Download Module 2 Chapter 8 Indefinite Integrals

8 Indefinite Integrals
Case Study
8.1
Concepts of Indefinite Integrals
8.2
Indefinite Integration of Functions
8.3
Integration by Substitution
8.4
Integration by Parts
8.5
Applications of Indefinite Integrals
Chapter Summary
Case Study
Can you estimate the number
of radioactive particles in the
sample from your readings?
By measuring the level of radioactivity
with a counter, it is estimated that the
number of radioactive particles, y, in
the sample is decreasing at a rate of
1000e–0.046t per hour, where t is
expressed in hours.
I have already recorded the
readings for the level of
radioactivity.
According to what we learnt in Section 7.5 (Rates of Change), we have
dy
 1000e –0.046 t .
dt
In order to express y in terms of t, we need to find a function y of t such
that its derivative is equal to –1000e–0.046t .
The process of finding a function from its derivative is called integration
and will be discussed in this chapter.
P. 2
8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals
In previous chapters, we learnt how to find the derivative
of a given function.
Suppose we are given a function x2, by differentiation, we have
d 2
( x )  2 x.
dx
2
As 2x is the derivative of x , we call x2 the primitive function (or
antiderivative) of 2x. Generally, for any differentiable function F(x),
we have the following definition:
Definition 8.1
d
If [ F ( x)]  f ( x), then F(x) is called a primitive function of f(x).
dx
Although x2 is a primitive function of 2x, it is not the unique primitive
function. If we add an arbitrary constant C to x2 and differentiate it, we
d 2
d 2
d
(
x

C
)

(
x
)

(C )  2 x  0  2 x.
have
dx
dx
dx
Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C.
P. 3
8.1 Concepts of Indefinite Integrals
A. Definition of Indefinite Integrals
In order to represent all the primitive functions of a function
f (x), we introduce the concept of indefinite integral as below:
Definition 8.2
d
If
F ( x)  f ( x) , then the indefinite integral of f(x), which is
dx
denoted by  f ( x)dx , is given by
 f ( x)dx  F ( x)  C, where C is an arbitrary constant.
Note:
1. In the notation of
 f ( x)dx , f (x) is called the integrand, and ‘ ’
is called the integral sign. The process of finding the primitive
function is called integration.
2. C is called the constant of integration (or integration constant).
P. 4
8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals
As integration is the reverse process of differentiation, the
basic formulas for integrations can be derived from the
differentiation formulas.
d
For example: Since ( x n1 )  (n  1) x n,
dx
d  1 n1 
i.e.,
x   xn ,

dx  n  1

8.1
n
x
 dx 
1 n1
x  C , for all real numbers n  1.
n 1
Note:
1. Formula 8.1 is also called the Power Rule for integration.
2. When n  0, the left hand side of the formula becomes x 0 dx   1dx .
For convenience we usually express  1dx as  dx .
P. 5
8.1 Concepts of Indefinite Integrals
B. Basic Formulas of Indefinite Integrals
In addition to the Power Rule, we can use the similar way
to derive the following integration formulas:
8.2
8.3
8.5
 kdx  kx  C , where k is a constant.
1
 x dx  ln x  C
 sin xdx   cos x  C
8.4
8.6
8.7
2
sec
 xdx  tan x  C
8.8
8.9
 sec x tan xdx  sec x  C
8.10
x
x
e
dx

e
C

 cos xdx  sin x  C
2
csc
xdx   cot x  C

 csc x cot xdx   csc x  C
P. 6
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Theorem 8.1
If k is a non-constant, then  kf ( x)dx  k  f ( x)dx .
Proof:
d
[ g ( x)]  f ( x).
dx
d
d

[kg ( x)]  k [ g ( x)]  kf ( x)
dx
dx
Let  f ( x)dx  g ( x)  C. Then
By definition,  kf ( x)dx  kg ( x)  C , where C  is an arbitrary constant.
On the other hand,
k  f ( x)dx  k[ g ( x)  C ]  k g ( x)  kC
Since C  and kC are arbitrary constants, the expressions kg(x)  C and
kg(x)  kC represent the same family of primitive functions.

 kf ( x)dx  k  f ( x)dx
P. 7
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Theorem 8.2
 [ f ( x)  g ( x)]dx   f ( x)dx   g ( x)dx
Proof:
Let  f ( x)dx  F ( x)  C1 and  g ( x)dx  G( x)  C2 , where C1 and C2 are
d
d
arbitrary constants. Then [ F ( x)]  f ( x) and [G ( x)]  g ( x).
dx
dx
d

