calculus - integration by substitution.pdf

8/14/2019 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.

Chapter 10 : Techniques of Integration page 1 of 26


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First Year Calculus c W W L Chen, 1994, 2008

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.


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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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.


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.


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.


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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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

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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.


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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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.


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.


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.


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)


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)


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)


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)


/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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/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.


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)


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.


sin3 x cos3 x dx.


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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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.


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.


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)


tan3 x dx.


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.


tan4 x dx.


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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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)


tan2 x sec x dx.


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)


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.


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)


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)


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.


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.


cot x cscn x dx= 1

ncscn x+C ifn= 0,



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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.


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.


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.


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).


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.


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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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

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