8.1 concepts of indefinite integrals
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
8.1 Concepts of Indefinite Integrals8.2 Indefinite Integration of Functions8.3 Integration by Substitution
Chapter Summary
Case Study
Indefinite Integrals8
8.4 Integration by Parts8.5 Applications of Indefinite Integrals
P. 2
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.
Case StudyCase Study
According to what we learnt in Section 7.5 (Rates of Change), we have
The process of finding a function from its derivative is called integration and will be discussed in this chapter.
I have already recorded the readings for the level of radioactivity.
Can you estimate the number of radioactive particles in the sample from your readings?
.0100 0.046– tedt
dy
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 .
P. 3
In previous chapters, we learnt how to find the derivative of a given function.
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Suppose we are given a function x2, by differentiation, we have
As 2x is the derivative of x2, we call x2 the primitive function (or antiderivative) of 2x.
Although x2 is a primitive function of 2x, it is not the unique primitive function.
A. A. Definition of Indefinite IntegralsDefinition of Indefinite Integrals
.2)( 2 xxdx
d
Definition 8.1
If , then F(x) is called a primitive function of f(x).
)()]([ xfxFdx
d
If we add an arbitrary constant C to x2 and differentiate it, we
have .202)()()( 22 xxCdx
dx
dx
dCx
dx
d
Thus, x2 + C is also a primitive function of 2x for an arbitrary constant C.
Generally, for any differentiable function F(x), we have the following definition:
P. 4
In order to represent all the primitive functions of a function f (x), we introduce the concept of indefinite integral as below:
Note: 1. In the notation of , f (x) is called the integrand, and ‘ ’
is called the integral sign. The process of finding the primitive function is called integration.
Definition 8.2
If , then the indefinite integral of f(x), which is
denoted by , is given by
, where C is an arbitrary constant.
)()( xfxFdx
d
dxxf )(
CxFdxxf )()(
2. C is called the constant of integration (or integration constant).
dxxf )(
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
A. A. Definition of Indefinite IntegralsDefinition of Indefinite Integrals
P. 5
Note: 1. Formula 8.1 is also called the Power Rule for integration.
B. B. Basic Formulas of Indefinite IntegralsBasic Formulas of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
As integration is the reverse process of differentiation, the basic formulas for integrations can be derived from the differentiation formulas.
For example: Since , nn xnxdx
d)1()( 1
8.1 , for all real numbers n 1.Cxn
dxx nn
1
1
1
2. When n 0, the left hand side of the formula becomes .
For convenience we usually express as .
dxdxx 10
dxdx1
,1
1i.e., 1 nn xx
ndx
d
P. 6
B. B. Basic Formulas of Indefinite IntegralsBasic Formulas of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
In addition to the Power Rule, we can use the similar way to derive the following integration formulas:
8.2 , where k is a constant.
8.3 8.4
8.5 8.6
8.7 8.8
8.9 8.10 Cxxdxx csccotcsc
Cxxdx cotcsc2
Cxxdx sincos
Cedxe xx
Cxxdxx sectansec
Cxxdx tansec2
Cxxdx cossin
Cxdxx
ln1
Ckxkdx
P. 7
Theorem 8.1If k is a non-constant, then .
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
dxxfkdxxkf )()(
Proof:
.)()(Let Cxgdxxf ).()]([Then xfxgdx
d
)()]([)]([ xkfxgdx
dkxkg
dx
d
kCxgkCxgkdxxfk )(])([)(
Since C and kC are arbitrary constants, the expressions kg(x) C and kg(x) kC represent the same family of primitive functions.
dxxfkdxxkf )()(
)()( xkgdxxkfBy definition, C , where C is an arbitrary constant. On the other hand,
P. 8
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Theorem 8.2
dxxgdxxfdxxgxf )()()]()([
Let and , where C1 and C2 are
arbitrary constants. ).()]([ and )()]([Then xgxGdx
dxfxF
dx
d
)()()]()([ xgxfxGxFdx
d
21)()( CCxGxF 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.
