Chapter 8
Techniques of Integration
1
8.1
Basic Integration Formulas
2
3
Example 1 Making a simplifying
substitution
u


2x  9
dx  
d ( x 2  9 x)
2
x
 9x  1
x  9x  1
du
d (u  1)
dv


 2v1/ 2  C
u 1
u 1
v
2
 2(u  1)
1/ 2
 C  2  x  9 x  1
2
1/ 2
C
4
Example 2 Completing the square
dx

8x  x

2

dx

16  ( x  4)
d ( x  4)
16  ( x  4)
2

2

du
42  u 2
u
 x4
 sin 1  C  sin 1 
C
4
 4 
5
Example 3 Expanding a power and using a
trigonometric identity
2
(sec
x

tan
x
)
dx

  (sec 2 x  tan 2 x  2sec x tan x)dx.
d
d
2
Racall:tan x  sec x  1; tan x  sec x; sec x  tan x sec x;
dx
dx
2
2
  (2sec 2 x  1  2sec x tan x)dx
 2 tan x   x  2sec x  C
6
Example 4 Eliminating a square root
 /4

1  cos 4xdx 
0
cos 4 x  cos 2(2 x)  2cos 2 (2 x)  1
 /4

 /4
1  cos 4 xdx 
0

0
 /4
2cos 2 2 xdx  2
 | cos 2 x | x
0
 /4
 2
 cos 2 xdx  ...
0
7
Example 5 Reducing an improper fraction
3x 2  7 x
 3x  2 dx
6
  x 3
dx
3x  2
2
  x 3
dx
x  2/3
1 2
2
 x  3x  2ln | x  | C
2
3
8
Example 6 Separating a fraction

 3
3x  2
1 x
x
2
dx
dx  
2
dx
1 x
1 x
1
d ( x2 )
1
2
 3
 2
dx
1  x2
1  x2
3
du
 
 2sin 1 x  C
2 1 u
3
 [2(1  u )1/ 2 ]  2sin 1 x  C ''
2
2
2
du
1/ 2


2(1

u
)
C'
 (1  u)1/ 2
 3 (1  x 2 )  2sin 1 x  C ''
9
Example 7 Integral of y = sec x
 sec xdx  ?
d sec x  sec x tan xdx
d tan x  sec xdx  sec x sec xdx
2
d (sec x  tan x)  sec x(sec x  tan x)dx
d (sec x  tan x)
sec xdx 
sec x  tan x
d (sec x  tan x)
 sec xdx   sec x  tan x  ln | sec x  tan x | C
10
11
12
8.2
Integration by Parts
13
Product rule in integral form
d
d
d
[ f ( x) g ( x)]  g ( x) [ f ( x)]  f ( x) [ g ( x)]
dx
dx
dx
d
d
d
 dx [ f ( x) g ( x)]dx  g ( x) dx [ f ( x)]dx   f ( x) dx [ g ( x)]dx
f ( x) g ( x)   g ( x) f '( x)dx   f ( x) g '( x)dx
Integration by parts formula
14
Alternative form of Eq. (1)

We write
dg  x 
g  x  
 h x    g  x dx   h x dx
dx
 f  x  g   x  dx  f  x  g  x   f   x  g  x  dx
  f  x  h  x  dx  f  x    h  x  dx     f   x    h  x  dx  dx




15
Alternative form of the integration by
parts formula
d
d
d
[ f ( x) g ( x)]  g ( x) [ f ( x)]  f ( x) [ g ( x)]
dx
dx
dx
d
d
d
 dx [ f ( x) g ( x)]dx   g ( x) dx [ f ( x)]dx   f ( x) dx [ g ( x)]dx
f ( x) g ( x)   g ( x)df ( x)   f ( x)dg ( x)
Let u  f ( x); v  g ( x).The above formular is recast into the form
uv   vdu   udv
16
Example 4 Repeated use of integration
by parts
2 x
x
 e dx  ?
17
Example 5 Solving for the unknown
integral
x
e
 cos xdx  ?
18
Evaluating by parts for definite
integrals
or, equivalently
 f  x  h  x  dx  f   x  h  x    f   x   h  x  dx  dx
b
b
b
a
a
a
19
Example 6 Finding area

Find the area of the region in Figure 8.1
20
Solution
4
x
xe
 dx  ...
0
21
Example 9 Using a reduction formula


