In class today, we went over two examples of integration of trigonometric functions. In this
note, I’ll show you the facts you need to know from this section, with examples for each technique.
1
R
Integrals of the form
odd natural number
sinm (x) cosn (x)dx, where m is an
Method:
1. Set aside one copy of sin(x), and place it next to the differential dx.
2. You have an even number (m − 1) of copies of sin(x) remaining. Since m − 1 = 2k for
some number k, write:
R
= sinm−1 (x) cosn (x)sin(x)dx
R
= sin2k (x) cosn (x)sin(x)dx
R
= (sin2 (x))k cosn (x)sin(x)dx
3. Rewrite sin2 (x) = 1 − cos2 (x) to get:
=
R
(1 − cos2 (x))k cosn (x)sin(x)dx
4. Make the substitution: u = − cos(x). Then cos(x) = −u, and du = sin(x)dx, so we get:
=
R
(1 − (−u)2 )k (−u)n du
5. Expand the integrand. You will get a polynomial function of u.
6. Integrate the resulting polynomial function of u, and substitute back for u.
Example:
R
sin5 (x) cos10 (x)dx
Solution:
1. =
R
sin4 (x) cos10 (x)sin(x)dx
2. =
R
sin2·2 (x) cos10 (x)sin(x)dx =
3. =
R
(1 − cos2 (x))2 cos10 (x)sin(x)dx
4. =
R
(1 − (−u)2 )2 (−u)10 du =
R
5. =
R
(1 + u4 − 2u2 )u10 du =
(u10 + u14 − 2u12 )du
6. =
u11
11
+
u15
15
13
− 2 u13 =
R
R
(sin2 (x))2 cos10 (x)sin(x)dx
(1 − u2 )2 u10 du
(− cos(x))11
11
+
(− cos(x))15
15
1
− 2 (− cos(x))
13
13
+C
2
Integrals of the form
odd natural number
R
sinm (x) cosn (x)dx, where n is an
Method:
1. Set aside one copy of cos(x), and place it next to the differential dx.
2. You have an even number (n − 1) of copies of cos(x) remaining. Since n − 1 = 2k for some
number k, write:
R
= R cosn−1 (x) sinm (x)cos(x)dx
= R cos2k (x) sinm (x)cos(x)dx
= (cos2 (x))k sinm (x)cos(x)dx
3. Rewrite cos2 (x) = 1 − sin2 (x) to get:
=
R
(1 − sin2 (x))k sinm (x)cos(x)dx
4. Make the substitution: u = sin(x). Then du = cos(x)dx, so we get:
=
R
(1 − (u)2 )k (u)m du
5. Expand the integrand. You will get a polynomial function of u.
6. Integrate the resulting polynomial function of u, and substitute back for u.
Example:
R
sin4 (x) cos5 (x)dx
Solution:
1. =
R
sin4 (x) cos4 (x)cos(x)dx
2. =
R
cos2·2 (x) sin4 (x)cos(x)dx =
3. =
R
(1 − sin2 (x))2 sin4 (x)cos(x)dx
4. =
R
(1 − (u)2 )2 (u)4 du =
5. =
R
(1 + u4 − 2u2 )u4 du =
6. =
u5
5
+
u9
9
7
− 2 u7 =
R
(cos2 (x))2 sin4 (x)cos(x)dx
(1 − u2 )2 u4 du
R
R
(sin(x))5
5
(u4 + u8 − 2u6 )du
+
(sin(x))9
9
7
− 2 (sin(x))
+C
7
2
3
Integrals of the form
odd natural number
R
secm (x) tann (x)dx, where n is an
Method:
1. Set aside one copy of sec(x) tan(x), and place it next to the differential dx.
2. You have an even number (n − 1) of copies of tan(x) remaining. Since n − 1 = 2k for some
number k, write:
R
= tann−1 (x) secm−1 (x)sec(x) tan(x)dx
R
= R tan2k (x) secm−1 (x)sec(x) tan(x)dx
= (tan2 (x))k secm−1 (x)sec(x) tan(x)dx
3. Rewrite tan2 (x) = sec2 (x) − 1 to get:
=
R
(sec2 (x) − 1)k secm−1 (x)sec(x) tan(x)dx
4. Make the substitution: u = sec(x). Then du = sec(x) tan(x)dx, so we get:
=
R
(u2 − 1)k um−1 du
5. Expand the integrand. You will get a polynomial function of u.
6. Integrate the resulting polynomial function of u, and substitute back for u.
Example:
R
sec4 (x) tan5 (x)dx
Solution:
1. =
R
sec3 (x) tan4 (x)sec(x) tan(x)dx
2. =
R
sec3 (x) tan2·2 (x)sec(x) tan(x)dx =
3. =
R
(sec2 (x) − 1)2 sec3 (x)sec(x) tan(x)dx
4. =
R
((u)2 − 1)2 (u)3 du =
5. =
R
(u4 + 1 − 2u2 )u3 du =
6. =
u8
8
+
u4
4
6
− 2 u6 =
(tan2 (x))2 sec3 (x)sec(x) tan(x)dx
R
(u2 − 1)2 u3 du
R
R
(sec(x))8
8
(u7 + u3 − 2u5 )du
+
(sec(x))4
4
6
− 2 (sec(x))
+C
6
3
4
Integrals of the form
even natural number
R
secm (x) tann (x)dx, where m is an
Method:
1. Set aside two copies of sec(x) (that is, set aside sec2 (x)), and place them next to the
differential dx.
