curs analiza matematica

MATHEMATICAL ANALYSIS II

INTEGRAL CALCULUS

SEVER ANGEL POPESCU

To my family...........................................................................................

To those who really try to put their lives in the Light ofTruth.

............................................... ...............................................Dedicated to the memory of my late Professor and

Mentor, Dr. Doc. Nicolae Popescu.

Contents

Preface 7

Chapter 1. Indefinite integrals (Primitives, Antiderivatives) 11. Definitions, some properties and basic formulas 12. Some results on polynomials 103. Primitives of rational functions 214. Primitives of irrational and trigonometric functions 275. Problems and exercises 47

Chapter 2. Definite integrals 491. The mass of a linear bar 492. Darboux sums and their applications 523. Lebesgue criterion and its applications 594. Mean theorem. Newton-Leibniz formula 635. The measure of a figure in Rn 686. Areas of plane figures bounded by graphics of functions 757. The volume of a rotational solid 858. The length of a curve in R3 879. Approximate computation of definite integrals. 9010. Problems and exercises 101

Chapter 3. Improper (generalized) integrals 1051. More on limits of functions of one variable 1052. Improper integrals of the first type 1123. Improper integrals of the second type 1224. Problems and exercises 135

Chapter 4. Integrals with parameters 1371. Proper integrals with parameters 1372. Improper integrals with parameters 1473. Eulers functions gamma and beta 1674. Problems and exercises 179

Chapter 5. Line integrals 1811. The mass of a wire. Line integrals of the first type. 1812. Line integrals of the second type. 194

3

4 CONTENTS

3. Independence on path. Conservative fields 2014. Computing plane areas with line integrals 2115. Supplementary remarks on line integrals 2166. Problems and exercises 221

Chapter 6. Double integrals 2251. Double integrals on rectangles. 2252. Double integrals on an arbitrary bounded domain 2333. Green formula. Applications. 2444. Change of variables in double integrals. Polar coordinates. 2515. Problems and exercises 263

Chapter 7. Triple integrals 2651. What is a triple integral on a parallelepiped? 2652. Triple integrals on a general domain. 2713. Iterative formulas for a general space domain 2784. Change of variables in a triple integral 2865. Problems and exercises 303

Chapter 8. Surface integrals 3051. Deformation of a 2D-domain into a space surface 3052. Surface integral of the first type. The mass of a 3D-lamina. 3133. Flux of a vector field through an oriented surface. 3194. Problems and exercises 335

Chapter 9. Gauss and Stokes formulas. Applications 3391. Gauss formula (divergence theorem) 3392. A mathematical model of the heat flow in a solid 3483. Stokes Theorem 3494. Problems and exercises 357

Chapter 10. Some remarks on complex functions integration 3591. Complex functions integration 3592. Applications of residues formula 3773. Problems and exercises 385

Appendix A. Exam samples 3871. June, 2010 3872. June, 2010 3873. June, 2010 3874. June, 2010 3885. September, 2010 3886. September, 2010 388

Appendix B. Basic antiderivatives 389

CONTENTS 5

Appendix. Index 391

Appendix. Bibliography 397

Preface

Here is the second volume "Mathematical Analysis II. Integral Cal-culus" of my course of Advanced Calculus for Engineers and beginningMathematicians. The first volume "Mathematical Analysis I. Differ-ential Calculus" appeared in 2009 (see [Po]). I invite the reader tocarefully meditate on the main ideas I discussed in the Preface of thisfirst volume. The matter is about the way I think Mathematical Analy-sis (or Advanced Calculus) have to be taught to an engineering student.First of all I insist on a strong and correct intuitive basis of any math-ematical notion and statement. Then I proceed step by step to a rig-orously mathematical model of this notion or statement. My aim inthis course (primarily intended for engineering students) is to create anintuitive thinking to the reader who must feel the simplicity and thenaturalness of the introduced notion or concept. Abstract Mathemat-ics need not destroy the intuition of the future engineer. Mathematicalcourses must make this intuition stronger, more mature and very solid.This is why this course appeared to be "a spoken course". I wanted itto be so! In general, I like a lot "live" Mathematics. A "dead" abstractnotion, without at least one small motivation, generally will remaindead and nonuseful for an engineering student.

I also blame the opposite side, that mathematical teaching whichconsists only on a sequence of examples, motivations, etc., without ageneral definition, without a mathematical correct reasoning, based on"stories", etc. A creative middle way teaching is to be preferred. All ofthese belong to something which is beyond Mathematics itself, namelyto the "Art of Teaching".

For instance, many times in this volume, the idea of the convergenceof a system A of numbers to a number I, was substituted with the ideaof a "well approximation" of I with numbers of A. Surely, I could usefrom the very beginning the definition "with and everything couldappeared formally rigorous for an "extremist" mathematician, but thiscourse is not intended for such a people! If I had done so, I wouldhave killed the intuitive power of the idea of approximation, extremelyuseful for a future engineer and not only for him!

7

8 PREFACE

In fact, in all my courses of Mathematical Analysis, I carried on thebasic idea of Prof. Dr. Gavril Paltineanu to make appropriate mathe-matical courses for engineers and not for mathematicians. I think thateven a student in abstract Mathematics could find a good opportunityto read such an "applicable" course.

This book contains ten chapters and an appendix with exam sam-ples and a table with some basic formulas for primitives (antideriva-tives). Since I successively introduced in the first volume ([Po]) basicelements of complex functions theory, I continued here with some ba-sic remarks on integrals of functions with a complex variable, residuestheory and some of their applications.

I started with a serious review of the calculus of primitives, usuallyassumed to be known from high school by the Romanian student.

I go on with definite or Riemann integrals, improper integrals, in-tegrals with parameters, line integrals, double and triple integrals, sur-face integrals, Gauss and Stokes formulas and I finish with complexfunctions integrals.

Any chapter contains many motivation-examples, worked exercisesand proposed exercises. I insisted on using instead of an abstract math-ematical notion more intuitive objects taken from Physics, Mechanics,Strains of Materials, etc. Dont say that the word "deformation" is notbetter for an engineering student then the corresponding mathemati-cal notion of "vector transformation", "parametric path", "parametricsheet", etc.

Many people contributed directly or indirectly to this volume.The late Professor Dr. Doc. Nicolae Popescu, my Professor and

my Master, was teaching me to do "live" Mathematics.My colleague and my friend Prof. Dr. Gavriil Paltineanu challenged

me to investigate the interesting way to teach Analysis for engineers.He gave me many deep advises on this fascinating domain which isMathematical Analysis. He also carried this difficult task to read thishuge volume and to push me in making many corrections.

Prof. Dr. Octav Olteanu and Prof. Dr. Nicolae Danet read care-fully the entire manuscript and made some very useful suggestions.

I am also grateful to my younger colleagues Dr. Emil Popescu, Dr.Viorel Petrehus, Dr. Marilena Jianu and Dr. Valentin Popescu forhelping me a lot with the proofreading.

At last, but not the least, I will always express my thanks to twopersons: to my wife Angela for encouraging me to "go on" with thisdifficult task and to my 11

2years old grand-daughter Monica Izabela

who succeeded to refresh a lot my tired mind before chapter 4. This

PREFACE 9

is why the reader must be grateful to her for the easy (but not short)way I succeeded to present the improper integrals with parameters.

Finally, I mention the special help that trees and birds of my closepark gave me during my silent walk around the lake in those hotevenings of August 2010. All of these are of a great importance whenone is thinking of mathematical affairs.

Prof. Dr. Sever Angel PopescuTechnical University of Civil Engineering BucharestDepartment of Mathematics and Computer ScienceB=dul Lacul Tei 124, Sector 2, Bucharest, [email protected], 2011

CHAPTER 1

Indefinite integrals (Primitives, Antiderivatives)

1. Definitions, some properties and basic formulas

We shall see later that in many problems coming from Geometry,Statistics, Mechanics, Physics, Chemistry, Engineering, Biology, Econ-omy, etc. we have to compute the area of a plane figure bounded by thelines x = a, x = b, y = 0 and the graphic of a nonnegative continuousfunction y = f(x), where x [a, b], a < b (see Fig.1).

Fusc 1.

Such a figure is usually called a curvilinear trapezoid. The greatEnglish scientist Sir Isaac Newton (1642-1727) discovered the mathe-matical relation between the function y = f(x) and this area.

T 1. (I. Newton) Let f : [a, b] R+ be a continuousfunction defined on a real closed interval [a, b] with nonnegative realvalues. Let x be a number in (a, b] and let F (x) be the area of thecurvilinear trapezoid [AxMD] (the area under the graphic of f "up tothe point x" (see Fig.2). Then this area function is differentiable andits derivative F (x) at the point x is exactly f(x), the value of the initialfunction f at the same point x.

1

2 1. INDEFINITE INTEGRALS (PRIMITIVES, ANTIDERIVATIVES)

Fusc 2.

P. (intuitively proof; you will see later a mathematically rig-orous proof) Let us fix a point x0 in (a, b]. We are going to prove thatthe derivative F (x0) exists and it is equal to f(x0). By definition

(1.1) F (x0) = limxx0

F (x) F (x0)x x0 ,

if this limit exists. Let us take for instance x (a, b], x > x0 (ifx0 = b, then take x < x0 and do a completely similar reasoning!). ButF (x)F (x0) is exactly the area of the curvilinear trapezoid [M0x0xM ](see also Fig.2). Since f is continuous, this area can be well approxi-mated by the area of the rectangle [M0x0xM

]. Indeed, in the case ofour figure (Fig.2), the least value of f on the interval [x0, x] is real-ized at the point x0 and the greatest is realized at the point x (usethe boundedness Weierstrass theorem ([Po], Theorem 32), for a moregeneral situation). So, the area of the curvilinear trapezoid [M0x0xM ]is a number between the area of the rectangle [M0x0xM

] and the areaof the rectangle [M 0x0xM ]. When M becomes closer and closer to M0,these last two areas tends to give the same number, so the area of thecurvilinear trapezoid can be well approximated by the area of the rec-tangle [M0x0xM

]. Thus, using now the corresponding mathematicalexpressions, we get:

(x x0)f(x0) F (x) F (x0) (x x0)f(x),

1. DEFINITIONS, SOME PROPERTIES AND BASIC FORMULAS 3

or

(1.2) f(x0) F (x) F (x0)(x x0) f(x).

