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Файл:Mathematics for Foreign Students integrals. Study Guide
.pdf
Example. To solve
xdx3sin
2
xdx3sin
2
=
−
=
dx
x26cos1
=−
xdxdx 6cos
2
1
2
1
Cxx +− 6sin
12
1
2
1
dx
x
2
cos
4
dx
x
2
cos
4
dx
x
2
2
2
cos
=
+
=
dx
x
2
2
cos1
( )
=++
dxxx2coscos21
4
1
=
+
++
dx
x
xdxdx
2
2cos141
cos
2
1
4
1
+++ xxx
8
1
sin
2
1
4
1
=++ Cx2sin
16
1
Cxxx +++ 2sin
16
1
sin
2
1
8
3
xdxx42cossin
= xdxx42cossin
=
+
−
dx
xx
2
2
2cos1
2
2cos1
( )
( )
=+−= dxxx 2cos12cos1
8
1
2
( )
=+− dxxx 2cos12sin
8
1
2
( )
=+
−
= dxxx2cos1
2
4cos181
( )
=−−+= dxxxxx 2cos4cos4cos2cos1
16
1
−
−+=
4
4sin
2
2sin
16
1 xx
x
Cxx+
+
2
2sin
6
6sin
2
1
Solution.
.
=
Example. To solve
Solution.
=
Example. To solve
Solution.
.
.
=
=
.
.
.
31
T H E U N I V E R S A L T R I G O N O M E T R I C S U B S T I T U T I O N
t
x
tg =
2
2
x
tg
arctgtx 2=
2
1
2
t
dt
dx
+
=
2
1
2
sin
t
t
x
+
=
2
2
1
1
cos
t
t
x
+
−
=
− x
dx
cos35
=
+
−
=
+
=
==
=
−
2
2
2
1
1
cos,
1
2
2,
2
cos35
t
t
x
t
dt
dx
arctgtxt
x
tg
x
dx
( )
=
+
−
−+
=
2
2
2
1
1
351
2
t
t
t
dt
( )
=
+
−
−+
2
2
2
1
33
51
2
t
t
t
dt
( )
=
+
+−+
+
=
2
22
2
1
3355
1
2
t
tt
t
dt
=
+
=
+
2
2
4
1
4
1
41
t
dt
t
dt
C
x
tgarctgCtarctg +
=+=
2
2
2
1
2
2
1
+− 2cossin xx
dx
For
we will use the substitution
is called the universal trigonometric substitution, so that by using that substitution, the functions sin x and cos x can be expressed through rational func-
, which
tions with respect to
Example. To solve
Solution.
and by using the double-angle formulas:
;
,
.
,
.
Example. Solve the following integral
.
.
32
33
Solution.
=
+− 2cossin xx
dx
=
+
−
=
−
=
+
==
2
2
2
2
1
1
cos
1
2
sin
1
2
2
t
t
x
t
t
x
t
dt
dx
x
tgt
( )
( )
=
+
+
+
−
−
+
+
+
=
2
2
2
2
2
2
12
1
1
1
2
1
1
2
t
t
t
t
t
t
t
dt
=
+++−
22
2212
2
ttt
dt
=
++ 123
2
2
tt
dt
=
++
=
3
1
3
2
3
2
2
t
t
dt
=
+
−
+
3
1
3
1
3
1
3
2
22
t
dt
=
=
+=
=
+
+
=
dtdz
tz
t
dt
3
1
2
9
2
3
1
3
2
=
+
9
2
3
2
2
z
dz
=+
+
=+= C
t
arctgC
z
arctg
2
13
2
2
3
3
2
1
3
2
C
x
tg
arctg +
+21
2
3
2
.
For
dxtgxR )(
or
dxctgxR )(
we will use the substitutions
ttgx =
or
tctgx =
.
Example. To solve
xdxtg
5
.
Solution.
xdxtg
5
=
+
+−=
+
=
+
=
=
=
= dt
t
t
ttdt
t
t
dt
t
dx
arctgtx
tgxt
2
3
2
5
2
11
1
1
34
=+++−−= Ct
tt
1ln
2
1
24
2
24
Cxtg
xtgxtg
+++− 1ln
2
1
24
2
24
.
