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Файл:Mathematics for Foreign Students integrals. Study Guide
.pdf
21
Examples: Solve the following integrals
1.
=
+
−
+−=
+
−+
dx
x
x
xxdx
x
xx
1
12
1
1
2
3
2
5
=
+
−
+
+−=
11
2
24
22
24
x
dx
x
xdxxx
Carctgxx
xx
+−++− 1ln
24
2
24
.
2.
=++=+−+=++=
++
+
1)2(54)2(54
54
25
222
2
xxxxdx
xx
x
=
=
++
+
dx
x
x
1)2(
25
2
=
+
−
=
+
+−
=
=
−=
+=
dt
t
t
dt
t
t
dtdx
tx
xt
1
85
1
2)2(5
2
2
22
=+−+=
+
−
+
= carctgtt
t
dt
t
tdt
81ln
2
1
5
1
8
1
5
2
22
( )
Сxarctgxx ++−++= 2854ln
2
5
2
.
Suppose we have
+−
+
dx
xx
x
23
1
2
. Clearly this is not one of our
standard types, and the numerator is not the differential coefficient of
the denominator. In such a case, we first of all have to express the rather
cumbersome algebraic fraction in terms of its partial fractions, i. e. a number of simpler algebraic fractions, that we shall be able to integrate separately without difficulty.
=
−
−
−
=
+−
+
dx
x
dx
x
dx
xx
x
1
2
2
3
23
1
2
3ln(x–2)–2ln(x–1) + C.
The method, of course, is based on the ability to express the given
function in terms of its partial fractions.
The rules of partial fractions are as follows:
The degree of numerator of the given function must be lower than
the degree of denominator. If it is not, then first of all we have to use
the method of dividing like in the first exercise of the previous examples.
22
We shall factorize the denominator into its prime factors, and this is
very important, since the obtained factors determine the shape of the partial
fractions.
A linear factor (ax + b) gives a partial fraction of the form
bax
A
+
.
Factors (ax + b)2 give partial fractions
2
)( bax
B
bax
A
+
+
+
.
Example.
+−
+
dx
xx
x
23
1
2
.
Solution:
21)2)(1(
1
23
1
2
−
+
−
=
−−
+
=
+−
+
x
B
x
A
xx
x
xx
x
.
Multiple both sides by the denominator (x – 1) (x – 2).
x+1=A(x–2) + B(x–1).
This is an identity and true for any value of x we like to substitute. Where
possible, choose a value of x which will make one of the brackets zero.
Let (x–1) =0, i.e. substitute x=1:
)0()1(2 BA +−=
2−= A
.
Let (x–2) =0, i.e. substitute x=2:
)1()0(3 BA +=
3= B
.
So the integral can now be written
−
−
−
=
+−
+
dx
x
dx
x
dx
xx
x
1
2
2
3
23
1
2
.
Now the rest is easy.
−
−
−
=
+−
+
dx
x
dx
x
dx
xx
x
1
1
2
2
1
3
23
1
2
=
Cxx +−−−= )1ln(2)2ln(3
(do not forget the constant of integration!).
23
Example.
−+
dx
xx
x
2
2
)1)(1(
.
Solution:
22
2
)1(
11
)1)(1( −
+
−
+
+
=
−+ x
C
x
B
x
A
xx
x
.
Clear the denominators x2 = A(x – 1)2 + B(x + 1)(x – 1) + C(x + 1)
Put (x – 1) =0, i.e. x = 1:
2001 ++= CBA
2
1
= C
.
Put (x+1) = 0, i.e. x = –1:
0041 ++= CBA
4
1
= A
.
When the crafty substitution has come to an end, we can find the remaining constants (in this case, just B) by comparing coefficients. Choose
the highest power involved, i.e. x2 in this example.
x2:
BA+=1
=−=−=
4
3
4
1
11 AB
C
x
xx +
−
−−++=
)1(2
1
|1|ln
4
3
|1|ln
4
1
.
Example. Find the following integral
+
+
dx
x
x
3
2
)2(
1
.
Solution:
3
2
)2(1++x
x
=
32
)2()2(
2
+
+
+
+
+
x
C
x
B
x
A
.