Since f is continuous, taking limits when x x0 in (1.2), we get thatF (x0) exists and that it is equal to f(x0).

The area function constructed above is called the Newton area func-tion of f.

C 1. Let f : [a, c) R be a continuous function whichdoes not change its sign on the interval [a, c) (here c may be even !).Then, there is a differentiable function F defined on this interval (a, c)such that F (x) = f(x) for any x (a, c).

P. It is sufficient to take f to be nonnegative (otherwise re-place f by f). Then we define like above F (x) = the area of thecurvilinear trapezoid [AxMD] (see Fig.2).

E 1. Try to substitute the above interval [a, c) for an openfinite or infinite interval (a, c). Moreover, try to substitute this lastinterval for any open subset of R.

All of the discussion above is a real motivation for the followinggeneral definition.

D 1. Let A be an open subset of the real line R. Let f :A R be a function defined on A with values in R. Any differentiablefunction F : A R such that its derivative F (x) is equal to f(x) forany x in A (simply if F = f) is called a primitive of f on A. It is alsocalled an antiderivative or an integral of f on A.

In the following all functions are defined only on subsets A of R,which have no isolated points. We shall simply note A for a subsetsatisfying this last property. We introduced this restriction becauseit is a nonsense to speak about the limit of a function at an isolatedpoint. In particular, about its derivative!

If we remember the definition of the notion of a differential dF (a)of a function F at a point a of A, namely dF (a) which is the linearmapping defined on R with values in the same R, such that dF (a)(h) =F (a)h for any h in R, or dF (a) = F (a)dx, where dx is the differentialof the variable x (in our case the identity mapping-see Analysis I, [Po]),we can easily prove the following very useful result.

T 2. Let f : A R be a function defined on A (as above).A function F : A R is a primitive of f on A if and only if it is


differentiable on A and dF = fdx, i.e. its differential dF (a) at anypoint a of A is equal to f(a)dx.

A primitive F of f is denoted by the notationf(x)dx and it is

also called an integral of f with respect to the variable x. Moreover, anexpression of the form fdx is called a differential form of order one inone variable x. A primitive of such a differential form is a function U :A R such that U is differentiable on A and dU = fdx, equivalentlyU (x) = f(x) for any x in A. Now, the notation

f(x)dx means such

a primitive U for the differential form fdx. It is useful for us in thefollowing to speak the primitives of differential forms language, insteadof using the language of primitives for usual functions.

To decide if a differential form has a primitive or not is in general avery difficult problem. We shall prove later a more general result thanthat one of exercise 1. Namely, one can rigorously prove (see Chapter2) that any continuous function f on an interval I has a primitive Fon I. This means that for a continuous function f on an interval I, thedifferential form fdx has a primitive F, i.e. dF = fdx.

For instance, a primitive of f(x) = cosx is F (x) = sinx becaused(sin x) = cosxdx. We see that a function of the form sin x+C, whereC is an arbitrary constant (does not depend on x. It may depend onany other variable distinct of x), is also a primitive of cosx. Indeed,

d(sin x+ C) = d(sin x) + dC = cosxdx+ 0 = cosxdx.

So, if a function f has a primitive F on an open subset A of R, then ithas an infinite number of primitives, namely, any function of the form:F + C, where C is a constant w.r.t. x (prove this as in the case off(x) = cosx).

T 3. (a structure theorem for all primitives) Let f : I Rbe a function defined on an interval I and let F be a fixed primitiveof it (on I). Then any other primitive G of f on the interval I is ofthe form G = F + C, where C is a constant number (which dependsonly on G). This is why the set of all primitives of f is denoted byf(x)dx+ C. Here

f(x)dx means any fixed primitive of f on I.

P. Since F (x) = f(x) = G(x) for any x in I, one can seethat the difference function H = GF has its derivative equal to zeroat any point of I. We are going to prove that in this case H must be aconstant, i.e. it cannot depend on the variable x. This means that forany points a and b of I we must have that H(a) = H(b). Indeed, let ussuppose that a < b and let us apply Lagrange theorem on the closedinterval [a, b] (which is included in I, why?):

H(b)H(a) = H (c)(b a),


for a point c in (a, b). Since H (x) = 0 for any x of I, one gets thatH(b) = H(a). Let us denote by C this common value of H(x). Hence,G F = C, or G = F + C. This means that for this fixed number C,we have that G(x) = F (x) + C for any x I.

R 1. If the set A is not an interval, then the above result isnot always true. For instance, let A = (0, 1) (2, 3) and f(x) = 2x forany x of A. It is easy to see that the following two functions

F (x) =

{x2, for x (0, 1)

x2 + 1, for x (2, 3) ,

G(x) =

{x2, for x (0, 1)x2, for x (2, 3)

are primitives of f. It is clear enough that there is no constant numberC such that G(x) = F (x) + C for any x in A. Indeed, for x (0, 1),C = 0 and for x (2, 3), C = 1, so we cannot have the same constantC on the entire set A.

E 1. Let us find a primitive for the following continuousfunction

f(x) =

x, if x [1, 1]x3, if x (1, 2)x2 + 4, if x [2, 3) ,f : [1, 3) R. Since

(x2

2

)= x,

(x4

4

)= x3 and

(x3

3+ 4x

)= x2+4,

any primitive of f is of the following form:

F (x) =

x2

2+ C1, if x [1, 1]

x4

4+ C2, if x (1, 2)

x3

3+ 4x+ C3, if x [2, 3)

,

Let us force now F to be continuous at x = 1 (F is differentiable, thusit must be continuous):

1

2+ C1 =

1

4+ C2,

so we get that C2 =14+ C1. The continuity of F at x = 2 implies that

24

4+

1

4+ C1 =

23

3+ 8 + C3,

thus C3 = 7712 + C1. Hence, all the primitives of f are:

F (x) =

x2

2, if x [1, 1]

x4

4+ 1

4, if x (1, 2)

x3

3+ 4x 77

12, if x [2, 3)

+ C1 ,


where C1 is an arbitrary constant (with respect to x) number. It is easyto see that F (x) is differentiable and F (x) = f(x) for any x [1, 3).

E 2. Here is an example of a discontinuous function whichhas no primitive at all. Let

f(x) =

{1, for x [0, 1]0, for x (1, 2] .

It is easy to see that a primitive of f must be of the form:

F (x) =

{x+ C1, for x [0, 1]C2, for x (1, 2] .

The continuity of F at x = 1 implies that C2 = 1 + C1, thus,

F (x) =

{x+ C1, for x [0, 1]1 + C1, for x (1, 2] .

But this last function is not differentiable at x = 1 for no value of C1(why?). Hence, F cannot be a primitive of f on [0, 2]!

But, do not you hurry to conclude that all the discontinuous func-tions have no primitive. Here is a counterexample!

E 3. The function

f(x) =

{2x sin 1

x cos 1

x, for x (0, 1]

0, for x = 0,

f : [0, 1] R. Since limx0

2x sin 1x= 0 (prove it!) and since lim

x0cos 1

x

does not exist (xn =1

2nand xn =

2(4n+1)

are convergent to 0, but

cos 1xn

= 1 1 and cos 1xn

= 0 0), we see that function f has nolimit at x = 0. In particular, it is not continuous at x = 0. At the sametime, it is easy to prove that the function

F (x) =

{x2 sin 1

x, for x (0, 1]

0, for x = 0

is a primitive for f on [0, 1].

In general, to decide if a given noncontinuous function has a prim-itive or not is a difficult task.

The main problem which is to be considered in the following isthe one of the effective construction of primitives for some classes ofcontinuous functions.

Here is a table with the primitives of the basic elementary functions,mostly used in Mathematics and in its applications. We write downonly one primitive (the immediate one, obtained by a direct process of


anti-differentiation). The others can be obtained by adding a constantnumber to this last one.

(1.3)

xdx =

x+1

+ 1, if = 1, (x > 0, if is not an integer)

(1.4)

1

xdx = lnx, if x I,

where I is any interval contained in (0, ).(1.5)

1

xdx = ln(x), if x J,

where J is any interval contained in (, 0).(1.6)

exdx = ex

(1.7)

axdx =

ax

ln a, if a > 0 and a = 1.

(1.8a)

1

x2 + 1dx = arctan x

(1.9)

1

x2 + a2dx =

1

aarctan

x

a, if a = 0.

(1.10)

sin xdx = cosx

(1.11)

cosxdx = sin x

(1.12)

1

1 x2dx = arcsin x, if x (1, 1).

(1.13)

1

a2 x2dx = arcsinx

a, if x (a, a).

(1.14)

1

x2 + dx = ln(x+

x2 + ),

if = 0, x2 + > 0 and x+x2 + > 0.(1.15)

1

cos2 xdx = tan x,


if x belongs to any interval which does not contain any number of theform n +

2, n Z.

(1.16)

1

sin2 xdx = cotx,

if x belongs to any interval which does not contain any number of theform n, n Z.

