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Файл:Mathematics for Foreign Students integrals. Study Guide
.pdf
11
22)
Cxxxx +−+−++ 9ln115ln3
22
;
23)
C
x
arctg +
77
4
; 24)
C
x
x
+
+
−
−
4
4
ln
16
5
;
25)
C
x
xxx +−−
8
6
3
2
; 26)
C
x
+
7
3
arcsin
3
3
;
27)
;23ln
2
Сxxx ++−
28)
Сxxxxх ++− 8410
4
3
10
3
.
Often we are required to integrate functions like those in the standard
list, but where x is replaced by a linear function of x, as a following example,
Find the integration,
+ dxx
8
)14(
, which is similar to
dxx
8
, except that
x is replaced by (4x+1), so that
=+
+
=+
C
x
dxx
94
)14(
)14(
9
8
C
x
+
+
36
)14(
9
.
Where the corresponding standard integral is
dxx
8
=
C
x
+
9
9
. We
note that, even if we replace x by (4x + 1), the ‘power’ rule will still apply,
i. e. (4x + 1) replaces the single x in the result as long as we divide by the co-
efficient of x, in this case is 4.
This will always happen when we integrate functions that consist of
a linear function of x.
e. g.
+= Cedxe
xx
+=
−
−
C
e
dxe
x
x
8
38
38
,
i. e. (8x –3) replaces x in the integral, then (8x –3) in the result too, provided
that we also have to divide by the coefficient of x.
Similarly, since
+= Cxxdx cossin
,
then
=+ dxx )56sin(
=+ dxx )56sin(
Cx+
+6)56cos(
.
12
So if a linear function of x replaces the single x in the standard integral,
the same linear function of x replaces the single x in the result, as long as we
divide by the coefficient of x in the result.
Examples. Solve the following integrals:
1.
;
3
32
32
C
e
dxe
x
x
+−=
−
−
2.
C
x
x
dx
+
+
=
+
2
52ln
52
;
3.
=dxe
x8
C
e
x
+
8
8
;
4.
=+ dxx
7
)6(
C
x
+
+
8
)6(
8
;
5.
=+ dxx 32
=+ dxx
2/1
)32(
C
x
+
+
2
3
2
)32(
2/3
C
x
+
+
=
3
)32(
3
;
6.
=
− 2
5
x
dx
Cx +− |2|ln5
;
7.
CxC
x
x
dx
+−=+
−
=
−
12
2
122
12
.
Do the following exercises and test Yourself
1.
− dxx
7
)15(
; 2.
+ dxx )24cos(
;
3.
−
dxe
x 32
; 4.
+113x
dx
;
5.
− dxх)35sin(
; 6.
+dxx25
4
;
13
7.
− dxxtg )15(
; 8.
− )53(cos
2
x
dx
;
Answers:
1)
C
x
+
−
40
)15(
8
; 2)
C
x
+
+
4
)24sin(
;
3)
32
2
1
−x
e
+C; 4)
C
x
+
+
3
)113ln(
;
5)
Cx+
−3)35cos(
; 6)
C
x
+
+
4ln
4
2
1
25
;
7)
Cx+
−
−
5
|)15cos(|ln
; 8)
C
xtg
+
−3)53(
.
Integration by Substitution
= dtttfdxxf )())(()(
.
Examples. Find the integral
1.
.
5
3
2
+x
dxx
To solve this example have to use the method of substitution as a following
Solution:
;5ln
3
1
ln
3
1
3
1
3
3
5
5
3
2
2
3
3
2
СxСt
t
dt
dt
dxx
dxxdt
xt
x
dxx
++=+==
=
=
+=
=
+
14
2.
;
26
1
22
1
3
1
4
3
1
3
3
4
3
2
2
2
3
6
2
С
x
arctgС
t
arcctg
t
dt
dt
dxx
dxxdt
xt
x
dxx
+=+=
+
=
=
=
=
=
+
3.
С
С
4.
=+==
=
=
= С
t
dtt
xdxdt
xt
xdxx
4
cos
sin
cossin
4
33
С
x
+
4
sin
4
;
5.
