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CHAPTER 7. TECHNIQUES OF INTEGRATION
7.2
33
Trigonometric Integrals
Example 1. Find
�
sin(x) cos18 (x) dx.
Solution. Let u = cos(x), du = − sin(x) dx and the integral becomes
�
− u18 du
Then the anti-derivative is
−
Example 2. Find
u19
cos19
=−
19
19
�
sin7 (x)
�
sin2 (x) cos2 (x) dx
�
cos(x) dx.
Solution. Let u = cos(x), du = sin(x) dx. We get rid of sin6 (x) by rewriting
it as (1 − cos2 (x))3 . Then we have:
�
�
�
7
sin (x) cos(x) dx = sin(x)(1 − cos2 (x))3 (cos(x))1/2 dx
�
= − (1 − u2 )3 u1/2 du
�
= − (1 − 3u2 + 3u4 − u6 )u1/2 du
�
= − u1/2 − 3u5/2 + 3u9/2 − u13/2 du
�
�
2 3/2
6 7/2
6 11/2
2 15/2
=−
u −
u +
u
−
u
3
7
11
15
�
�
2
6
6
2
3/2
7/2
11/2
15/2
=−
(cos(x)) −
(cos(x)) +
(cos(x))
−
(cos(x))
3
7
11
15
Example 3. Find
Solution. We apply the identities mentioned in Case 3 above to get
�
�
1
1
1
(1 − cos(2x)) (1 + cos(2x)) dx =
1 − cos2 (2x) dx
2
2
4
�
1
1
1 − (1 + cos(4x)) dx
=
4
2
�
1
1 1
=
− cos(4x) dx
4
2 2
�
1
=
1 − cos(4x) dx
8
�
�
1
sin(4x)
=
x−
8
4
CHAPTER 7. TECHNIQUES OF INTEGRATION
Example 4. Find
�
34
tan3 (x) sec3 (x) dx.
Solution. Note that the power of tangent is odd, so we use
u = sec(x)
du = sec(x) tan(x) dx
If we apply this we can translate part of the integral already
�
�
2
2
tan (x) tan(x) sec(x) sec (x) ()dx = tan2 (x)u2 du
�
��
� � �� � ����
u2
du
du
Now to finish we need to get rid of the last functions of u. We do this with
the identity
�
�
2
2
tan (x)u du = (sec2 (x) − 1)u2 du
�
= (u2 − 1)u2 du
�
= u4 − u2 du
u5 u3
−
5
3
sec5 (x) sec3 (x)
=
−
5
3
=
Example 5. Find
�
sec3 (x) dx.
Solution. We have an even number of tangents (zero) and an odd number
of secants. The integral is already written in terms of secants, so we apply
integration by parts. To do this, we need to write sec3 (x) as a product, namely
as sec(x) sec2 (x).
�
� �� � � �
��
� � �� �
sec(x) sec (x) dx = sec(x) tan(x) − sec(x) tan(x) tan(x) dx
�
= sec(x) tan(x) − sec(x) tan2 (x) dx
�
= sec(x) tan(x) − sec(x)(sec2 (x) − 1) dx
�
�
3
= sec(x) tan(x) − sec (x) + sec(x) dx
�
�
3
sec (x) dx = sec(x) tan(x) − sec3 (x) dx + ln | sec(x) + tan(x)|
�
2 sec3 (x) dx = sec(x) tan(x) + ln | sec(x) + tan(x)|
anti-derivative
2
derivative
anti-derivative
CHAPTER 7. TECHNIQUES OF INTEGRATION
�
sec3 (x) dx =
Example 6. Find
�
35
�
1�
sec(x) tan(x) + ln | sec(x) + tan(x)|
2
tan2 (x) sec(x) dx.
Solution. Since we have an even number of tangents, and an odd number of
secants, we start by getting rid of all powers of tangent
�
�
�
2
2
tan (x) sec(x) dx = (sec (x) − 1) sec(x) dx = sec3 (x) − sec(x) dx.
�
The integrals are now known:
sec(x) dx was given above, and the previous
�
example solved sec3 (x) dx. We put this in, simplify, and we are done:
�
1�
sec(x) tan(x) + ln | sec(x) + tan(x)| − ln | sec(x) + tan(x)|
2
�
1�
=
sec(x) tan(x) − ln | sec(x) + tan(x)|
2