7
TECHNIQUES OF INTEGRATION
TECHNIQUES OF INTEGRATION
7.2
Trigonometric Integrals
In this section, we will learn:
How to use trigonometric identities to integrate
certain combinations of trigonometric functions.
TRIGONOMETRIC INTEGRALS
We start with
powers of sine and cosine.
SINE & COSINE INTEGRALS
Example 1
Evaluate ∫ cos3x dx
 Simply substituting u = cos x isn’t helpful,
since then du = -sin x dx.
 In order to integrate powers of cosine, we
would need an extra sin x factor.
 Similarly, a power of sine would require
an extra cos x factor.
SINE & COSINE INTEGRALS
Example 1
Thus, here we can separate one cosine factor
and convert the remaining cos2x factor to
an expression involving sine using the identity
sin2x + cos2x = 1:
cos3x = cos2x . cosx = (1 - sin2x) cosx
Example 1
SINE & COSINE INTEGRALS
We can then evaluate the integral by
substituting u = sin x.
So, du = cos x dx and
 cos x dx 
3
 cos x  cos x dx
2

 (1  sin

 (1  u
2
2
x ) cos x dx
) du  u  u  C
1
3
 sin x  sin x  C
1
3
3
3
SINE & COSINE INTEGRALS
In general, we try to write an integrand
involving powers of sine and cosine in
a form where we have only one sine factor.
 The remainder of the expression can be
in terms of cosine.
SINE & COSINE INTEGRALS
We could also try only one cosine
factor.
 The remainder of the expression can be
in terms of sine.
SINE & COSINE INTEGRALS
The identity
sin2x + cos2x = 1
enables us to convert back and forth
between even powers of sine and cosine.
SINE & COSINE INTEGRALS
Example 2
Find ∫ sin5x cos2x dx
 We could convert cos2x to 1 – sin2x.
 However, we would be left with an expression
in terms of sin x with no extra cos x factor.
Example 2
SINE & COSINE INTEGRALS
Instead, we separate a single sine factor
and rewrite the remaining sin4x factor in
terms of cos x.
So, we have:
sin x cos x  (sin x ) cos x sin x
5
2
2
2
2
 (1  cos x ) cos x sin x
2
2
2
Example 2
SINE & COSINE INTEGRALS
Substituting u = cos x, we have du = sin x dx.
So,
 sin

5
 (sin
x co s x d x 
2
2
2
x ) co s x sin x d x
 (1  co s x ) co s x sin x d x 
2
2
2
2
 (1  u ) u (  d u )
2
2
2
3
5
7

u
u
u 
2
4
6
   (u  2 u  u ) d u   
2

C
5
7 
 3
  13 co s x 
3
2
5
co s x 
5
1
7
co s x  C
7
SINE & COSINE INTEGRALS
The figure shows the graphs of the integrand
sin5x cos2x in Example 2 and its indefinite
integral (with C = 0).
SINE & COSINE INTEGRALS
In the preceding examples, an odd power of
sine or cosine enabled us to separate a single
factor and convert the remaining even power.
 If the integrand contains even powers of
both sine and cosine, this strategy fails.
SINE & COSINE INTEGRALS
In that case, we can take advantage
of the following half-angle identities:
sin x 
1
2
(1  co s 2 x )
co s x 
1
2
(1  co s 2 x )
2
2
SINE & COSINE INTEGRALS

Example 3
Evaluate  sin x d x
2
0
 If we write sin2x = 1 - cos2x, the integral
is no simpler to evaluate.
SINE & COSINE INTEGRALS
Example 3
However, using the half-angle formula
for sin2x, we have:



1
2

0

1
2
( x  sin 2 x )  0

1
2
(  12 sin 2  ) 

1
2

sin x dx 
2
0

(1  cos 2 x ) dx
1
2

1
2
(0  12 sin 0)
SINE & COSINE INTEGRALS
Example 3
Notice that we mentally made
the substitution u = 2x when integrating
cos 2x.
 Another method for evaluating this integral
was given in Exercise 43 in Section 7.1
SINE & COSINE INTEGRALS
Example 4
Find ∫ sin4x dx
 We could evaluate this integral using the reduction
formula for ∫ sinnx dx (Equation 7 in Section 7.1)
together with Example 3.
Example 4
SINE & COSINE INTEGRALS
However, a better method is to write and use a
half-angle formula:
 sin
4
x dx 
 (sin
2
2
x) dx
2
 1  co s 2 x 
 
 dx
2



1
4
 (1  2 co s 2 x  co s
2
2 x) dx
SINE & COSINE INTEGRALS
Example 4
As cos2 2x occurs, we must use another
half-angle formula:
cos 2 x 
2
1
2
(1  cos 4 x )
SINE & COSINE INTEGRALS
Example 4
This gives:
 sin x dx 
1
4
 1  2 cos 2 x 