+
+
−
=
−
=
=
=
=
−
;
3
3
ln
32
1
33
22
С
t
t
t
dt
dxedt
et
e
dxe
x
x
x
x
6.
=+−===
=
=
=
= Ctdtt
dt
t
dt
dx
dtdx
tx
xdx cos
3
1
sin
3
1
3
sin
3
3
3
3sin
Cx +−= 3cos
3
1
;
7.
+=+==
=
=
=
= CeCe
dt
e
dt
dx
dtxdx
tx
dxex
xttx
22
2
1
2
1
2
2
2
2
;
8.
=++=+=
=
=
=
+
Ct
t
dtt
dx
x
dt
xt
x
dxx
3
)1(
1
ln
)1(ln
3
2
2
Cx
x
++= ln
3
ln
3
;
15
9.
( )
( )
==
=+
=+
=+
+
dt
dtdxx
txx
dxx
txx
2
13
132
2
3
2
3
CC
xxt
+=+=
+
2ln
2
2ln
2
3
;
10.
===
=
−
=
=
−
dttdtt
dtdx
x
tx
dx
x
x
3
1
3
2
2
3
1
1
arcsin
1
arcsin
CxC
t
+=+=
3
4
3
4
arcsin
4
3
3
4
;
11.
====
=
=
=+
=
+
−
3
2
3
1
3
1
3
1
3
3
1
1
3
2
3
1
3
2
2
3
3
3
2
t
dtt
dt
t
dt
dxx
dtdxx
tx
dx
x
x
( )
Cx+
+
=
2
1
3
2
3
;
12.
+=+=
−
=
=
=
=
=
−
CxCt
t
dt
dt
xdx
dtxdx
tx
x
xdx
2
2
2
4
arcsin
2
1
arcsin
2
1
1
2
1
2
2
1
.
Do the following exercises and test Yourself
1.
dxxx
43
sin
; 2.
+ 9
4
x
xdx
;
3.
( )
xdxx cossin1
3
2
−
; 4.
dx
x
e
x
;
16
5.
( )
− xx
dx
32
arcsin1
; 6.
−
−
dx
xx
x
2
)sin(
cos1
;
7.
+11
3
2
x
xdx
; 8.
x
xdx
3
ln
;
9.
+
dx
x
x
5
8
7
; 10.
xdx
x
cos3
sin
;
11.
dxxe
x
2sin
2
sin
; 12.
( )
+ dxxx
10
52
.
Answers:
1)
Cx +−4cos
4
1
; 2)
C
x
arctg +
36
1
2
;
3)
( )
C
x
+
−
−
5
sin13
3
5
; 4)
Ce
x
+2
;
5)
C
x
+
−
arcsin
2
; 6)
C
xx
+
−
−
sin
1
;
7)
Cx ++11ln
2
3
2
; 8)
C
x
+
4
ln
4
;
9)
Cx ++ 5ln
8
1
8
; 10)
C
x
+
3ln
3
sin
;
11)
Ce
x
+
2
sin
; 12)
( ) ( )
C
xx
+
+
−
+
44
525
48
52
1112
.
Integration by parts
The formula for integration by parts states as a following
−= vduuvudv
.
Which is fulfilled for any differentiable functions u(x) and v(x).
17
Example.
xdxx ln
2
.
We have two factors, they are x2 and ln x, and we have to decide which
one better to take as u and which one as dv. If we choose x2 to be u and ln x
to be dv, then we shall have to integrate ln x in order to find v. Unfortunately,
xdxln
is not in our basic list of standard integrals, therefore we must allo-
cate u and dv by another way, i. e. let ln x = u and x2 =dv.
dx
x
x)
x
x(
x
v =dv x
du dx
x
x = u
xdxx
−=
=
=
=
1
3
1
3
ln
3
1
ln
ln
3
3
3
2
2
.
Notice that we can tidy up - the writing of the second integral by writing the constant factors involved, outside the integral.
=−=−=
dxxx
x
dx
x
x)
x
x(xdxx
2
3
3
3
2
3
1
ln
3
1
3
1
3
lnln
Cx
x
C
x
x
x
+
−=+−=
3
1
ln
333
1
ln
3
333
.