Evaluate
 cos
3
xdx
Use
u
 cos
n
xdx   cos
n 1
dv
x  cos xdx
cos n 1 x sin x n  1
n2


cos
xdx

n
n
22
8.3
Integration of Rational Functions by
Partial Fractions
23
General description of the method



A rational function f(x)/g(x) can be written as a sum
of partial fractions. To do so:
(a) The degree of f(x) must be less than the degree
of g(x). That is, the fraction must be proper. If it isn’t,
divide f(x) by g(x) and work with the remainder term.
We must know the factors of g(x). In theory, any
polynomial with real coefficients can be written as a
product of real linear factors and real quadratic
factors.
24
Reducibility of a polynomial




1.
2.
A polynomial is said to be reducible if it is the product
of two polynomials of lower degree.
A polynomial is irreducible if it is not the product of
two polynomials of lower degree.
THEOREM (Ayers, Schaum’s series, pg. 305)
Consider a polynomial g(x) of order n ≥ 2 (with leading
coefficient 1). Two possibilities:
g(x) = (x-r) h1(x), where h1(x) is a polynomial of degree
n-1, or
g(x) = (x2+px+q) h2(x), where h2(x) is a polynomial of
degree n-2, and (x2+px+q) is the irreducible quadratic
factor.
25
Example
g ( x)  x 3  4 x  ( x  2)  x( x  2)
linear factor poly. of degree 2
g ( x)  x 3  4 x 
( x 2  4)

irreducible quadratic factor
g ( x)  x 4  9 
( x 2  3)
irreducible quadratic factor
x
poly. of degree 1
 ( x  3)( x  3)
poly. or degree 2
g ( x)  x 3  3x 2  x  3  ( x  1) ( x  2) 2
linear factor poly. or degree 2
26
Quadratic polynomial




A quadratic polynomial (polynomial or order n
= 2) is either reducible or not reducible.
Consider: g(x)= x2+px+q.
If (p2-4q) ≥ 0, g(x) is reducible, i.e.
g(x)
= (x+r1)(x+r2).
If (p2-4q) < 0, g(x) is irreducible.
27

In general, a polynomial of degree n can
always be expressed as the product of
linear factors and irreducible quadratic
factors:
Pn ( x)  ( x  r1 ) n1 ( x  r2 ) n2 ...( x  rl ) nl 
( x 2  p1 x  q1 ) m1 ( x 2  p2 x  q2 ) m2 ...( x 2  pk x  qk ) mk
n  (n1  n2  ...  nl )  2(m1  m2  ...  ml )
28
Integration of rational functions by
partial fractions
29
Example 1 Distinct linear factors
x2  4 x  1
 ( x  1)( x  1)( x  3) dx  ...
x2  4 x  1
A
B
C



 ...
( x  1)( x  1)( x  3) ( x  1) ( x  1) ( x  3)
30
Example 2 A repeated linear factor
6x  7
 ( x  2)2 dx  ...
6x  7
A
B


2
( x  2)
( x  2) ( x  2) 2
31
Example 3 Integrating an improper
fraction
2 x3  4 x 2  x  3
 x 2  2 x  3 dx  ...
2 x3  4 x 2  x  3
5x  3
 2x  2
2
x  2x  3
x  2x  3
5x  3
5x  3
A
B



 ...
2
x  2 x  3 ( x  3)( x  1) ( x  3) ( x  1)
32
Example 4 Integrating with an irreducible
quadratic factor in the denominator
2 x  4
 ( x 2  1)( x  1)2 dx  ...
2 x  4
Ax  B
C
D
 2


 ...
2
2
2
( x  1)( x  1)
( x  1) ( x  1) ( x  1)
33
Example 5 A repeated irreducible
quadratic factor
1
 x( x 2  1)2 dx  ?
1
A Bx  C Dx  E
  2
 2
 ...
2
2
2
x( x  1)
x ( x  1) ( x  1)
34
Other ways to determine the
coefficients


Example 8 Using
differentiation
Find A, B and C in the
equation
x 1
A
B
C



( x  1)3 ( x  1) ( x  1) 2 ( x  1)3
A( x  1) 2  B ( x  1)  C
x 1

( x  1)3
( x  1)3
 A( x  1) 2  B ( x  1)  C  x  1
x  1  C  2
 A( x  1) 2  B ( x  1)  x  1
 A( x  1)  B  1
d
d
[ A( x  1)  B]  (1)  0
dx
dx
A0
B 1
35
Example 9 Assigning numerical values to
x

