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Chapter 4: Methods of integrals
Definition
Section 4.1: Indefinite Integrals
The process of finding the indefinite integral of f(x) is called integration of f(x) or integrating f(x). If we need to be specific about the integration variable we will say that we are integrating f(x) with respect to x.
Properties of the Indefinite Integral:
need. weas functionsmany as to
extended becan rule Thisintegrals. individual theof difference
or sum theis functions of differenceor sum a of integral the
s,other wordIn 2 .)()()()( dxxgdxxfdxxgxf
integrals. indefinite ofout constants tivemultiplicafactor can
weSo, number.any iswhere1 ,)()(. kdxxfkdxxfk
.)()()()( dxxgdxxfdxxgxf
.)(
)(
)(
)(
dxxg
dxxfdx
xg
xf
!!!!! Warning !!!!!
.1,1
1
ncn
xdxx
nn
.constants areand, kcckxkdx
.cossin cxxdx .sincos cxxdx
.tansec2 cxxdx .sectansec cxxdxx
The first integral that we will look at is the integral of a power of x.
The general rule when integrating a power of x we add one onto the exponent and then divide by the new exponent. It is clear that we will need to avoid n = -1 in this formula. If we allow in this formula we will end up with division by zero.
Next is one of the easier integrals but always seems to cause problems for students.
Let us now take a look at the trigonometric functions:
Example:
.cedxe xx .||ln1 1 cxdxxdxx
.4105)( 63 dttta
.)( 88 dxxxb
.424
54
510
45
4105
5454
63
cttt
cttt
dtdttdtt
.7979
797988 c
xxc
xxdxxdxx
.1 cydy .)( dyc
Now, let us take care of exponential and logarithm functions.
.6
773)(
54 3 dx
xxxd dx
xdx
xdxx
1
6
7173
54 3
dxxdxxdxx 2
154
13
6
773
cxxx
1216
7
157
143
3
12
1151
4
3
cxxx
3
7
4
7
7
12 2
144
7
.4)( 23 dwwwwe
.1524
)(3
2610
dxx
xxxf
.44 3233 dwwwwww
dxxdxxdxx 137 1524
.44 3
7
3
13
dwwwww
.
310
34
442
43
10
3
442
cwwww
.10
33
42
3
10
3
442 c
ww
ww
cxxx
||ln154
28
448
cxxx
||ln1522
48
Integration by substitution:
dyewdwdtt
dxx ycos13
4
dyeydwww
wyy2418lncos
11
dttt
tdxxx
34
34 32
)2(
125618
After the last section we now know how to do the following integrals
However, we can’t do the following integrals:
Let us start with the first one
dxxx 4 32 5618
In this case let us notice
duudxxxdxxx 4
124
134 32 18565618
56 3 xu
and we compute the differential dxxdxudu 218'
Now, we go back to our integral and notice that we can eliminate every x that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential
Evaluating the integral gives,
c
xc
uduudxxx
5
564
5
45618
4
534
5
4
14 32
Example:
Solution:
dwww
wa lncos
11)(
dxxxc432 103)(
dyeyb yy24183)(
dx
x
xd
241)(
(a) In this case let us take wwu lnand we compute the differential
dww
dwudu
11'
Now, we go back to our integral and notice that we can eliminate every w that exists in the integral and write the integral completely in terms of u using both the definition of u and its differential
cwwcuduu
dww
wwdwwww
lnsin)sin(cos
11lncoslncos
11
dyeyb yy24183)( In this case let us take yyu 24
and we compute the differential dyydyudu 18'
Thus we get
ceceduedyyedyey yyuuyyyy 222 444 333183183
In this case let us take3103 xu
So, dxxdxudu 230'
Thus we get
dxxxdxxx )30(103
30
1103 243432