dxxgdxxfdxxgxf )()()]()([
1)()( CxFdxxf 2)()( CxGdxxg
.)()()]()([ CxGxFdxxgxf By definition,
])([])([)()( 21 CxGCxFdxxgdxxf On the other hand,
Proof:
P. 9
Example 8.1T
Solution:
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Find .)543( 5 dxxx
dxxx )543( 5 dxdxxdxx 543 2
5
2
1
322
7
12
3
57
24
3
23 CxCxCx
Cxxx 57
82 73 Use C to express the sum
of all constants
P. 10
Example 8.2T
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Solution:
dxx
x
1
13
dxx
xxx
1
1)()1(
3
3233
dxxx )1( 3
1
3
2
dxdxxdxx 3
1
3
2
Cxxx 3
4
3
5
4
3
5
3
.1
13
dx
x
x
Cancel the common factor
a3 b3 (a b)(a2 ab b2)
Find
P. 11
Example 8.3T
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Solution:
. 2
23 dxx
x
dx
xx
223 dx
xdxxdx
1223
Cxxx ln222
32
Cxxx ln222
3 2
Find
P. 12
Example 8.4T
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Solution:
dxe
xx x2
53 2
.25
3 2
dxe
xx x
dxedxx
dxx x21
53 2
Cexx x 2ln53
33
Cexx x 2ln53
Find
P. 13
Example 8.5T
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Solution:
.)tan1( 2 dxx
dxx)tan1( 2 dxx 2sec
Cx tan
Find
P. 14
Example 8.6T
C. C. Basic Properties of Indefinite IntegralsBasic Properties of Indefinite Integrals
8.1 8.1 Concepts of Indefinite IntegralsConcepts of Indefinite Integrals
Let y ln x – ln (x 1).
(a) Find .
(b) Hence find .
Solution:
dx
dy
dxxx )1(
1
)1(1
11
x
dx
d
xxdx
dy(a)
1
11
xx
(b) By (a),
Cxxdx
xx1lnln
1
11
Cxxdxxx
xx1lnln
)1(
1
Cxxdxxx
1lnln)1(
1
P. 15
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)?
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
Using the same method, we can obtain the following integration formulas:
.))(1()(Consider 1 nn baxnabaxdx
d
8.11 , where n –1 and a 0.Cna
baxdxbax
nn
)1(
)()(
1
Suppose a and b are real numbers with a 0.
8.12
8.13
8.14
8.15
Cbaxa
dxbax
ln
11
Cea
dxe baxbax 1
Cbaxa
dxbax )cos(1
)sin(
Cbaxa
dxbax )sin(1
)cos(
P. 16
Example 8.7T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
Solution:
.2
93
dxx
dx
x3 2
9
dxx 3
1
)2(9
Cx
1
3
1
)2(1
3
19
Cx 3
2
)2(2
27
Find
P. 17
Example 8.8T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
Solution:
.5232
1
dxxx
dx
xx 5232
1
dxxx
xx
xx 5232
5232
5232
1
dx
xx
xx
)52()32(
5232
dxxx )5232(8
1
Cxx
23
23
)52(1
21
28
1)32(
121
28
1
Cxx ])52()32[(24
1 2
3
2
3
Find
Rationalize the denominator
P. 18
Example 8.9T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
Solution:
.456
432
dx
xx
x
dxxx
x
456
432
dxxx
x
)12)(43(
43
dx
x 12
1
Cx 12ln2
1
Find
Cancel the common factor
P. 19
Example 8.10T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
.3
2
14
dte
eet
tt
dte
eet
tt
2
14 3
Cee tt
1
3
2
12
Cee tt 32
1 12
Solution:
Find
dtee tt )3( 12
dt
e
e
e
et
t
t
t
22
14
3
P. 20
Example 8.11T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
Solution:
.)]23sin(2[ 2 dttt
dttt )]23sin(2[ 2
Ctt )23cos(31
32 3
Ctt
3
)23cos(
3
2 3
dttdtt )23sin(2 2
Find
P. 21
Example 8.12T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
Solution:
.52 dxxx
dxxx 52 dxx10
dxe x )( 10ln
Ce x
10ln
)10(ln
Cx
10ln
10
A. Integration of Functions Involving the ExpressionA. Integration of Functions Involving the Expression (ax + b)(ax + b)
If y ln 10, then ey 10 by definition,i.e., eln 10 10.
