Calculus - Integration by Substitution.pdf
Transcript of Calculus - Integration by Substitution.pdf
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FIRST YEAR CALCULUS
W W L CHEN
c W W L Chen, 1994, 2008.This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain,
and may be downloaded and/or photocopied, with or without permission from the author.
However, this document may not be kept on any information storage and retrieval system without permission
from the author, unless such system is not accessible to any individuals other than its owners.
Chapter 10
TECHNIQUES OF INTEGRATION
10.1. Integration by Substitution
In this section, we discuss how we can use the Chain rule in differentiation to help solve problems inintegration. This technique is usually called integration by substitution. As we shall not prove any resulthere, our discussion will be only heuristic.
We emphasize that the technique does not always work. First of all, we have little or no knowledge ofthe antiderivatives of many functions. Secondly, there is no simple routine that we can describe to helpus find a suitable substitution even in the cases where the technique works. On the other hand, whenthe technique does work, there may well be more than one suitable substitution!
Occasionally, the possibility of substitution may not be immediately obvious, and a certain amount oftrial and error does occur. The fact that one substitution does not appear to work does not mean that
the method fails. It may very well be the case that we have used a bad substitution.INTEGRATION BY SUBSTITUTION VERSION 1.If we make a substitutionx = g(u), thendx= g(u) du, and
f(x) dx=
f(g(u))g(u) du.
Example 10.1.1.Consider the indefinite integral 1
1 x2 dx.
If we make a substitution x = sin u, then dx= cos u du, and
11 x2dx=
cos u
1 sin2 udu=
du= u+C= sin1 x+C.
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On the other hand, if we make a substitution x = cos v, then dx=sin v dv, and
1
1 x2
dx=
sin v
1 cos2
v
dv=
dv=
v+C=
cos1 x+C.
Example 10.1.2.Consider the indefinite integral
1
1 +x2dx.
If we make a substitution x = tan u, then dx= sec2 u du, and
1
1 +x2dx=
sec2 u
1 + tan2 udu=
du= u+C= tan1 x+C.
On the other hand, if we make a substitution x = cot v, then dx=csc2
v dv, and 1
1 +x2dx=
csc2 v
1 + cot2 vdv=
dv=v+C=cot1 x+C.
Example 10.1.3.Consider the indefinite integral
x
x+ 1 dx.
If we make a substitution x = u2 1, then dx= 2u du, and
xx+ 1 dx= 2(u2 1)u2 du= 2 u4 du 2 u2 du
=2
5u5 2
3u3 +C=
2
5(x+ 1)5/2 2
3(x+ 1)3/2 +C.
On the other hand, if we make a substitution x = v 1, then dx= dv, and
x
x+ 1 dx=
(v 1)v1/2 dv=
v3/2 dv
v1/2 dv
=2
5v5/2 2
3v3/2 +C=
2
5(x+ 1)5/2 2
3(x+ 1)3/2 +C.
We can confirm that the indefinite integral is correct by checking that
ddx
25
(x+ 1)5/2 23
(x+ 1)3/2 +C
=xx+ 1.
INTEGRATION BY SUBSTITUTION VERSION 2. Suppose that a function f(x) can bewritten in the formf(x) =g(h(x))h(x). If we make a substitutionu= h(x), thendu= h(x) dx, and
f(x) dx=
g(h(x))h(x) dx=
g(u) du.
Remark.Note that in Version 1, the variable x is initially written as a function of the new variable u,whereas in Version 2, the new variable u is written as a function of x. The difference, however, is
minimal, as the substitution x= g(u) in Version 1 has to be invertible to enable us to return from thenew variableu to the original variable x at the end of the process.
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Example 10.1.4.Consider the indefinite integral
x2ex3 dx.
Note first of all that the derivative of the function x3 is equal to 3x2, so it is convenient to make thesubstitutionu = x3. Then du= 3x2 dx, and
x2ex
3
dx= 1
3
3x2ex
3
dx=1
3
eu du=
1
3eu +C=
1
3ex
3
+C.
A somewhat more complicated alternative is to note that the derivative of the function e x3
is equal to3x2ex
3
, so it is convenient to make the substitution v = ex3
. Then dv= 3x2ex3
dx, and
x
2
e
x3
dx=
1
3
3x
2
e
x3
dx=
1
3
dv=
1
3 v+C=
1
3 e
x3
+C.
Example 10.1.5.Consider the indefinite integral
x(x2 + 3)4 dx.