=−+
−
+
+
=
−+
−
dxxdx
x
dx
x
xx
x
2
2
2
)1(
2
1
1
1
4
3
1
1
4
1
)1)(1(
24
Now by multiplying both sides of the previous equality by (x + 2)3, we get
x2+1 =A (x +2)2 + B (x + 2) + C.
We now put (x + 2) = 0, i.e. x = –2:
CBA ++=+ 0014
5= C
.
There are no other brackets in this identity so by comparing the coefficients, starting with the highest power, i.e. x2.
x2+ 1 = A(x + 2)2 + B(x + 2) + C.
x2:
A=1
We now go to another part and by comparing the coefficients of lowest
power i.e. the constant terms (or absolute terms) on each side.
x0:
CBA ++= 241
5241 ++= B
82 −=B
4=B
.
3
2
)2(1++x
x
32
)2(
5
)2(
4
2
1
+
+
+
−
+
=
xx
x
=
+
+
dx
x
x
3
2
)2(
1
C
xx
x +
−
+
+
−
+
−+
−−
2
)2(
5
1
)2(
4)2ln(
21
=
C
x
x
x +
+
−
+
−+=
2
)2(2
5
2
4
)2ln(
.
Example. Solve the following integral
+−
dx
xx
x
)1)(2(
2
2
.
Solution:
In this example, we have a quadratic factor which we cannot factorize
to prime factors, so that we will use the following procedure,
1
2
)1)(2(
22
2
+
+
+
−
=
+− x
CBx
x
A
xx
x
.
x2 = A (x2 +1) + (x – 2) (Bx +C).
054 += A
5
4
= A
BA +=1
5
1
5
4
11 =−=−= AB
=
2
A
C
5
2
=C
1
1
5
2
1
5
1
2
1
5
4
1
5
2
5
1
2
1
5
4
)1)(2(
2222
2
+
+
+
+
−
=
+
+
+
−
=
+− xx
x
x
x
x
x
xx
x
Carctgxxxdx
xx
x
++++−=
+−
5
2
)1ln(
10
1
)2ln(
5
4
)1)(2(
2
2
2
( )
+1
2
xx
dx
( ) ( )
111
1
2
22
22
+
+++
=
+
+
+=
+ xx
CxBxAAx
x
CBx
x
A
xx
CxBxAAx +++=
22
1
0=+ BA
0=C
1=A
1−=B
( )
++−=
+
−=
+
Cxxdx
x
x
dx
x
xx
dx
1ln
2
1
ln
1
1
1
2
22
−+
+−−
dx
xx
xxx
2
742
2
23
Put (x – 2) =0, i.e. x = 2:
x2:
x0: 0=A–2C
Here is one for you to do on your own.
Example.
Solution:
.
.
.
.
.
x2:
x1:
x0:
1)
.
.
Do the following exercises and test Yourself
.
.
;
25
2)
+−
dx
xxx
x652
23
2
++
−
dx
xx
x623
2
Cxxx
x
+−++− )1)(2(ln
3
1
3
2
2
Cxx +−−− 2ln43ln6
( )
C
x
arctgx +
+
−++
5
1
5
4
51ln
2
1
2
;
3)
Answers.
1)
2)
3)
.
;
;
.
26
I N T E G R A T I O N O F T R I G O N O M E T R I C A L F U N C T I O N S
+−=−= C
xx
dxxxdx
4
2sin
2
)2cos1(
2
1
sin
2
++=+= C
xx
dxxxdx
4
2sin
2
)2cos1(
2
1
cos
2
=−== xdxxxdxxxdx sin)cos1(sinsinsin
223
−= xdxxxdx sincossin
2
C
x
x ++−=
3
cos
cos
3
=
−
== dx
x
dxxxdx
4
)2cos1(
)(sinsin
22
224
=
+−
= dx
xx
4
2cos2cos21
2
)2cos1(
2
1
cos
2
xx +=
=++− dxxx )4cos
2
1
2
1
2cos21(
4
1
To integrate sin2 x and cos2 x, we express the function in terms of
the cosine of the double angle:
cos2x = 1 – 2 sin2x and cos2x = 2 cos2x – 1
Examples.