[ F ( x)  G ( x)]  f ( x)  g ( x)
dx
By definition,  [ f ( x)  g ( x)]dx  F ( x)  G( x)  C.
On the other hand,  f ( x)dx   g ( x)dx  [ F ( x)  C1 ]  [G( x)  C2 ]
 F ( x)  G ( x)  C1  C2
Since C1  C2 is an arbitrary constant, the expressions F(x)  G(x)  C
and F(x)  G(x)  C1  C2 represent the same family of primitive
functions.
  [ f ( x)  g ( x)]dx   f ( x)dx   g ( x)dx
P. 8
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.1T
Find  (3 x  4 x 5  5)dx.
Solution:
5
(
3
x

4
x
 5)dx  3

1
x 2 dx  4

5
x 2 dx 
 5dx
2 3
 2 7

 3 x 2  C1   4 x 2  C 2   5 x  C3
3
 7


 

8 7
Use C to express the sum
 2 x3 
x  5x  C
7
of all constants
P. 9
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.2T
Find  3
x 1
dx.
x 1
Solution:
 (3 x ) 2  3 x  1
x 1
3
 3 x  1 dx   ( x  1) 3 x  1  dx





2
(x 3
2
x 3 dx 
5
3 3
x
5

1
x3


 1)dx
1
x 3 dx 
4
3 3
x
4
a3  b3  (a  b)(a2  ab  b2)
Cancel the common factor
 dx
 xC
P. 10
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.3T
2

Find   3x  2   dx.
x

Solution:
2
1

3
x

2

dx

3
xdx

2
dx

2
dx






x
x
2
x
 3  2 x  2 ln x  C
2
3
 x 2  2 x  2 ln x  C
2
P. 11
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.4T
5
Find   3 x 2   2e x dx.
x


Solution:
1
 2 5
x
2
x
3
x


2
e
dx

3
x
dx

5
dx

2
e
dx






x
x

3
x
 3   5 ln x  2e x  C
3
 x 3  5 ln x  2e x  C
P. 12
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.5T
Find  (1  tan 2 x)dx.
Solution:
 (1  tan
2
x)dx   sec 2 xdx
 tan x  C
P. 13
8.1 Concepts of Indefinite Integrals
C. Basic Properties of Indefinite Integrals
Example 8.6T
Let y  ln x – ln (x  1).
dy
(a) Find .
dx
1
dx .
(b) Hence find 
x( x  1)
Solution:
dy 1
1 d
 
( x  1)
dx x x  1 dx
1
1
 
x x 1
1
1 
(b) By (a),   
dx  ln x  ln x  1  C
 x x  1
x 1 x
 x( x  1) dx  ln x  ln x  1  C
1
 
dx  ln x  ln x  1  C
x( x  1)
(a)
P. 14
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
The integration formulas mentioned in Section 8.1 enable us
to find the indefinite integrals of simple functions such as ex,
sin x and cos x. But how about e2x, sin 4x and cos (7x + 5)?
d
Consider (ax  b) n1  a(n  1)(ax  b) n .
dx
n 1
(
ax

b
)
8.11  (ax  b) n dx 
 C, where n  –1 and a  0.
a(n  1)
Using the same method, we can obtain the following integration formulas:
Suppose a and b are real numbers with a  0.
1
1
8.12 
dx  ln ax  b  C
ax  b
a
1
8.13  e axb dx  e axb  C
a
1
8.14  sin( ax  b)dx   cos(ax  b)  C
a
1
8.15  cos(ax  b)dx  sin( ax  b)  C
a
P. 15
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.7T
9
dx.
Find  3
2 x
Solution:
3
1

9
dx  9  (2  x) 3 dx
2 x
1
 1
9

(2  x) 3  C
 1 
    1
 3 
2
27
  (2  x) 3  C
2
P. 16
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.8T
1
dx.
Find 
2x  3  2x  5
Solution:

