, can you just show how we got the 1/8 circled in blue ? What steps did we take to get that 1/8. I thought we only took 1/2 out and it confused me.
![1+
8.3.25
sin? x cos x dx
Cos 2.x
1+ cos 2x
(1-cos² 2x)(1+ cos 2x) dx
8.
dx
COS
cos 2x- cos 2a - cos 2x da =
1+ cos 4x
1.
CoS 2x dx =
1
+cos 2x
COS
1+ cos 2x
3
COs 4x dx
-
sin 2x
sin 4x
(1-sin2 2x) cos 2x d.x. To compute this last integral,
8.
2.
(1-sin 2x) cos 2x da
16
%3D
16
64
we let u = sin 2x, so that du = 2 cos 2x dx. Then
%3D
sin3 2x
Ja-
|(1- sin 2r) cos 2x da
1
(1-u²) du:
+C
sin 2x
+C.
3.
Thus, our original given integral is equal to
sin 2x
+C.
3
sin 2x
)
sin 2x
sin 4.x
sin 4x
+C =
16
sin 2x
%3D
-
16
16
64
8.
64
48](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F98b00e30-c6af-4a76-ae37-440d63ad066d%2F78i5pl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1+
8.3.25
sin? x cos x dx
Cos 2.x
1+ cos 2x
(1-cos² 2x)(1+ cos 2x) dx
8.
dx
COS
cos 2x- cos 2a - cos 2x da =
1+ cos 4x
1.
CoS 2x dx =
1
+cos 2x
COS
1+ cos 2x
3
COs 4x dx
-
sin 2x
sin 4x
(1-sin2 2x) cos 2x d.x. To compute this last integral,
8.
2.
(1-sin 2x) cos 2x da
16
%3D
16
64
we let u = sin 2x, so that du = 2 cos 2x dx. Then
%3D
sin3 2x
Ja-
|(1- sin 2r) cos 2x da
1
(1-u²) du:
+C
sin 2x
+C.
3.
Thus, our original given integral is equal to
sin 2x
+C.
3
sin 2x
)
sin 2x
sin 4.x
sin 4x
+C =
16
sin 2x
%3D
-
16
16
64
8.
64
48
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