8
Integration Techniques, L’Hôpital’s Rule,
and Improper Integrals
Copyright © Cengage Learning. All rights reserved.
8.3
Trigonometric Integrals
Copyright © Cengage Learning. All rights reserved.
Objectives
 Solve trigonometric integrals involving
powers of sine and cosine.
 Solve trigonometric integrals involving
powers of secant and tangent.
 Solve trigonometric integrals involving
sine-cosine products with different angles.
3
Integrals Involving Powers of
Sine and Cosine
4
Integrals Involving Powers of Sine and Cosine
In this section you will study techniques for evaluating integrals of the form
where either m or n is a positive integer.
To find antiderivatives for these forms, try to break them into combinations of
trigonometric integrals to which you can apply the Power Rule.
For instance, you can evaluate  sin5 x cos x dx with the Power Rule by letting u = sin x.
Then, du = cos x dx and you have
To break up  sinm x cosn x dx into forms to which you can apply the Power Rule, use the
following identities.
5
Integrals Involving Powers of Sine and Cosine
6
Example 1 – Power of Sine Is Odd and Positive
Find
Solution:
Because you expect to use the Power Rule with u = cos x, save one sine factor
to form du and convert the remaining sine factors to cosines.
My Example
3
6
2
6
2
6
cos
(
x
)
sin
(
x
)
dx

cos
(
x
)
sin
(
x
)
cos(
x
)
dx

(
1

sin
(
x
))
sin
( x) cos( x)dx 



u 7 u 9 sin 7 ( x) sin 9 ( x) 7
[u  sin( x)  du  cos( x)dx]   (1  u )u du   (u  u )du   

7 9
7
9
2
6
6
8
My example

sin 3 ( x)
(1  cos 2 ( x)) sin( x)
dx  
dx 
cos( x)
cos( x)
(1  u 2 )
[u  cos( x)  du   sin( x)dx]   
du 
u
 1/ 2 u 5 / 2 
u1/ 2 u 5 / 2
 1
3/ 2 
 
 
 u du  

 2  u 
1
5
5 
 u


2
2

cos 5 / 2 ( x) 
1/ 2

2  cos ( x) 
5


Note that in CAS we also found
the area under the graph of f.
8
9
Integrals Involving Powers of Sine and Cosine
These formulas are also valid if cosn x is replaced by sinn x.
 /2
In 
 cos
 /2
n
( x)dx 
0
 cos n 1 ( x) sin( x)
 cos
n 1
( x) cos( x)dx  [u  cos n 1 ( x), v'  cos( x)  v  sin( x)] 
0
 /2
0
 /2

n2
 (n  1) cos ( x)( sin( x)) sin( x)dx 
0
 (n  1) I n  2  (n  1) I n 
In 
 /2
 (n  1) cos
n2
( x)(1  cos 2 ( x)) dx 
0
n 1
In2
n
1)n  odd , n  3
 /2
n 1 n  3
n 1 n  3 n  5
n 1 n  3 n  5
n 1 n  3 n  5
 /2
In 
In4 
I n  6  ... 
....  cos( x)dx 
....sin( x) 0 
n n2
n n2 n4
n n2 n4 0
n n2 n4
246
n 1
In  1
.....
357
n
n=1
10
1)n  even, n  2
n 1
In 
I n2
n
 /2
 /2
n 1 n  3
n 1 n  3 n  5
n 1 n  3 n  5
n 1 n  3 n  5
1  cos( 2 x)
2
In 
I n4 
I n 6  ... 
....  cos ( x)dx 
.... 
dx 
n n2
n n2 n4
n n2 n4 0
n n2 n4 0
2
 /2

n 1 n  3 n  5  1
sin( 2 x) 
.... x 

n n2 n4 2
4 0
In 

n 1 n  3 n  5   
.... 
n n2 n4  4 
n=2
135
n 1 
.....
246
n 2
So in particular for n  4 :
I4 
1 3  3

2 4 2 16
- - as we got in Ex 3
Maxima can NOT do this
integral for general n
11
Integrals Involving Powers of
Secant and Tangent
12
The following guidelines can help you evaluate integrals of the form
then in 1st integral u=tan(x) and in the
second repeat if necessary
13
Example 4 – Power of Tangent Is Odd and Positive
Find
Do my example instead, next slide
Solution:
Because you expect to use the Power Rule with u = sec x,
save a factor of (sec x tan x) to form du and convert the
remaining tangent factors to secants.
14
My example

tan 5 ( x)
3
dx  [power of tangent is odd and positive  convert to secants] 
sec ( x)
5 / 2
4
sec
(
x
)
tan
( x) sec( x) tan( x)dx  [u  sec( x)  du  sec( x) tan( x)dx] 



5 / 2
2
5 / 2
4
2
3/ 2
1 / 2
5 / 2
u
u

1
du

u
(
u

2
u

1
)
du

(
u

2
u

u
)du 



2
u5/ 2
u1/ 2 u 3 / 2 2u 5 / 2
2u 3 / 2 2 sec 5 / 2 ( x)
2 sec 3 / 2 ( x)
1/ 2
1/ 2

2


 4u 

 4 sec ( x) 
5/ 2
1/ 2  3 / 2
5
3
5
3
15
My example
4
5
sec
(
2
x
)
tan
(2 x)dx  [power of secant is even and positive  convert to tangents] 

2
5
2
2
sec
(
2
x
)
tan
(
2
x
)
sec
(
2
x
)
dx

[
u

tan(
2
x
)

du

2
sec
(2 x)dx] 

6
8
6
8
1
1
1
u
1
u
tan
(
2
x
)
tan
(2 x)
2
5
5
7




1

u
u
du

u

u
du




2
2
2 6 2 8
12
16
16
My example
 tan
6
( x)dx  [power of tan is even and positive, no secant (case 3)  convert ta n 2 to sec 2 ] 
tan 5 ( x)
2
2
tan
(
x
)
(sec
(
x
)

1
)
dx

tan
(
x
)
sec
(
x
)
dx

tan
(
x
)
dx


tan
(
x
)
(sec
( x)  1)dx




5
tan 5 ( x)
tan 5 ( x) tan 3 ( x)
2
2
2

  tan ( x) sec ( x)dx   tan ( x)dx 

  sec 2 ( x)  1dx
5
5
3
tan 5 ( x) tan 3 ( x)


 tan( x)  x
5
3
4
2
4
2
4
Also computed area under graph:
17
18
19
Integrals Involving Sine-Cosine
Products with Different Angles
20
Integrals Involving Sine-Cosine Products with Different Angles
Integrals involving the products of sines and cosines of two
different angles occur in many applications (Fourier Series).
In such instances you can use the following product-to-sum
identities.
21
Example 8 – Using Product-to-Sum Identities
Find
Solution:
Considering the second product-to-sum identity above,
you can write
22
My example
 cos(3x) cos(2 x)dx 
1
cos( x)  cos(5 x)dx 

2
1
1
 sin( x)  sin( 5 x)
2
10
23