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Chapter 7
Techniques of Integration
7.1 Integration by parts
Question:

How to integrate


x ln  xdx, x sin  xdx, ex sin  xdx , where the
integrands are the product of two kinds of functions?
Every differentiation rule has a corresponding
integration rule:
Differentiation
Integration
the Chain Rule
the Substitution Rule
the Product Rule
the Rule for Integration by Parts
The formula for integration by parts
(1)
 udv  uv  vdu
Let u = f (x) and v = g(x) are both differentiable,
then du= f’(x) dx and dv= g’(x) dx
(2)
 f ( x) g( x)dx  f ( x) g ( x)  f ( x) g ( x)dx
Example 1. Find

x cos xdx
Example 2. Evaluate  ln xdx
Example 3. Find
2 x
x
 e dx
x
e
Example 4. Evaluate  sin xdx
The formula for definite integration by parts
(3)
b
b
a f ( x) g( x)dx  f ( x) g ( x)  a f ( x) g ( x)dx
Example 5. Calculate
1
0
tan 1 xdx
Example 6. Prove the reduction formula
1
n 1
n 1
n2
sin
xdx


cos
x
sin
x

sin
xdx


n
n
n
where
n2
is an integer.
Summarize
•When the integrands are the product of two kinds of
functions and neither of them is derivative of the other,
we use the integration by parts. We can compare:
the integratio n by parts
 f ( x) g ( x)dx  f ( x) g ( x)  f ( x) g ( x)dx
the substituti on rule


f ( g ( x)) g ( x)dx  f ( g ( x))dg ( x) 
 f (t )dt

•First we recognize u and v, then confirm udv to be
more easily integrated than vdu . For example

x ln  xdx,


x sin  xdx,

ex sin  xdx
7.2 Trigonometric integrals
Question 1: How to evaluate  sin m x cos n xdx ？
(a) If the power of cosine is odd ( n=2k+1 ).
• Save one cosine factor to cos xdx  d sin x
• Use cos 2 x  1  sin 2 x to express the remaining factors
in terms of sine:
 sin
m
2 k 1
x cos
xdx   sin x(cos x) cos xdx
m
2
k
  sin x(1  sin x) d sin x
m
•Then substitute u=sinx.
2
k
(b) If the power of sine is odd ( m=2k+1 ).
• Save one sine factor to sin xdx  d cos x
• Use sin 2 x  1  cos 2 x to express the remaining
factors in terms of sine:
2 k 1
n
2
k
n
sin
x
cos
xdx

(sin
x
)
cos
x sin xdx


  (1  cos2 x) k cosn xd cos x
•Then substitute u=cosx.
(c) If the powers of both sine and cosine are even,
use the half-angle identities
1
sin x  (1  cos2 x)
2
2
1
cos x  (1  cos2 x)
2
2
It is sometimes helpful to use the identity
sin x cos x  12 sin 2 x
5
2
sin
x
cos
xdx

Example 1. Evaluate
Example 2. Evaluate
 4

0
Example 3. Find
 sin
4
4
2
sin x cos xdx
xdx
Question 2: How to evaluate  tan m x sec n xdx ？
(a) If the power of secant is even ( n=2k).
2
2
sec
x
to
sec
xdx  d tan x
• Save a factor of
• Use sec x  1  tan x to express the remaining
factors in terms of tanx:
2
2
m
2k
m
2
k 1
2
tan
x
sec
xdx

tan
x
(sec
x
)
sec
xdx


  tan m x(1  tan 2 x) k 1 d tan x
•Then substitute u=tanx.
6
4
Example 4. Find  tan x sec xdx
(b) If the power of tangent is odd ( m=2k+1).
•Save a factor of
sec x tan x to sec x tan xdx  d sec x
2
2
tan
x

sec
x  1 to express the remaining
•Use
factors in terms of secx:
2 k 1
n
2
k
n 1
tan
x
sec
xdx