(1.17)

sinhxdx = cosh x,

coshxdx = sinh x,

where sinh x = exex

2and cosh x = e

x+ex2

.All of the above formulas can be directly verified by differentiat-

ing the function from the right side of each equality. Lets verify forinstance formula (1.14):(

ln(x+x2 + )

)=

1

x+x2 + )

(1 +

2x

2x2 +

)=

1x2 +

.

T 4. (linearity) Let f, g : I R be two functions definedon the same interval I and let , be two real numbers. Let

f(x)dx

andg(x)dx be two primitives of f and g respectively (on I). Then

f(x)dx+

g(x)dx is a primitive of f + g on I.

P. Since(

f(x)dx)

= f and(

g(x)dx)

= g (use only thedefinition!), we get that(

f(x)dx+

g(x)dx

)=

=

(f(x)dx

)+

(g(x)dx

)= f + g

and the proof is complete (why?). Here we used the linearity of thedifferential operator h h.

This last result says that the "mapping" f f(x)dx is linear.Here is a concrete example of the way we can work with this lastproperty and with the above basic formulas.

E 4. Let us compute (

x5 3x+ 2 3x 75x3

)dx. The

linearity of the integral operator(see theorem 4) and formula (1.3)

imply that (x5 3x+ 2 3x 7

5x3

)dx =

x5dx 3

xdx+


+2

x

13dx 7

x

35dx =

x6

6 3x

2

2+ 2

x43

43

7x25

25

=

=x6

6 3

2x2 +

3

2x

43 35

2x

25 .

Let J be another interval and let u : J I, x = u(t), be a functionof class C1(J). Then dx = u(t)dt (see Analysis I, [Po]). Let f : I Rbe a continuous function of the variable x I. Let F (x) be a primitiveof f on I. Then F (u(t)) is a primitive of the function f(u(t))u(t) of t,i.e.

(1.18)

f(u(t))d(u(t)) =

(f(x)dx

)(u(t)).

Indeed,

[F (u(t))] = F (u(t))u(t) = f(u(t))u(t).

Formula (1.18) is called the change of variable formula for integralcomputation. This formula says that if in an integral

h(x)dx one can

put in evidence an expression u = u(x) such that h(x)dx = f(u(x))duand if one can compute

f(u)du then, put instead of u, u(x) in this

last primitive and we obtain a primitiveh(x)dx for the differential

form hdx.

E 5. Let us compute

13x+2

dx, where 3x + 2 < 0. If wedenote u = 3x+ 2 then, du = 3dx (why?). Thus, our integral becomes13

1

3x+2d(3x+2). Since u = 3x+2 < 0, a primitive for

1udu is ln(u)

(see formula (1.5)). So, a primitive of 13x+2

dx is 13ln(3x 2).

Sometimes it is easier to directly compute dx as a function of t anddt. For instance, let us compute

tan3 xdx. Let us change the variable:

t = tan x, x (2, 2) (why this restriction?). Since

dt =1

cos2 xdx =

(1 + tan2 x

)dx = (1 + t2)dx,

we get thattan3 xdx =

t3

1 + t2dt =

t3 + t tt2 + 1

dt =

tdt

tdt

t2 + 1=

=t2

2 1

2

d(t2 + 1)

t2 + 1=t2

2 1

2ln(t2 + 1) =

=1

2

[tan2 x ln(tan2 x+ 1)] .


Let f, g : I R be two functions of class C1(I), where I is a realinterval. We know the "product formula" :

(fg) = f g + fg.

Applying the integral operator to both sides, we get:

(1.19)

f(x)g(x)dx = fg

f (x)g(x)dx.

or, in language of differentials,

(1.20)

fdg = fg

gdf

These last two formulas are called the integral by parts formulas.When do we use them? If one has to compute the integral

h(x)dx

and if we can write h(x) = f(x)g(x) and if the integralf (x)g(x)dx

can be easier computed, then formula (1.19) works. For instance, if wedifferentiate the function f(x) = ln x, we get 1

xwhich is an "easier"

function from the point of view of integral calculus. Practically, let ususe this philosophy to compute the integral: In =

xn ln xdx, where n

is a natural number. For n = 0 we get

I0 =

ln xdx =

(lnx)(x)dx.

In formula (1.19) one can put f(x) = lnx and g(x) = x. Thus,

I0 = x ln x

1

x xdx = x ln x x.

Let us use the same "trick" for computing In, n > 0.

In =

xn ln xdx =

(ln x)

(xn+1

n+ 1

)dx =

xn+1

n+ 1ln x

1

x

xn+1

n+ 1dx =

=xn+1

n+ 1ln x 1

n+ 1

xndx =

xn+1

n+ 1lnx x

n+1

(n+ 1)2.

2. Some results on polynomials

We begin with some facts on polynomial functions.A polynomial function ( or simply a polynomial) is a function P of

one variable x, defined on R by the following formula

P (x) = a0 + a1x+ a2x2 + ...+ anx

n, x R,where a0, ..., an are fixed real numbers. These numbers are said to bethe coefficients of P. We shortly write that P (x), or P R[x], whereR[x] is the set (a ring!) of all polynomials with real coefficients. For

2. SOME RESULTS ON POLYNOMIALS 11

instance, P (x) = 2x 1 is a polynomial. Here n = 1, a0 = 1 anda1 = 2. If an = 0 (except the case P = 0 this means that ALL itscoefficients are zero, we always assume that the dominant coefficient anis not zero) we say that the degree of P is n and write this as degP = n.If P (x) = a0 = 0, a nonzero constant, then degP = 0. If P (x) = 0for any x (this is equivalent to saying that all its coefficients are zero),then the degree of P is by definition . This value was chosen topreserve the formula degPQ = degP + degQ for P = 0 and Q = 0.

If an = 1 we say that the polynomial P is monic. For instance,P (x) = x 1 is monic, but Q(x) = x 1 is not monic. If P has thedegree greater or equal to 1 and if it cannot be written as a productof two polynomials of degrees greater or equal to 1, it is said to beirreducible. For instance, P (x) = x2+2, P (x) = 2x1 are irreducible,but Q(x) = x2 1 or Q(x) = x3 are reducible, i.e. they can bedecomposed into at least two factors of degrees greater than zero.

The following result is fundamental in Algebra.

T 5. (Euclids division algorithm) Let P and Q = 0 be twopolynomials. Then there are another two polynomials C and R suchthat P = CQ + R and degR < degQ. Here the polynomials C and Rare uniquely defined by P and Q.

P. An "abstract" general proof can be found in [La1]. But,...nothing is "abstract" here! We simply use the "division algorithm" forpolynomials, which is very known even from elementary school Algebra.If degP < degQ, then take C = 0 and R = P. Assume that n =degP m = degQ. We use mathematical induction with respect ton. If n = 0, then m = 0 (Q = 0) and P = k, Q = h, where k and h aretwo real numbers and h = 0 (why?). Since

k =k

hh+ 0,

one can take C = khand R = 0 (degR = < degQ = 0). Let

now n be grater than 0 and let us assume that we have just proved thetheorem for any k = 0, 1, ..., n 1. Let us prove it for k = n. SupposeP (x) = anx

n+an1xn1+...+a0 andQ(x) = bmxm+bm1xm1+...+b0,where an = 0 and bm = 0. Then, it is easy to see that(2.1) P (x) = Q an

bmxnm + P1(x),

where

P1(x) =

(an1 anbm1

bm

)xn1 + ...+

(anm anb0

bm

)xnm+


+anm1xnm1 + ...+ a0is a polynomial of degree at most n 1. Let us use now the inductionhypothesis and write

P1(x) = C1Q+R,

where C1 and R are polynomials, degR < degQ. Let us come back toformula (2.1) with this expression of P1(x) and find

P (x) =

(anbm

xnm + C1

)Q+R.

If we put now C = anbmxnm+C1, we just obtained the statement of the

theorem for n. Hence the proof of the theorem is complete.The uniqueness can be derived as follows. Let P = DQ+ S, where

deg S < degQ. Then, RS = (DC)Q. If R = S, then the inequality,deg(RS) = deg(DC)Q degQ, give rise to a contradiction (why?).Thus, R = S and so (D C)Q = 0 implies D = C (Q = 0).

C 2. (the P -expansion of Q) Let P be a nonconstantpolynomial and let Q be another arbitrary polynomial. Then Q can beuniquely written as

(2.2) Q = A0 + A1P + ... + AnPn,

where A0, A1, ..., An are polynomials of degrees at most degP 1. Wesay that "we write Q in base P". In particular, for any nonzero naturalnumber m, one get:

(2.3)Q

Pm=

A0Pm

+A1

Pm1+ ...+

AnPmn

,

if m > n and

(2.4)Q

Pm=

A0Pm

+A1

Pm1+ ...+

Am1P

+ S,

where S is a polynomial, if n m.P. We apply again mathematical induction on the degree of

Q. If Q is a nonzero constant (degree zero) polynomial, then Q = Q(A0 = Q). Assume that degQ > 0 and that the statement is true forany polynomial Q1 of degree < degQ. Let us apply Euclids divisionalgorithm for Q and P :

(2.5) Q = CP + A0,

where degA0 < degP. Since degC is less than degQ (degP > 0),using the induction hypothesis, we get C = A1 + A2P + ...+ AnP

n1,where degAj < degP, j = 1, 2, ..., n. Coming back to formula (2.5)with this expression of C, we obtain exactly formula (2.2).