1
1
2x  3  2x  5
dx  

dx
2x  3  2x  5
2x  3  2x  5 2x  3  2x  5
Rationalize the denominator
2x  3  2x  5

dx
(2 x  3)  (2 x  5)
1
  ( 2 x  3  2 x  5 )dx
8
3
3
1
1

(2 x  3) 2 
(2 x  5) 2  C
1 
1 
8  2  1
8  2  1
2 
2 
3
3
1
 [(2 x  3) 2  (2 x  5) 2 ]  C
24
P. 17
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.9T
3x  4
dx.
Find  2
6 x  5x  4
Solution:
3x  4
3x  4

dx
 (3x  4)(2 x  1) dx
 6 x 2  5x  4
1

dx
2x 1
Cancel the common factor
1
 ln 2 x  1  C
2
P. 18
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example
8.10T
4t 1
e
 3e t
dt.
Find 
2t
e
Solution:
 e 4t 1
e 4t 1  3e t
et 
 e 2t dt    e2t  3 e2t dt


  (e 2t 1  3e t )dt
e 2t 1 3e t


C
2
1
1
 e 2t 1  3e t  C
2
P. 19
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.11T
Find  [2t 2  sin(3t  2)]dt.
Solution:
 [2t
2
 sin(3t  2)]dt   2t 2 dt   sin(3t  2)dt
2t 3  cos(3t  2)


C
3
3
2 3 1
 t  cos(3t  2)  C
3
3
P. 20
8.2 Indefinite Integration of Functions
A. Integration of Functions Involving the Expression
(ax + b)
Example 8.12T
Find  2 x  5 x dx.
Solution:
x
x
x
2

5
dx

10

 dx
  (e ln 10 ) x dx
If y  ln 10, then ey  10 by definition,
i.e., eln 10  10.
e (ln 10 ) x

C
ln10
10 x

C
ln 10
P. 21
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
To find integrals where the integrand is the product or
power of trigonometric functions, we can first use
double angle formulas and product-to-sum formulas to
express the integrand in the sum of trigonometric functions.
cos 2A  2 cos2 A  1 or cos 2A  1  2 sin2 A
Product-to-sum Formulas
1
sin A cos B  [sin (A  B)  sin (A  B)]
2
1
cos A sin B  [sin (A  B)  sin (A  B)]
2
1
cos A cos B  [cos (A  B)  cos (A  B)]
2
1
sin A sin B   [cos (A  B)  cos (A  B)]
2
P. 22
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
Example 8.13T


6
sin
cos
d .
Find 
2
2
Solution:


 6 sin 2 cos 2 d   3 sin  d
  3 cos   C
sin 2A  2 sin A cos A
P. 23
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
Example 8.14T
2 x
2 x
sin
cos
dx.
Find 
4
4
Solution:
2
x
1
2 x
2 x
sin
cos
dx

sin

 dx
 4

4
2
2
1 2x
  sin dx
4
2
1  1  cos x 
 
dx
4
2 
1
  (1  cos x)dx
8
1
 ( x  sin x)  C
8
sin 2A  2 sin A cos A
cos 2A  1  2 sin2 A
P. 24
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
Example 8.15T
Find  sin 8t sin 7tdt.
Solution:
1


sin
8
t
sin
7
tdt
 2 [cos(8  7)t  cos(8  7)t ]dt

1
  (cos t  cos15t )dt
2
Product-to-sum formula
1
sin 15t 
  sin t 
C
2
15 
sin t sin 15t


C
2
30
P. 25
8.2 Indefinite Integration of Functions
B. Integration of Trigonometric Functions
Example 8.16T
2 cos
Find 
d .
cos 2  1
Solution:
2 cos
cos
d


 cos 2  1
 cos 2  1 d
2
cos

d
2
 sin 
  ( cot  csc )d
 csc   C
cos 2A  1  2 sin2 A
cot A 
cos A
sin A
P. 26
8.3 Integration by Substitution
A. Change of Variables
In Section 8.2, we learnt some basic formulas to find the
indefinite integrals of functions.
However, not all functions can be integrated directly using these formulas.
In this case, we have to use the method of integration by substitution.
The following shows the basic principle of this method.
Let  f (u )du  F (u )  C and u  g (x).
d
d
du
Since [ F (u )]  [ F (u )] 
dx
du
dx
 f (u) g(x)
 f [g(x)] g(x)
By the definition of integration,  f [ g ( x)]g' ( x)dx  F (u)  C.