Find A, B and C in
x2  1
( x  1)( x  2)( x  3)
A
B
C



( x  1) ( x  2) ( x  3)
A( x  2)( x  3)  B( x  1)( x  3)  C ( x  1)( x  2)  f ( x)
 x2  1
f (1)  2 A  12  1  2  A  1
f (2)   B  22  1  5;  B  5
f (3)  2C  32  1  10;  C  5
36
8.4
Trigonometric Integrals
37
38
Example 1 m is odd
 sin
3
x cos x dx  ?
2
 sin x cos x dx   sin x cos x d  cos x 
  (cos x  1)cos x d  cos x 
  (u  1)u du  ...
3
2
2
2
2
2
2
2
39
Example 2 m is even and n is odd
 cos x
5
dx  ?
 cos x dx   cos x cosx dx    cos x  d  sin x  
  (1-sin x) d sin x
5
4
2
2
2
2
u
  (1-u 2 ) 2 du  1+u 4  2u 2 du  ...
40
Example 3 m and n are both even
 cos x sin
2
 cos x sin
2
4
4
x dx  ?
x dx 
2
 1-cos2x  1+cos2x 
 dx
  2 
2


1
2
  1-cos2x 1+cos2x  dx
4
1
  1  cos 2 2 x  cos 2 x  cos 3 2 x  dx  ...
4
41
Example 6 Integrals of powers of tan x
and sec x
3
sec xdx  ?
Use integration by parts.

3
2
sec
xdx

sec
x

sec
xdx ;


u
dv
dv  sec 2 xdx  v   sec 2 xdx  tan x
u  sec x  du  sec x tan xdx
2
sec
x

sec
xdx

u
dv
 sec x tan x   tan x  sec x tan xdx
(tan x  sec x)
dx
tan x  sec x
(sec x tan x  sec 2 x)

dx
tan x  sec x
d (sec x  tan x)

tan x  sec x
 ln | sec x  tan x | C
 sec xdx   sec x
du
 sec x tan x   tan 2 x sec xdx
 sec x tan x   (sec 2 x  1)sec xdx
3
3
sec
xdx

sec
x
tan
x

sec

 xdx   sec xdx...
42
Example 7 Products of sines and
cosines
 cos5 x sin 3xdx  ?
1
cos(m  n) x  cos(m  n) x ;
2
1
sin mx cos nx  sin( m  n) x  sin( m  n) x ;
2
1
cos mx cos nx   cos( m  n) x  cos( m  n) x 
2
sin mx sin nx 
 cos5x sin 3xdx
1
  [sin(2 x)  sin8 x]dx
2
 ...
43
8.5
Trigonometric Substitutions
44
Three basic substitutions
Useful for integrals involving
a2  x2 , a2  x2 , x2  a2
45
Example 1 Using the substitution x=atanq

dx
4 x
2
?
x  2tan y  dx  2sec2 ydy  2(tan 2 y  1)dy


dx
4  4 tan y
2
(tan 2 y  1)
1  tan 2 y

2(tan 2 y  1)
4  4 tan y
2
dy
dy   sec 2 ydy   | sec y | dy
 ln | sec y  tan y | C
46
Example 2 Using the substitution x =
asinq

x 2 dx
9 x
2
?
x  3sin y  dx  3cos y dy

x 2 dx
9 x
 9
2

9sin 2 y  3cos y dy
9  9sin y
2

sin 2 y  cos y dy
1  sin 2 y
 9  sin 2 ydy  ...
47
Example 3 Using the substitution x =
asecq
dx

25 x  4
2
?
2
2
x  sec y  dx  sec y tan y dy
5
5

dx
2 sec y tan y dy 1 sec y tan y dy
 
 
2
2
25 x  4 5
4sec y  4 5
sec 2 y  1
1 sec y tan y dy 1
 
  sec y dy
2
5
5
sec y  1
1
 ln | sec y  tan y | C  ...
5
48
Example 4 Finding the volume of a
solid of revolution
2
V  16 
0
dx
x
2
 4
2
?
49
Solution
2
V  16 
0
dx
x
2
 4
2
?
Let x  2tan y  dx  2sec2 ydy
V 
 /4