Find
P. 22
To find integrals where the integrand is the product or power of trigonometric functions, we can first usedouble angle formulas and product-to-sum formulas to express the integrand in the sum of trigonometric functions.
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions
Product-to-sum Formulas
sin A cos B [sin (A B) sin (A B)]
cos A sin B [sin (A B) sin (A B)]
cos A cos B [cos (A B) cos (A B)]
sin A sin B [cos (A B) cos (A B)]
2
1
2
1
2
1
2
1
cos 2A 2 cos2 A 1 or cos 2A 1 2 sin2 A
P. 23
Example 8.13T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
Solution:
B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions
.2
cos2
sin6 d
d
2cos
2sin6 dsin3
C cos3
sin 2A 2 sin A cos A
Find
P. 24
Example 8.14T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
Solution:
B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions
dxxx
4cos
4sin 22
.4
cos4
sin 22 dxxx
dx
x 2
2sin
2
1
dx
x
2
cos1
4
1
dxx)cos1(8
1
Cxx )sin(8
1
Find
dxx
2sin
4
1 2
sin 2A 2 sin A cos A
cos 2A 1 2 sin2 A
P. 25
Example 8.15T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
Find
Solution:
B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions
tdtt 7sin8sin
.7sin8sin tdtt
dttt ])78cos()78[cos(2
1
dttt )15cos(cos2
1
Ct
t
15
15sinsin
2
1
Ctt
30
15sin
2
sin
Product-to-sum formula
P. 26
Example 8.16T
8.8.22 Indefinite Integration of FunctionsIndefinite Integration of Functions
Solution:
B. Integration of Trigonometric FunctionsB. Integration of Trigonometric Functions
.12cos
cos2
d
d12cos
cos2
d
2
12coscos
d2sin
cos
d)csccot(
C csc
Find
cos 2A 1 2 sin2 A
cot A A
A
sin
cos
P. 27
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. 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 and u g (x). CuFduuf )()(
dx
duuF
du
duF
dx
d )]([)]([ Since
f (u) g(x) f [g(x)] g(x)
By the definition of integration, . CuFdxxg'xgf )()()]([
duufdxxg'xgf )()()]([
P. 28
8.8.33 Integration by SubstitutionIntegration by Substitution
For an integral , we can transform it into
a simpler integral , by the following steps.
dxxg'xgf )()]([
duuf )(
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 together. Note that , xdxxdxxx 2112 22
so we let u x2 + 1, such that 2x. dx
du
duuxdxx 212
Cu 2
3
3
2
Cx 23
2 )1(3
2
A. Change of VariablesA. Change of Variables
dxxx 12 2
P. 29
Example 8.17T
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. Change of Variables
Solution:Let u 1 – x2. Then .
.12 3 2 dxxx
xdx
du2
dxxx 3 212 dxxx )2(13 2
duu3
Cu 3
4
4
3
Cx 3
42 )1(
4
3
Find
Express the answer in terms of x
P. 30
Example 8.18T
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. Change of Variables
Solution:Let u x2 – 7. Then . x
dx
du2
.)7( 1023 dxxx
dxxx 1023 )7( xdxxx 22
1)7( 1022
duuu2
1)7( 10
duuu )7(2
1 1011
Cuu
1112
11
7
12
1
2
1
Cxx 112122 )7(22
7)7(
24
1
Find
Rewrite x2 as (u 7)
P. 31
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. 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. 32
Example 8.19T
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. Change of Variables
Let u csc x – cot x.