Note first of all that the derivative of the function x2 + 3 is equal to 2x, so it is convenient to make thesubstitutionu = x2 + 3. Then du= 2x dx, and
x(x
2
+ 3)4
dx=1
2
2x(x2
+ 3)4
dx= 1
2
u4
du= 1
10 u5
+C= 1
10 (x2
+ 3)5
+C.
Example 10.1.6.Consider the indefinite integral
1
x log xdx.
Note first of all that the derivative of the function log x is equal to 1/x, so it is convenient to make thesubstitutionu = log x. Then du= (1/x) dx, and
1x log xdx=
1udu= log |u| +C= log | log x| +C.
Example 10.1.7.Consider the indefinite integral
tan3 x sec2 x dx.
Note first of all that the derivative of the function tan x is equal to sec2 x, so it is convenient to makethe substitution u = tan x. Then du= sec2 x dx, and
tan3 x sec2 x dx=
u3 du= 1
4u4 +C=1
4tan4 x+C.
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Example 10.1.8.Consider the indefinite integral
sin3 x cos3 x dx.Note first of all that the derivative of the function sin x is equal to cos x, so it is perhaps convenient tomake the substitution u = sin x. Then du= cos x dx, and
sin3 x cos3 x dx=
u3(1 u2) du=
(u3 u5) du= u
4
4 u
6
6 +C=
sin4 x
4 sin
6 x
6 +C.
Alternatively, note that the derivative of the function cos xis equal tosin x, so it is convenient to makethe substitution v = cos x. Then dv=sin x dx, and
sin3 x cos3 x dx=
(1 v2)v3 dv=
(v5 v3) dv= v
6
6 v
4
4 +C=
cos6 x
6 cos
4 x
4 +C.
It can be checked that
sin4 x
4 sin
6 x
6 =
cos6 x
6 cos
4 x
4 +
1
12.
Example 10.1.9.Recall Example 10.1.1. Since
1
1 x2dx= sin1 x+C,
we have
1/20
11 x2dx=
sin1 x
1/20
= sin11
2 sin1 0 =
6.
Note that we have in fact used the substitution x= sin u to show that
1
1 x2dx=
du= u+C,
followed by an inverse substitution u = sin1 x. Here, we need to make the extra step of substituting thevaluesx = 0 andx = 1/2 to the indefinite integral sin1 x. Observe, however, that with the substitutionx= sin u, the variablex increases from 0 to 1/2 as the variable u increases from 0 to /6. But then
/60
du=
u
/6
0
=
6 =
1/20
11 x2dx,
so it appears that we do not need the inverse substitutionu = sin1 x. Perhaps we can directly substituteu= 0 and u= /6 to the indefinite integral u.
DEFINITE INTEGRAL BY SUBSTITUTION VERSION 1. Suppose that a substitutionx= g(u) satisfies the following conditions:(a) There exist, R such thatg() = A andg() = B.(b) The derivativeg(u)> 0 for everyu satisfying < u < .
Thendx= g (u) du, and
B
A
f(x) dx=
f(g(u))g(u) du.
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Remark.If condition (b) above is replaced by the condition that the derivative g(u) < 0 for every usatisfying < u < , then the same conclusion holds if we adopt the convention that
f(g(u))g(u) du=
f(g(u))g(u) du.
Example 10.1.10.To calculate the definite integral
10
1
1 +x2dx,
we can use the substitution x = tan u, so that dx = sec2 u du. Note that tan 0 = 0 and tan(/4) = 1,and that sec2 u >0 whenever 0< u < /4. It follows that
10
1
1 +x2dx=
/40
sec2 u
1 + tan2 udu=
/40
du=
u
/40
=
4 0 =
4.
We can compare this to first observing Example 10.1.2, so that
10
1
1 +x2dx=
tan1 x
10
= tan1 1 tan1 0 = 4 0 =
4.
Example 10.1.11.To calculate the definite integral
30
xx+ 1 dx,
we can use the substitution x = g (u) =u2 1, so that dx = 2u du. Note that g(1) = 0 and g(2) = 3,and thatg (u) = 2u >0 whenever 1< u 0 for everyx satisfyingA < x < B.
Thendu= h(x) dx, and
BA
f(x) dx=
BA
g(h(x))h(x) dx=
h(B)h(A)
g(u) du.