Let us apply this theory to the polynomials Q(x) = x3 + 2 andP (x) = x 1. Since

x3 + 2 = C(x)(x 1) +R,putting x = 1 we get R = 3. Thus,

x3 1 = C(x)(x 1).Hence, C(x) = x2 + x+ 1. Now,

x2 + x+ 1 = C1(x)(x 1) +R1.Making x = 1, we get R1 = 3. So,

x2 + x+ 1 3 = x2 + x 2 = C1(x)(x 1).Thus,

C1(x) = x+ 2 = 1 (x 1) + 3.Coming back, step by step, we obtain

x3 + 2 = 3 + (x 1)[{3 + 1 (x 1)}(x 1) + 3] == 3 + 3(x 1) + 3(x 1)2 + (x 1)3.

Let C = {a+ bi : a, b R}, i = 1, be the field of complex num-bers. A fundamental result in Algebra (C. F. Gauss) [La1] says thatthe roots of any polynomial P R[x] are complex numbers, namelyelements of the form a + bi, with a, b real numbers (see also theorem100 in this book).

T 6. (Bezouts theorem) Let P = P (x) be a nonconstantpolynomial and let C be a root of P (P () = 0). Then P (x) =A(x )Q(x), where A is a constant number and Q(x) is a monicpolynomial. Moreover, if 1, ..., s are all the distinct roots of P, then

(2.6) P (x) = A(x 1)n1(x 2)n2 ...(x s)ns,where A is the dominant coefficient of P (A = an) and n1, ..., ns arethe algebraic multiplicities of the roots 1, ..., s. All of these numbersare uniquely determined only by P.

P. Let P (x) = a0+a1x+a2x2+ ...+anx

n, be our polynomialand let A = an. Then

(2.7) P (x) = AT (x)

where T (x) is the monic polynomial xn + an1an

xn1 + ... + a0an. Let us

use Euclids division algorithm for dividing T (x) by x :(2.8) T (x) = (x )Q(x) +R,


where Q(x) is a polynomial and R is a constant number. Since T () =0 (why?), making x = in (2.8) one gets R = 0. Thus, T (x) =(x )Q(x) and, coming back to formula (2.7), we finally obtain that

P (x) = A(x )Q(x).Now we apply the same procedure to the monic polynomial Q(x) andanother root of it:

P (x) = A(x )(x )Q1(x),where Q1 is a polynomial with degP > degQ > degQ1. We continuein this way and finally obtain

P (x) = A(x )(x )...(x ),where , , ..., are all the n roots of P. Let us put together the equalroots and so we get the required expression (2.6).

A bijection : C C which is a field morphism, i.e.(2.9) (z + w) = (z) + (w), (zw) = (z)(w)

is called an automorphism of C. An R-automorphism of C is anautomorphism of C which does not change the elements of R, i.e. (a+0i) = a, if a R. For instance, it is easy to see (prove it!) thatthe identity e : C C, e(z) = z and the conjugation e : C C,e(a+ bi) = a bi are R-automorphisms of C. We also denote e(z) byz, the conjugate of z. For instance, 3 4i = 3 + 4i and 3.5 = 3.5.

L 1. Let

P (x) = a0 + a1x+ a2x2 + ...+ anx

n

be a polynomial with real coefficients and let C be a root of it, i.e.P () = a0 + a1 + a2

2 + ...+ ann = 0.

Let be a R-automorphism of C. Then, the complex number () isalso a root of the polynomial P.

P. Let us apply to the equality

a0 + a1+ a22 + ...+ an

n = 0.

Taking into account the algebraic properties (2.9) of and the factthat it preserves the real coefficients a0, ..., an, we get

a0 + a1() + a2()2 + ...+ an()

n = 0,

i.e. the complex number () verifies the equality: P (x) = 0, thus() is another root of P.


This lemma says that any R-automorphism of C permutes theroots between them. For instance, since i is a root of the equationX2 + 1 = 0, (i) = i or (i) = i. If (i) = i, (a + bi) = a+ bi, i.e. = e, the identity. If (i) = i, then (a+ bi) = abi, i.e. = e, theconjugation automorphism of C. These considerations lead us directlyto the following basic result.

T 7. Any monic irreducible polynomial P with real coef-ficients is either of the form P (x) = x a, or of the form P (x) =x2 + bx+ c, where a, b, c are real numbers.

P. If our polynomial P is of degree 1 everything is clear. IfdegP > 1, let be one of its root in C. Use lemma 1 and find thatits conjugate is also a root of P (x). If = , i.e. if is a realnumber, then P (x) is divisible by the polynomial Q(x) = x (seetheorem 6). Since degP > 1 and since Q(x) is a factor of P (x), weobtain a contradiction (P was considered to be irreducible). Hence,the root is not real. Let us denote b = ( + ) and c = . Since+ = + and = we see that b, c R. The polynomialH(x) = (x)(x) = x2+bx+c is a divisor of P. Indeed, let us divideP by H. Thus, P = HU + V, where deg V < 2. Moreover, V () = 0and V () = 0, i.e. V is a polynomial of degree at most 1 which has twodistinct roots! The only possibility is that V be identically with zero,so P = HU. Since P is irreducible, U must be a constant polynomialand, since P and H are both monic, this last constant polynomial mustbe 1. Thus,

P = H = x2 + bx+ c.

The following theorem (see [La1] for instance) is a basic result inAlgebra.

T 8. (factorial theorem). Let P be a nonconstant polyno-mial with real coefficients. Then P can be uniquely written as

(2.10) P = AP n11 Pn22 ...P

nkk ,

where A is a constant number and P1, P2, ..., Pk are distinct monicirreducible polynomials of degrees 1 or 2. "Uniquely" here means thatif

P = BQm11 Qm22 ...Q

mhh ,

is a decomposition of P of the same type (B constant and Q1, Q2, ..., Qhare monic and irreducible polynomials), then B = A, k = h and foreach Q

mjj , there is a Pi such that Qj = Pi and mj = ni.


P. In formula (2.6) some s could be real numbers and somecould be complex numbers. If for instance 1 is a real number, thenP1 = (x 1)n1 . If i is a complex number (nonreal), then there isexactly an j between the roots 1, ..., s of P such that j = i (seelemma 1) and ni = nj (why?). Thus, the polynomial (x i)(x j)is a monic irreducible polynomial with real coefficients (why?). Let usdenote it by P2. Hence, the contribution of i and j to P is exactlyP n22 , where n2 is the common value of ni and nj . We continue thisreasoning on the roots of P up to obtaining the formula (2.10). SinceA = B = an, the dominant coefficient of P , since any root h of P isa root of a Pt and of a Qv at the same time and since Pt and Qv aremonic irreducible polynomials, we see that Pt = Qv. Indeed, if h is areal number, then Pt(x) = x h = Qv(x). If h is a complex nonrealnumber, since Pt and Qt are monic irreducible polynomials with realcoefficients, we see that both of them must be equal to (x h)(xh)(why?). So, we have the required uniquenees.

D 2. (the greatest common divisor) Let P and Q be twononzero polynomials. The monic polynomial D of maximum degreesuch that D is a divisor of P and of Q is said to be the greatest commondivisor of P and Q. We denote D by (P,Q).

For instance, if P = x4(x + 1)2(x2 + x + 1) and Q = x2(x + 1)5,then D = (P,Q) = x2(x+ 1)2. In general, if

P = AP n11 Pn22 ...P

nkk ,

andQ = BQm11 Q

m22 ...Q

mtt ,

are the decompositions of P and respectively of Q, of type (2.10), suchthat P1 = Q1, P2 = Q2, ..., Pr = Qr and all the others are distinct oneto each other, then (why?)

D = Pmin(n1,m1)1 P

min(n2,m2)2 ...P

min(nr ,mr)r .

Moreover, the roots of D are exactly the common roots of P and Qwith the least multiplicities (look at the above example!). If D is 1,we say that P and Q are coprime. For instance, P = 2x2 + 2 andQ = x3+1 are coprime, but P = x+1 and Q = x3+1 are not coprime(why?).

T 9. D is the greatest common divisor of P and Q (P andQ are not zero) if and only if there exists two polynomials U0 and V0such that

(2.11) D = PU0 +QV0


and D is monic with the least degree such that D can be written asin (2.11). In particular, if P and Q are coprime, then there are twopolynomials U0 and V0 such that

(2.12) 1 = PU0 +QV0

P. Let us define the following set of polynomials(2.13)S = {H = PU +QV : U and V are arbitrary polynomials in R[x]}.It is easy to see that the sum between two polynomials of S is also apolynomial in S.

If we multiply a polynomial of S by an arbitrary polynomial of R[x],we also get a polynomial of S (prove these two last statements). LetD be a (monic) nonzero polynomial of S of the least degree and letM be another polynomial of S. Let us apply the Euclids algorithm fordividing M by D.We get

M = CD +R,

where degR < degD. Since M,D S, then R = M CD S. Thus,R = 0 and so M = CD. Therefore D is a divisor of any polynomial ofS. In particular, D divides P and Q (P,Q S!) Since D S, thereare two polynomials U0 and V0 such that D = PU0 +QV0.

=) Let now G be the greatest common divisor of P and Q. SinceG divides P and Q, G also divides D. But D also divides P and Q,thus the degree of D is at most equal to degG. Since G divides D, wehave that degG = degD. Since D and G are both monic, we musthave that D = G. Thus, for the greatest common divisor we have therelation (2.11).

=) Let D0 be a monic nonzero polynomial of minimal degree suchthat

D0 = PU1 +QV1 S.Exactly like above we prove that D0 is the greatest common divisor ofP and Q.

The last statement is obvious.