1
4

3
2

1
4

x  sin 2 x  81 sin 4 x   C
4
3
2
 2 cos 2 x 
1
2
1
2
(1  cos 4 x )  dx
cos 4 x  dx
SINE & COSINE INTEGRALS
To summarize, we list guidelines
to follow when evaluating integrals
of the form
∫ sinmx cosnx dx
where m ≥ 0 and n ≥ 0 are integers.
STRATEGY A
If the power of cosine is odd (n = 2k + 1),
save one cosine factor.
 Use cos2x = 1 - sin2x to express the remaining
factors in terms of sine:
 sin
m
x cos
2 k 1
x dx 
 sin
m
x (cos x ) cos x dx

 sin
m
x (1  sin x ) cos x dx
 Then, substitute u = sin x.
2
k
2
k
STRATEGY B
If the power of sine is odd (m = 2k + 1),
save one sine factor.
 Use sin2x = 1 - cos2x to express the remaining
factors in terms of cosine:
 sin
2 k 1
x cos x dx 
n

 (sin x ) cos x sin x dx
2
k
n
 (1  cos x ) cos x sin x dx
 Then, substitute u = cos x.
2
k
n
STRATEGIES
Note that, if the powers of both sine
and cosine are odd, either (A) or (B)
can be used.
STRATEGY C
If the powers of both sine and cosine are
even, use the half-angle identities
sin x 
1
2
(1  cos 2 x )
cos x 
1
2
(1  cos 2 x )
2
2
 Sometimes, it is helpful to use the identity
sin x cos x 
1
2
sin 2 x
TANGENT & SECANT INTEGRALS
We can use a similar strategy to
evaluate integrals of the form
∫ tanmx secnx dx
TANGENT & SECANT INTEGRALS
As (d/dx)tan x = sec2x, we can separate
a sec2x factor.
 Then, we convert the remaining (even) power
of secant to an expression involving tangent using
the identity sec2x = 1 + tan2x.
TANGENT & SECANT INTEGRALS
Alternately, as (d/dx) sec x = sec x tan x,
we can separate a sec x tan x factor
and convert the remaining (even) power
of tangent to secant.
TANGENT & SECANT INTEGRALS Example 5
Evaluate ∫ tan6x sec4x dx
 If we separate one sec2x factor, we can express
the remaining sec2x factor in terms of tangent
using the identity sec2x = 1 + tan2x.
 Then, we can evaluate the integral by substituting
u = tan x so that du = sec2x dx.
TANGENT & SECANT INTEGRALS Example 5
We have:
 tan x sec x dx   tan x sec x sec x dx
6
4
6

 tan

u

u
6
2

x (1  tan x ) sec x dx
6
2
(1  u ) du 
7

2
u
7
1
7
2
2
 (u
6
9
C
9
tan x 
7
1
9
tan x  C
9
 u ) du
8
TANGENT & SECANT INTEGRALS Example 6
Find ∫ tan5 θ sec7θ
 If we separate a sec2θ factor, as in the preceding
example, we are left with a sec5θ factor.
 This isn’t easily converted to tangent.
TANGENT & SECANT INTEGRALS Example 6
However, if we separate a sec θ tan θ factor,
we can convert the remaining power of
tangent to an expression involving only
secant.
 We can use the identity tan2θ = sec2θ – 1.
TANGENT & SECANT INTEGRALS Example 6
We can then evaluate the integral by
substituting u = sec θ, so du = sec θ tan θ dθ:
 tan  sec    tan  sec  sec  tan  d 
5
7
4
6

 (sec   1) sec  sec  tan  d 

 (u

u
2
2

6
 1) u du 
2
11
2
11
1
11
2
u
6
9

9
u
 (u
 2 u  u ) du
8
7
C
7
sec   sec  
11
10
2
9
9
1
7
sec   C
7
6
TANGENT & SECANT INTEGRALS
The preceding examples demonstrate
strategies for evaluating integrals in the form
∫ tanmx secnx for two cases—which we
summarize here.
STRATEGY A
If the power of secant is even (n = 2k, k ≥ 2)
save sec2x.
 Then, use tan2x = 1 + sec2x to express
the remaining factors in terms of tan x:
 tan
m
x sec
2k
k 1
x dx 
 tan
m
x (sec x )

 tan
m
x (1  tan x )
 Then, substitute u = tan x.
2
2
2
sec x dx
k 1
2
sec x dx
STRATEGY B
If the power of tangent is odd (m = 2k + 1),
save sec x tan x.
 Then, use tan2x = sec2x – 1 to express
the remaining factors in terms of sec x:
 tan
2 k 1
x sec x dx 
n