Remark.If condition (b) above is replaced by the condition that the derivative h(x)< 0 for every xsatisfyingA < x < B, then the same conclusion holds if we adopt the convention that
h(B)h(A)
g(u) du= h(A)h(B)
g(u) du.
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Example 10.1.12.To calculate the definite integral
1
0
x(x2 + 3)4 dx,
we can use the substitution u = h(x) =x2 + 3, so that du = 2x dx. Note that h(0) = 3 and h(1) = 4,and thath (x) = 2x >0 whenever 0< x 0 whenever2< x
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Furthermore, v = ex and u = 1. It follows that
uv
vu dx= xex
ex dx= xex
ex +C.
Hence xex dx= xex ex +C.
Example 10.2.2.Consider the indefinite integral
log x dx.
Writingu = log x and v = 1, we have
uv dx=
log x dx.
Furthermore,
v= x and u= 1
x.
It follows that
uv
vu dx= x log x
x1
xdx= x log x x+C.
Hence log x dx= x log x x+C.
Example 10.2.3.Consider the indefinite integral
ex sin x dx.
Writingu = ex andv = sin x, we have
uv dx=
ex
sin x dx.
Furthermore, v =cos x and u= ex. It follows that
uv
vu dx=ex cos x+
ex cos x dx.
Hence ex sin x dx=ex cos x+
ex cos x dx. (2)
We now need to study the indefinite integral
ex cos x dx.
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Writingu = ex andv = cos x, we have
uv dx= ex cos x dx.Furthermore, v = sin x and u= ex. It follows that
uv
vu dx= ex sin x
ex sin x dx.
Hence
ex cos x dx= ex sin x
ex sin x dx. (3)
It looks like we are back to the same old problem. However, if we combine (2) and (3), then we obtain
ex sin x dx=ex cos x+ ex sin x ex sin x dx,
so that
2
ex sin x dx= ex sin x ex cos x= ex(sin x cos x).
Adding an arbitrary constant, which we may in view of Proposition 9C, we have
ex sin x dx=
1
2ex(sin x cos x) +C.
Example 10.2.4.Consider the indefinite integral
x3 cos x dx.
Writingu = x3 and v = cos x, we have
uv dx=
x3 cos x dx.
Furthermore, v = sin x and u= 3x2. It follows that
uv vu dx= x3
sin x 3 x2
sin x dx.
Hence x3 cos x dx= x3 sin x 3
x2 sin x dx. (4)
We now need to study the indefinite integral
x2 sin x dx.
Writingu = x2 and v = sin x, we have
uv dx=
x2 sin x dx.
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Furthermore, v =cos x and u= 2x. It follows that
uv
vu dx=
x2 cos x+ 2 x cos x dx.
Hence x2 sin x dx=x2 cos x+ 2
x cos x dx. (5)
Combining (4) and (5), we have
x3 cos x dx= x3 sin x+ 3x2 cos x 6
x cos x dx. (6)
We now need to study the indefinite integral
x cos x dx.
Writingu = x and v = cos x, we have
uv dx=
x cos x dx.
Furthermore, v = sin x and u= 1. It follows that
uv
vu dx= x sin x
sin x dx.
Hence x cos x dx= x sin x
sin x dx. (7)
Combining (6) and (7), we have
x3 cos x dx= x3 sin x+ 3x2 cos x 6x sin x+ 6
sin x dx
=x3 sin x+ 3x2 cos x 6x sin x 6cos x+C.
The technique is also valid for definite integrals, in view of the first Fundamental theorem of integralcalculus. For definite integrals over the interval [A, B], we have
BA
uv dx=
uv
x=Bx=A
BA
vu dx. (8)
Equation (8) is called the formula for integration by parts for definite integrals.
Example 10.2.5.Consider the definite integral
/20
x3 cos x dx.
Writingu = x3 and v = cos x, we have
/2
0
uv dx= /2
0
x3 cos x dx.
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Furthermore, v = sin x and u= 3x2. It follows that
uvx=/2
x=0 /2
0
vu
dx= x3 sin x/2
0 3
/2
0
x2 sin x dx.
Hence
/20
x3 cos x dx=
x3 sin x
/20
3 /2
0
x2 sin x dx= 3
8 3
/20
x2 sin x dx. (9)
We now need to study the definite integral
/20
x2 sin x dx.
Writingu = x2
and v = sin x, we have /20
uv dx=
/20
x2 sin x dx.