Practically, we can find the greatest common divisor and its ex-pression (2.11) by transferring the problem from polynomials P andQ to another pair of polynomials Q and R, where degR < degQ (ifdegP degQ). Then to another pair R and R1, where degR1 degP, change Pwith Q, etc.) and divide P by Q. There are polynomials C and R suchthat P = CQ+R and degR < degQ. It is easy to see that the greatest


common divisor of P and Q is equal to the greatest common divisor ofQ and R (prove slowly!). Now divide Q by R and get Q = C1R + R1,with degQ > degR > degR1. The greatest common divisor of Q andR is equal to the greatest common divisor of R and R1, and so on... upto a Rj = 0. Then Rj1 is exactly the greatest common divisor of Pand Q eventually multiplied by a constant (in our above definition thegreatest common divisor is a monic polynomial!) (prove all of these!).

Let for instance P (x) = x4 x and Q(x) = x2 1. It is not neces-sarily to use Euclids division algorithm, but to successively diminishthe degrees of P or of Q. Let us write:

x4 x = x2(x2 1) + x2 x = x2(x2 1)++(x2 1) (x 1) = (x2 + 1)(x2 1) (x 1).

Since (P,Q) = (x2 1, x 1) we go on with division:x2 1 = (x+ 1)(x 1),

the greatest common divisor of P and Q is x 1 andx 1 = (x4 x)(1) + (x2 1)(x2 + 1).

Thus, the polynomials U0 and V0 are

U0(x) = (1), and V0(x) = x2 + 1and they can be chosen such that degU0 < degQ and deg V0 < P.

A representation ofD as in (2.11) with degU0 < degQ and deg V0

1 anddenote P n22 P

n33 ...P

nkk by L. Suppose that the statement of the theorem

is true for any nonzero natural number less than k and we shall prove it


for k. Since P n11 , Pn22 , ... , P

nkk are monic irreducible and distinct one to

each other, P n11 and L have no common roots. Indeed, if is a root ofP n11 and of L, then there is a j = 1 such that is a root of P1 and of Pj;if is real, P1 and Pj are both equal to x, a contradiction! (why?).If is not real, let be the conjugate of . Then P1 and Pj are bothequal to (x )(x ), again a contradiction! (why?). Thus P n11 andL are coprime and we can apply theorem 9, formula (2.12). So thereare two uniquely defined polynomials V and U with deg V < degL anddegU < degP n11 such that

P n11 V + LU = 1

(see theorem 10). Let us multiply both sides by QP= Q

APn11 L

. We get:

(2.19)Q

P=

1

A

[QV

L+QU

P n11

].

Let us put Q1 instead of QU and let us see that the number of irre-ducible factors of L is k 1. Applying the induction hypothesis weobtain the representation (L is monic, so A = 1 in this case!):

(2.20)QV

L=

Q2P n22

+ ...+QkP nkk

,

where Q2, Q3, ..., Qk are uniquely defined polynomials. If we come backto formula (2.19) with this expression of QV

L, we get exactly formula

(2.17). The other statement is a direct computational consequence ofthis last formula.

E 6. Let us find the decomposition into simple fractionsof the rational function Q(x)

P (x)= x+1

x4+x2. Since P (x) = x2(x2 + 1), we

have two blocks of simple fractions, one corresponding to the irreduciblepolynomial P1(x) = x and the other corresponding to the irreduciblepolynomial P2(x) = x

2 + 1. Hence,

(2.21)x+ 1

x4 + x2=

A

x2+B

x+Cx+D

x2 + 1.

Thus,

x+ 1

x4 + x2=A(x2 + 1) +Bx(x2 + 1) + (Cx+D)x2

x4 + x2,

orx+ 1 = (B + C)x3 + (A+D)x2 +Bx+ A.

We identify the coefficients of both sides and obtain:

1 = A,

1 = B,

3. PRIMITIVES OF RATIONAL FUNCTIONS 21

0 = A+D,

0 = B + C,

so A = 1, B = 1, C = 1 and D = 1. Therefore our decompositionis

(2.22)x+ 1

x4 + x2=

1

x2+

1

x+x 1x2 + 1

.

3. Primitives of rational functions

We shall use now the basic properties of the integrals (primitives)and the above sketchy theory of rational functions in order to computetheir primitives.

Let us compute for instance a primitive for the rational functionwhich appeared in example 6, namely

x+1x4+x2

dx. We use the linearity

ofin formula (2.22):

x+ 1

x4 + x2dx =

1

x2dx+

1

xdx+

x 1x2 + 1

dx =

=

x2dx+

x1dx 1

2

2x

x2 + 1dx

1

x2 + 1dx =

=x3

3 + ln |x| 1

2ln(x2 + 1) arctanx,

because2x

x2 + 1dx =

d(x2 + 1)

x2 + 1=

du

u= ln |u| = ln(x2 + 1).

Since any simple fraction has one of the following forms:

(3.1)A

x , where A, R,

(3.2)A

(x )n , where A, R, n N, n 2,

(3.3)Ax+B

x2 + bx+ c, where A, b, c R, = b2 4c < 0,

(3.4)Ax+B

(x2 + bx+ c)n, where A, b, c R, = b2 4c < 0, n 2,

we must show how to find a primitive for each of these cases.Case 1

(3.5)

A

x dx = A

1

x dx = A ln |x | .


Case 2 A

(x )ndx = A

1

(x )ndx =(3.6)

A

(x )ndx = A(x )

n+1

n+ 1 , if n 2.

Case 3 SinceAx+B

x2 + bx+ cdx =

A

2

2x+ b

x2 + bx+ cdx+

+

(Ab

2+B

)1

x2 + bx+ cdx

and since 2x+ b

x2 + bx+ cdx = ln(x2 + bx+ c),

one can reduce everything to the computation of a primitive of thefollowing type:

A

x2 + bx+ cdx = A

1

(x+ b2)2 + a2

dx,

where a =c b2

4. Thus,A

x2 + bx+ cdx = A

d(x+ b

2)

(x+ b2)2 + a2

=(3.7)

= A1

aarctan

x+ b2

a.

Here we use the basic formula

1x2+a2

dx = 1aarctan x

a(see formula

(1.9)).Case 4 Using a similar reasoning as in Case 3 one can reduce the

computation of the primitive of the simple fraction of formula (3.4) tothe following primitive:

1

(x2 + bx+ c)ndx =

d(x+ b

2)[

(x+ b2)2 + a2

]n .Denoting x + b

2by a new variable u we must compute the primitive

du(u2+a2)n

, for n 2. For convenience we use the same variable x. Letus denote In =

dx

(x2+a2)nfor any n > 0 this time (a = 0). We know

that

I1 =1

aarctan

x

a.


Let n > 1. We shall construct now a recurrence formula for In :

In =1

a2

x2 + a2 x2(x2 + a2)n

dx =1

a2

1

(x2 + a2)n1dx

1a2

x2

(x2 + a2)ndx =

1

a2In1 1

2a2

x

(x2 + a2)nd(x2 + a2) =

=1

a2In1 1

2a2

[xd

((x2 + a2)n+1

n+ 1)]

.

Let us use now the formula of integrating by parts (see (1.20)) andcompute

xd

((x2 + a2)n+1

n+ 1)

= x(x2 + a2)n+1

n+ 1 1

n+ 1

1

(x2 + a2)n1dx,

Thus,

In =1

a2In1 1

2a2

[x(x2 + a2)n+1

n+ 1 1

n+ 1In1],

or

(3.8) In =

[1

a2 1

2a2(n 1)]In1 +

x

2a2(n 1)(x2 + a2)n1 .

For instance, let us compute I2 =

dx(x2+a2)2

.

(3.9) I2 =1

2a3tan1

x

a+

x

2a2(x2 + a2).

Here tan1 x is another notation for arctan x. Formula (3.8) can be usedto compute In "from up to down", i.e. the computation of In reducesto the computation of In1. The computation of In1 reduces to thecomputation of In2, etc., up to the computation of I1 = 1a arctan

xa.

In the following examples we use the ideas and experience just ex-posed in the above four cases.

E 7. What is a primitive of 34x+5

, where x runs over an

interval I which does not contain x = 54.

3

4x+ 5dx =

3

4

d(4x+ 5)

4x+ 5=

3

4

du

u=

3

4ln |u| = 3

4ln |4x+ 5| .

E 8. Let us compute a primitive of f(x) = 1(2x+5)5

, where

x belongs to an interval I which does not contain x = 52.

1

(2x+ 5)5dx =

1

2

(2x+ 5)5d(2x+ 5) =

1

2

u5du =


=1

2

u5+1

5 + 1 =(2x+ 5)4

8 .

E 9. What is the set of all primitives of the function f(x) =3x+2

x2+x+1, x R?

3x+ 2

x2 + x+ 1dx = 3

x+ 2

3

x2 + x+ 1dx =

3

2

2x+ 4

3

x2 + x+ 1dx =

=3

2

2x+ 1 + 1

3

x2 + x+ 1dx =

3

2

(x2 + x+ 1)

x2 + x+ 1dx+

1

2

1

x2 + x+ 1dx =

=3

2

d (x2 + x+ 1)

x2 + x+ 1+

1

2

1(

x+ 12

)2+ 3

4

d

(x+

1

2

)=

=3

2ln(x2 + x+ 1) +

1

2

1

u2 +(

32

)2du = 32 ln(x2 + x+ 1)++1

2

132

arctanu32

=3

2ln(x2 + x+ 1) +

13arctan

2(x+ 1

2

)3

=

3

2ln(x2 + x+ 1) +

13arctan

2x+ 13

.

So the set of all primitives of f(x) = 3x+2x2+x+1

is{3

2ln(x2 + x+ 1) +

13arctan

2x+ 13

+ C

},

where C is an arbitrary constant.

E 10. Let us find a primitive for f(x) = x+3(x2+4x+6)3

, x R.

