Blank B.E., Krantz S.G. Calculus: Single Variable
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⁄ EXAMPLE 6 Show that
Z
sec
3
ðxÞdx 5
1
2
secðxÞtanðxÞ1
1
2
ln
jsecðxÞ1 tanðxÞj
1 C: ð6:1:8Þ
Solution Be
cause sec
2
(x) is the derivative of tan(x), we write sec
3
(x) dx 5 sec(x)
sec
2
(x) dx and set u 5 sec(x)anddv 5 sec
2
(x) dx.Wethenhavedu 5 sec(x)tan(x) dx
and v 5 tan(x). The integration by parts formula gives us
Z
se
c
3
ðxÞdx 5 secðxÞtanðxÞ2
Z
tanðxÞsecðxÞtanðxÞdx 5 secðxÞtanðxÞ2
Z
tan
2
ðxÞsecðxÞdx:
We now use the identity 1 1 tan
2
(x) 5 sec
2
(x), or tan
2
(x) 5 sec
2
(x) 2 1, and sub-
stitute for tan
2
(x) in the integral on the right. With S denoting
R
sec
3
ðxÞdx; we have
S 5 sec ðxÞtanðxÞ2
Z
sec
2
ðxÞ2 1
secðxÞdx
5 secðxÞtanðxÞ2
Z
sec
3
ðxÞ2 secðxÞ
dx
5 secðxÞtanðxÞ2 S 1
Z
secðxÞdx
5 secðxÞtanðxÞ2 S 1 ln
jsecðxÞ1 tanðxÞj
1 C:
After bringing S to the left side (and changing its coefficient from 21to11 in doing
so), we obtain
2S 5 secðxÞtanðxÞ1 ln
jsecðxÞ1 tanðxÞj
1 C;
from which formula (6.1.8) follows.
¥
Reduction Formulas In Example 4, we evaluated the integral
R
r
2
e
22r
dr by applying the integration by
parts formula twice. The first application results in an integral of the form
R
re
22r
dr; which is just like the original integral except that the power of r has been
reduced by one. When we perform this second integration by parts, we reduce it to
R
r
0
e
22r
dr; or
R
e
22r
dr: This last integral is a standard one that does not require
further reduction. In summary, for n 5 2 and n 5 1, applying integration by parts to
R
r
n
e
22r
dr results in a formula involving the integral
R
r
n21
e
22r
dr, which is just like
the original except for the lower power. Such a formula is known as a reduction
formula.
⁄ EX
AMPLE 7 Let a be a nonzero constant. Derive the reduction formula
Z
x
n
e
ax
dx 5
1
a
x
n
e
ax
2
n
a
Z
x
n21
e
ax
dx: ð6:1:9Þ
Solution We
let u 5 x
n
and dv 5 e
ax
dx. Then du 5 nx
n21
dx and v 5 e
ax
/a. The
integration by parts formula gives us
Z
x
n
e
ax
dx 5 x
n
1
a
e
ax
2
Z
1
a
e
ax
nx
n21
dx;
which, after a little rearrangement, is equation (6.1.9).
¥
6.1 Integration by Parts 475
⁄ EXAMPLE 8 Use formula (6.1.9) to evaluate
R
x
3
e
2x
dx:
Solution Note
that a 521. Setting n 5 3 in equation (6.1.9), we obtain
Z
x
3
e
ax
dx 5
1
a
x
3
e
ax
2
3
a
Z
x
2
e
ax
dx: ð6:1:10Þ
We now use equation (6.1.9) with n 5 2 to reduce the integral on the right side
of (6.1.10):
Z
x
3
e
ax
dx 5
1
a
x
3
e
ax
2
3
a
1
a
x
2
e
ax
2
2
a
Z
xe
ax
dx
5
1
a
x
3
e
ax
2
3
a
2
x
2
e
ax
1
6
a
2
Z
xe
ax
dx:
The integral on the right side of the last equation can be reduced by applyin g
equation (6.1.9) with n 5 1:
R
x
3
e
ax
dx 5
1
a
x
3
e
ax
2
3
a
2
x
2
e
ax
1
6
a
2
1
a
xe
ax
2
1
a
Z
x
0
e
ax
dx
5
1
a
x
3
e
ax
2
3
a
2
x
2
e
ax
1
6
a
3
xe
ax
2
6
a
3
Z
e
ax
dx
5
1
a
x
3
e
ax
2
3
a
2
x
2
e
ax
1
6
a
3
xe
ax
2
6
a
4
e
ax
1 C
5
1
a
x
3
2
3
a
2
x
2
1
6
a
3
x 2
6
a
4
e
ax
1 C:
We now set a 521 to obtain
Z
x
3
e
2x
dx 52ðx
3
1 3x
2
1 6x 1 6Þe
2x
1 C: ¥
QUICK QUIZ
1. The integration by parts formula is another way of looking at what formula for
the derivative?