Furthermore, v =cos x and u= 2x. It follows that
uv
x=/2x=0
/2
0
vu dx=
x2 cos x
/20
+ 2
/20
x cos x dx.
Hence
/20
x2 sin x dx=x2 cos x
/2
0+ 2
/20
x cos x dx= 2 /2
0x cos x dx. (10)
Combining (9) and (10), we have
/20
x3 cos x dx=3
8 6
/20
x cos x dx. (11)
We now need to study the definite integral
/20
x cos x dx.
Writingu = x and v = cos x, we have
/20
uv dx=
/20
x cos x dx.
Furthermore, v = sin x and u= 1. It follows that
uv
x=/2x=0
/2
0
vu dx=
x sin x
/20
/2
0
sin x dx.
Hence
/2
0
x cos x dx=
x sin x
/2
0
/2
0
sin x dx= 2 /2
0
sin x dx. (12)
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Combining (11) and (12), we have
/2
0
x3 cos x dx= 3
8 3+ 6
/2
0
sin x dx=3
8 3+ 6 cos x
/2
0
=3
8 3+ 6.
10.3. Trigonometric Integrals
In this section, we consider integrals involving the six trigonometric functions sin x, cos x, tan x, cot x,sec x and csc x. If we consider differentiation formulas involving these functions, then we can dividethese into three groups: (a) sin x and cos x; (b) tan x and sec x; and (c) cot x and csc x. Note that thederivative of any of these functions can be expressed in terms of the two functions in the group to whichit b elongs. This division is also substantiated by integral formulas.
It follows that given any indefinite integral
f(x) dx,
where the integrand f(x) involves trigonometric functions, it may be beneficial to try first to expressf(x) in terms of trigonometric functions from only one of these three groups.
Example 10.3.1.Consider the indefinite integral
tan x+ sec3 x cot x
cos2 x dx=
tan x
cos2 x+
sec3 x cot x
cos2 x
dx
=
sin xcos3 x
+ sec5 xtan x
dx=
sin xcos3 x
dx+
sec5 xtan x
dx.
Note that we can also write
sin x
cos3 xdx=
tan x sec2 x dx=
1
2tan2 x+C.
However, the indefinite integral
sec5 x
tan x dx
does not appear to be so simple.
Let us consider first integrals involving sin x and cos x. Consider an integral of the form
sinm x cosn x dx.
Whenm = 1, the integral is simple to evaluate. Clearly
sin x cosn x dx= 1
n+ 1cosn+1 x+C ifn=1,
and
sin x cos1 x dx= log | cos x| +C.
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Whenn = 1, the integral is also simple to evaluate. Clearly
sinm x cos x dx= 1
m+ 1
sinm+1 x+C ifm
=
1,
and sin1 x cos x dx= log | sin x| +C.
In the general case, we may use standard trigonometric formulas like
sin2 x+ cos2 x= 1, (13)
sin2x= 2 sin x cos x, (14)
cos2x= cos2 x sin2 x. (15)
Note also that combining (13) and (15), we have
cos2x= 2 cos2 x 1 = 1 2sin2 x. (16)
Example 10.3.2.Consider the indefinite integral
sin5 x dx.
Using (13), we can write
sin5 x= sin4 x sin x= (1
cos2 x)2 sin x= (1
2cos2 x+ cos4 x)sin x,
so that sin5 x dx=
(1 2cos2 x+ cos4 x)sin x dx
=
sin x dx 2
cos2 x sin x dx+
cos4 x sin x dx
=cos x+23
cos3 x 15
cos5 x+C.
Example 10.3.3.Consider the indefinite integral
sin3 x cos3 x dx.
Using (13), we can write
sin3 x cos3 x= cos2 x sin3 x cos x= (1 sin2 x)sin3 x cos x= sin3 x cos x sin5 x cos x,
so that
sin3 x cos3 x dx=
(sin3 x cos x sin5 x cos x) dx
= sin3 x cos x dx sin
5 x cos x dx
=1
4sin4 x 1
6sin6 x+C.
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Example 10.3.4.Consider the indefinite integral
sin4 4x dx.Using (16), we can write
sin4 4x= 1
4(1 cos8x)2 = 1
4(1 2cos8x+ cos2 8x)
= 1
4
1 2cos8x+1
2(1 + cos 16x)
=
3
8 1
2cos 8x+
1
8cos16x,
so that
sin4 4x dx=
3
8 1
2cos 8x+
1
8cos16x
dx
= 38
dx 1
2
cos8x dx+1
8
cos16x dx
= 3
8x 1
16sin8x+
1
128sin 16x+C.