 f [ g ( x)]g' ( x)dx   f (u)du
P. 27
8.3 Integration by Substitution
A. Change of Variables
For an integral  f [ g ( x)]g' ( x)dx, we can transform it into
a simpler integral
 f (u)du , by the following steps.
Step 1: Separate the integrand into two parts: f [g(x)] and g(x)dx.
Step 2: Replace every occurrence of g(x) in the integrand by u.
Step 3: Replace the expression ‘g(x)dx’ by ‘du’.
Let us use this method to find the integral  2 x x 2  1dx together.
Note that  2 x x 2  1dx   x 2  1  2 xdx ,
du
so we let u  x2 + 1, such that
 2x.
dx
2
  x  1  2xdx   udu
3
2
 u2 C
3
3
2 2
 ( x  1) 2  C
3
P. 28
8.3 Integration by Substitution
A. Change of Variables
Example 8.17T
3
Find  2 x 1  x 2 dx.
Solution:
Let u  1 – x2. Then

du
 2 x .
dx
3
3
2
2


1

x
(2 x)dx
2
x
1

x
dx


   3 udu
4
3
  u3 C
4
4
3
  (1  x 2 ) 3  C
4
Express the answer in
terms of x
P. 29
8.3 Integration by Substitution
A. Change of Variables
Example 8.18T
Find  x 3 ( x 2  7)10 dx.
Solution:
Let u  x2 – 7. Then

du
 2x.
dx
3 2
10
2
2
10 1
x
(
x

7
)
dx

x
(
x

7
)
  2 xdx


2
1
Rewrite x2 as (u  7)
  (u  7)u10  du
2
1
  (u11  7u10 )du
2
1 1
7

  u12  u11   C
2  12
11 
1
7
 ( x 2  7)12  ( x 2  7)11  C
24
22
P. 30
8.3 Integration by Substitution
A. Change of Variables
With the method of integration by substitution, we can
find the integrals of trigonometric functions other than
sine and cosine, as shown in the following example.
P. 31
8.3 Integration by Substitution
A. Change of Variables
Example 8.19T
Find  csc xdx.
Solution:
csc x  cot x
dx
csc x  cot x
csc 2 x  csc x cot x

dx
csc x  cot x
 csc xdx   csc x 
Let u  csc x – cot x.
du
Then
 csc 2 x  csc x cot x.
dx
1
  csc xdx   du
u
 ln u  C
 ln csc x  cot x  C
Alternative Solution:
d
csc x   csc x cot x
dx
d
and
cot x   csc 2 x
dx
d

(csc x  cot x)
dx
 csc x(csc x  cot x)

Let u  csc x – cot x.
du
Then
 u csc x.
dx
1
  csc xdx   du
u
 ln csc x  cot x  C
P. 32
8.3 Integration by Substitution
A. Change of Variables
It is tedious to write u and du every time when finding
the integrals by substitution, as shown in the previous
examples.
After becoming familiar with the method of integration by substitution,
the working steps can be simplified by omitting the use of the variable u.
Let us study the following example.
P. 33
8.3 Integration by Substitution
A. Change of Variables
Example 8.20T
sin(ln x)
dx.
Find 
x
Solution:
sin(ln x)
1

sin(ln
x
)

dx
dx

 x
x
  sin(ln x)d (ln x)
  cos(ln x)  C
P. 34
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Sometimes we need to handle indefinite integrals that involve
the products of powers of trigonometric functions, such as
m
n
m
n
or
sin
x
cos
xdx
tan
x
sec
xdx , where m and n are integers.


In the following discussion, we will see how to apply different strategies
according to different values of m and n.
Strategies for finding integrals in the form  sin m x cosn xdx
Case 1: m is an odd number.
Use sin x dx  –d(cos x) and express all the other sine terms
as cosine terms.
Case 2: n is an odd number.
Use cos x dx  d(sin x) and express all the other cosine terms
as sine terms.
Case 3: both m and n are even numbers.
Use the double-angle formula to reduce the powers of the
functions.
P. 35
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.21T
Find  sin 3 x cos5 xdx.
Solution:
3
5
sin
x
cos
xdx

   sin 2 x cos5  ( sin x)dx
   (1  cos2 x) cos5 xd (cos x)
  (cos7 x  cos5 x)d (cos x)
cos8 x cos6 x


C
8
6
P. 36
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.22T
Find  3 cos x sin 3 xdx.
Solution:

3
cos x sin 3 xdx
   3 cos x sin 2 x  ( sin x)dx
   3 cos x (1  cos2 x)d (cos x)
7
1