.csc xdx
dxxx
xxx
cotcsc
cotcsccsc xdxcsc
dx
xx
xxx
cotcsc
cotcsccsc2
.cotcsccscThen 2 xxxdx
du
duu
xdx1
csc
Cu ln
Cxx cotcscln
Solution:
Find
Alternative Solution:xxx
dx
dcotcsccsc
xxdx
d 2csccotand
)cot(csccsc xxx )cot(csc xx
dx
d
Let u csc x – cot x.
duu
xdx1
cscCxx cotcscln
.cscThen xudx
du
P. 33
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. 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. 34
Example 8.20T
8.8.33 Integration by SubstitutionIntegration by Substitution
A. Change of VariablesA. Change of Variables
dxx
x)sin(ln
.)sin(ln
dxx
x
dxx
x1
)sin(ln
)(ln)sin(ln xdx
Cx )cos(ln
Solution:
Find
P. 35
Strategies for finding integrals in the form
Sometimes we need to handle indefinite integrals that involvethe products of powers of trigonometric functions, such as
or , where m and n are integers.
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
xdxx nm cossin xdxx nm sectan
In the following discussion, we will see how to apply different strategies according to different values of m and n.
xdxx nm cossin
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. 36
Example 8.21T
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
Find
Solution: xdxx 53 cossin
.cossin 53 xdxx
dxxx )sin(cossin 52
)(coscos)cos1( 52 xxdx
)(cos)cos(cos 57 xdxx
Cxx
6
cos
8
cos 68
P. 37
Example 8.22T
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
Find
Solution: xdxx 33 sincos
.sincos 33 xdxx
dxxxx )sin(sincos 23
)(cos)cos1(cos 23 xdxx
)(cos)(cos)(cos 3
1
3
7
xdxx
Cxx 3
4
3
10
)(cos4
3)(cos
10
3
P. 38
Example 8.23T
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
Find
Solution: xdxx 42 cossin
.cossin 42 xdxx
xdx
x 22
cos2
2sin
xdxx 22 cos2sin4
1
dxxx )2cos1)(4cos1(16
1
dxxxxx )4cos2cos4cos2cos1(16
1
dxxxxx 2cos
2
16cos
2
14cos2cos1
16
1
dxxxx 6cos
2
14cos2cos
2
11
16
1
Cxxxx 6sin
192
14sin
64
12sin
64
1
16
P. 39
Similarly, integrals in the form may be
found by using the method of integration by substitution.
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
xdxx nm sectan
Strategies for finding integrals in the form xdxx nm sectan
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. 40
Example 8.24T
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
Find
Solution: xdxx 53 sectan
.sectan 53 xdxx
xdxxxx sectansectan 42
)(secsec)1(sec 42 xxdx
)(sec)sec(sec 46 xdxx
Cxx
5
sec
7
sec 57
P. 41
Example 8.25T
8.8.33 Integration by SubstitutionIntegration by Substitution
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
Find
Solution:
.csccot 42 xdxx
xdxx 42 csccot
dxxxx )csc(csccot 222
)(cot)1(cotcot 22 xdxx
)(cot)cotcot( 24 xdxx
Cxx
3
cot
5
cot 35
P. 42
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.
8.8.33 Integration by SubstitutionIntegration by Substitution
For example, to find (m 0 and n 1), we may follow the
method in Example 8.19.
xdxsec
In some other cases, such as (m 2 and n 1), we may
need to use the technique ‘integration by parts’, which will be discussed later in this chapter.
xdxx sectan2
B. Integrals Involving Powers of TrigonometricB. Integrals Involving Powers of Trigonometric FunctionsFunctions
P. 43
If an indefinite integral involves radicals in the form
, or , we can use the method of integration by substitution to eliminate the radicals.
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
The following three trigonometric identities are very useful for the elimination:
For example, if we substitute x a sin into the expression ,
we have
22 ax 22 xa 22 xa
22 xa
22222 sinaaxa
coscos22
aa
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.
cos2 1 sin2 , sec2 1 tan2 , tan2 sec2 1
P. 44
In order to express in terms of x, let us first introduce the following notations:
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
Inverse of Trigonometric FunctionsLet x be a real number. 1. sin–1x is defined as the angle such that sin x (where –1 x
1)
and .