Example.
xdxctg
4
.
Solution.
=
+
−=
+
−=
=
=
= dt
t
t
dt
t
dx
arcctgtx
tctgx
xdxctg
1
1
1
2
4
2
4
+++−=
+
+−−= Carctgtt
t
dt
t
t
3
1
1
1
3
2
2
.
Do the following exercises and test Yourself
1.
xdxx cossin
;
2.
+
xdx
x
x
cos
sin
2sin
2
;
3.
x
xdx
6
3
cos
sin
;
4.
++ xx
dx
cos3sin5
;
5.
− x
dx
sin1
;
6.
dx
x
x
6
3
sin
cos
;
35
Answers.
1)
C
x
+
2
sin
2
; 2)
Cxx++ sinln2
2
sin
2
; 3)
C
xx
+−
35
cos3
1
cos5
1
;
4)
C
x
tg
arctg +
+
15
2
21
15
2
; 5)
C
x
tg
+
−
2
2
;
6)
Cxx+−
53
sin5
1
sin3
1
.
Integrals Involving Rational Exponents.
1) Integrals with the rational exponents
n
x
1
can be transformed to in-
tegrals of rational functions by using the substitution x = un, which implies
ux
n
=
and dx=nu
n
-1
du.
2) Integrals with a several rational exponents can be evaluated by using the substitution x=un, where n is the least common multiple of the denominators of exponents.
3) Integrals involving expressions of the form
n
dcx
bax++
can be evalu-
ated by the substitution
n
u
dcx
bax
=
+
+
.
4) R(x,
( ) ( )
)ax+b,...,ax+b
l
l
n
m
n
m
1
1
dx, by using the substitution
ax+b=tk, dx=
dtkt
1
1k−
a
,
Example. To solve
+dxx
x
1
.
36
Solution.
+
dx
x
x
1
=
=
=
=
xt
tdtdx
tx
2
2
=
=
+
dt
t
tt
1
2
=
=
+
dt
t
t
1
2
2
=
+
+−
dt
t
t
1
11
2
2
( )( )
=
+
+
+
+−
t
dt
dt
t
tt
1
2
1
11
2
=
( )
=+++− Ctlndtt 1212
Ctlnt
t
+++− 122
2
2
2
=
=
Cxlnxx +++− 122
.
Example. To solve
+ x
dx
21
.
Solution.
+ x
dx
21
=
=
=
=
xt
tdtdx
tx
2
2
=
=
+
dt
t
t
21
2
=
=
+
−+
dt
t
t
21
112
=
+
−
t
dt
dt
21
Ctt ++− 12ln
2
1
=
=
Cxx ++− 12ln
2
1
.
Example. To solve
( )
+ xx
dx
3
1
.
Solution.
( )
+ xx
dx
3
1
=
=
=
=
6
5
6
6
xt
dttdx
tx
=
( )
=
+
dt
tt
t
32
5
1
6
=
=
+
dt
t
t
2
2
1
6
=
+
−+
dt
t
t
2
2
1
11
6
=
+
−
2
166t
dt
dt
=
Carctgtt +− 66
=
Cxarctgx +−6666
.
Example. To solve
dx
xx
x
+
+41
.
37
Solution.
dx
xx
x
+
+41
=
=
=
=
4
3
4
4
xt
dttdx
tx
=
=
+
+
dtt
tt
t
3
24
4
1
=
( )
( )
=
+
+
dt
tt
tt
1
1
4
22
3
=
+
+
dt
t
tt
2
2
1
4
=
+
+−+
dt
t
tt
2
2
1
11
4
=
=
+
−
+
+
22
1
4
1
44
t
dt
t
tdt
dt
Carctgt
t
td
t +−
+
+
+
4
1
)1(
24
2
2
=
=
Carctgttt +−++ 41ln24
2
=
=
Cxarctgxx +−++
44
41ln24
.
Example.
dxxx
−3
.
