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Файл:Mathematics for Foreign Students integrals. Study Guide
.pdf
51
Example:
dxex
x
+
1
0
2
)1(
.
Solution:
dxex
x
+
1
0
2
)1(
=
==
=+=
xx
evdxedv
dxduxu
22
2
1
,
,1
=
=
−+
1
0
2
)1(
2
1
x
ex
−
1
0
2
2
1
dxe
x
=
1
0
202
4
1
)2(
2
1
x
eee −−
=
=
4
1
4
1
2
1
22
+−− ee
=
4
1
4
3
2
−e
.
Example:
dxx
e
+
1
2
)ln1(
.
Solution:
dxx
e
+
1
2
)ln1(
=
=
==
+=+=
xvdxdv
x
dx
xdu,x)(u
,
)ln1(2ln1
2
=
−+exx
1
2
)ln1(
+−
e
x
dx
xx
1
)ln1(2
=
+−+−+
e
dxxee
1
22
)ln1(2)1ln1()ln1(
=
=
==
=+=
xvdxdv
x
dx
dux,u
,
ln1
=
−+−−
e
e
x
dx
xxxe
1
1
)ln1(214
=
=
( )
e
xee
1
12214 −−−−
=
222414 −++−− eee
=2e-1.
Example:
dxx
+
3
0
2
1
.
Solution:
J=
=
=
==
+
=+=
xvdxdv
x
xdx
du,xu
,
1
1
2
2
dx
x
x
xxx
+
−+
3
0
2
3
0
2
1
1
dx
x
x
+
−+
−
3
0
2
2
1
11
32
+
+
+
+
−
3
0
2
3
0
2
2
11
1
32
x
dx
dx
x
x
3
0
2
3
0
2
|1|ln132 xxdxx ++++−
)23ln(32 ++− J
)23ln(322 ++=J
)23ln(
2
1
3 ++=J
=
=
=
=
=
=
=
;
.
52
I N T E G R A T I O N O F R A T I O N A L F R A C T I O N S
b
a
dx
xQ
xP
)(
)(
dx
xx
x
++
+
1
0
2
342
32
=
+−+=
++=++
2
3
1)1(2
2
3
22342
222
xxxxx
++
2
1
)1(22x
2
1
1
0
t
x
dx
xx
x
++
+
1
0
2
342
32
dx
x
x
++
+
1
0
2
2
1
)1(2
32
dx
x
x
++
+
1
0
2
2
1
)1(
32
2
1
dt
t
t
+
+−
2
1
2
2
1
3)1(2
2
1
dt
t
t
+
+
2
1
2
2
1
12
2
1
+
+
+
2
1
2
2
1
2
2
1
2
1
2
2
1
t
dt
dt
t
t
+
+
+
+
2
1
2
2
2
1
2
2
2
1
2
1
2
1
2
1
t
dt
t
td
++
2
1
2
1
2
2121
1
2
1
ln
2
1 t
arctgt
this integral is calculated by the Newton-Leibniz formula.
We can find the primitive in this case in the same way as in the case of an
indefinite integral.
Example:
Solution:
(D=16-24=-8<0).
х+1=t, dx=dt, x=t-1,
=
=
.
, we get
=
=
=
=
=
=
=
=
=
53
54
=
−+− 22222
2
3
ln
2
9
ln
2
1
arctgarctg
=
=
222223ln
2
1
arctgarctg −+
.
Example:
dx
x
x
−
−
1
0
4
16
12
.
Solution:
)4)(2)(2(
12
)4)(4(
12
16
12
2224
++−
−
=
+−
−
=
−
−
xxx
x
xx
x
x
x
)2(
1
−х
А
.
)2(
2
+х
А
.
(D=0-16=-16<0)
4
2
+
+
х
NMx
.
We get the following amount
4)2()2()4)(2)(2(
12
2
21
2
+
+
+
+
+
−
=
++−
−
x
NМx
x
А
x
А
xxx
x
.
)2)(2)((
)4)(2()4)(2(12
2
2
2
1
+−++
++−+++=−
xxNMx
xxAxxAx
х1=2 and х2= -2, , х3=0 и х4=1.