And notice that if one of the factors of the product which to be integrated is a log function, this must be chosen as u.
Example.
dxex
x32
.
=
=
=
=
xx
x
ev dv=e
xdx duu=x
dxex
33
2
32
3
1
2
=−=
xdxe
e
x
xx3
3
2
3
2
3
=
=
=
xx
ev dv=e
dx duu=x
33
3
1
=++−=
−−=
C
exeex
dxe
e
x
ex
xxx
x
xx
39
2
9
2
33
1
33
2
3
3332
3
332
C
x
x
e
x
+
+−=
9
2
3
2
3
2
3
.
18
In Example 1 we saw that if one of the factors is a log function, this
function must be taken as u.
In Example 2 we saw that, there is not log function, and the power of
x is taken as u. (By the way, this method is good choice only for positive
whole-number powers of x. For other powers, a different method must be
applied.)
Example.
xdxe
x
sin
3
.
Here we have neither a log factor nor a power of x.
=
=
=
=
x-v x dv =
dxeduu = e
xdxe
xx
x
cossin
3
sin
33
3
=+−=−−−= dxxexedxexx)(e
xxxx 3333
cos3cos)cos(3cos
=
=
=
=
xv x dv =
dxeduu = e
xx
sincos
3
33
( )
−+−= dxexxexe
xxx 333
sin3sin3cos
,
and it looks like we are back where we started. So to solve this problem, let’s
I =
xdxe
x
sin
3
, we have the following equality
IxexeI
xx
9sin3cos
33
−+−=
.
Then, treating this as a simple equation, we get
1
3
)cossin3(10 CxxeI
x
+−=
;
1
3
)cossin3(
10
Cxx
e
I
x
+−=
.
Whenever we integrate functions of the form ekx sin x or ekx cos x, we get
similar types of results after applying the rule of integration by parts twice.
Examples: Solve the following integral
1.
.
19
Solution:
( )
( )
=
+=+=
=
+=
=
=+
x
x
dxxv
dx
x
du
dxxdv
xu
xdxx
3
3
3
1
)3(
ln
ln3
3
2
2
2
=
=
+−
+
dx
x
x
x
x
x
x
1
3
3
3
3
ln
33
;3
9
3
3
ln
33
Cx
x
x
x
x +−−
+
2.
=
==
==
=
xvxdv
dxduxu
xdxx
3sin
3
1
3cos
3cos
=−
xdxxx 3sin
3
1
3sin
3
1
C
xxx
Cxxx +−=+−=
9
3cos
3
3sin
3cos
3
1
3
1
3sin
3
1
.
Do the following exercises and test Yourself
1.
xdxx2ln
;
2.
;
3.
xdxx 2sin
2
.
Answers:
1)
;
2)
3)
Cxx
x
xx ++− 2cos
4
1
2sin
2
2cos
2
1
2
.
20
Integration by partial fractions
Let
)()(xQ
xP
be a rational function, and let degree of the polynomial P(x)
be greater than or equal to degree of the polynomial Q(x). Then there exist
the uniquely determined polynomials S(x) and R(x), such that the rational
function
)()(xQ
xP
can be represented in the form
)(
)(
xQ
xP
= S(x) +
)(
)(
xQ
xR
,
where
)(
)(
xQ
xR
is a proper fraction. The polynomial S(x) is called the quotient,
where the term Q(x) is the divisor and the expression R(x) is called the remainder.
Example. Solve the following integral
dx
x
xx
+
++
4
69
4
.
Solution: We will solve it by the following method,
226
22055
655
6416
916
164
04
55164 4
4 6900
2
2
23
23
2334
234
−−
+−
+
+
−−
+−
−+−+
+++++
x
x
xx
xx
xx
xx
xxxxx
хxxxx
=
+
++
dx
x
xx
4
69
4
=
+
+−+− dx
x
xxx
4
226
55164
23
4ln226558
3
4
4
2
34
++−+−= xxx
xx
+С.
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