(tan
x
)
sec
x sec x tan xdx


  (sec 2 x  1) k sec n 1 xd sec x
•Then substitute u=secx.
Example 5. Find
5
7
tan
x
sec
xdx

(c) If n = 0, only tanx occurs. Use tan x  sec x  1
and, if necessary, the formula
2
2
 tan xdx  ln sec x  C
Example 6. Find  tan 3 xdx
(d) If n is odd and m is even, we express the integrand
completely in term of secx.
Power of secx may require integration by parts.
Example 7. Find
 sec xdx
Example 8. Find
3
sec
 xdx
Question 3: How to evaluate  sin mxcos nxdx ？
 sin mx cosnxdx ,
To evaluate the integrals (a)
(b)
 sin mxsin nxdx,
(c)
 cosmx cosnxdx
use the corresponding identity:
1
sin( A  B)  sin( A  B)
2
1
(b) sin A sin B  cos(A  B)  cos(A  B)
2
1
(c) cos A cos B  cos(A  B)  cos(A  B)
2
(a) sin A cos B 
Example 9. Evaluate  sin 4 x cos5xdx
7.3 Trigonometric substitution
•How to find the area of a circle or an ellipse?
•How to integrate

a 2  x2 dx,

a2  x2 dx and

x2  a 2 dx ?
In general we can make a substitution of the form x=g(t) by
using the Substitution Rule in reverse(called inverse
substitution):
Assume that g has an inverse function, that is, g
is one-to-one, we obtain
x  g (t )
 f ( x)dx 
t  g 1 ( x )
1

f
(
g
(
t
))
g
(
t
)
dt

F
(
t
)

F
(
g
( x))

One kind of inverse substitution is trigonometric
substitution.
Table of trigonometric substitution
Expression
a2  x2
a2  x2
x2  a2
Substitution



2
2


x  a tan ,    
2
2

x  a sec , 0    or
2
3

2
x  a sin , 
Identity
1  sin 2   cos2 
1  tan 2   sec2 
sec2   1  tan 2 
Example 1. Evaluate
Example 2. Find

Example 3. Evaluate
Example 4. Evaluate
9  x2
dx
2
x

1
x2 x2  4


dx
dx
x a
2
2
, where a  0.
x
3  2x  x
2
dx
Inverse Substitution Formula For Definite Integral
Let x=g( t ) and g has an inverse function, we have

b
a

f ( x)dx   f ( g (t )) g (t )dt

where
  g 1 (a)
Example 5. Find

a
0
  g 1 (b)
a 2  x 2 dx (a  0)
Example 6. Find the area enclosed by the ellipse
x2 y 2
 2 1
2
a
b
3 3 2
Example 7. Find 0
x3
dx
2
32
(4 x  9)
7.4 Integration of rational functions
by partial fraction
In this section we show how to integrate any rational
function (a ratio of polynomials) by expressing it as a
sum of simpler fraction(called partial fraction).
Consider a rational function
P( x)
f ( x) 
Q( x )
where P and Q are polynomials. If
P( x)  an x n  an1x n1    a1x  a0
where an  0 ,then the degree of P is n and we write
deg(P) = n
We can integrate rational functions according to 3 steps:
Step 1. First express f as a sum of simpler fractions.
Provide that the degree of P is less than the degree
of Q. Such a rational function is called proper.
If f is improper, that is, deg(P)  deg(Q), we can
divide Q into P by long division until a remainder R(x) is
obtained such that deg(R)<deg(Q). The division
statement is
P( x)
R( x)
f ( x) 
 S ( x) 
(1)
Q( x)
Q( x)
where S and R are also polynomials.
Step 2. Second factor the denominator Q(x) as far as
possible. It can be shown that any polynomial Q can be
factored as a product of linear factors(of the form ax+b)
and irreducible quadratic factors (of the form
ax2  bx  c, where b2  4ac  0).
Step 3. Finally express the proper rational function
R(x)/Q(x) as a sum of partial fractions of the form
A
(ax  b)i
or
Ax  B
(ax2  bx  c) j
A theorem in algebra guarantees that it is always possible
to do it.
We explain the details for the 4 cases that occur.
Case 1. The denominator Q(x) is a product of
distinct linear factors.
Q( x)  (a1x  b1 )( a2 x  b2 )(ak x  bk )
The partial fraction theorem states there exist constants
A1, A2 , Ak
(2)
such that
R( x)
A1
A2
Ak