  (cos x) 3  (cos x) 3  d (cos x)


10
4
3
3
 (cos x) 3  (cos x) 3  C
10
4
P. 37
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.23T
2
4
Find  sin x cos xdx.
Solution:
2
 sin 2 x 
2

cos
xdx
sin
x
cos
xdx



 2 
1
  sin 2 2 x cos 2 xdx
4
1
  (1  cos 4 x)(1  cos 2 x)dx
16
1
  (1  cos 2 x  cos 4 x  cos 2 x cos 4 x)dx
16
1 
1
1

  1  cos 2 x  cos 4 x   cos 6 x  cos 2 x  dx
16 
2
2

1  1
1

  1  cos 2 x  cos 4 x  cos 6 x dx
16  2
2

x 1
1
1
  sin 2 x  sin 4 x 
sin 6 x  C
16 64
64
192
2
4
P. 38
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Similarly, integrals in the form  tan m x sec n xdx may be
found by using the method of integration by substitution.
Strategies for finding integrals in the form  tan m x sec n xdx
Case 1: m is an odd number.
Use tan x sec x dx  d(sec x) and then express all other
tangent terms as secant terms.
Case 2: n is an even number.
Use sec2x dx  d(tan x) and then express all other secant
terms as tangent terms.
P. 39
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.24T
3
5
Find  tan x sec xdx.
Solution:
3
5
tan
x
sec
xdx

  tan 2 x sec 4 x  tan x sec xdx
  (sec 2 x  1) sec 4 xd (sec x)
  (sec 6 x  sec 4 x)d (sec x)
sec 7 x sec 5 x


C
7
5
P. 40
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
Example 8.25T
Find  cot 2 x csc 4 xdx.
Solution:
2
4
cot
x
csc
xdx

   cot 2 x csc 2 x( csc 2 x)dx
   cot 2 x(cot 2 x  1)d (cot x)
  ( cot 4 x  cot 2 x)d (cot x)
cot 5 x cot 3 x


C
5
3
P. 41
8.3 Integration by Substitution
B. Integrals Involving Powers of Trigonometric
Functions
In the above examples, the case that m is even while n is odd is
not considered. This is because there is no standard technique
and the method varies from case to case.
For example, to find  sec xdx (m  0 and n  1), we may follow the
method in Example 8.19.
In some other cases, such as  tan 2 x sec xdx (m  2 and n  1), we may
need to use the technique ‘integration by parts’, which will be discussed
later in this chapter.
P. 42
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
If an indefinite integral involves radicals in the form
a 2  x 2 , a 2  x 2 or x 2  a 2 , we can use the method
of integration by substitution to eliminate the radicals.
The following three trigonometric identities are very useful for the
elimination:
cos2  1  sin2 , sec2  1  tan2 , tan2  sec2  1
For example, if we substitute x  a sin  into the expression
a 2  x 2,
we have a 2  x 2  a 2  a 2 sin 2 
 a 2 cos2 
 a cos
Then we can express the integrand in terms of .
After finding the indefinite integral in terms of  (say, 3 + C), the final
answer should be expressed in terms of the original variable, say, x.
P. 43
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
In order to express  in terms of x, let us first introduce
the following notations:
Inverse of Trigonometric Functions
Let x be a real number.
1. sin–1x is defined as the angle  such that sin   x (where –1  x  1)
π
π
and     .
2
2
2. cos–1x is defined as the angle  such that cos   x (where –1  x  1)
and 0    p.
π
π
3. tan–1x is defined as the angle  such that tan   x and     .
2
2
P. 44
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
Example 8.26T
dx
.
Find  2
2
x 1 x
Solution:
Let x  sin.Then dx  cos d.
dx
cos d

 x 2 1  x 2  sin 2  1  sin 2 
cos d
 2
sin  cos
  csc 2  d
  cot   C
Since sin   x,
cot  csc2   1
1
1
2
sin 
1