dxxxc432 103)(
c
xcuduu
150
103
150
1
30
153
54
dx
x
xd
241)(
In this case let us take 241 xu
So, dxxdxudu 8'
Thus,
dxxxxdxxdx
x
x8)41(
8
1)41(
412
122
12
2
cu
duu
218
1
8
1 2
1
2
1
cx 2
12 )41(
4
1
Integration by parts:
Example 1:
Solution: dxxex Find
dxex x2 FindExample 2: Solution:
.,., Then Let xx evdxdudxedvxu
c. xxxxx exedxexedxxe
.,2., Then Let 2 xx evxdxdudxedvxu
dxxeexdxex xxx 222
c22 xxx exeex
C222 xxx exeex
Inserting this integral in the first one yields
Bringing the last term to the left hand side and dividing by 2 gives
Example 3:
Solution:
xdxex cos Find
.sin,.cos, Then Let xvdxeduxdxdveu xx
.sinsincos Thus, xdxexexdxe xxx
So,Then Let
. integral theagain partsby integrate Now
.cos,.sin,
sin
xvdxeduxdxdveu
xdxe
xx
x
.coscos)cos(cossin xdxexedxxexexdxe xxxxx
.coscossincos xdxexexexdxe xxxx
xexexdxe xxx cossincos2 .cossin
2
1cos cxexexdxe xxx
Example 4:
Solution:
.5
,1
.,ln 5
4 Then Letx
vdxx
dudxxdvxu
.1
5ln
5ln Thus,
554 dx
x
xx
xxdxx
.ln Find 4 xdxx
dxxxx 4
5
5
1ln
5
c55
1ln
5
55
x
xx
c.25
ln5
55
x
xx
Example 5:
Solution:.,
1..1,ln Then Let xvdx
xdudxdvxu
.1
lnln Thus, dxx
xxxxdx
.ln Find xdx
dxxx .1ln
cln xxx
Integration by partial fractions:
Main Rules:
Example 1:
First we write
Thus
Integrating, we get
Distinct roots:
)2)(1( Integrate
xx
xdx
Solution:
21)2)(1(
x
B
x
A
xx
x
getting ,)2)(1(by equation hismultiply t Now x-x-
)1()2( xBxAx
;1 so ,)21(1get we,1 substitute weIf -AAx
.2 so ,)12(2get we,2 letting Now BBx
2
2
1
1
)2)(1(
xxxx
x
.|2|ln2|1|ln)2(
2)1()2)(1(
cxxx
dx
x
dx
xx
xdx
Integrating, we obtain
Therefore
Example 2:
)3)(1(
)3( Integrate
2
2
xx
dxx
Solution:
)3)(1)(1( getting ,by equation hismultiply t Now xxx
;4/1 so ,)4)(2(3-1get we,1 Substitute -AAx
and writeFirst we ),1)(1()1( 2 xxx
)3()1()1()3)(1)(1(
)3(
)3)(1(
)3( 2
2
2
x
D
x
B
x
A
xxx
x
xx
x
)1)(1()3)(1()3)(1()3( 2 xxDxxBxxAx
;2/1 so ,)2)(2(3-1get we,1 Substitute BBx
;4/38/6 so ,)2)(4(3-9get we,3 Substitute CCx
)3(
1
4
3
)1(
1
2
1
)1(
1
4
1
)3)(1(
)3(2
2
xxxxx
x
.|3|ln4
3|1|ln
2
1|1|ln
4
1
)3)(1(
)3(2
2
cxxxxx
dxx
which we can integrate term by term
Thus
Therefore
Multiple roots:Example 3:
First we write
)1(
)1( Integrate
2
2
xx
dxx
Solution:
1)1(
)1(22
2
x
D
x
B
x
A
xx
x
1
211
)1(
)1(22
2
xxxxx
x
22 )1()1(1 DxxBxAxx
.1 so ,)1(1get we,0 ngSubstituti BBx
; of eother valuany take we find To .2get we,1 ngSubstituti xADx
.1soand,22-
so ,2)2)(1()2)(1(2get we,1 Substitute
AA
Ax
.|1|ln21
||ln)1(
)1(2
2
cxx
xxx
dxx
Example 4:
Solution:
First we write
Multiply by we get
Thus
Integrating, we obtain
8
3,So.
8
9
8
85121
8
5
2
33
,8
5
2
33331get we0 Substitute
AA
ADBA x
3)1(1)1)(3(
)1(22
2
x
D
x
B
x
A
xx
x
22 )1()3()1)(3(1 xDxBxxAx
.8
5,)4(19get we,3 Substitute 2 DsoD x
.2
1),4(2get we,1 Substitute BsoB x
3
1
8
5
)1(
1
2
1
1
1
8
3
)1)(3(
)1(22
2
xxxxx
x
.|3|ln8
5
)1(2
1|1|ln
8
3
)1)(3(
)1(2
2
cxx
xxx
dxx
Solution:
Example 5:
No roots:
Main Rule:
Thus
cx
x
dx
x
dx
)
3(tan
3
1
391
222
ca
x
aax
dx
)(tan
1 122
9 Evaluate
2x
dx
Solution:
Example 6:
1)2(45)44(54 222 xxxxx
54 Evaluate
2 xx
dx
Here we can’t find real factors, because the roots are complex. But we can complete the square:
.)2(tan1)2(54
122
cxx
dx
xx
dx