1)
2)
To integrate sin3x, we release one of the factors sin x from the power
and convert the remaining sin2 x into (1 – cos2 x), thus:
To integrate sin4 x and cos4 x.
and
;
.
.
.
N.B.
=
27
28
N.B.
)4cos1(
2
1
2cos
2
xx +=
=+−= dxxx )4cos
2
1
2cos2
2
3
(
4
1
C
xxx
C
xxx
++−=+
+−=
32
4sin
4
2sin
8
3
)
4
4sin
2
1
2
2sin2
2
3
(
4
1
.
Remember not this result, but the method.
Now you find
xdx
4
cos
in much the same way.
C
xxx
xdx +++=
32
4sin
4
2sin
8
3
cos
4
.
Finally, while we are dealing with the integrals of trigonometrical
functions, let us consider one further type.
( )
)sin()sin(
2
1
cossin
++−=
,
( )
)cos()cos(
2
1
sinsin
+−−=
,
( )
)cos()cos(
2
1
coscos
++−=
,
Example.
xdxx 2cos4sin
.
Solution:
( )
( )
xx
xxxxxx
2sin6sin
2
1
24sin()24sin(
2
1
2cos4sin
+=
=−++=
C
xx
dxxxxdxx +−−=+=
4
2cos
12
6cos
)2sin6(sin
2
1
2cos4sin
.
Example. Solve the integral
xdxx 4cos6cos
.
( ))
( )
xx
xxxxxx
10cos2cos
2
1
46cos()46cos(
2
1
4cos6cos
+=
=++−=
29
C
xx
dxxxxdxx ++=+=
20
10sin
4
2sin
)10cos2(cos
2
1
4cos6cos
.
They are all done in the same basic way.
Example.
=−=
=−
=
=
dtt
dtxdx
tx
xdxx
55
sin
cos
sincos
C
x
C
t
+−=+−
6
cos
6
66
.
Example. To solve
x
xdx
3
sin
cos
.
Solution.
===
=
=
=
−
dtt
t
dt
dtxdx
tx
x
xdx
3
33
cos
sin
sin
cos
C
x
C
t
+−=+
−
=
−22
sin2
1
2
.
Example. To solve
xdxx sincos
3
2
.
Solution.
=
=−
=
=
dtxdx
tx
xdxx
sin
cos
sincos
3
2
−= dtt
3
2
C
x
C
t
dtt +−=+−=−=
5
cos3
3
5
3
5
3
5
3
2
.
Example. To solve
xdx2cos
3
.
Solution.
==
xdxxxdx 2cos2cos2cos
23
=
( )
=−
xdxx 2cos2sin1
2
=
=
=
2
2cos
2cos2
2sin
dt
xdx
dtxdx
tx
=
( )
=−
dtt
2
1
2
1
=
C
x
xC
t
t +−=+−
6
2sin
2sin
2
1
62
1
33
.
30
Example. To solve
xdxx
35
sincos
.
Solution.
== xdxxxxdxx cossincossincos
2535
=
( )
=− xdxxx sincos1cos
25
−=
−=
=
22
1sin
sin
cos
tx
xdxdt
xt
( )
( )
=−−= dttt
25
1
=
+− dttdtt
75
=
C
xx
C
tt
++−=++−
8
cos
6
cos
86
8686
.
Example.
xdxx
23
cossin
= xdxx
23
cossin
( )
=− xdxxx sincos1cos
22
=
−=
−=
=
22
1sin
sin
cos
tx
xdxdt
xt
( )
=−−= dttt
22
1
( )
=++−=−−=
C
tt
dxtt
53
53
42
C
xx
++−=
5
cos
3
cos
53
.
Example. To solve
xdxx22sincos
.
Solution.
= xdxx
22
sincos
N.B.
2
21
2
xcos
xcos
+
=
and
2
21
2
xcos
xsin
−
=
.
=
=
−
+
dx
xcosxcos
2
21
2
21
( )
=−
dxxcos 21
4
1
2
=
=
+
−
dx
xcos
dx
2
4141
4
1
=−−
xdxcosdxx 4
8
1
8
1
4
1
=
=+−− Cxsinxx 4
418
1
8
1
4
1
Cxsinx +− 4
32
1
8
1
.
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