2. An application of integration by parts leads to an equation of the form
Z
x
5
e
23x
dx 5 Ax
5
e
23x
1 B
Z
x
4
e
23x
dx:
What are A and B?
3. An application of integration by parts leads to an equation of the form
Z
x
5
lnðxÞdx 5
1
6
x
6
lnðxÞ2
Z
ψðxÞdx:
What is ψ(x)?
4. Evaluate
R
lnðxÞdx:
Answers
1. Product Rule 2. A 521/3; B 5 5/3
3. x
5
/6 4. x ln(x) 2 x 1 C
476 Chapter 6 Techniques of Integration
EXERCISES
Problems for Practice
c In Exercises 1210, integrate by parts to evaluate the given
indefinite integral. b
1.
R
xe
x
dx
2.
R
xe
2x
dx
3.
R
ð2x 1 5Þe
x=3
dx
4.
R
x sinðxÞdx
5.
R
ð4x 1 2Þsinð2xÞdx
6.
R
9x cosð3xÞdx
7.
R
x lnð4xÞdx
8.
R
lnðxÞx
2
dx
9.
R
9x
2
lnðxÞdx
10.
R
16x
3
lnð2xÞdx
c In Exercises 11220, integrate by parts to evaluate the
given
definite integral. b
11.
R
1
0
xe
x
dx
12.
R
2
0
xe
2x=2
dx
13.
R
π
0
xcosðxÞdx
14.
R
π=4
0
xcosð2xÞdx
15.
R
π=2
0
4xsinðx=3Þdx
16.
R
e
1
lnðxÞdx
17.
R
e
1
x lnðxÞdx
18.
R
1
1=3
x lnð 3xÞdx
19.
Z
4
1
lnðxÞ
ffiffiffi
x
p
dx
20.
R
1
0
x3
x
dx
c In Exercises 21230, integrate by parts successively to
evaluate
the given indefinite integral. b
21.
R
x
2
e
x
dx
22.
R
x
2
e
2x
dx
23.
R
x
2
cosðxÞdx
24.
R
x
2
sinðxÞdx
25.
R
4x
2
cosð2xÞdx
26.
R
27x
2
sinð3xÞdx
27.
R
x
3
e
x
dx
28.
R
x
3
sinðxÞdx
29.
R
16x
3
ln
2
ðxÞdx
30.
R
ln
3
ðxÞdx
c In Exercises 31236, integrate by parts to evaluate the
given
definite integral. b
31.
R
1
0
2arctanðxÞdx
32.
R
1
0
arcsinðxÞdx
33.
R
ffiffi
3
p
0
arcsinðx=2Þdx
34.
R
2
0
arccotðx=2Þdx
35.
R
1
1=2
2arccosðxÞdx
36.
R
2
ffiffi
2
p
2xarcsecðxÞdx
c In Exercises 37240, use reduction formula (6.1.9) to
evaluate
the given integral. b
37.
R
1
0
x
2
e
x=2
dx
38.
R
1
0
4x
2
e
22x
dx
39.
R
1
0
x
3
e
x
dx
40.
R
1
0
8x
3
e
2x
dx
Further Theory and Practice
c In each of Exercises 41244, simplify the integrand before
integrating by parts. b
41.
R
lnð
ffiffi
ffi
x
p
Þdx
42.
R
xlnðx
5
Þdx
43.
R
x
3
lnð1=xÞdx
44.