I =

x+ 3

(x2 + 4x+ 6)3dx =

1

2

2x+ 4

(x2 + 4x+ 6)3dx+

(3.10) +

1

(x2 + 4x+ 6)3dx =

1

2

(x2 + 4x+ 6)

(x2 + 4x+ 6)3dx+

+

1

[(x+ 2)2 + 2]3d(x+ 2) =

1

2

u3du+

dv[

v2 + (2)2]3 ,

where u = x2 + 4x+ 6 and v = x+ 2. Let

I3 =

dv[

v2 + (2)2]3


and let us use successively the recurrence formula (3.8) (put v insteadof x, 3 instead of n and

2 instead of a):

I3 =

[1

2 1

4(3 1)]I2 +

v

4(3 1)(v2 + 2)2 =3

8I2 +

v

8(v2 + 2)2.

But, from formula (3.9) we get

I2 =1

42arctan

v2+

v

4(v2 + 2),

thus

I3 =3

322arctan

v2+

3v

32(v2 + 2)+

v

8(v2 + 2)2.

Let us come back with this value of I3 in (3.10) and find:

I =1

2

u3du+

dv[

v2 + (2)2]3 =

1

2

u2

2 +3

322arctan

v2+

3v

32(v2 + 2)+

v

8(v2 + 2)2.

Finally, we make u = x2 + 4x+ 6 and v = x+ 2 :

I = 14 (x2 + 4x+ 6)2

+3

322tan1

x+ 22

+

+3(x+ 2)

32 [(x+ 2)2 + 2]+

x+ 2

8 [(x+ 2)2 + 2]2.

E 11. Compute I =

dx(x2+a2)(x2+b2)

for all values of a and

b.Case 1. a = 0 = b. Then

I =

x4dx =

x3

3 .

Case 2. a = 0, b = 0. Then we must decompose into simple frac-tions the integrand function: 1

x2(x2+b2). Let us denote x2 by u :

1

u(u+ b2)=A

u+

B

u+ b2,

or

1 = A(u+ b2) +Bu = (A+B)u+ Ab2.

So A = b2 and B = b2. Coming back to x, we get1

x2(x2 + b2)=b2

x2 b

2

x2 + b2.


Thus,

(3.11)

1

x2(x2 + b2)dx = b2

x2dx b2

1

x2 + b2dx =

= b2x1

1 b21barctan

x

b.

Case 3. a = 0, b = 0. Then I = 1x2(x2+a2)

dx. Use now formula

(3.11) with a instead of b and find1

x2(x2 + a2)dx = a2

x1

1 a2 1aarctan

x

a.

Case 4. a = 0, b = 0, a = b. Let us take the fraction 1(x2+a2)(x2+b2)

and

decompose it into simple fractions. Since this last expression is in facta rational function of x2, let us put u = x2 (it it easier to work withfactors of degree one at denominator-why?). So

1

(x2 + a2)(x2 + b2)=

A

u+ a2+

B

u+ b2.

Thus,

1 = A(u+ b2) +B(u+ a2) = (A+B)u+ Ab2 +Ba2.

Hence, A+ B = 0 and Ab2 + Ba2 = 1, or A = 1b2a2 and B = 1b2a2 .

Finally, 1

(x2 + a2)(x2 + b2)dx =

=1

b2 a2[

1

x2 + a2dx

1

x2 + b2dx

]=

=1

b2 a2[1

aarctan

x

a 1barctan

x

b.

]Case 5. a = b = 0. Then I = 1

(x2+a2)2= I2, with the notation of

formula (3.9). We use this last formula and find

I =1

2a3arctan

x

a+

x

2a2(x2 + a2).

4. PRIMITIVES OF IRRATIONAL AND TRIGONOMETRIC FUNCTIONS 27

4. Primitives of irrational and trigonometric functions

What is an irrational function? The big temptation were to say thatsuch a function is a function "which is not rational"! But, it is not so!For instance f(x) = exp(x) is not a rational function (why?-prove it!)but it is also not an irrational one. We say that it is a transcendentalfunction. "Easily" speaking, a function is irrational if it is "rational"and contains some radicals...The exact definition is the following. Afunction of two variables

P (x, y) =ni=0

mj=0

aijxiyj = a00 + a10x+ a01y + ... + anmx

nym,

where aij R, is called a polynomial of two variables. A ratio-nal function of two variables R(x, y) is a quotient of two polynomi-

als P (x, y), Q(x, y) of two variables: R(x, y) = P (x,y)Q(x,y)

. An algebraic

function of one variable is a root y = f(x) of a nonzero polynomialP (x, y) = a0(x) + a1(x)y + ...+ am(x)y

m, where ai(x) are polynomialsof one variable with coefficients in R, i.e. P (x, f(x)) = 0 for any realnumber x in the definition domain of f. For instance, y =

x2 + 1

is algebraic because it is a root of the equation 1 + x2 y2 = 0. Anintegral of the form

R(x, y)dx, where R(x, y) is a rational function

and y is an algebraic function is called an abelian integral ("abelian"comes from the name of the great mathematician Niels Abel who sys-tematically studied such integrals). If the algebraic function y = f(x)can be expressed by radicals we say that R(x, y) is an irrational func-

tion. For instance, R(x,x2 + 1) = x+

x2+1

1+3x2+1

is an irrational function.

Since primitives of some classes of irrational functions appears in manyapplications, we present here some modalities to compute them.

A. Primitives of the formR(x,

ax+ b)dx, a = 0

We can reduce the computation of such a primitive to the compu-tation of a primitive of a rational function by the following change ofvariable: u2 = ax + b for ax + b 0, a = 0 (a, b are fixed constants).Our primitive becomes

F (u) =

R

(u2 ba

, u

)2u

adu.

Then G(x) = F(

ax+ b)is a primitive of R(x,

ax+ b) (see formula

(1.18)).


E 12. Compute I =

11+2x+3

dx. Write u2 = 2x + 3,

2udu = 2dx = dx = udu, so

I =

u

1 + udu =

u+ 1 1u+ 1

du =

1du

1

1 + udu =

= u ln |u+ 1| = 2x+ 3 ln(

2x+ 3 + 1).

B. Primitives of the formR(x,

ax2 + bx+ c)dx, a = 0,where

ax2 + bx+ c is not a perfect squareAssume that a > 0. Then

ax2 + bx+ c =

(ax+

b

2a

)2+ c b

2

4a.

Let us make the following convention: if c b24a> 0, we denote it by 2

(any positive real number is a square!!); if c b24a< 0, we denote it by

2. Since ax2 + bx+ c is not a perfect square, c b24a

cannot be zero.

Thus, if we denoteax + b

2a= u, then du =

adx and our integral

becomes

(4.1)

R

(u b

2a

a,u2 2

)1adu.

Assume now that a < 0. Then

ax2 + bx+ c = (ax b

2a

)2+ c b

2

4a.

Now, always c b24a

is positive because ax2 + bx+ c 0 (otherwise wecannot define the square root of it!). So this time we put c b2

4a= 2.

Making the change of variable u =ax b

2a , du =

adx, weget

R

(u+ b

2aa ,

2 u2

)1adu.

Hence, we have to study the following three types of such integrals:

B1 :R(x,

x2 + 2)dx

B2 :R(x,

x2 2)dx and

B3 :R(x,

2 x2)dx.

In order to eliminate the radical in B1 and in B2 one can use the(Euler) substitution

(4.2)x2 2 = x+ t,


where t is a new variable. Squaring in (4.2) we get

x =2 t2

2t,x2 2 =

2 + t2

2tand dx =

2 t22t2

dt.

Thus our primitives B1 and B2 becomeR

(2 t22t

,2 + t2

2t

) 2 t22t2

dt

which is a primitive of a rational function in the new variable t. Com-pute it by the methods described in the previous section and finally

put t =x2 2 x (see formula (4.2)).

E 13. Let us compute I =

3x2 4x+ 1dx. First of allwe must reduce the computation of I to one of the form B1 or B2. Since

3x2 4x+ 1 =(

3x 23

)2 1

3.

Thus for u =3x 2

3, dx = 1

3du and our primitive becomes

13

u2 1

3du.

Let us make now an Euler substitution

(4.3)

u2 1

3= u+ t, u =

13+ t2

2t, du = t

2 13

2t2dt,

so13

u2 1

3du = 1

43

t4 2

3t2 + 1

9

t3dt =

= 143

[t2

2 2

3ln |t| t

2

18

].

Coming back to the initial variable x we get (t =u2 1

3 u, see

formula (4.3)):

I = 143[

(u2 1

3 u

)22

23ln

u2 1

3 u

(u2 1

3 u

)218

].

To find the expression of the primitive I in x we must put in this lastexpression u =

3x 2

3, etc. (go on with computations up to the

end).


In order to eliminate the radical in B3 we can make the substitution:

(4.4)2 x2 = xt+ .

Squaring and dividing by x we get:

x = xt2 + 2t = x = 2tt2 + 1

, dx =2t2 2(t2 + 1)2

dt,(4.5)2 x2 = t

2 +

t2 + 1.(4.6)

So

(4.7)

R(x,

2 x2)dx =

R

( 2tt2 + 1

,t2 + t2 + 1

)2t2 2(t2 + 1)2

dt,

which is a primitive of a rational function in the variable t. After finding

it we put instead of t its expression from (4.4), t =

2x2x

.

E 14. Let us use formula (4.7) to find the primitive ofI =

1

x+4x2dx.

For this we change the variable x with a new one t (like in (4.4)):

(4.8)22 x2 = xt+ 2.

So, using (4.5) we get1

x+4 x2dx =

1

4tt2+1

+ 2t2+2

t2+1

4t2 4(t2 + 1)2

dt =

2

t2 1(t2 + 2t 1)(t2 + 1)dt.