2. cos–1x is defined as the angle such that cos x (where –1 x 1)and 0 .
3. tan–1x is defined as the angle such that tan x and .
2
π
2
π
2
π
2
π
P. 45
Example 8.26TFind
Solution:
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
.1 22 xx
dx
Let x sinThen dx cos d
2222 sin1sin
cos
1
d
xx
dx
cossin
cos2
d
d2csc
C cot
Since sin x,
1csccot 2
1sin
12
112
x
x
x21
Cx
x
xx
dx 2
22
1
1
P. 46
Example 8.27T
Solution:
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
Let x sinThen dx cosd.
Since sin x,
.1 22 dxxx
dxxx 22 1
d22 cossin
d
2
2
2sin
d2sin4
1 2
C
4
4sin
8
1
,1cos 2x
2sin2cos24sin )sincos2)(sin21(2 2
23 1)2(4 xxx
dxxx 22 1
Cxxx
x
4
1)2(4sin
8
1 231
Cxxxx ]sin1)2[(8
1 123
d2
4cos1
4
1
dcossin1sin 22
)1)(21(4 22 xxx
Find
P. 47
Example 8.28T
Solution:
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
Let x 3tanThen dx 3sec2 d.
.9 22 xx
dx
22 9 xx
dx
sectan27
sec32
2 d
dcot
sin
cos
cos
1
9
1
Since ,3
tanx
2cot1csc x
2tan
11
Cx
x
xx
dx
9
9
9
2
22
22
2
tan99tan9
sec3 d
d2tan
sec
9
1
dcotcsc9
1
C csc9
1 2
91
x
x
x 92
Find
P. 48
Example 8.29T
Solution:
8.8.33 Integration by SubstitutionIntegration by Substitution
C. Integration by Trigonometric SubstitutionC. Integration by Trigonometric Substitution
.342
xx
dx
1)2(34 22 x
dx
xx
dx
Let x + 2 sec
342 xx
dx
tan
tansec d
d
tansec
tansecsec
tansec
)tan(secd
C tansecln
Since sec x + 2, 341)2(1sectan 222 xxx
Cxxxxx
dx
342ln34
2
2
1sec
tansec2
d
dsec
Find
1
x + 2 __________(x 2)2 1
Then dx sectan d.
P. 49
Some indefinite integrals such as , and
cannot be found by using the techniques we have
learnt so far.
8.8.44 Integration by PartsIntegration by Parts
xdxln xdxx sin
dxxe x
To evaluate them, we need to introduce another method called integration by parts.
Theorem 8.3 Integration by PartsIf u(x) and v(x) are two differentiable functions, then
In other words, .
. vu'dxuvuv'dx
vduuvudv
Proof:Suppose u and v are two differentiable functions.
Since (uv) uv + vu, by definition, .dx
dCuvdxvu'uv' )(
Cuvvu'dxuv'dx vu'dxuvuv'dx
P. 50
From Theorem 8.3, we can see that the problem of finding
can be transformed into the problem of finding instead.
8.8.44 Integration by PartsIntegration by Parts
If the integral is much simpler than , then the original integralcan be found easily.
If we want to apply the technique of integration by parts to an integral,
such as , we need to transform the integral into the form
first, such as or .
udv vdu
udvvdu
udv xdxe x sin
)()(sin xedx )cos( xde x
P. 51
Example 8.30TFind
Solution:
8.8.44 Integration by PartsIntegration by Parts
.ln xdxx
2lnln
2xxdxdxx
)(ln
22)(ln
22
xdxx
x
xdxxx
2
1
2
ln2
dxx
xxx 1
22
ln 22
Cxxx 42
ln 22
P. 52
Example 8.31T
Solution:
8.8.44 Integration by PartsIntegration by Parts
.log52 xdxx
dx
xx
xdxx
5ln
ln
log2
52
3ln
5ln
1 3xxd
)(ln
33)(ln
5ln
1 33
xdxx
x
dxx
xxx 1
3
1
3
ln
5ln
1 33
dxx
xx 23
3
1
3
ln
5ln
1
Cx
xx
9ln
35ln
1 33
Find
P. 53
In some cases, there may be more than one choice for u and v.