Q( x) a1x  b1 a2 x  b2
ak x  bk
These constants need to be determined.
x2  2x 1
Example 1. Evaluate  3
dx.
2
2 x  3x  2 x
Example 2. Find
dx
 2 2 , where a  0.
x a
Case 2. Q(x) is a product of linear factors, some of
which are repeated.
Suppose the first linear factor (a1 x  b1 ) is repeated r
times; that is, (a1 x  b1 ) r occurs in the factorization of
Q(x). Then instead of the single term A1 (a1 x  b1 )
in Equation 2, we would use
(3)
A1
A2
Ar


2
a1 x  b1 (a1 x  b1 )
(a1 x  b1 ) r
x4  2x2  4x  1
dx.
Example 3. Find  3
2
x  x  x 1
Case 3. Q(x) contains irreducible quadratic factors,
none of which is repeated.
If Q(x) has the factor (ax2  bx  c), where b2  4ac  0,
then, in addition to the partial fraction in Equation 2 and 3,
the expression for R(x)/Q(x) will have a term of the form
(4)
Ax  B
ax2  bx  c
where A and B are constants to be determined. We can
integrate (4) by completing the square and using the
formula
dx
1 1 x
 x 2  a 2  a tan ( a )  C
2 x2  x  4
dx
Example 3. Find  3
x  4x
(5)
4 x 2  3x  2
dx
Example 4. Evaluate  2
4x  4x  3
Case 4. Q(x) contains a repeated irreducible
quadratic factors.
2
r
2
(
ax

bx

c
)
,
where
b
 4ac  0,
If Q(x) has the factor
then instead of the single partial fraction(4), the sum
(6)
A1x  B1
A2 x  B2
Ar x  Br


2
2
2
ax  bx  c (ax  bx  c)
(ax2  bx  c) r
occurs in the partial fraction decomposition of R(x)/Q(x).
Each of the term in (6) can be integrated by completed the
square and making a tangent substitution.
1  3x  2 x 2  x 3
Example 5. Evaluate 
dx
2
2
x( x  1)
7.5 Rationalizing substitutions
By means of appropriate substitutions, some functions
can be changed into rational functions. In particular, when
an integrand contains an expression of the form n g ( x) ,
then the substitution u = n g ( x) may be effective.
Example 1. Evaluate 
Let
Example 2. Find
Let
x4
dx
x
u x4

u6 x
dx
x 3 x
The substitution t = tan(x/2) will convert any rational
function of sinx and cosx into an ordinary rational
function. This is called Weierstrass substitution.
Let
Then
 x
t  tan  
 x
2
1
1
1
 x
cos  


2
 2  sec x 
1

t
2 x 
 
1  tan  
2
 
2
t
 x
 x  x
sin    cos  tan   
2
2 2
1 t2
Therefore
t
1
2t
 x  x
sin x  2 sin   cos   2

