Integrating
∫
cos
j
x
sin
k
x
d
x Where
k Is odd
Evaluate
∫
cos
8
x
sin
5
x
d
x
.
Since the power on
sin
x is odd, use strategy 1. Thus,
∫
cos
8
x
sin
5
x
d
x
=
∫
cos
8
x
sin
4
x
sin
x
d
x
Break off
sin
x
.
=
∫
cos
8
x
(
sin
2
x
)
2
sin
x
d
x
Rewrite
sin
4
x
=
(
sin
2
x
)
2
.
=
∫
cos
8
x
(
1
−
cos
2
x
)
2
sin
x
d
x
Substitute
sin
2
x
=
1
−
cos
2
x
.
=
∫
u
8
(
1
−
u
2
)
2
(
−
d
u
)
Let
u
=
cos
x
and
d
u
=
−
sin
x
d
x
.
=
∫
(
−
u
8
+
2
u
10
−
u
12
)
d
u
Expand
.
=
−
1
9
u
9
+
2
11
u
11
−
1
13
u
13
+
C
Evaluate the integral
.
=
−
1
9
cos
9
x
+
2
11
cos
11
x
−
1
13
cos
13
x
+
C
.
Substitute
u
=
cos
x
.
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Integrating
∫
cos
j
x
sin
k
x
d
x Where
k And
j Are even
Evaluate
∫
sin
4
x
d
x
.
Since the power on
sin
x is even
(
k
=
4
) and the power on
cos
x is even
(
j
=
0
)
, we must use strategy 3. Thus,
∫
sin
4
x
d
x
=
∫
(
sin
2
x
)
2
d
x
Rewrite
sin
4
x
=
(
sin
2
x
)
2
.
=
∫
(
1
2
−
1
2
cos
(
2
x
)
)
2
d
x
Substitute
sin
2
x
=
1
2
−
1
2
cos
(
2
x
)
.
=
∫
(
1
4
−
1
2
cos
(
2
x
)
+
1
4
cos
2
(
2
x
)
)
d
x
Expand
(
1
2
−
1
2
cos
(
2
x
)
)
2
.
=
∫
(
1
4
−
1
2
cos
(
2
x
)
+
1
4
(
1
2
+
1
2
cos
(
4
x
)
)
d
x
.
Since
cos
2
(
2
x
) has an even power, substitute
cos
2
(
2
x
)
=
1
2
+
1
2
cos
(
4
x
)
:
=
∫
(
3
8
−
1
2
cos
(
2
x
)
+
1
8
cos
(
4
x
)
)
d
x
Simplify
.
=
3
8
x
−
1
4
sin
(
2
x
)
+
1
32
sin
(
4
x
)
+
C
Evaluate the integral
.
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In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include
sin
(
a
x
)
,
sin
(
b
x
)
,
cos
(
a
x
)
, and
cos
(
b
x
)
. These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.
Rule: integrating products of sines and cosines of different angles
To integrate products involving
sin
(
a
x
)
,
sin
(
b
x
)
,
cos
(
a
x
)
, and
cos
(
b
x
)
, use the substitutions
sin
(
a
x
)
sin
(
b
x
)
=
1
2
cos
(
(
a
−
b
)
x
)
−
1
2
cos
(
(
a
+
b
)
x
)
sin
(
a
x
)
cos
(
b
x
)
=
1
2
sin
(
(
a
−
b
)
x
)
+
1
2
sin
(
(
a
+
b
)
x
)
cos
(
a
x
)
cos
(
b
x
)
=
1
2
cos
(
(
a
−
b
)
x
)
+
1
2
cos
(
(
a
+
b
)
x
)
These formulas may be derived from the sum-of-angle formulas for sine and cosine.
Integrating products and powers of tan
x And sec
x
Before discussing the integration of products and powers of
tan
x and
sec
x
, it is useful to recall the integrals involving
tan
x and
sec
x we have already learned:
∫
sec
2
x
d
x
=
tan
x
+
C
∫
sec
x
tan
x
d
x
=
sec
x
+
C
∫
tan
x
d
x
=
ln
|
sec
x
|
+
C
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
.
For most integrals of products and powers of
tan
x and
sec
x
, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form
∫
tan
j
x
sec
2
x
d
x or
∫
sec
j
x
tan
x
d
x
. As we see in the following example, we can evaluate these new integrals by using
u -substitution.
We now take a look at the various strategies for integrating products and powers of
sec
x and
tan
x
.
Problem-solving strategy: integrating
∫
tan
k
x
sec
j
x
d
x
To integrate
∫
tan
k
x
sec
j
x
d
x
, use the following strategies:
If
j is even and
j
≥
2
, rewrite
sec
j
x
=
sec
j
−
2
x
sec
2
x and use
sec
2
x
=
tan
2
x
+
1 to rewrite
sc
j
−
2
x in terms of
tan
x
. Let
u
=
tan
x and
d
u
=
sec
2
x
.
If
k is odd and
j
≥
1
, rewrite
tan
k
x
sec
j
x
=
tan
k
−
1
x
sec
j
−
1
x
sec
x
tan
x and use
tan
2
x
=
sec
2
x
−
1 to rewrite
tan
k
−
1
x in terms of
sec
x
. Let
u
=
sec
x and
d
u
=
sec
x
tan
x
d
x
. (
Note : If
j is even and
k is odd, then either strategy 1 or strategy 2 may be used.)
If
k is odd where
k
≥
3 and
j
=
0
, rewrite
tan
k
x
=
tan
k
−
2
x
tan
2
x
=
tan
k
−
2
x
(
sec
2
x
−
1
)
=
tan
k
−
2
x
sec
2
x
−
tan
k
−
2
x
. It may be necessary to repeat this process on the
tan
k
−
2
x term.
If
k is even and
j is odd, then use
tan
2
x
=
sec
2
x
−
1 to express
tan
k
x in terms of
sec
x
. Use integration by parts to integrate odd powers of
sec
x
.