Example 10.3.5.Consider the indefinite integral
sin2 x cos4 x dx.
Using (14) and (16), we can write
sin2 x cos4 x= cos2 x(sin x cos x)2 =18
(1 + cos 2x)sin2 2x= 18
sin2 2x+18
cos 2x sin2 2x
= 1
16(1 cos4x) +1
8cos 2x sin2 2x=
1
16 1
16cos 4x+
1
8cos 2x sin2 2x,
so that
sin2 x cos4 x dx=
1
16 1
16cos 4x+
1
8cos 2x sin2 2x
dx
= 1
16
dx 1
16
cos4x dx+
1
8
cos2x sin2 2x dx
= 1
16x 1
64sin4x+
1
48sin3 2x+C.
Let us consider next integrals involving tan x and sec x. Consider an integral of the form
tanm x secn x dx.
Whenm = 1, the integral is simple to evaluate. Clearly
tan x secn x dx=
1
nsecn x+C ifn= 0,
and
tan x dx= log | cos x| +C.
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Whenn = 2, the integral is also simple to evaluate. Clearly
tanm x sec2 x dx= 1m+ 1
tanm+1 x+C ifm=
1,
and
tan1 x sec2 x dx= log | tan x| +C.
In the general case, we may use standard trigonometric formulas like
1 + tan2 x= sec2 x. (17)
Example 10.3.6.Consider the indefinite integral
tan3 x dx.
Using (17), we can write
tan3 x= tan2 x tan x= (sec2 x 1) tan x= sec2 x tan x tan x,
so that
tan3 x dx=
(sec2 x tan x tan x) dx
=
sec2 x tan x dx
tan x dx
=1
2tan2 x+ log | cos x| +C.
Example 10.3.7.Consider the indefinite integral
tan4 x dx.
Using (17), we can write
tan4 x= tan2 x tan2 x= (sec2 x 1) tan2 x= sec2 x tan2 x tan2 x= sec2 x tan2 x sec2 x+ 1,
so that
tan4 x dx=
(sec2 x tan2 x sec2 x+ 1) dx
=
sec2 x tan2 x dx
sec2 x dx+
dx
= 13
tan3 x tan x+x+C.
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Example 10.3.8.Consider the indefinite integral
sec3 x dx.Writingu = sec x andv = sec2 x, we have
uv dx=
sec3 x dx.
Furthermore, v = tan x andu = tan x sec x. It follows that
uv
vu dx= sec x tan x
tan2 x sec x dx.
Hence
sec3 x dx= sec x tan x
tan2 x sec x dx. (18)
We now need to study the indefinite integral
tan2 x sec x dx.
Using (17), we can write
tan2 x sec x= (sec2 x 1) sec x= sec3 x sec x,
so that
tan2 x sec x dx=
(sec3 x sec x) dx=
sec3 x dx
sec x dx. (19)
Combining (18) and (19), we have
sec3 x dx= sec x tan x
sec3 x dx+
sec x dx,
so that
sec3 x dx=1
2sec x tan x+
1
2
sec x dx=
1
2sec x tan x+
1
2log | sec x+ tan x| +C.
Example 10.3.9.Consider the indefinite integral
tan2 x sec3 x dx.
Writingu = tan2 sec x and v = sec2 x, we have
uv dx=
tan2 x sec3 x dx.
Furthermore, v = tan x andu = 2 tan x sec3 x+ tan3 x sec x. It follows that
uv
vu dx= tan3 x sec x
(2tan2 x sec3 x+ tan4 x sec x) dx.
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Hence
tan2 x sec3 x dx= tan3 x sec x
(2 tan2 x sec3 x+ tan4 x sec x) dx
= tan3 x sec x 2
tan2 x sec3 x dx
tan4 x sec x dx. (20)
We now need to study the indefinite integral
tan4 x sec x dx.
Using (17), we can write (the reader should check this)
tan4 x sec x= tan2 x sec3 x sec3 x+ sec x,
so that tan4 x sec x dx=
tan2 x sec3 x dx
sec3 x dx+
sec x dx. (21)
Combining (20) and (21), we have
tan2 x sec3 x dx=
1
4tan3 x sec x+
1
4
sec3 x dx 1
4
sec x dx
= 1
4tan3 x sec x+
1
8tan x sec x 1
8log | tan x+ sec x| +C.