8.8.44 Integration by PartsIntegration by Parts
For example, we can transform into
However, if we choose the former, then
xdxx sin
)(sin
22)(sin
2sin
222
xdxx
xx
xd
xdxxxx
cos2
1
2
sin 22
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.
2
sin2x
xd )(cos xxdor .
P. 54
Example 8.32T
Solution:
8.8.44 Integration by PartsIntegration by Parts
.csc2 xdxx
xdxx 2csc
)cot( xxd
xdxxx cot)cot(
dxx
xxx
sin
coscot
)(sinsin
1cot xd
xxx
Cxxx sinlncot
Find
P. 55
Example 8.33T
Solution:
8.8.44 Integration by PartsIntegration by Parts
(a) Show that
(b) Hence find .
.1cos
1
2cot
x
x
dx
d
dx
x
x
1cos
22csc
2cot 2 x
dx
dxx
dx
d
2csc
2
1 2 x
2sin2
1
2 x
2
cos12
1x
1cos
1
x
(a) (b)
2cot
1cos
xxddx
x
x
dxxx
x2
cot2
cot
22
sin
2cos
22
cotx
dx
xx
x
2sin
2sin
12
2cot
xd
xx
x
Cxx
x 2
sinln22
cot
P. 56
Example 8.34T
Solution:
8.8.44 Integration by PartsIntegration by Parts
.2 dxex x
)(2 xedx
])([ 22 xdeex xx
)2( 2 dxxeex xx
)(22 xx exdex
])2(2[2 xdexeex xxx
dxexeex xxx 222
Cexeex xxx 222
Cexx x )22( 2
dxex x2
Find
P. 57
Example 8.35T
Solution:
8.8.44 Integration by PartsIntegration by Parts
dxx)cos(ln
.)cos(ln dxx
)][cos(ln)][cos(ln xxdxx
dxx
xxxx1
)sin(ln)cos(ln
dxxxx )sin(ln)cos(ln
)][sin(ln)][sin(ln)cos(ln xxdxxxx
dxx
xxxxxx1
)cos(ln)sin(ln)cos(ln
dxxxxxx )cos(ln)sin(ln)cos(ln
Therefore, Cxxxxdxx )sin(ln)cos(ln)cos(ln2
Cxxx
dxx )]cos(ln)[sin(ln2
)cos(ln
Find
P. 58
In previous chapters, we learnt that of a curve y F(x) is the slope function of the curve.
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
Since integration is the reverse process of differentiation,
if we let f(x), then by the definition of integration, we have
where C is an arbitrary constant.
Thus we can see that the equation of a family of curves y F(x) + C
can be found by integration, providing that the slope function of the curve is known.
A. Geometrical ApplicationsA. Geometrical Applications
dx
dy
dx
dy
dxxfy )(
CxF )(
dx
dy
P. 59
Example 8.36TThe equation of the slope of a curve at the point (x, y) is given by
. If the curve passes through (2, 2), find the equation of
the curve. Solution:
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
A. Geometrical ApplicationsA. Geometrical Applications
6)32( xdx
dy
6)32( xdx
dy
dxxy 6)32( ,72
)32( 7
Cx
where C is an arbitrary constant
Cx 14
)32( 7
When x 2, y 2, we have
14
2714
)322(2
7
C
C
∴ The equation of the curve is .14
27
14
)32( 7
xy
P. 60
Example 8.37TIt is given that at each point (x, y) on a certain curve, . If
the curve passes through and , find the equation of that curve. Solution:
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
A. Geometrical ApplicationsA. Geometrical Applications
xdx
yd 22
2
sin2
xdx
yd 22
2
sin2
dxxdx
dy 2sin2
dxx)2cos1(
dxC
xxy 12
2sin
21
2
4
2cos
2CxC
xx
Since and lie on the curve, we have
4
5 ,0
4
5 ,
14
10
4
5
2
2
C
C
12
2sinC
xx
1
2
1
1
2
2
14
1
24
5
C
C
C
∴ The equation of the curve is
.124
2cos
2
2
xxxy
4
5 ,0
4
5 ,
P. 61
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
where s is the displacement of the particle at time t.