R
4xsinðxÞcosðxÞdx
c Evaluate each of the integrals in Exercises 45254. b
45.
R
ðx=2
x
Þdx
46.
Z
lnðxÞ
ffiffiffi
x
p
dx
47.
Z
lnðxÞ
x
2
dx
48.
R
9
ffiffiffi
x
p
lnðxÞdx
49.
R
x secðxÞ tanðxÞdx
50.
R
x sec
2
ðxÞdx
51.
R
x cosðx 1 π=4Þdx
52.
R
9 x sinð3x 1 4Þdx
53.
R
2 cos
lnðxÞ
dx
54.
Z
lnðxÞ
x
1
lnðxÞ
x
2
dx
55. Evaluate
R
1
22
xðx13Þ
21=2
dx: For the first step, integrate by
parts with u 5 x.
56. Evaluate
R
2
1
xðx21Þ
1=2
dx: For the first step, integrate by
parts with u 5 x.
c In Exercises 57260, make an appropriate substitution
before
integrating by parts. b
57.
R
sinð
ffiffi
ffi
x
p
Þdx
58.
R
expð
ffiffiffi
x
p
Þdx
59.
R
2x
3
expðx
2
Þdx
6.1 Integration by Parts 477
60.
R
3x
5
sinðx
3
Þdx
61. Evaluate
R
1
0
ðe
x
2 xÞ
2
dx:
62. Evaluate
R
2
1
3ðx 2 2Þ
2
lnðxÞdx:
63. Evaluate
R
x
p
lnðxÞdx:
c In each of Exercises 64269, integrate by parts. This will
result
in an integrand of the form P(x)/Q(x) where P(x) and
Q(x) are polynomials with the degree of P(x) greater than or
equal to the degree of Q(x). Such an integrand is handled by
performing polynomial division to put P(x)/Q(x) into the
form r(x) 1 s(x)/Q(x) where r(x) and s(x) are polynomials
with the degree of s(x) less than the degree of Q(x). b
64.
R
1
0
x arctanðxÞdx
65.
R
1
0
2x arctanðx=
ffiffiffi
3
p
Þdx
66.
R
1
0
6x
2
arctanðxÞdx
67.
R
1
0
lnð1 1 x
2
Þdx
68.
R
1
0
x lnð1 1 xÞdx
69.
R
0
21
9x
2
lnð2 1 xÞdx
70. Suppose that a is a nonzero constant. Calculate
R
sec
3
ðaxÞdx:
71. Derive the reduction formula
Z
ln
n
ðxÞdx 5 x ln
n
ðxÞ2 n
Z
ln
n21
ðxÞdx:
72. Suppose that a is a nonzero constant. Derive the formula
Z
lnðax 1 bÞdx 5
1
a
ðax 1 bÞlnðax 1 bÞ2 x 1 C:
73. Use the formula of Exercise 72 to evaluate
R
lnðx
2
2 1Þdx:
74. Use the formula of Exercise 72 to evaluate
R
lnðx
2
1 7x 1 10Þdx:
c In each of Exercises 75278 a region R is
described.
Calculate its area. b
75. R is
the region that is bounded above by y 5 πx/2 and
below by y 5 arcsin(x) for 0 # x # 1.
76. R is the region that is bounded above by y 5 2 arctan(x)
and below by y 5 πx/2 for 0 # x # 1.
77. R is the region that is bounded above by y 5 cos(πx/2)
and below by y 5 x cos(πx/2) for 21 # x # 1.
78. R is the region that is bounded above by y 5 ln(1 1 x) and
below by y 5 ln(1 1 x
2
) for 0 # x # 1.
79. Let α 5
R
1
0
ð11x
4
Þ
21
dx: Express
R
1
0
lnð1 1 x
4
Þdx in terms
of α. (It is known that α 5
ffiffiffi
2
p
(π 1 2 ln(1 1
ffiffiffi
2
p
))/8, but do
not attempt this evaluation.)