Occasionally, it may be necessary to convert an expression involving tan x and sec x to one involvingsin x and cos x instead.
Example 10.3.10.Consider the indefinite integral
tan4 7x
sec5 7xdx.
Here the identity (17) does not help very much. However, we have
tan4 7x
sec5 7x= sin4 7x cos7x,
so that
tan4 7x
sec5 7xdx=
sin4 7x cos7x dx=
1
35sin5 7x+C.
Let us consider finally integrals involving cot x and csc x. Consider an integral of the form
cotm x cscn x dx.
Whenm = 1, the integral is simple to evaluate. Clearly
cot x cscn x dx= 1
ncscn x+C ifn= 0,
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Example 10.4.2.Consider the indefinite integral
x3
9 4x2 dx.
If we use the substitution x = 32
sin , then
9 4x2 = 3| cos | and dx= 3
2cos d.
Suppose that cos >0. Then
x3
9 4x2 dx= 243
16
sin3 cos2 d
= 243
16
(1 cos2 )sin cos2 d
= 243
16
sin cos2 d 243
16
sin cos4 d
=8116
cos3 +243
80 cos5 +C.
Next, note that cos2 = 1 sin2 = 1 49
x2, so that
x3
9 4x2 dx=81
16
1 4
9x23/2
+243
80
1 4
9x25/2
+C.
Let us consider next the case
a2 +b2x2. If we use the substitution
x=a
b tan ,
then
a2 +b2x2 =
a2(1 + tan2 ) =
a2 sec2 = a| sec |,
while
dx=a
bsec2 d.
Example 10.4.3.Consider the indefinite integral
x2
1 +x2 dx.
If we use the substitution x = tan , then
1 +x2 =|sec | and dx= sec2 d.
Suppose that sec >0. Then
x2
1 +x2 dx=
tan2 sec3 d.
We have shown earlier that tan2 sec3 d=
1
4tan3 sec +
1
8tan sec 1
8log | tan + sec | +C.
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Next, note that sec2 = 1 + tan2 = 1 +x2, so that
x21 +x2 dx=1
4
x3(1 +x2)1/2 +1
8
x(1 +x2)1/2
1
8
log
|x+ (1 +x2)1/2
|+C.
Let us consider finally the case
b2x2 a2. If we use the substitution
x=a
bsec ,
then
b2x2 a2 =
a2(sec2 1) =
a2 tan2 = a| tan |,
while
dx= ab
tan sec d.
Example 10.4.4.Consider the indefinite integral
x2 4
x dx.
If we use the substitution x = 2 sec , then
x2 4 = 2| tan | and dx= 2 tan sec d.
Suppose that tan >0. Then
x2 4
x dx= 2
tan2 d= 2
(sec2 1) d= 2
sec2 d 2
d= 2 tan 2+C.
Next, note that
tan2 = sec2 1 =14
x2 1 and = sec1x
2
,
so that
x2 4x
dx= 214
x2
1
1/2
2sec1 x
2+C= x2 4 2sec
1 x2+C.
10.5. Completing Squares
In this section, we shall consider techniques to handle integrals involving square roots of the formx2 +x +, where = 0. Our task is to show that such integrals can be reduced to integrals
discussed in the previous section.
Note that
x2 +x +=
x2 +
x+
=
x2 +
x+
2
42
+
b2
4
=
x+
2
2
+
2
4
.
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Suppose first of all that we use a substitution
y= x+
2
.
Then dy= dx and
x2 +x += y2 +,
where
= 2
4.
It now follows that
x2 +x + is of the form
a2
b2y2 if 0,
a2 +b2y2 if >0 and >0,b2y2 a2 if >0 and 0. Then 13 2x x2 dx=
1
4 y2 dx.
We have shown earlier that 1
4 y2 dy= sin1y
2
+C.
It follows that
1
3 2x x2
dx= sin1x+ 1
2+C.
Example 10.5.2.Consider the indefinite integral
x2 4xx 2 dx.
We have
x2 4x= (x2 4x+ 4) 4 = (x 2)2 4 =y2 4,
where we use the substitution y = x 2. Note that = 1> 0 and =4< 0. Thenx2 4x
x 2 dx=
y2 4y
dy.
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We have shown earlier that
y2 4
y
dy= y2 4
2sec1
y
2+C.