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
Since integration is the reverse process of differentiation, we have
B. Applications in PhysicsB. Applications in Physics
,and2
2
dt
sd
dt
dva
dt
dsv
P. 62
Example 8.38TA particle moves along a straight line such that its velocity v at time t is given by . Find the displacement s of the particle at time t, given that s 6 when t 2.
Solution:
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
B. Applications in PhysicsB. Applications in Physics
52 ttv
dttts 52 )5(52
1 22 tdt
Ct 2
32 )5(
3
2
2
1Ct 2
32 )5(
3
1
When t 2, s 6, we have
3
)52(3
16 2
32
C
C
3)5(3
1 2
32 ts
P. 63
Example 8.39TA particle moves along a straight line so that its acceleration a at time t is
given by a for t > 0. When t 1, the velocity of the particle is 8 and
its displacement is 24. Find the displacement of the particle at time t. Solution:
8.8.55 Applications of Indefinite Applications of Indefinite IntegralsIntegrals
B. Applications in PhysicsB. Applications in Physics
t
1
dtt
v1
1ln Ct
When t 1, v 8, we have
81ln8
1
1
C
C
8ln tv
ttdt
dtts
8ln
)8(ln
tdtt
ttt
tttdtt
81
ln
8)(lnln
27ln8ln
Cttttdttt
When t 1, s 24, we have
17)1(71ln124
2
2
C
C
177ln ttts
P. 64
8.1 Concepts of Indefinite Integrals
Chapter Chapter SummarySummary
)()]([ xfxFdx
d
CxFdxxf )()(
1. If , then the indefinite integral of f(x)
is defined by , where C is an arbitrary constant.
2. 1. where,1
1 1
nCxn
dxx nn
Cedxe xx
Cxdxx
ln1
constant. a is where, kCkxkdx 3.
4.
5.
P. 65
8.1 Concepts of Indefinite Integrals
Chapter Chapter SummarySummary
6.
7.
Cxxdxx csccotcsc
Cxxdx cotcsc2
Cxxdx sincos
Cxxdxx sectansec
Cxxdx tansec2
Cxxdx cossin
8.
9.
10.
11.
12.
13. dxxgdxxfdxxgxf )()()()(
constant. zero-non a is where,)()( kdxxfkdxxkf
P. 66
8.2 Indefinite Integration of Functions
Chapter Chapter SummarySummary
Let a and b be real numbers with a 0.
1.
2.
3.
4.
.1 where,)1(
)()(
1
nCna
baxdxbax
nn
Cbaxa
dxbax
ln
11
Cea
dxe baxbax 1
Cbaxa
dxbax )cos(1
)sin(
Cbaxa
dxbax )sin(1
)cos(5.
P. 67
1. Let u g(x) be a differentiable function. Then, 8.3 Integration by Substitution
Chapter Chapter SummarySummary
.)()()]([ duufdxxg'xgf
2. If the integrated involves terms like , and , we can simplify the integrand by substituting x a sin , x a tan or x a sec respectively.
22 ax 22 xa 22 xa
P. 68
If u(x) and v(x) are two differentiable functions, then8.4 Integration by Parts
Chapter Chapter SummarySummary
. vduuvudv
P. 69
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
where F (x) f(x).
8.5 Applications of Indefinite Integrals
Chapter Chapter SummarySummary
CxFdxxf )()(
2. Let s, v and a be the displacement, velocity and acceleration of a particle moving along a straight line respectively, then
. and adtvvdts