х1=2
0)2(084122
21
+++=− NMAA
,
)4)(2)(2(
)2)(2)(()4)(2()4)(2(
)4)(2)(2(
12
2
2
2
2
1
2
++−
+−+++−+++
=
=
++−
−
xxx
xxNMxxxAxxA
xxx
x
55
1
323 A=
,
32
3
1
=A
.
х2=-2
0))2((8)4(01)2(2
21
+−+−+=−− NMAA
,
2
325 A−=−
,
32
5
2
=A
.
х3=0
2)2()0(4)2(42102
21
−++−+=− NMAA
.
32
3
1
=A
,
32
5
2
=A
.
N416
32
5
16
32
3
1 −−=−
, 4N=0, N=0.
х4=1
3)1()1(5)1(53112
21
−++−+=− NMAA
.
By substituting
32
3
1
=A
,
32
5
2
=A
, N=0, we get
М3
32
25
32
45
1 −−=
,
М3
32
12
−=
,
8
1
−=М
.
Therefore,
4
8
1
)2(
32
5
)2(
32
3
16
12
24
+
−
+
+
−
=
−
−
x
x
xxx
x
.
So that we get sum of integrals as a following
dx
x
x
−
−
1
0
4
16
12
=
=
+
−
+
+
−
1
0
2
4
8
1
)2(
32
5
)2(
32
3
dx
x
x
xx
=
+
−
+
+
−
1
0
1
0
2
1
0
48
1
)2(32
5
)2(32
3
x
xdx
x
dx
x
dx
=
=
1
0
1
0
2ln
32
5
2ln
32
3
++− xx
( )
+
+
−
1
0
2
2
4
4
2
1
8
1
x
xd
56
=
1
0
2
1
0
1
0
4ln
16
1
2ln
32
5
2ln
32
3
+−++− xxx
=
−− 21ln
32
3
20ln
32
3
−−
++−+−++ 41ln
16
1
20ln
32
5
21ln
32
5
40ln
16
1
++
=
+−−+− 5ln
16
1
2ln
32
5
3ln
32
5
2ln
32
3
1ln
32
3
4ln
16
1
+
=
5ln
16
1
4ln
16
1
2ln
32
8
3ln
32
5
++−
.
Calculation of definite integrals involving trigonometric expressions
To calculate definite integrals involving trigonometric expressions, we
will use the same integration methods were used for finding indefinite integrals
Example:
+
2
0
2
cos1
sin
x
xdx
.
Solution:
+
2
0
2
cos1
sin
x
xdx
=
0
2
1
0
sin
cos
t
x
dtxdx
tx
=−
=
=
+
−
0
1
2
1 t
dt
= =
=
)1(0 arctgarctg −−−
=
4
.
Example:
2
0
2
2sinsin
xdxx
.
Solution:
2
0
2
2sinsin
xdxx
=
2
0
2
cossin2sin
xdxxx
=
2
0
3
cossin2
xdxx
=
0
1
arctgt−
57
=
1
2
0
0
cos
sin
t
x
dtxdx
tx
=
=
=
1
0
3
2 dtt
=
1
0
4
4
2
t
=
2
1
4
1
2 =
.
Example:
4
0
2
2cos
xdx
.
Solution:
4
0
2
2cosxdx
==
+
4
0
2
4cos1
dx
x
=
+
4
0
4
0
4cos
2
1
2
1
xdxdx
=
=
4
0
4sin
4
1
2
1
2
1
+ xx
=
( )
0sinsin
8
1
42
1
−+
=
8
.
Example:
3
0
3
cos
sin
x
xdx
.
Solution:
3
0
3
cos
sin
x
xdx
=
3
0
2
cos
sinsin
x
xdxx
=
( )
−
3
0
2
cos
sincos1
x
xdxx
=
=
2131
0
sin
cos
t
x
dtxdxtx=−
=
=
−
−
2
1
1
2
1
dt
t
t
=
+−
2
1
1
2
1
1
1
tdtdt
t
=
2
1
1
2
2
||ln
+−tt
=
=
+−−
+−
2
1
1ln
2
2
1
2
1
ln
2
=
8
3
2ln −
.