80. Integrate by parts repeatedly, each time with u equal to a
power of 1 2 x, to prove that
Z
1
0
ffiffiffi
x
p
ð1 2 xÞ
4
dx 5
8
3
6
5
4
7
2
9
2
11
:
81. Derive the reduction formula
Z
1
ðx
2
1 a
2
Þ
n11
dx 5
1
2a
2
n
x
ðx
2
1 a
2
Þ
n
1
2n 2 1
2a
2
n
Z
1
ðx
2
1 a
2
Þ
n
dx:
82. Derive formula (6.1.5).
83. Derive formula (6.1.6).
84. Derive formula (6.1.7).
85. Suppose that a and b are constants, not both 0. Derive the
formula
Z
e
ax
cosðbxÞdx 5
e
ax
a cosðbxÞ1 b sinðbxÞ
a
2
1 b
2
1 C:
86. Suppose that a and b are constants, not both 0. Derive the
formula
Z
e
ax
sinðbxÞdx 5
e
ax
a sinðbxÞ2 b cosðbxÞ
a
2
1 b
2
1 C:
87. The Fresnel sine and Fresnel cosine functions, denoted by
FresnelS and FresnelC, are important in the theory of
optics. They are defined by
FresnelSðxÞ5
Z
x
0
sin
π
2
t
2
dt
and
FresnelCðxÞ5
Z
x
0
cos
π
2
t
2
dt:
a. Integrate by parts to calculate
R
FresnelSðxÞ dx: (Your
answer will involve FresnelS(x) and elementary
functions.)
b. Integrate by parts to calculate
R
FresnelCðxÞdx: (Your
answer will involve FresnelC(x) and elementary
functions.)
c. Integrate by parts to calculate
R
2xFresnelSðxÞdx:
(Your answer will involve both Fresnel functions and
elementary functions.)
d. Integrate by parts to calculate
R
2xFresnelCðxÞdx:
(Your answer will involve both Fresnel functions and
elementary functions.)
88. The error function (denoted by erf) is central to the
subjects of probability and statistics. It is defined by
erfðxÞ5
2
ffiffiffi
π
p
Z
x
0
expð2t
2
Þdt:
a. Integrate by parts to calculate
R
erf ð xÞdx: (Your
answer will involve erf (x) and elementary functions.)
b. Integrate by parts to calculate
R
x
2
expð2x
2
Þdx : (Take
u 5 x, dv 5 x exp(2x
2
) dx. Your answer will involve
erf (x) and elementary functions.)
c. Integrate by parts to calculate
R
x
2
erfðxÞdx: (You will
need the result of part b. Your answer will involve
erf (x) and elementary functions.)
478 Chapter 6 Techniques of Integration
89. The sine integral Si (x) is defined by
SiðxÞ5
Z
x
0
sinðtÞ
t
dt:
(The integrand has a continuous extension at 0.)
a. Integrate by parts to calculate
R
SiðxÞdx: (Your answer
will involve Si(x) and elementary functions.)
b. Integrate by parts to calculate
R
xSiðxÞdx: (Your
answer will involve Si(x) and elementary functions.)
Calculator/Computer Exercises
90. Refer to Example 4. For approximately what value ρ is
the probability of finding the hydrogen electron within a
distance ρ α
0
of the nucleus equal to 1/2?
91. Find the area of the region that is bounded above by
y 5 1 1 2x and below by y 5 xe
x
.
92. Find the total area of the regions that are bounded above
and below by y 5 x sin(πx)andy 5 x sin(πx
2
)for0# x # 1.
93. Plot arctan(x)21 and ln(x)/x for 0.7 # x # 5. Find the two
points of intersection. Calculate the area of the region
enclosed by the two curves.
94. Suppose that f (x) is an odd function: f (2x) 52f (x). Let
b
n
5
1
π
Z
π
2π
f ðxÞsinðnxÞdx ðn $ 1Þ:
The function
X
N
n51
b
n
sinðnxÞ
is said to be the order N Fourier sine-polynomial
approximation of f. Using the viewing window [22, 2] 3
[22, 2], plot the order N Fourier sine-polynomial
approximation of f (x) 5 x for N 5 3 and 5.