2
2
2
1

t
2 2
1 t 1 t
2
x
x
1

t
 
 
cos x  cos2    sin 2   
2
2
 2  1 t
Since t = tan(x/2), we have x  2 tan 1 t
, so
2
dx 
dt
2
1 t
Thus if we make the substitution t = tan(x/2), then we
have
(1) sin x  2t
1 t2
1 t2
cos x 
1 t2
1
dx
Example 3. Find 
3sin x  4 cos x
2
dx 
dt
2
1 t
7.6 Strategy For Integration
Integration is more challenging than differentiation.
In finding the derivative of a function it is obvious which
differentiation formula we should apply. But when
integrating a given function, it may not be obvious which
techniques we should use.
First it is useful to be familiar with the basic integration
formulas.
Table of integration formulas
Constants of integration have been omitted.
n1
x
1.  x n dx 
(n  1)
n 1
1
2.  dx  ln x
x
3.
e
x
dx  e
x
 sin xdx   cos x
7.  sec2 xdx  tan x
9.  sec x tan xdx  sec x
11.  sec xdx  ln sec x  tan x
13.  tan xdx  ln sec x
15.  sinh xdx  cosh x
5.
ax
4.  a dx 
ln a
6.  cos xdx  sin x
x
8.
2
csc
xdx   cot x

10.  csc x cot xdx   csc x
12.  csc xdx  ln csc x  cot x
14.  cot xdx  ln sin x
16.  cosh xdx  sinh x
dx
1 1 x
17.  2
 tan ( )
2
a
a
x a
18.

dx
1
xa
19.  2
 ln
2
2a x  a
x a
20.

dx
x
 sin ( )
2
2
a
a x
dx
 ln x  x 2  a 2
x2  a2
1
Secondly if you do not immediately see how to attack a
given integral, you might try the following four-step
strategy.
1. Simplify the integrand if possible.
2. Look for an obvious substitution.
3. Classify the integrand according to its form.
4. Try again.
(a) Try substitution.
(b) Try parts.
(c) Manipulate the integrand.
(d) Relate the problem to previous problems.
(e) Use several methods.
Example 1.
tan 3 x
 cos3 x dx
Example 2.
x
e
 dx
Example 3.
x5  1
 x3  3x2  10 x dx
Example 4.
dx
 x ln x
Example 5.
1 x
dx
1 x

7.7 Using Tables of Integrals and
Computer Algebra Systems
7.8 Approximation Integration
•How to integrate
x2
 e dx or
1
1

1  x 3 dx ?
It is difficult, or even impossible, to find an
antiderivative.
•When the function is determined from a scientific
experiment through instrument readings, how to
integrate such discrete function?
In both cases we need to find approximate values of
definite integrals.
Using Riemann sums
The left endpoint approximation
(1)
n
 f ( x)dx  Ln   f ( xi 1 )x
i 1
The right endpoint approximation
b
a
(2)
n
 f ( x)dx  Rn   f ( xi )x
i 1
(3) Midpoint rule
b
a
 f ( x)dx  M n  x[ f ( x1 )  f ( x2 )    f ( xn )]
b
a
where
and
x  b  a
n
1
xi  ( xi 1  xi )  midpoit of
2
[ xi 1 , xi ]
(4) Trapezoidal rule
x
b
a f ( x)dx  Tn  [ f ( x0 )  2 f ( x1 )  2 f ( x2 )  
2
 2 f ( xn 1 )  f ( xn )]
ba
where
x 
and
xi  a  ix
n
Example 1. Use (a) the Trapezoidal Rule
(b) the Midpoint Rule with n=5
to approximate the integral 12 (1 x)dx.
Notice
.
 f ( x)dx  approximation  error
b
a
The error in using an approximation is defined to be the
amount that needs to be added to the approximation to
make it exact.
In general, we have
ET   f ( x)dx  Tn
b
a
(5) Error bounds
and
EM   f ( x)dx  TM
b
a
Suppose f ( x)  K for a  x  b. If
ET and EM are the errors in the Trapezoidal and
Midpoint Rules, then
K (b  a)3
K (b  a)3
ET 
and
EM 
2
2
12n
24n
Example 2. (a) Use the Midpoint Rule with n=10 to
1 x
approximate the integral 0 e dx.
2
(b) Give an upper bound for the error
involved in this approximation.
(6) Simpson’s Rule
x
 f ( x)dx  S n  [ f ( x0 )  4 f ( x1 )  2 f ( x2 )  4 f ( x3 )  
3
 2 f ( xn  2 )  4 f ( xn 1 )  f ( xn )]
b
a
where n is even and
x  (b  a) n .
Example 3. Use Simpson’s Rule with n=10 to
approximation 2 (1 x)dx.
1
(7) Error bound for Simpson’s Rule Suppose
that
If
ES
then
f ( 4) ( x)  K for a  x  b.
is the error involved in using Simpson’s Rule,
K (b  a)3
ES 
4
180n
Example 4. (a) Use Simpson’s Rule with n=10 to
approximate the integral 1 e x dx.
0
2
(b) Estimate the error involved in this approximation.
7.9 Improper Integrals
In defining a definite integral ab f ( x)dx we deal with
(1) the function f defined on a finite interval [a, b];
(2) f is a bounded function.
Question: How to integrate a definite integral when
the interval is infinite or when f is unbounded ?
Type 1 Infinite Intervals
y  1 x2
Consider the infinite region that lies under the
curve
, above the x-axis and from the line
x=1 to infinite, can this area A be infinite?
Notice
A(t )  
t
1
and
t
1
1
1
dx    1 
2
x
x1
t
1
lim A(t )  lim (1  )  1
t 
t 
t
So the area of the infinite region is equal to 1 and
we write