0
 2
2sec2 ydy
 tan
2
y  1
2

 /4

0
2sec 2 ydy
 sec y 
2
2
 /4
 cos
2
ydy ...
0
50
8.6
Integral Tables
51
Integral tables is provided at the back of
Thomas’


T-4 A brief tables of integrals
Integration can be evaluated using the tables
of integral.
52
53
54
55
56
57
8.8
Improper Integrals
58
59
Infinite limits of integration
A(a)  lim A(b)  lim 2  2e
b 
b 
b / 2
2
b
A(b)   e x / 2dx  ...  2  2eb / 2
0
60
61
Example 1 Evaluating an improper
integral on [1,∞]

Is the area under the curve y=(ln x)/x2 from
1 to ∞ finite? If so, what is it?
b
ln x
lim  2 dx  ?
b 
x
1
62
Solution
b
b
ln b
ln x dx
ln x
u
1 x x  1 x d (ln x)  ln1 eu du
ln b
ln b
u
u
u
e
du

u
(

e
)

0
 ue
dw
u 0
ln b
w
ln b

e
u
ln b

0
du  ue

;u  ln x, x  eu
(e  u ) du
0
w
u 0
ln b
e
 u ln b
0
0
1
1
  ln b  e  ln b  (e  ln b  1)   ln b   1
b
b
ln x
1 
 1
lim  2 dx  lim   ln b   1  1
b 
b 
x
b 
 b
1
b
63
Example 2 Evaluating an integral on
[-∞,∞]

dx
 1  x2  ?

0
b
dx
dx
dx
 lim 
2
 1  x 2  lim

b 
b  1  x 2
1

x
b
0
b
dx
 2lim 
2
b 
1

x
0
64
Solution
Using the integral table (Eq. 16)
dx
1
1 x
 a2  x2  a tan a  C
b
b
dx
1
1
1
1



tan
x

tan
b

tan
0

tan


 b .
0 1  x 2 
0

dx

1
tan b  2   
 1  x2  2lim
b 
2
b
y
1
y  tan 1 b  b  tan y
lim tan 1 b 
b 

2
65
66
Example 3 Integrands with
vertical asymptotes
67
Example 4 A divergent improper
integral

Investigate the
convergence of
1
dx
0 1  x
68
Solution
1
b
dx
dx
b
  lim  ln | x  1|0
0 1  x  blim
 
1
b 1
1 x
0
  lim  ln | b  1|  ln | 0  1|
b 1
  lim  ln | b  1|  ln | 0  1|  lim ln | b  1|1 
b 1
b 1
 1
 lim ln   
 0
 
69
Example 5 Vertical asymptote at an
interior point
dx
 ( x  1)  ?
3
2/3
0
70
Example 5 Vertical asymptote at an
interior point  dx   dx   dx
3
0
1
( x  1) 2 / 3
1
0
3
( x  1) 2 / 3
1
( x  1) 2 / 3
b
dx
dx
1/ 3 b
 lim 3( x  1)  
2/3
0 ( x  1)2 / 3  blim
0
1  ( x  1)
b 1
0
lim 3(b  1)1/ 3  3( 1)1/ 3   lim 0  3  3;
b 1
3
b 1
3
dx
dx
1/ 3 3
 lim 3( x  1) 
1 ( x  1)2 / 3  clim
c
1  ( x  1) 2 / 3
c 1
c
 lim 3(3  1)1/ 3  3(c  1)1/ 3   3  22 / 3
c 1
3
dx
2/3

3(1

2
)
2/3
( x  1)
0

71
Example 7 Finding the volume of an
infinite solid

The cross section of
the solid in Figure
8.24 perpendicular to
the x-axis are circular
disks with diameters
reaching from the xaxis to the curve y =
ex, -∞ < x < ln 2. Find
the volume of the
horn.
72
Example 7 Finding the volume of an
infinite solid
volume of a slice of disk of thickness dx,diameter y
V
1
V   dV  lim
4 b
0

1
lim
4 b
ln 2
2

y
(
x
)
dx

dV   ( y / 2)2 dx
b
ln 2
2x

e
 dx
b
1
2 x ln 2
 lim  e 
b
8 b
1
 lim  4   e 2b 
8 b
1

2b
  lim (4  e ) 
8 b
2
y
dx
73