95. Suppose that f (x) is an even function: f (2x) 5 f (x). Let
a
n
5
1
π
Z
π
2π
f ððxÞcosðnxÞdx ðn $ 0Þ:
The function
a
0
2
1
X
N
n51
a
n
cosðnxÞ
is said to be the order N Fourier cosine-polynomial
approximation of f. Using the viewing window [22, 2] 3
[20.2, 4], plot the order N Fourier cosine-polynomial
approximation of f (x) 5 x
2
for N 5 3 and 5.
96. Let the Fourier coeffcients a
n
(n $ 0) and b
n
(n $ 1) of a
function f be defined as in Exercises 94 and 95. If f is a
differentiable function on the interval (2π, π ), then it can
be shown that the degree N Fourier polynomial
a
0
2
1
X
N
n51
a
n
cosðnxÞ1 b
n
sinðnxÞ
converges to f (x)asN tends to infinity. Calculate the
degrees 2 and 3 Fourier polynomials of f (x) 5 exp(x). Plot
these Fourier polynomials and the function f in the
viewing window [22.6, 2.6] 3 [22.25, 14.25]. (With these
small values of N, you can only begin to see the Fourier
polynomials taking on the shape of f. In Figure 1, we have
plotted f and the approximating Fourier polynomials of
degree 5 and 20.)
6.2 Powers and Products of Trigonometric Functions
This section introduces several procedures for treating integrals involving powers
of sines and cosines. The techniques we use are a combination of trigonometric
identities and substitutions. We will group the problems by type. You will want
to study the technique for each type of problem. There are no new formulas to
memorize in this section. To master this material you must learn how and when
to apply each technique.
y e
x
x
y
15
10
5
2.6
2.6
m Figure 1a The order 5 Fourier polynomial of
the exponential function.
y e
x
x
y
10
5
2.62.6
m Figure 1b The order 20 Fourier polynomial of
the exponential function.
6.2 Powers and Products of Trigonometric Functions 479
Squares of Sine,
Cosine, Secant, and
Tangent
In this subsection, you will learn how to integrate the expressions sin
2
(θ) and
cos
2
(θ) by using the eq uations
sin
2
ðθÞ5
1 2 cosð2θÞ
2
ð6:2:1Þ
and
cos
2
ðθÞ5
1 1 cosð2θÞ
2
; ð6:2:2Þ
which are obtained by replacing θ with 2θ in Half-Angle Formulas (1.6.10) and
(1.6.11). Notice that the right sides of identities (6.2.1) and (6.2.2) differ only in the
sign that precedes the term cos(2θ). In particular, formula (6.2.1) for sin
2
(θ)
involves cos(2θ) and not sin(2θ).
⁄ EX
AMPLE 1 Show that
Z
sin
2
ðxÞdx 5
1
2
x 2
1
4
sinð2xÞ1 C ð6:2:3Þ
and
Z
cos
2
ðxÞdx 5
1
2
x 1
1
4
sinð2xÞ1 C: ð6:2:4Þ
Solution Identiti
es (6.2.1) and (6.2.2) are used to derive formulas (6.2.3) and
(6.2.4), respectively. Let us illustrate with the second formula:
R
cos
2
ðxÞdx 5
Z
1 1 cos ð2xÞ
2
dx
5
Z
1
2
dx 1
Z
cosð2xÞ
2
dx
5
1
2
x 1
1
4
sinð2xÞ1 C: ¥
INSIGHT
In the next section, you will find that it is sometimes best to express the
antiderivatives of sin
2
(x) and cos
2
(x) in terms of trigonometric functions of the argument
x (instead of 2x). Using the identity sin(2x) 5 2 sin(x) cos(x), we may rewrite formulas
(6.2.3) and (6.2.4) as
Z
sin
2
ðxÞdx 5
1
2
x 2 sinðxÞcosðxÞ
1 C ð6:2:5Þ
and
Z
cos
2
ðxÞdx 5
1
2
x 1 sinðxÞcosðxÞ
1 C: ð6:2:6Þ
We conclude our discussion of sin
2
(x) and cos
2
(x) by using equations (6.2.5)
and (6.2.6) to evaluate some definite integrals that arise frequently:
Z
π=2
0
cos
2
ðxÞdx 5
Z
π=2
0
sin
2
ðxÞdx 5
π
4
; ð6:2:7Þ
480 Chapter 6 Techniques of Integration
Z
π
0
cos
2
ðxÞdx 5
Z
π
0
sin
2
ðxÞdx 5
π
2
; ð6:2:8Þ
and
Z
2π
0
cos
2
ðxÞdx 5
Z
2π
0
sin
2
ðxÞdx 5 π: ð6:2:9Þ
We now turn our attention to
R
sec
2
ðxÞdx and
R
tan
2
ðxÞdx: The first of these is
a basic indefinite integral that we already know:
R
sec
2
ðxÞdx 5 tanðxÞ1 C: In the
next example, we will see how this formula can be used to obtain an antiderivative
of tan
2
(x).