1
1
t 1
dx  lim 1 2 dx  1
2
t  x
x
(1)Definition of An Improper Integral of Type 1
(a) If t f ( x)dx exists for every number t  a , then
a

t
a f ( x)dx  lim a f ( x)dx
t 
provide this limit exists(as a finite number).
(b) If b f ( x)dx exists for every number t  b , then
t
b

f ( x)dx  lim
t 
b
t
f ( x)dx
provide this limit exists(as a finite number).
The improper integrals in (a) and (b) are called convergent
if the limit exists and divergent if the limit does not exist.
(c) If both   f ( x)dx and  a f ( x)dx are convergent,
a

then we define

a

 f ( x)dx   f ( x)dx  a f ( x)dx
Example 1. Determine whether the integral
1
1 dx
is convergent or divergent.
x
0
x
Example 2. Evaluate  xe dx.

Example 3. Evaluate


1
dx
2
1 x
Example 4. For what value of p is the integral

1
1
dx
p
x
convergent?
(2)

1
1
dx is convergent if p >1 and divergent if p  1.
p
x
Type 2 Discontinuous Integrands
(3) Definition of An Improper Integral of Type 2
(a) If f is continuous on [a,b) and lim f ( x)   , then
b
a
f ( x)dx  lim
t b

t b
t
a
f ( x)dx
if this limit exists(as a finite number).
(b) If f is continuous on (a,b] and lim f ( x)   , then
b
a
f ( x)dx  lim
t a

t a
b
t
f ( x)dx
if this limit exists(as a finite number).
The improper integrals in (a) and (b) are called convergent
if the limit exists and divergent if the limit does not exist.
(c) If lim f ( x)   , where a<c<b, and both
t c
c
a
f ( x)dx and cb f ( x)dx are convergent, then we define

b
a
c
b
a
c
f ( x)dx   f ( x)dx   f ( x)dx
t
b
 lim  f ( x)dx  lim  f ( x)dx
t c
Example 5. Find
5
2
t c
a
t
1
dx.
x2
Example 6. Determine whether 1 2 sec x dx converges or
diverges.
3 dx
Example 7. Evaluate 0
x

1
1
Example 8. Find 0 ln xdx.
if possible.
A Comparison Test For Improper Integrals
(4)Comparison Theorem Suppose that f and g are
continuous functions with f ( x)  g ( x)  0

a
f ( x)dx
(a) If
convergent.

a g ( x)dx
(b) If
divergent.
for
x  a.
is convergent, then

a g ( x)dx
is divergent, then

a
Example 9. Show that (a)
(b)
  x2
0 e dx
x
1

e

dx
1
x
f ( x)dx is
is convergent.
is divergent.
is