Example:
4
0
3
xdxtg
.
58
Solution:
4
0
3
xdxtg
=
1
4
0
0
1
2
t
x
t
dt
dx
ttgx
+
=
=
=
+
1
0
2
3
1
dt
t
t
=
=
+
−+
1
0
2
3
1
dt
t
ttt
=
+
+
1
0
2
2
1
)1(
dt
t
tt
+
−
1
0
2
1
dt
t
t
=
1
0
tdt
+
−
1
0
2
1
dt
t
t
=
=
1
0
2
2
1ln
2
1
2
+− t
t
=
2ln
2
1
2
1
−
=
)2ln1(
2
1
−
.
Example:
+
2
0
3cos2
x
dx
.
Solution:
t
x
tg =
2
.
+
2
0
3cos2
x
dx
=
+
−
=
+
==
1
2
0
0
1
1
cos
1
2
2
2
2
2
t
x
t
t
x
t
dt
dxt
x
tg
=
+
+
−
+
1
0
2
2
2
3
1
1
2
1
2
t
t
t
dt
=
=
+
++−
+
1
0
2
22
2
1
3322
1
2
t
tt
t
dt
=
+
1
0
2
5
2tdt
=
+
1
0
22
)5(
2
t
dt
1
0
55
2 t
arctg=
=
5
1
5
2
0
5
1
5
2
arctgarctgarctg =
−=
.
Calculation of definite integrals involving irrational expressions
When we need to calculate the definite integrals involving irrational
expressions, we use the same integration techniques were used for finding
indefinite integrals.
59
Example:
dx
x
x
+
4
1
2
1
.
Solution:
dx
x
x
+
4
1
2
1
=
2
4
1
1
2
2
t
x
tdtdx
tx
=
=
=
dtt
t
t
+
2
1
4
2
1
=
=
dt
t
t
+
2
1
3
1
2
=
( )
dttt
−−
+
2
1
23
2
=
2
1
12
1
2
2
2
−
+
−
−−
tt
=
2
1
2
21
−−
tt
=
=
211
4
1
++−−
=
4
7
.
Example:
+
1
0
1 dxxx
.
Solution:
+
1
0
1 dxxx
=
2
1
1
0
2
1
1
2
2
t
x
tdtdx
tx
tx
=
−=
=+
=
( )
−
2
1
2
21 tdttt
=
=
( )
−
2
1
24
2 dttt
=
2
1
35
35
2
−
tt
=
+−−
3
1
5
1
3
8
5
32
2
=
15
11
7
.
Example:
++
1
0
2
58хх
dx
.
Solution:
++
1
0
2
58хх
dx
=
+−++
1
0
2
5161642 хх
dx
=
−+
1
0
2
9)4(х
dx
=
=
51 4
0
4
t
x
dxdtxt=
+=
−
5
4
2
9t
dt
5
4
2
9ln −+ tt
74ln165ln +−+
74
9
ln
+
−+
−
1
0
2
247
)56(
хx
dxx
−+
−
1
0
2
247
)56(
хx
dxx
−+
−
1
0
2
)2
2
7
(2
)56(
хx
dxx
−−
−
1
0
2
)1(
2
9
)56(
2
1
х
dxх
01 1
0
1
1
−
=
+=
−=
t
x
dxdt
tх
xt
−
−
−+
0
1
2
2
9
)5)1(6(
2
1
t
dtt
+
−
−01
2
2
9
2
2
3
t
tdt
−
−
0
1
2
2
92
1
t
dt
290 27
1
2
2
9
2
−
−=
−=
z
t
tdtdz
tz
+−
2
9
2
7
2
3zdz
0
1
3
2
arcsin
2
1
−
t
+−
2
9
2
7
2
2
3
z
+
3
2
arcsin0arcsin
2
1
( )
3
2
arcsin
2
1
733 +−−
=
=
=
=
=
=
=
.
Example:
Solution:
=
=
.
=
=
=
=
=
=
+
=
.
60
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