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 AND COSINE INTEGRALS
Example 1
Evaluate ∫ cos3x dx
 Simply substituting u = cos x is not 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 AND 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 AND COSINE INTEGRALS
We can then evaluate the integral by substituting
u = sin x. So, du = cos x dx and
 cos
3
x dx   cos x  cos x dx
2
  (1  sin x) cos x dx
2
  (1  u ) du  u  u  C
2
1
3
 sin x  sin x  C
1
3
3
3
SINE AND 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 AND COSINE INTEGRALS
We could also try only one cosine factor.
 The remainder of the expression can be in terms of sine.
SINE AND COSINE INTEGRALS
The identity
sin2x + cos2x = 1
enables us to convert back and forth between even
powers of sine and cosine.
SINE AND 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 AND 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 AND COSINE INTEGRALS
Substituting u=cos x, we have du =-sin x dx. So,
 sin x cos
  (1  cos
5
2
2
x dx   (sin x) cos x sin x dx
2
2
2
x) 2 cos 2 x sin x dx   (1  u 2 ) 2 u 2 (du )
3
5
7

u
u u 
2
4
6
   (u  2u  u )du     2    C
5 7 
 3
  13 cos3 x  52 cos5 x  17 cos 7 x  C
SINE AND COSINE INTEGRALS
The figure shows the graphs of the integrand sin5x
cos2x in Example 2 and its indefinite integral (with
C = 0).
SINE AND 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 AND COSINE INTEGRALS
In that case, we can take advantage of the
following half-angle identities:
sin x  12 (1  cos 2 x)
2
cos x  12 (1  cos 2 x)
2
SINE AND COSINE INTEGRALS
Evaluate


0
Example 3
2
sin x dx
 If we write sin2x = 1 - cos2x, the integral
is no simpler to evaluate.
SINE AND COSINE INTEGRALS
Example 3
However, using the half-angle formula for sin2x,
we have:


0
sin x dx 
2
1
2


0
(1  cos 2 x) dx
  ( x  sin 2 x) 0
1
2
1
2

 12 (  12 sin 2 )  12 (0  12 sin 0)
 12 
SINE AND 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 AND COSINE INTEGRALS
Find
Example 4
sin
x
dx

4
 We could evaluate this integral using the reduction formula for
∫ sinn x dx (Equation 7 in Section 7.1) together with Example 3.
Example 4
SINE AND COSINE INTEGRALS
However, a better method is to write and use a
half-angle formula:
4
2
2
sin
x
dx

(sin
x
)
dx


 1  cos 2 x 
 
 dx
2


2

1
4
 (1  2 cos 2 x  cos
2
2 x) dx
SINE AND COSINE INTEGRALS
Example 4
As cos2 2x occurs, we must use another half-angle
formula:
cos 2 x  12 (1  cos 4 x)
2
SINE AND COSINE INTEGRALS
Example 4
This gives:
 sin x dx 
1
4
1
1

2
cos
2
x

2 (1  cos 4 x )  dx


1
4


1
4
4

3
2
3
2
 2 cos 2 x  12 cos 4 x  dx
x  sin 2 x  81 sin 4 x   C
SINE AND COSINE INTEGRALS
To summarize, we list guidelines to follow when
evaluating integrals of the form

m
n
sin x cos x 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:
m
2 k 1
m
2
k
sin
x
cos
x
dx

sin
x
(cos
x
)
cos x dx


  sin m x(1  sin 2 x) k cos x dx
 Then, substitute u = sin x.
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:
2 k 1
n
2
k
n
sin
x
cos
x
dx

(sin
x
)
cos
x sin x dx


  (1  cos 2 x) k cos n x sin x dx
 Then, substitute u = cos x.
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  12 (1  cos 2 x)
2
cos x  12 (1  cos 2 x)
2
 Sometimes, it is helpful to use the identity
sin x cos x  12 sin 2 x
TANGENT & SECANT INTEGRALS
We can use a similar strategy to evaluate integrals
of the form

m
n
tan x sec x 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
6
x sec x dx   tan x sec x sec x dx
4
6
2
2
  tan x(1  tan x) sec x dx
6
2
2
  u 6 (1  u 2 ) du   (u 6  u 8 ) du
7
9
u u
  C
7 9
7
9
1
1
 7 tan x  9 tan x  C
TANGENT & SECANT INTEGRALS
Find
Example 6
∫ tan5 θ sec7θ d θ
 If we separate a sec2θ factor, as in the preceding example,
we are left with a sec5θ factor.
 This is not 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.
Example 6
TANGENT & SECANT INTEGRALS
We then evaluate the integral by substituting u =
sec θ, so du = sec θ tan θ dθ:
5
7
4
6
tan

sec


tan

sec
 sec  tan  d


  (sec   1) sec  sec  tan  d
2
2
6
  (u  1) u du   (u  2u  u ) du
2
2
6
10
8
u11
u9 u7

2  C
11
9 7
11
9
7
1
2
1
 11 sec   9 sec   7 sec   C
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:
m
2k
m
2
k 1
2
tan
x
sec
x
dx

tan
x
(sec
x
)
sec
x dx


  tan m x(1  tan 2 x) k 1 sec 2 x dx
 Then, substitute u = tan x.
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:
2 k 1
n
2
k
n 1
tan
x
sec
x
dx

(tan
x
)
sec
x sec x tan x dx


  (sec 2 x  1) k sec n 1 x sec x tan x dx
 Then, substitute u = sec x.
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
a substitution,
 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
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
2
sec x  sec x tan x

dx
sec x  tan x
TANGENT & SECANT INTEGRALS
If we substitute u = sec x + tan x, then
du = (sec x tan x + sec2x) dx.
 The integral becomes: ∫ (1/u) du = ln |u| + C
 Thus, we have:
 sec x dx  ln | sec x  tan x |  C
TANGENT & SECANT INTEGRALS
Example 7
Find  tan 3 x 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 x dx   tan x(sec x  1) dx
2
2
  tan x sec x dx   tan x dx
2
tan 2 x

 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  sec3 x dx
 Here, we integrate by parts with
u  sec x
dv  sec2 x dx
du  sec x tan x dx
v  tan x
Example 8
TANGENT & SECANT INTEGRALS
Then,
 sec
3
x dx  sec x tan x   sec x tan x dx
2
 sec x tan x   sec x(sec 2 x  1) dx
 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
3
x dx
 12 (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.
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
Identity
a
∫ sin mx cos nx dx
sin A cos B
b
∫ sin mx sin nx dx
sin A sin B
c
∫ cos mx cos nx dx
cos A cos B
 12 sin( A  B)  sin( A  B) 
 12  cos( A  B)  cos( A  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 is easier to use the identity in Equation 2(a):
1
sin
4
x
cos
5
x
dx


 2 sin( x)  sin 9 x
 12  ( sin x  sin 9 x) dx
 12 (cos x  91 cos 9 x)  C