⁄ EX
AMPLE 2 Show that
Z
tan
2
ðxÞdx 5 tanðxÞ2 x 1 C: ð6:2:10Þ
Solution As
in Example 6 of Section 1, we use the identity tan
2
(x) 5 sec
2
(x)21.
We have
Z
tan
2
ðxÞdx 5
Z
sec
2
ðxÞ2 1
dx 5
Z
sec
2
ðxÞdx 1
Z
ð21Þdx 5 tanðxÞ2 x 1 C:
Higher Powers of Sine,
Cosine, Secant, and
Tangent
We now know how to evaluate
R
f
n
ðxÞdx where f (x) is one of the trigonometric
functions sin(x), cos(x), sec (x), tan(x), and n 5 0, 1, or 2. If n is a higher integer
power, then we can use a reduction formula, repeatedly if necessary, to reduce the
calculation to one we already know. Here, for future reference, are the reduction
formulas we need:
Z
sin
n
ðxÞdx 52
1
n
sin
n21
ðxÞcosðxÞ1
n 2 1
n
Z
sin
n22
ðxÞdx ðn 6¼ 0Þð6:2:11Þ
Z
cos
n
ðxÞdx 5
1
n
sinðxÞcos
n21
ðxÞ1
n 2 1
n
Z
cos
n22
ðxÞdx ðn 6¼ 0Þð6:2:12Þ
Z
sec
n
ðxÞdx 5
1
n 2 1
sec
n22
ðxÞtanðxÞ1
n 2 2
n 2 1
Z
sec
n22
ðxÞdx ðn 6¼ 1Þð6:2:13Þ
Z
tan
n
ðxÞdx 5
1
n 2 1
tan
n21
ðxÞ2
Z
tan
n22
ðxÞdx ðn 6¼ 1Þ: ð6:2:14Þ
⁄ EX
AMPLE 3 Derive reduction formula (6.2.11).
Solution Let I
n
5
R
sin
n
ðxÞdx denote the left side of formula (6.2.11). We apply the
integration by parts formula as follows:
I
n
5
Z
sin
n21
ðxÞ
|fflfflfflfflffl{zfflfflfflfflffl}
u
sinðxÞdx
|fflfflfflfflffl{zfflfflfflfflffl}
dv
5 sin
n21
ðxÞ
|fflfflfflfflffl{zfflfflfflfflffl}
u
2cosðxÞ
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
v
2
Z
2cosðxÞ
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
v
ðn 2 1Þsin
n22
ðxÞcosðxÞdx
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
du
:
This equation simplifies to
I
n
52sin
ðn21Þ
ðxÞcosðxÞ1 ðn 2 1Þ
Z
sin
ðn22Þ
cos
2
ðxÞdx:
6.2 Powers and Products of Trigonometric Functions 481
We replace cos
2
(x) with 1 2 sin
2
(x) and expand the integral:
I
n
52sin
ðn21Þ
ðxÞcosðxÞþðn 2 1Þ
Z
sin
ðn22Þ
ð1 2 sin
2
ðxÞÞdx
52sin
ðn21Þ
ðxÞcosðxÞþðn 2 1Þ
Z
sin
ðn22Þ
dx 2 ðn 2 1Þ
Z
sin
n
ðxÞdx
52sin
ðn21Þ
ðxÞcosðxÞþðn 2 1Þ
Z
sin
ðn22Þ
dx 2 ðn 2 1ÞI
n
:
We now bring the unknown I
n
that appears on the right side of the equation over to
the left side. We obtain
I
n
1 ðn 2 1ÞI
n
52sin
ðn21Þ
ðxÞcosðxÞ1 ðn 2 1Þ
Z
sin
ðn22Þ
dx:
Because the left side of this equation simplifies to n I
n
, dividing through by n
results in formula (6.2.11).
b
In practice, reduction formulas (6.2.11) through (6.2.14) are used successively
until the power of the trigonometric function has been reduced to 0 or 1.