Since t2 + 2t 1 = (t + 1 + 2)(t + 1 2), we have the followingdecomposition into simple fractions

t2 1(t2 + 2t 1)(t2 + 1) =

A

t+ 1 +2+

B

t+ 12 +Ct+D

t2 + 1.

An ugly computation leads to

(4.9) A = 1 +2

4 + 22, B = 1

4 + 22, C = D =

1

2.

Thus

I = 2

t2 1(t2 + 2t 1)(t2 + 1)dt = 2A

1

t+ 1 +2dt

2B

1

t+ 12dt C

d(t2 + 1)

t2 + 1 2D

1

t2 + 1dt =


(4.10)

= 2A lnt+ 1 +22B ln t+ 12C ln(t2+1)2D arctan t,

where A,B,C,D have the numerical values from formula (4.9). From(4.8) we get

t =

4 x2 2

x.

With this last value of t we come in (4.10) and obtain

I = 2A ln4 x2 2x + 1 +2

2B ln

4 x2 2x + 12

C ln([

4 x2 2x

]2+ 1

) 2D arctan

4 x2 2

x.

R 2. To compute the primitive B1 we also can use trigono-

metric substitutions. For instance, if in I =R(x,

x2 + 2)dx we

put x = tan t, we get

dx =

cos2 tdt, I =

R( tan t,

|cos t|)1

cos2 tdt.

This last primitive is a rational function of sin x and cosx. We shellsee later how to integrate a rational function of sin x and cosx.

E 15. Let us make x = a tan t (a > 0) and compute

I =

x2 + a2dx = a2

1

cos3 tdt = a2

d(sin t)

cos4 t= a2

du

(1 u2)2 ,

where u = sin t. Since1

(1 u2)2 =A

(1 u)2 +B

1 u +C

(1 + u)2+

D

1 + u,

we find A = B = C = D = 14. So

I =a2

4

[du

(1 u)2 +

du

1 u +

du

(1 + u)2+

du

1 + u

]=

a2

4

[1

1 u ln(1 u)1

1 + u+ ln(1 + u)

]=a2

4

(2u

1 u2 + ln1 + u

1 u).

Coming back to t we obtain

I =a2

4

(2 sin t

cos2 t+ ln

1 + sin t

1 sin t).


But t = arctan xa, thus I finally becomes:

I =a2

4

(2 sin

[arctan x

a

]cos2

[arctan x

a

] + ln 1 + sin [arctan xa]1 sin [arctan x

a

]) .R 3. To compute

R(x,

x2 2)dx we can use sinhx and

cosh x. Recall that sinhx = expxexp(x)2

and coshx = expx+exp(x)2

. It iseasy to see that

(4.11) cosh2 x sinh2 x = 1.

Hence, if we put x = cosh t we get dx = sinh tdt andx2 2 =

|sinh t| . Thus our integral becomes a rational integral in sinh x andcosh x.

To computeR(x,

2 x2)dx we can use the trigonometric sub-

stitution x = sin t. Thus dx = cos tdt and2 x2 = |sin t| . So

our integral reduces to an integral of a rational function in sinx andcosx.

C. Primitives of the formxm(axn+ b)pdx, where m,n, p are

nonzero rational numbers and a, b = 0Let m = m1

m2, n = n1

n2and p = p1

p2be three nonzero rational numbers

(fractions are simplified) withm1, n1, p1,m2, n2, p2 natural numbers andm2, n2, p2 > 0. The expression x

m(axn + b)pdx is called a binomialdifferential form. If there is a differential function F (x) such thatdF (x) = xm(axn + b)pdx we say that the binomial differential is exactand the function F (x) is a primitive of it. A substitution or a change ofthe variable x with a new variable t is an equation of the type g(x) = t,where g is a diffeomorphism on a fixed interval J (g is of class C1 andhas an inverse x = g1(t) also of class C1). A substitution can beinterpreted as a plane curve h(x, t) = g(x) t = 0. The substitutionis called "rational" if one can find a rational parameterization of thecurve h(x, t) = 0, i.e. if one can find two rational functions R1(u),R2(u), where u L, a real interval, such that h(R1(u), R2(u)) = 0for any u L. For instance, (3x+ 5) 53 = t is a rational substitutionbecause for x = R1(u) =

u353, and t = R2(u) = u

5, rational functions

of u, one has: (3R1(u) + 5)53 = R2(u). At the end of 19-th century, the

great Russian mathematician P. L. Cebyshev proved a basic results onsuch binomial differential forms (on the existence of their primitives).


T 12. (Cebyshev) One can find an integral of a rationalfunction starting from

xm(axn + b)pdx, by making "rational" substi-

tutions, if and only if m,n and p are in one of the following threesituations:

Case 1. p is an integer (p2 = 1).Case 2. p is not an integer but m+1

nis an integer.

Case 3. p is not an integer, m+1n

is not an integer, but m+1n

+ p isan integer.

P. We prove only the fact that if we are in one of the threecases above, then the integral can be "rationalized", i.e. there is asequence of "rational substitutions" of variables such that our binomialdifferential form xm(axn+b)pdx becomes a rational differential form, i.e.a differential form of the type Q(u)du, where Q is a rational function ina new variable u. The reverse part of the theorem cannot be proved byelementary tools. It involves deep knowledge of Algebraic Geometry.

First of all let us make the natural substitution: xn = t (if n = 1; ifn = 1 we pass directly to the second step of the following reasoning).

Thus x = t1n and dx = 1

nt1n1dt. Hence, our binomial integral becomes

(4.12)1

n

tq(at+ b)pdt,

the canonical form of the binomial integral. Here we denoted m+1n 1

by q. In general, since q = m+1n 1 and p are rational numbers we

have in the expression tm+1n1(at + b)p two radicals. If we had only

one radical, it could be easy to change the variable t with another oneu such that the new differential form becomes rational. For instance,int4(3t + 4)

13dt we have only one radical, so we "kill" it by the

obvious substitution 3t+ 4 = u3, or t = u343

and dt = u2du. Thus ourlast integral becomes

34 (

u3 4)4 u1u2du = 81 u(u3 4)4du,

which is a primitive of a rational function. We decompose u(u34)4 into

simple fractions, etc.Let us come back to our general case of the primitive (from formula

(4.12)).Case 1. p is an integer.If q is also an integer, we have nothing to do, the integrand tq(at+b)p

being rational.


If q = m+1n 1 is not an integer, i.e. if m+1

nis not an integer, q =

q1q2, simplified and q1, q2 integers with q2 = 1, then the obvious substitu-

tion t = uq2 makes the binomial differential form tq(at+ b)pdt rational.Indeed,

tq(at+ b)pdt = uq1 (auq2 + b)p q2uq21du

is rational because q1, q2 and p are integers.Case 2. p is not an integer, but m+1

nis an integer. Let p = p1

p2,

simplified and p1, p2 integers, with p2 = 1. Then the substitutionat+ b = up2 give rise to a rational differential form. Indeed, this timeq = m+1

n 1 is an integer, t = up2b

aand dt =

p2

aup21du, so

tq(at+ b)pdt =

(up2 b

a

)qup1

p2

aup21du

is rational because p1, p2 and q are integers.Case 3. p = p1

p2is not an integer and m+1

nis not an integer too.

In this case we make the following "trick" in the canonical differ-ential form:

(4.13) tq(at+ b)pdt = tq+p(at+ b

t

)pdt.

We apply now the same idea like above.If q + p = m+1

n 1 + p is an integer, i.e. if m+1

n+ p is an integer,

then the obvious substitution at+bt

= up2 , where p2 is the denominator

of p, leads to a rational differential form. Indeed, this time

t =b

up2 a = b (up2 a)1 ,

dt = bp2up21 (up2 a)2 du,so

tq+p(at+ b

t

)pdt = bq+p (up2 a)(q+p) up1 (bp2)up21 (up2 a)2 du.

Since q + p, p1 and p2 are integers, then this last differential formis rational and the implication "=" of the theorem is completelyproved.

R 4. The case 3 which naturally appeared during the proofof the above theorem 12 can also be manipulated in the following way.Instead of the "trick" used in formula (4.13) we can use the followingone:

(4.14) tq(at+ b)pdt =

(t

at+ b

)q(at+ b)p+qdt.


Since q + p is an integer, and since q = q1q2is simplified with q2 = 1

we make the natural substitution tat+b

= uq2 in order to "kill" the onlyradical which appears in (4.14) (because q is not an integer!!). Now,

t = b (uq2 a)1 and dt = bq2uq21 (uq2 a)2 du. Thus formula(4.14) becomes(

t

at+ b

)q(at+b)p+qdt = uq1 (1 auq2)(q+p) bq2uq21

(uq2 a)2 du,

which is a rational differential form because q+p, q1 and q2 are integers.

We give now some examples in which we practically use the theoryexposed above.

E 16. Reduce to an integral of a rational function and thencompute the following integral:

3x (2 + 3

x)2dx. Here m = 1

3, n =

12and p = 2. Thus we are in Case 1 of the Cebyshev theorem. Let us

make the canonical substitution

t = x12 or x = t2, dx = 2tdt.

So 3x(2 + 3

x)2

dx = 2

t53 (2 + 3t)2 dt.

Now we "kill" the radical t53 by making the new substitution t = u3.

Thus

2

t53 (2 + 3t)2 dt = 6

u5(2 + 3u3)3u2du

by parts=

13

u5[(2 + 3u3)2

]du =

13

{u5(2 + 3u3)2 5

u4(2 + 3u3)2du

}=

13

{u5(2 + 3u3)2 +

5

9

u2[(2 + 3u3)1

]du

}by parts=

(4.15) 13

{u5(2 + 3u3)2 +

5

9

[u2(2 + 3u3)1 2

u

2 + 3u3du

]}.