1
2
x

1 x 2

x
dx
1  x2
 

C
2
2
x
x 1 x
P. 45
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
Example 8.27T
Find  x 2 1  x 2 dx.
Solution:
Since sin   x, cos   1  x 2 ,
Let x  sin.Then dx  cosd.
sin 4  2 cos 2 sin 2
 2(1  2 sin 2  )(2 cos sin  )
2
2
x
1

x
dx

  sin 2  1  sin 2   cos d
  sin  cos  d
2
sin
2



 
 d
 2 
1
  sin 2 2 d
4
1  1  cos 4 
 
 d
4 
2

1
sin 4 
  
C
8
4 
2
 4(1  2 x 2 )( x 1  x 2 )
2
 4( x  2 x 3 ) 1  x 2

2
2
x
1

x
dx

1  1
4( x  2 x 3 ) 1  x 2 
 sin x 
C
8 
4

1
 [(2 x3  x) 1  x 2  sin 1 x]  C
8
P. 46
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
Example 8.28T
dx
.
Find  2
2
x 9 x
Solution:
x
Since tan   ,
3
Let x  3tan.Then dx  3sec2 d.
dx
 x2

csc x  1  cot 2 
9 x
3 sec 2  d
2
9 tan 2  9  9 tan 2 
3 sec 2  d

27 tan 2  sec
1 sec
  2 d
9 tan 
1 1 cos
 

 cot d
9 cos sin 
1
tan 2 
9
 1 2
x
 1
1
csc cot d

9
1
  csc  C
9


 x2

x2  9
x
9  x2

C
2
9x
9 x
dx
P. 47
8.3 Integration by Substitution
C. Integration by Trigonometric Substitution
Example 8.29T
dx
.
Find  2
x  4x  3
Solution:

dx

dx
x+2
__________
(x  2)2  1
x  4x  3
( x  2) 2  1

Let x + 2  sec. Then dx  sectan d.
1
dx
sec tan d
sec   tan 


sec


d

 x 2  4 x  3  sec2   1
sec   tan 
sec tan d
d (sec  tan  )


tan
sec   tan 
 ln sec   tan   C
  sec d
2
Since sec  x + 2, tan  sec 2   1  ( x  2) 2  1  x 2  4 x  3
dx
 
 ln x  2  x 2  4 x  3  C
x2  4x  3
P. 48
8.4 Integration by Parts
Some indefinite integrals such as  ln xdx,  xsin xdx and
x
xe
 dx cannot be found by using the techniques we have
learnt so far. To evaluate them, we need to introduce another method
called integration by parts.
Theorem 8.3 Integration by Parts
If u(x) and v(x) are two differentiable functions, then
 uv'dx  uv   vu'dx.
In other words,  udv  uv   vdu .
Proof:
Suppose u and v are two differentiable functions.
d
Since (uv)  uv + vu, by definition,  (uv'  vu' )dx  uv  C.
dx
 uv'dx   vu'dx  uv  C

 uv'dx  uv   vu'dx
P. 49
8.4 Integration by Parts
From Theorem 8.3, we can see that the problem of finding
 udv can be transformed into the problem of finding  vdu instead.
If the integral  vdu is much simpler than  udv, then the original integral
can be found easily.
If we want to apply the technique of integration by parts to an integral,
such as  e x sin xdx, we need to transform the integral into the form  udv
first, such as  (sin x)d (e x ) or  e x d ( cos x).
P. 50
8.4 Integration by Parts
Example 8.30T
Find
 x ln xdx.
Solution:
 x2 
 x ln xdx   ln xd  2 
 
 x2 
x2
 (ln x)    d (ln x)
2
 2
x 2 ln x
x2 1

   dx
2
2 x
x 2 ln x 1

  xdx
2
2
x 2 ln x x 2

 C
2
4
P. 51
8.4 Integration by Parts
Example 8.31T
2
Find  x log5 xdx.
Solution:
2
x
 log5 xdx
x 2 ln x

dx
ln 5
 x3 
1

ln xd  

ln 5
 3

 x3 
1 
x3



(ln x)    d (ln x)
ln 5 
3
 3

1  x 3 ln x 1 3 1 


  x  dx 

ln 5  3
3
x 
1  x 3 ln x 1 2 


  x dx 

ln 5  3
3

1  x3
x3 
 ln x    C

ln 5  3
9 
P. 52
8.4 Integration by Parts
In some cases, there may be more than one choice for u and v.
For example, we can transform  xsin xdx into
 x2 
 sin xd  2 
 
or
  xd (cos x).
However, if we choose the former, then
 x2 
 x2 
x2
 sin xd  2   (sin x) 2    2 d (sin x)
 