 (tan x ) sec
2
k
n 1
x sec x tan x dx
 (sec x  1) sec
2
 Then, substitute u = sec x.
k
n 1
x sec x tan x dx
OTHER INTEGRALS
For other cases, the guidelines are not
as clear-cut.
We may need to use:
 Identities
 Integration by parts
 A little ingenuity
TANGENT & SECANT INTEGRALS
We will need to be able to integrate tan x
by using Formula 5 from Section 5.5 :
 tan x dx  ln | sec x |  C
TANGENT & SECANT INTEGRALS Formula 1
We will also need the indefinite integral
of secant:
 sec x dx  ln | sec x  tan x |  C
TANGENT & SECANT INTEGRALS
We could verify Formula 1
by differentiating the right side,
or as follows.
TANGENT & SECANT INTEGRALS
First, we multiply numerator and denominator
by sec x + tan x:
sec x  tan x
 sec x dx   sec x sec x  tan x dx
sec x  sec x tan x
2


sec x  tan x
dx
TANGENT & SECANT INTEGRALS
If we substitute u = sec x + tan x,
then du = (sec x tan x + sec2x).
 The integral becomes: ∫ (1/u) du = ln |u| + C
TANGENT & SECANT INTEGRALS
Thus, we have:
sec
x
dx

ln
|
sec
x

tan
x
|

C

TANGENT & SECANT INTEGRALS Example 7
Find ∫ tan3x dx
 Here, only tan x occurs.
 So, we rewrite a tan2x factor in terms of sec2x.
TANGENT & SECANT INTEGRALS Example 7
Hence, we use tan2x - sec2x = 1.
 tan
3
x dx 
 tan x tan
2

 tan x sec
2

tan x
x dx 
 tan x (sec
x dx 
 tan x dx
2
x  1) dx
2
 ln | sec x |  C
2
 In the first integral, we mentally substituted u = tan x
so that du = sec2x dx.
TANGENT & SECANT INTEGRALS
If an even power of tangent appears with an
odd power of secant, it is helpful to express
the integrand completely in terms of sec x.
 Powers of sec x may require integration by parts,
as shown in the following example.
TANGENT & SECANT INTEGRALS Example 8
Find ∫ sec3x dx
 Here, we integrate by parts with
u  sec x
dv  sec x dx
du  sec x tan x dx
v  tan x
2
TANGENT & SECANT INTEGRALS Example 8
Then,
 sec
3
x dx  sec x tan x   sec x tan x dx
2
 sec x tan x   sec x (sec x  1) dx
2
 sec x tan x   sec x dx   sec x dx
3
TANGENT & SECANT INTEGRALS Example 8
Using Formula 1 and
solving for the required integral,
we get:
sec
x
dx

3

1
2
(sec x tan x  ln | sec x  tan x |)  C
TANGENT & SECANT INTEGRALS
Integrals such as the one in the example
may seem very special.
 However, they occur frequently in applications of
integration.
 We will see this in Chapter 8.
COTANGENT & COSECANT INTEGRALS
Integrals of the form ∫ cotmx cscnx dx
can be found by similar methods.
 We have to make use of the identity
1 + cot2x = csc2x
OTHER INTEGRALS
Finally, we can make use of
another set of trigonometric identities,
as follows.
Equation 2
OTHER INTEGRALS
In order to evaluate the integral, use
the corresponding identity.
Integral
a
∫ sin mx cos nx dx
Identity
sin A cos B

b
∫ sin mx sin nx dx
∫ cos mx cos nx dx
 sin( A  B )  sin( A  B ) 
sin A sin B

c
1
2
1
2
 cos( A  B )  cos( A  B ) 
cos A cos B

1
2
 cos( A  B )  cos( A  B ) 
TRIGONOMETRIC INTEGRALS
Example 9
Evaluate ∫ sin 4x cos 5x dx
 This could be evaluated using integration by parts.
 It’s easier to use the identity in Equation 2(a):
 sin 4 x cos 5 x dx 
  sin(  x )  sin 9 x 
1
2

1
2
 (  sin x  sin 9 x ) dx

1
2
(cos x  91 cos 9 x )  C