It follows that
x2 4xx 2 dx=
(x 2)2 4 2sec1
x 2
2
+C=
x2 4x 2sec1
x 2
2
+C.
10.6. Partial Fractions
In this section, we shall consider techniques to handle integrals of the form
p(x)
q(x)dx,
wherep(x) and q(x) are polynomials in x.
If the degree ofp(x) is not smaller than the degree ofq(x), then we can always find polynomials a(x)and r(x) such that
p(x)
q(x) =a(x) +
r (x)
q(x),
wherer(x) = 0 or has degree smaller than the degree ofq(x).
Example 10.6.1.Consider the indefinite integral
x5 + 2x4 + 4x3 +x+ 1
x2 +x+ 1 dx.
Note that
x5 + 2x4 + 4x3 +x+ 1
x2 +x+ 1 = (x3 +x2 + 2x 3) + 2x+ 4
x2 +x+ 1,
so that
x5 + 2x4 + 4x3 +x+ 1
x2 +x+ 1 dx=
(x3 +x2 + 2x 3) dx+
2x+ 4x2 +x+ 1
dx.
It does not take a genius to work out the indefinite integral
(x3 +x2 + 2x 3) dx.
We can therefore restrict our attention to the case when the polynomial p(x) is of lower degree thanthe polynomial q(x).
The first step is to factorize the polynomial q(x) into a product of irreducible factors. It is a funda-
mental result in algebra that a real polynomial q(x) can be factorized into a product of irreducible linearfactors and quadratic factors with real coefficients.
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Example 10.6.2. Suppose that q(x) = x4 4x3 + 5x2 4x+ 4. Thenq(x) can be factorized into aproduct of irreducible linear factors in the form (x 2)2(x2 + 1).
Suppose that a linear factor (ax + b) occursn times in the factorization ofq(x). Then we write downa decomposition
A1ax+b
+ A2
(ax+b)2+. . .+
An(ax+b)n
,
where the constantsA1, . . . , An will be determined later. Suppose that a quadratic factor (ax2 + bx + c)
occurs n times in the factorization ofq(x). Then we write down a decomposition
A1x+B1ax2 +bx+c
+ A2x+B2
(ax2 +bx+c)2+. . .+
Anx+Bn(ax2 +bx+c)n
,
where the constants A1, . . . , An and B1, . . . , Bn will be determined later. We proceed to add all thedecompositions and equate their sum to
p(x)
q(x),
and then calculate all the constants by equating coefficients.
Example 10.6.3.Suppose that
p(x)
q(x) =
2x3 11x2 + 17x 16x4 4x3 + 5x2 4x+ 4 =
2x3 11x2 + 17x 16(x 2)2(x2 + 1) .
We now write
2x3 11x2 + 17x 16(x 2)2(x2 + 1) =
c1x 2+
c2(x 2)2 +
c3x+c4x2 + 1
.
Now
c1x 2+
c2(x 2)2 +
c3x+c4x2 + 1
= c1(x 2)(x2 + 1) +c2(x2 + 1) + (c3x+c4)(x 2)2
(x 2)2(x2 + 1) ,
so that
c1(x
2)(x2 + 1) +c2(x
2 + 1) + (c3x+c4)(x
2)2 = 2x3
11x2 + 17x
16.
Note now that
c1(x 2)(x2 + 1) +c2(x2 + 1) + (c3x+c4)(x 2)2=c1(x
3 2x2 +x 2) +c2(x2 + 1) +c3(x3 4x2 + 4x) +c4(x2 4x+ 4)= (c1+c3)x
3 + (2c1+c24c3+c4)x2 + (c1+ 4c3 4c4)x+ (2c1+c2+ 4c4).
Equating coefficients, we have
c1 + c3 = 2,
2c1+c24c3+ c4=11,c1 + 4c3 4c4= 17,
2c1+c2 + 4c4=16.
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Example 10.6.4.Let us continue the discussion of Example 10.6.3. Note that
1
x 2dx= log
|x
2
|+C and
1
(x 2)2
dx=
1
x 2+C.
On the other hand, we have
x 3x2 + 1
dx= 1
2
2x
x2 + 1dx 3
1
x2 + 1dx.
Clearly
2x
x2 + 1dx= log |x2 + 1| +C.
Using the substitution x = tan , we have
1
x2 + 1dx=
sec2
1 + tan2 d=
d= +C= tan1 x+C.