 
2
x sin x 1 2

  x cos xdx
2
2
As a result, we get an integrand x2 cos x which is more complicated
than the original one x sin x.
Thus we should try sin x dx  d(–cos x) instead.
P. 53
8.4 Integration by Parts
Example 8.32T
Find  x csc 2 xdx.
Solution:
2
x
csc
xdx

  xd ( cot x)
 x( cot x)    cot xdx
cos x
  x cot x  
dx
sin x
1
  x cot x  
d (sin x)
sin x
  x cot x  ln sin x  C
P. 54
8.4 Integration by Parts
Example 8.33T
d 
x
1
cot

.


dx 
2  cos x  1
x
(b) Hence find 
dx .
cos x  1
(a) Show that
Solution:
(a)
d 
x
2 x d  x
cot


csc


 
dx 
2
2 dx  2 
1
2 x
  csc
2
2
1

2 x
2 sin
2
1

 1  cos x 
2

 2 
1

cos x  1
x
x

dx

xd
cot
(b) 
  2 
cos x  1
x
x
 x cot   cot dx
2
2x
cos
x
2 d x 
 x cot  2 
 
x 2
2
sin
2
x
1 
x
 x cot  2
d  sin 
x  2
2
sin
2
x
x
 x cot  2 ln sin  C
2
2
P. 55
8.4 Integration by Parts
Example 8.34T
Find
x
2 x
e dx.
Solution:
2 x
   x 2 d (e  x )
x
e
dx

 [ x 2 e  x   e  x d ( x 2 )]
 ( x 2 e  x   2 xe  x dx)
  x 2 e  x   2 xd (e  x )
  x 2 e  x  [2 xe  x   e  x d (2 x)]
  x 2 e  x  2 xe  x  2 e  x dx
  x 2 e  x  2 xe  x  2e  x  C
  ( x 2  2 x  2)e  x  C
P. 56
8.4 Integration by Parts
Example 8.35T
Find  cos(ln x)dx.
Solution:
 cos(ln x)dx  [cos(ln x)]x   xd[cos(ln x)]
1
 x cos(ln x)   x sin(ln x)  dx
x
 x cos(ln x)   sin(ln x)dx
 x cos(ln x)  [sin(ln x)]x   xd[sin(ln x)]
1
 x cos(ln x)  x sin(ln x)   x  cos(ln x)  dx
x
 x cos(ln x)  x sin(ln x)   cos(ln x)dx
Therefore, 2 cos(ln x)dx  x cos(ln x)  x sin(ln x)  C
x
  cos(ln x)dx  [sin(ln x)  cos(ln x)]  C
2
P. 57
8.5 Applications of Indefinite Integrals
A. Geometrical Applications
dy
In previous chapters, we learnt that of a curve y  F(x)
is the slope function of the curve. dx
Since integration is the reverse process of differentiation,
dy
if we let  f(x), then by the definition of integration, we have
dx
y   f ( x)dx
 F ( x)  C
where C is an arbitrary constant.
Thus we can see that the equation of a family of curves y  F(x) + C
dy
can be found by integration, providing that the slope function
of the
dx
curve is known.
P. 58
8.5 Applications of Indefinite Integrals
A. Geometrical Applications
Example 8.36T
The equation of the slope of a curve at the point (x, y) is given by
dy
 (2 x  3) 6. If the curve passes through (2, 2), find the equation of
dx
the curve.
Solution:
dy
 (2 x  3) 6
dx
(2 x  3) 7
6
 C , where C is an arbitrary constant
y   (2 x  3) dx 
27
(2 x  3) 7

C
14
7
(
2

2

3
)
When x  2, y  2, we have 2 
C
14
27
C
14 (2 x  3) 7 27
 .
∴ The equation of the curve is y 
14
14
P. 59
8.5 Applications of Indefinite Integrals
A. Geometrical Applications
Example 8.37T
d2y
2
It is given that at each point (x, y) on a certain curve,

2
sin
x . If
2
dx
5
5




the curve passes through  0,  and  p,  , find the equation of that curve.
 4
 4
Solution:
 5
 5
0
,
Since
 and  p,  lie on the curve, we have
d2y
2
 4
 4