Solution.
dxxx
−3
=
−=
−=
=−
−=
tdtdx
tx
tx
xt
2
3
3
3
2
2
=
=
( )
dtttt
−− 2 3
2
=
( )
dttt 26
42
−−
=
C
tt
++−
5
2
3
6
53
=
=
( ) ( )
Cxx +−+−−
53
3
5
2
32
.
Example.
dx
x
x
−1
.
Solution.
dx
x
x−1
=
=
+=
=−
−=
tdtdx
tx
tx
xt
2
1
1
1
2
2
=
=
+
dt
t
tt
2
1
2
38
=
=
+
dt
t
t
2
2
1
2
=
+
−+
dt
t
t
2
2
1
11
2
=
+
−
2
1
22
t
dt
dt
=
Carctgtt +−22
= .
Example. To solve
+−
+
dx
x
x
315
2
.
Solution.
=
+
+
+
=
=
+
=
−=
−=
=
+−
+
dt
t
t
t
dt
t
dx
t
x
xt
xt
dx
x
x
5
2
3
2
5
1
5
2
5
1
15
15
315
2
2
2
2
( )
=
+
+
=
+
+
= dt
t
tt
dt
t
tt
3
11
25
2
3
11
25
2
32
=+
+−+−
−
=
+
−
++−=
Ctt
tt
dt
t
tt 3ln60 20
2
3
325
2
3
60
203
25
2
23
2
( ) )
.315ln60152015
2
3
3
15
25
2
Cxxx
x
++−
−−+−
−
=
❖ The integration of functions which contain a radical expression of
the form
cbxax ++
2
.
To calculate such an integral, first of all we need to complete the square in
the quadratic expression:
=
++=++
a
c
x
a
b
xacbxax
22
−+
+==
+
−
++=
2222
2
2222 a
b
a
c
a
b
xa
a
c
a
b
a
b
x
a
b
xa
39
Then we will use the substitution:
a
b
xt2+=
dtdx =
.
Example. To solve
++ 104
2
xx
dx
.
Solution.
++ 104
2
xx
dx
=
+−++ 10444
2
xx
dx
=
=
( )
++ 62
2
x
dx
=
=
=+
dtdx
tx
2
=
+ 6
2
t
dt
=
=
Ctt +++ 6ln
2
=
Cxxx +++++ 1042ln
2
.
Example. To solve
+− 18
2
xx
dx
.
Solution.
+− 18
2
xx
dx
=
+−+− 116168
2
xx
dx
=
=
( )
−− 154
2
x
dx
=
=
=−
dtdx
tx
4
=
−15
2
t
dt
=
=
Ctt +−+ 15ln
2
=
Cxxx ++−+− 184ln
2
.
Example. To solve
+ xxdx6
2
.
Solution.
+ xx
dx
6
2
=
−++ 996
2
xx
dx
=
=
( )
−+ 93
2
x
dx
=
==+dtdxtx
3
=
−9
2
t
dt
=
40
=
Ctt +−+ 9ln
2
=
Cxxx ++++ 63ln
2
.
Example. To solve
−−
2
23 xx
dx
.
Solution.
−−
2
23 xx
dx
=
−−++− )3112(
2
xx
dx
=
=
( )
+−
2
14 x
dx
=
dtdxtx=
=+ 1
=
−24 t
dt
=
Ct+
2
arcsin
=
Cx+
+21
arcsin
.
Example. To solve
++ 52
2
xx
xdx
.
Solution.
++ 52
2
xx
xdx
=
+−++ 5112
2
xx
xdx
=
=
( )
++ 41
2
x
xdx
=
=
−=
=+
dtdx
tx
tx
1
1
=
+
−
4
)1(
2
t
dtt
=
=
+
−
+ 44
22
t
dt
t
tdt
=
=
=
=+
2
2
4
2
dz
tdt
dztdt
zt
=
=
Ctt
z
dz
+++−
4ln
2
1
2
=
Cttz +++− 4ln2
2
1
2
=
=
Cxxxxx +++++−++ 521ln52
22
.
Example. To solve
+− 542xx
xdx
.
Solution.
( )
=
+
+
=
=
+=
−=
=
+−
=
+−
dt
t
t
dtdx
tx
xt
x
xdx
xx
xdx
1
2
2
2
12
54
222
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