It follows that
2x3 11x2 + 17x 16x4 4x3 + 5x2 4x+ 4dx=
1
x 2dx
2
(x 2)2dx+
x 3x2 + 1
dx
= log |x 2| + 2x 2+
1
2log |x2 + 1| 3tan1 x+C.
Example 10.6.5.Consider the indefinite integral
x2 +x 3
x3 2x2 x+ 2dx.
Note first of all that
x3 2x2 x+ 2 = (x 2)(x+ 1)(x 1),
so we consider partial fractions of the form
x2 +x 3(x 2)(x+ 1)(x 1) =
c1x 2+
c2x+ 1
+ c3x 1
= c1(x+ 1)(x 1) +c2(x 2)(x 1) +c3(x 2)(x+ 1)(x 2)(x+ 1)(x 1) .
It follows that
c1(x+ 1)(x 1) +c2(x 2)(x 1) +c3(x 2)(x+ 1) =x2 +x 3. (24)
We may equate coefficients and solve forc1, c2, c3. Alternatively, substituting x = 2,1, 1 into equation(24), we get respectively 3c1 = 3, 6c2 =3 and2c3 =1, so that c1 = 1, c2 =1/2 and c3 = 1/2.Hence
x2 +x 3
x3
2x2
x+ 2
dx= 1
x
2dx 1
2 1
x+ 1dx+
1
2 1
x
1dx
= log |x 2| 12
log |x+ 1| +12
log |x 1| +C.
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Example 10.6.6.Consider the indefinite integral
x6 2x4 +x2
dx.
Note that
x6 2x4 +x2
=x2 1 + x2 2
x4 +x2,
so that
x6 2x4 +x2
dx=
(x2 1) dx+
x2 2x4 +x2
dx=1
3x3 x+
x2 2x4 +x2
dx. (25)
Next, we study the integral
x2 2x4 +x2
dx.
Note first of all that
x4 +x2 =x2(x2 + 1),
so we consider partial fractions of the form
x2 2x2(x2 + 1)
=c1
x +
c2x2
+c3x+c4
x2 + 1 =
c1x(x2 + 1) +c2(x
2 + 1) + (c3x+c4)x2
x2(x2 + 1) .
It follows that
c1x(x2 + 1) +c2(x
2 + 1) + (c3x+c4)x2 =x2 2.
Equating coefficients, we have
c1 +c3 = 0,
c2 +c4= 1,
c1 = 0,
c2 =2.
This system has solution c1= 0, c2=2, c3= 0 and c4= 3. Hence x2 2x4 +x2
dx=2
1
x2dx+ 3
1
x2 + 1dx=
2
x+ 3 tan1 x+C. (26)
Combining (25) and (26), we obtain
x6 2x4 +x2
dx= 1
3x3 x+2
x+ 3 tan1 x+C.
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Problems for Chapter 10
1. Evaluate each of the following indefinite integrals:
a)
sin x cos7x dx b)
e2x cos3x dx c)
x2 log x dx
d)
cos2x
1 sin2xdx e)
116 3x+x2dx f)
x sec2 x dx
g)
1
x2 + 4x 4dx h)
x2
x3 + 3x2 + 3x+ 1dx i)
cot x csc4 x dx
j)
x2 + 3x 1x4 +x3 +x2 +x
dx k)
log(x6) dx l)
sin2 3x dx
m)
1
x2 5x+ 4dx n)
e2x cos x dx o)
(x3 +
x) dx
p) x4 +x
x+ 1
x dx q) x2
x
1 dx r) xx2 + 4 dxs)
e4x+2 dx t)
xex
2
dx u)
log x
x dx
v)
(log x)5
x dx w)
2x+ 3
x2 + 3x 4dx x)
sin1 x1 x2 dx
y)
1
a2 x2dx z)
(
x+ 1)10x
dx aa)
1
x2 +a2dx
bb)
x5ex dx cc)
1
x2 4x+ 3dx dd)
xex dx
ee)
x 4
(x2 + 4)(x+ 1)dx
2. Evaluate each of the following definite integrals:
a)
32
x(1 + 2x2)4 dx b)
41
x+ 1
x
dx c)
21
x2 + 1
(x+ 1)4dx
d)
/40
cos x
(1 + sin x)2dx e)
10
x
x2 + 1 dx f)
41
ex
x
dx
g)
/40
cos2 2x dx h)
/20
2 2cos x dx i)
/40
x cos2x dx
j)
1/20
x1 x2 dx