2. You have an even number (m − 2) of copies of sec(x) remaining. Since m − 2 = 2k for
some number k, write:
R
= R tann (x) secm−2 (x)sec2 (x)dx
= R tann (x) sec2k (x)sec2 (x)dx
= (sec2 (x))k tann (x)sec2 (x)dx
3. Rewrite sec2 (x) = tan2 (x) + 1 to get:
=
R
(tan2 (x) + 1)k tann (x)sec2 (x)dx
4. Make the substitution: u = tan(x). Then du = sec2 (x)dx, so we get:
=
R
(u2 + 1)k un du
5. Expand the integrand. You will get a polynomial function of u.
6. Integrate the resulting polynomial function of u, and substitute back for u.
Example:
R
sec4 (x) tan5 (x)dx
Solution:
1. =
R
sec2 (x) tan5 (x)sec2 (x)dx
2. =
R
sec2 (x) tan5 (x)sec2 (x)dx =
3. =
R
(tan2 (x) + 1)1 tan5 (x)sec2 (x)dx
4. =
R
(u2 + 1)u5 du =
5. =
u8
8
+
u6
6
=
R
(tan(x))8
8
R
(sec2 (x))1 tan5 (x)sec2 (x)dx
u7 + u5 du
+
(tan(x))6
6
+C
4
R
Integrals of the form sinm (x) cosn (x)dx, where both m
and n are even natural numbers
5
This time, we aren’t able to use a substitution immediately. Instead, we’ll use the double angle
formulae to transform the integrand.
Method 1:
Remember that
cos(2x) = cos2 (x) − sin2 (x) = 2 cos2 (x) − 1 = 1 − 2 sin2 (x).
Rearranging this, we get:
cos2 (x) =
1 + cos(2x)
2
sin2 (x) =
1 − cos(2x)
.
2
and
Example:
R
cos2 (x) sin2 (x)dx.
Solution:
=
R
)( 1−cos(2x)
)dx
( 1+cos(2x)
2
2
=
1
4
R
(1 + cos(2x))(1 − cos(2x))dx
=
1
4
R
1 − cos2 (2x)dx
=
1
4
R
1dx −
= 14 x −
1
4
R
1
4
cos2 (2x)dx
R
cos2 (2x)dx
At this step, we can apply the formula above again, to get:
= 14 x −
1
4
R
1+cos(4x)
dx
2
= 41 x −
1
8
R
1dx −
= 14 x − 18 x −
1
8
R
1
8
R
cos(4x)dx
cos(4x)dx
Finally, we can use a substitution u = 4x to get:
= 14 x − 18 x −
1
32
sin(4x) + C
5
Method 2:
Remember that
sin(2x) = 2 sin(x) cos(x).
Rearranging this, we get:
sin(x) cos(x) =
Example:
R
cos2 (x) sin2 (x)dx.
Solution:
=
R
(cos(x) sin(x))2 dx
=
R
( 21 sin(2x))2 dx
=
1
4
R
sin2 (2x)dx
Using the formulas from Method 1, we get:
=
1
4
R
1−cos(4x)
dx
2
=
1
8
R
1 − cos(4x)dx
=
1
8
R
1dx −
= 18 x −
1
32
1
8
R
cos(4x)dx
sin(4x) + C
6
1
sin(2x).
2
6
The integral
Fact:
R
R
tan(x)dx
tan(x)dx = − ln | cos(x)| + C = ln | sec(x)| + C
Proof:
R
tan(x)dx
=
R
sin(x)
cos(x) dx
Using the substitution u = cos(x), we get that du = − sin(x)dx, so dx =
=
R
=−
sin(x) −1
u sin(x) du
R
1
u du
= − ln |u| + C
= − ln | cos(x)| + C
giving the first result. Next,
= ln |(cos(x))−1 | + C
1
= ln | (cos(x))
|+C
= ln | sec(x)| + C
giving the second result.
7
−1
sin(x) du,
and
7
The integral
Fact:
R
R
sec(x)dx
sec(x)dx = ln | sec(x) + tan(x)| + C
Proof:
This proof is clearer if we go backwards:
d
dx [ln | sec(x)
+ tan(x)|]
=
1
d
sec(x)+tan(x) dx [sec(x)
=
1
sec(x)+tan(x) [sec(x) tan(x)
=
1
sec(x)+tan(x) [tan(x)
+ tan(x)]
+ sec2 (x)]
+ sec(x)] sec(x)
= sec(x)
Therefore, integrating both sides of the equation:
sec(x) =
d
[ln | sec(x) + tan(x)|],
dx
we get:
Z
sec(x)dx = ln | sec(x) + tan(x)| + C
which is the result.
8