2
sin
x
5
1
dx 2

0

 C2
dy
2
4
4
  2 sin xdx
C2  1
dx
  (1  cos 2 x)dx
5 p2 1

  pC1  1
4 2 4
sin 2 x
2
 x
 C1
p
2
pC1  
sin 2 x
2


y  x 
 C1 dx
p
2
C1  



x 2 cos 2 x
∴
The
equation
of the curve is


 C1 x  C2
2
4
x 2 cos 2 x px
y

  1.
2
4
2
P. 60
8.5 Applications of Indefinite Integrals
B. Applications in Physics
In Section 7.5 of Book 1, we learnt that for a particle moving
along a straight line, its velocity v and acceleration a at time
t are given by
ds
dv d 2 s
v
and a 
 2,
dt
dt dt
where s is the displacement of the particle at time t.
Since integration is the reverse process of differentiation, we have
s   vdt and v   adt.
P. 61
8.5 Applications of Indefinite Integrals
B. Applications in Physics
Example 8.38T
A particle moves along a straight line such that its velocity v at time t is
given by v  t t 2  5 . Find the displacement s of the particle at time t,
given that s  6 when t  2.
Solution:
1
2
2
t

5
d
(
t
 5)

2
3
3
1
1 2 2
  (t  5) 2  C  (t 2  5) 2  C
3
2 3
When t  2, s  6, we have
s   t t  5dt 
2
3
1
6  (2 2  5) 2  C
3
C  3
3
1
 s  (t 2  5) 2  3
3
P. 62
8.5 Applications of Indefinite Integrals
B. Applications in Physics
Example 8.39T
A particle moves along a straight line so that its acceleration a at time t is
1
given by a  for t > 0. When t  1, the velocity of the particle is 8 and
t
its displacement is 24. Find the displacement of the particle at time t.
Solution:
1
v   dt  ln t  C1
t
When t  1, v  8, we have
8  ln1  C1
C1  8

v  ln t  8
s   (ln t  8)dt
  ln tdt  8t
 t ln t   td (ln t )  8t
1
 t ln t   t  dt  8t
t
 t ln t   dt  8t
 t ln t  7t  C2
When t  1, s  24, we have
24  1 ln1  7(1)  C 2
C 2  17

s  t ln t  7t  17
P. 63
Chapter Summary
8.1 Concepts of Indefinite Integrals
d
[ F ( x)]  f ( x), then the indefinite integral of f(x)
dx
is defined by  f ( x)dx  F ( x)  C , where C is an arbitrary constant.
1.
If
2.
n
x
 dx 
3.
 kdx  kx  C, where k is a constant.
4.
1
 x dx  ln x  C
5.
x
x
e
dx

e
C

1 n1
x  C , where n   1.
n 1
P. 64
Chapter Summary
8.1 Concepts of Indefinite Integrals
6.
7.
8.
9.
10.
11.
12.
13.
 sin xdx   cos x  C
 cos xdx  sin x  C
2
sec
 xdx  tan x  C
2
csc
xdx   cot x  C

 sec x tan xdx  sec x  C
 csc x cot xdx   csc x  C
 kf ( x)dx  k  f ( x)dx, where k is a non - zero constant.
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
P. 65
Chapter Summary
8.2 Indefinite Integration of Functions
Let a and b be real numbers with a  0.
(ax  b) n1
n
1.  (ax  b) dx 
 C , where n  1.
a(n  1)
1
1
dx  ln ax  b  C
2. 
ax  b
a
1
3.  e axb dx  e axb  C
a
1
4.  sin( ax  b)dx   cos(ax  b)  C
a
1
cos(
ax

b
)
dx

sin( ax  b)  C
5. 
a
P. 66
Chapter Summary
8.3 Integration by Substitution
1.
Let u  g(x) be a differentiable function. Then,
 f [ g ( x)]g' ( x)dx   f (u)du.
2.
If the integrated involves terms like a 2  x 2 , a 2  x 2 and x 2  a 2 ,
we can simplify the integrand by substituting x  a sin , x  a tan 
or x  a sec  respectively.
P. 67
Chapter Summary
8.4 Integration by Parts
If u(x) and v(x) are two differentiable functions, then
 udv  uv   vdu.
P. 68
Chapter Summary
8.5 Applications of Indefinite Integrals
1.
If the slope of a curve at point (x, y) is f(x), then the
equation of the family of curves is given by
 f ( x)dx  F ( x)  C
where F (x)  f(x).
2.
Let s, v and a be the displacement, velocity and acceleration of
a particle moving along a straight line respectively, then
s   vdt and v   adt.
P. 69