⁄ EX
AMPLE 4 Evaluate
R
cos
6
ðxÞdx:
Solution After
one application of formula (6.2.12), we obtain
Z
cos
6
ðxÞdx 5
1
6
sinðxÞcos
5
ðxÞ1
5
6
Z
cos
4
ðxÞdx:
When we apply formula (6.2.12) to
R
cos
4
ðxÞdx, we obtain
Z
cos
6
ðxÞdx 5
1
6
sinðxÞcos
5
ðxÞ1
5
6
1
4
sinðxÞcos
3
ðxÞ1
3
4
Z
cos
2
ðxÞdx
or
Z
cos
6
ðxÞdx 5
1
6
sinðxÞcos
5
ðxÞ1
5
24
sinðxÞcos
3
ðxÞ1
5
8
Z
cos
2
ðxÞdx:
At this point, we can exploit equation (6.2.6). Alternatively, reduction formula
(6.2.12) can be applied to
R
cos
2
ðxÞdx; resulting in
Z
cos
2
ðxÞdx 5
1
2
sinðxÞcosðxÞ1
1
2
Z
cos
0
ðxÞdx 5
1
2
sinðxÞcosðxÞ1
1
2
x 1 C:
Thus
Z
cos
6
ðxÞdx 5
1
6
sinðxÞcos
5
ðxÞ1
5
24
sinðxÞcos
3
ðxÞ1
5
16
sinðxÞcosðxÞ1
5
16
x 1 C: ¥
Odd Powers of Sine
and Cosine
Although reduction formulas (6.2.11) and (6.2.12) can be used to evaluate
R
sin
n
ðxÞdx and
R
cos
n
ðxÞdx when n is any positive integer, there is a simpler, less
tedious method available when n is odd. We will use the identities
cos
2
ðxÞ5 1 2 sin
2
ðxÞ and sin
2
ðxÞ5 1 2 cos
2
ðxÞ:
482 Chapter 6 Techniques of Integration
Suppose that k and ‘ are positive integers. We integrat e an odd power sin
2k11
(x)of
sin(x) by writing
sin
2k11
ðxÞ5
sin
2
ðxÞ
k
sinðxÞ5
12cos
2
ðxÞ
k
sinðxÞ: ð6:2:15Þ
We integrate an odd power cos
2‘11
(x) of cos(x) by writing
cos
2‘11
ðxÞ5
cos
2
ðxÞ
‘
cosðxÞ5
12sin
2
ðxÞ
‘
cosðxÞ: ð6:2:16Þ
Let us see how these formulas help us in practice:
⁄ EX
AMPLE 5 Use formula (6.2.16) to calculate
R
cos
3
ðxÞdx:
Solution Formul
a (6.2.16) with ‘ 5 1 tells us that
Z
cos
3
ðxÞdx 5
Z
ðcos
2
ðxÞÞ cosðxÞdx 5
Z
ð1 2 sin
2
ðxÞÞ cosðxÞdx:
We make the substitution u 5 sin(x ), du 5 cos(x ) dx, which results in
Z
ð1 2 u
2
Þdu 5 u 2
1
3
u
3
1 C:
Resubstituting the x variable, we have
Z
cos
3
ðxÞdx 5 sinðxÞ2
1
3
sin
3
ðxÞ1 C: ¥
Here is a slightly more complicated example:
⁄ EX
AMPLE 6 Calculate
R
sin
5
ðxÞdx:
Solution Using
formula (6.2.15) wi th k 5 2, we write
Z
sin
5
ðxÞdx 5
Z
sin
2
ðxÞ
2
sinðxÞdx 5
Z
1 2 cos
2
ðxÞ
2
sinðxÞdx:
The substitution u 5 cos(x), du52sin(x) dx results in
2
Z
1 2 u
2
2
du 5
Z
21 1 2u
2
2 u
4
du 52u 1
2
3
u
3
2
1
5
u
5
1 C:
Resubstituting the expressions in x yields the final solution:
Z
sin
5
ðxÞdx 52cosðxÞ1
2
3
cos
3
ðxÞ2
1
5
cos
5
ðxÞ1 C: ¥
Integrals That Involve
Both Sine and Cosine
In the preceding subsections, we saw how to integrate positive powers of sin(x)and
cos(x). Next we consider integrals
Z
sin
m
ðxÞcos
n
ðxÞdx
that involve posit ive integer powers of both sin(x) and cos(x). There are two
possibilities. Either at least one of the numbers m and n is odd, or both m and n are
even. Here is what we do in each of these circumstances:
6.2 Powers and Products of Trigonometric Functions 483
If at least one of m or n is odd, then we apply the odd-power technique that we
discussed in the preceding subsection. (If both are odd, then we apply the
technique to the smaller of the two.)
If both m and n are even, then we use the identity cos
2
(x) 1 sin
2
(x) 5 1to
convert the integrand to a sum of even powers of sine or of cosine. Then we
apply the appropriate reduction formula.
Our next two exampl es illustrate these techniques.
⁄ EX
AMPLE 7 Evaluate the integral
R
cos
3
ðxÞsin
4
ðxÞdx:
Solution Because
cosine is raised to an odd power, we apply formula (6.2.16).
We have
Z
cos
3
ðxÞsin
4
ðxÞdx 5
Z
sin
4
ðxÞcos
2
ðxÞcosðxÞdx 5
Z
sin
4
ðxÞ
1 2 sin
2
ðxÞ
cosðxÞdx:
The substitution u 5 sin(x), du 5 cos(x) dx converts the last integral to
Z
u
4
ð1 2 u
2
Þdu 5
Z
ðu
4
2 u
6
Þdu 5
1
5
u
5
2
1
7
u
7
1 C:
Therefore after resubstituting the x variable, we have
Z
cos
3
ðxÞsin
4
ðxÞdx 5
1
5
sin
5
ðxÞ2
1
7
sin
7
ðxÞ1 C: ¥
⁄ EXAMPLE 8 Evaluate
R
sin
4
ðxÞcos
6
ðxÞdx:
Solution Here
both powers are even. For the first step, it is more efficient to work
with the smaller power. We write
Z
sin
4
ðxÞcos
6
ðxÞdx 5
Z
sin
2
ðxÞ
2
cos
6
ðxÞdx
5
Z
12cos
2
ðxÞ
2
cos
6
ðxÞdx
5
Z
1 2 2cos
2
ðxÞþcos
4
ðxÞ
cos
6
ðxÞdx
5
Z
cos
6
ðxÞdx 2 2
Z
cos
8
ðxÞdx þ
Z
cos
10
ðxÞdx:
ð6:2:17Þ
We can apply reduction formula (6.2.12) to each of the three integrals in the last
line, but here it is most efficient to work with the highest power of cosine. Thus
Z
cos
10
ðxÞdx 5
1
10
sinðxÞcos
9
ðxÞ1
9
10
Z
cos
8
ðxÞdx ð6:2:18Þ
and, by substituting equation (6.2.18) into the right side of (6.2.17), we obtain
R
sin
4
ðxÞcos
6
ðxÞdx 5
Z
cos
6
ðxÞdx 2 2
Z
cos
8
ðxÞdx þ
1
10
sinðxÞcos
9
ðxÞþ
9
10
Z
cos
8
ðxÞdx
5
1
10
sinðxÞcos
9
ðxÞþ
Z
cos
6
ðxÞdx þ
22 þ
9
10
Z
cos
8
ðxÞdx
5
1
10
sinðxÞcos
9
ðxÞþ
Z
cos
6
ðxÞdx 2
11
10
Z
cos
8
ðxÞdx: ð6:2:19Þ
484 Chapter 6 Techniques of Integration