This last integralu

2 + 3u3du =

1

3

u

23+ u3

du =1

3

u

a3 + u3du,


where a = 3

23, is an integral of a rational function. We can integrate it

by the usual methods of decomposing of the integrand ua3+u3

into simplefractions:

u

a3 + u3= 1

3a

1

u+ a+

1

3a

u+ a

u2 ua+ a2 .Hence,

1

3

u

a3 + u3du = 1

9a

1

u+ adu+

1

18a

2u a+ 3au2 ua+ a2du =

(4.16) 19a

ln |u+ a|+ 118a

lnu2 ua+ a2+ 1

6

1

u2 ua+ a2du.

Let us evaluate this last integral:(4.17)

1

u2 ua+ a2du =

d(u a2)(

u a2

)2+(a3

2

)2 = 2a3 arctan 2u aa3 .Coming back with this result to formula (4.16) we get:

1

3

u

a3 + u3du = 1

9aln |u+ a|+ 1

18alnu2 ua+ a2+

+1

3a3arctan

2u aa3.

Now we go back to formula (4.15) and find:

2

t53 (2 + 3t)2 dt = 1

3{u5(2 + 3u3)2+

+5

9[u2(2 + 3u3)1 2

3 1

9aln |u+ a|+ 1

18alnu2 ua+ a2+

+1

3a3tan1

2u aa3

]}.

Now, instead of u let us put t13 and obtain:

2

t53 (2 + 3t)2 dt = 1

3{t 53 (2 + 3t)2 + 5

9[t

23 (2 + 3t)1

23 1

9alnt 13 + a+ 1

18alnt 23 t 13a+ a2+ 1

3a3tan1

2t13 aa3

]}.

But t = x12 , so our initial integral is:

3x(2 + 3

x)2

dx = 13{x 56 (2 + 3x 12 )2 + 5

9[x

13 (2 + 3x

12 )1


23 1

9alnx 16 + a+ 1

18alnx 13 x 16a+ a2+ 1

3a3tan1

2x16 aa3

]}.

where a = 3

23.

R 5. The last example 16 is a very particular example of amore general situation. Let x1, x2, ..., xn be n independent variables. Amonomial in these variables x1, x2, ..., xn is an expression of the form

ak1,k2,...,knxk11 x

k22 ...x

knn ,

where ak1,k2,...,kn is a real number and k1, k2, ..., kn are nonnegative inte-gers (they may be also zero). A finite sum of such expressions is calleda polynomial P = P (x1, x2, ..., xn) in the n variables x1, x2, ..., xn. Arational function

R = R(x1, x2, ..., xn)

in n variables is a quotient R = PQof two polynomials in n variables.

A rational expression of simple radicals is obtained from a rationalfunction

R(x1, x2, ..., xn) in n variables by substituting the variable xi with a

radical expression of the form(ax+bcx+d

)piqi , where x is our "old variable

of integration", a, b, c, d are real numbers with det

(a bc d

)= 0 and

pi, qi are nonzero natural numbers. Here i goes from 1 up to n. Such arational expression of simple radicals usually appears like:

(4.18) R

((ax+ b

cx+ d

)p1q1

,

(ax+ b

cx+ d

)p2q2

, ...,

(ax+ b

cx+ d

)pnqn

).

To integrate such a function of x we make the natural substitution

(4.19)ax+ b

cx+ d= tq,

where q is the least common multiple (lcm) of all the denominatorsq1, q2, ..., qn of the powers of

ax+bcx+d

. It is clear that the new obtaineddifferential form in t is a rational one. Indeed, starting from

R

((ax+ b

cx+ d

)p1q1

,

(ax+ b

cx+ d

) p2q2

, ...,

(ax+ b

cx+ d

)pnqn

)dx,

after substitution, we get:

q (ad bc)

R (ts1, ts2 , ..., tsn)tq1

(ctq1 a)2dt,


because, from (4.19) x = tqdbatqc and dx = q (ad bc) t

q1

(ctq1a)2dt. Here

si =piqiq are natural numbers (why?). Thus, the obtained integral is an

integral of a rational function in the variable t.For instance, in our example 16 the common expression ax+b

cx+dun-

der radicals is simply x. Now, the rational function is R(x1, x2) =x1 (2 + 3x2)

2 , p1q1

= 13, p2

q2= 1

2and q = lcm(3, 2) = 6. So the nat-

ural substitution is x = t6 and we get:3x(2 + 3

x)2

dx = 6

t2(2 + 3t3

)2t5dt,

etc.

E 17. Let us compute the integral I =x3 31 + x2dx.

Since m = 3, n = 2 and p = 13, p is not an integer and m+1

n= 2

is an integer, we are in Case 2. We can see that it is possible to putdirectly

(4.20) 1 + x2 = u3 , u =31 + x2

(without making first of all x2 = t, why?). Thus,

2xdx = 3u2du = dx = 3u2

2xdu.

Let us come back to our differential form x3 31 + x2dx and perform

these new substitutions:

x331 + x2dx = x3u

3u2

2xdu =

3

2x2u3du =

3

2(u3 1)u3du.

So3

2(u3 1)u3du = 3

2

(u7

7 u

4

4

)=

3

14

(1 + x2

) 73 3

8

(1 + x2

) 43 .

E 18. Let us compute now I =

1+x2

x4dx. This is a bino-

mial integral with m = 4, n = 2 and p = 12. Since p is not an integer,

m+1n

= 32is also not an integer, but m+1

n+p = 1 is an integer (Case

3), we can directly make the substitution 1+x2

x2= u2 (because t = x2 in

formula (4.13)). So

1 + x2 = u2x2, x2 =1

u2 1 , 2xdx = 2x2udu+ 2u2xdx,

dx =x2u

x u2xdu =xu

1 u2du.


Thus, our differential form becomes (we consider only the case x > 0and u (1,)):

1 + x2

x4dx =

ux

x4xu

1 u2du =u2

x2(1 u2)du =

=u2(

1u21

)(1 u2)du = u

2du.

Finally we get: 1 + x2

x4dx =

u2du = u

3

3= 1

3

(1 + x2

x2

) 32

.

The best idea is to make "formal" computations (not taking count ofthe definition domains of different expressions which appear during sub-stitutions!) and, in the end, to verify by a direct differentiation theobtained result. In our example,[

13

(1 + x2

x2

) 32

]=

1 + x2

x4, x = 0.

Hence, in spite of the required limitations which appeared during a par-

ticular computation, a primitive of the differential form1+x2

x4dx, on

each interval which does not contain 0, is 13

(1+x2

x2

) 32

.

R 6. In general, a primitive of the formR(x,

anxn + an1xn1 + ... + a0)dx,

where an = 0 and n > 2 is not an elementary function. By an ele-mentary function we understand a function which is a composition ofrational functions, trigonometric functions, exponential functions andlogarithm functions. If n = 3 our primitive is called an elliptic inte-gral. In general, if the integrand function is a rational expression whichcontains radicals of polynomials, then it is called an abelian integral (inmemory of the great Norway mathematician, Niels Abel). For instance,

x2(2 + 5x3

) 35 dx =

1

15

(2 + 5x3

) 35 d(2 + 5x3) =

=1

15

u

35du =

1

15

u35+1

35+ 1

=1

24

(2 + 5x3

) 85 ,

is an elementary function. But,

1 + x4dx is not an elementary

function. Indeed, let us make x4 = t. So, x = t14 , dx = 1

4t

34dt and the


integral becomes:1

4

t

34 (1 + t)

12 dt.

The last integral is a binomial integral with m = 34, n = 1 and p = 1

2.

Since p is not an integer, m+1n

= 14is also not an integer and m+1

n+ p

= 34is not an integer, we are in no one of the three cases of the Cebysev

theorem. So we cannot reduce this primitive to a rational primitive byrational substitutions. O. K., but the question is still alive! Is therean elementary differentiable function F (x) such that F (x) =

1 + x4?

The answer is no, but we need a lot of higher Mathematics to prove it!

D. Primitives of the formR(cosx, sin x)dx, where R(x1, x2)

is a rational functionThe general method consists in the following change of variable.

Assume that x belongs to an interval J on which the function x tan x2

is invertible and its inverse is of class C1. Since

cos x =1 tan2 x

2

1 + tan2 x2

, sinx =2 tan x

2

1 + tan2 x2

,

it is naturally to put t = tan x2, x = 2arctan t, dx = 2

1+t2dt, so

R(cosx, sin x)dx =

R

(1 t21 + t2

,2t

1 + t2

)2

1 + t2dt,

which is an integral of a rational function, etc.

E 19. Let us use this substitution to compute

I =

1

1 + sin x+ cosxdx.

I =

1

1 + 2t1+t2

+ 1t2

1+t2

2

1 + t2dt =

1

t+ 1dt =

= ln |t+ 1| = lntan x

2+ 1

.R 7. Sometimes this substitution is not indicated because it

leads to a very complicated computation.If R(cosx, sin x) can be written as S(cosx) sin x or as T (sin x) cosx,

where S(y) and T (y) are rational functions, then our integral becomeseither

(4.21)

R(cosx, sinx)dx =

S(cosx) sin xdx =

=

S(cosx)d(cosx) =

S(u)du,


where u = cosx, or

(4.22)

R(cosx, sin x)dx =

T (sin x) cosxdx =

=

T (sin x)d(sin x) =

T (v)dv,

where v = sin x.

E 20. If we want to compute I =sin4 x cos3 xdx by the

general substitution t = tan x2, we get a very complicated integral of a

rational function in t (why is it so complicated?)

I =

(2t

1 + t2

)4(1 t21 + t2

)32

1 + t2dt.

But, if we use the substitution described in (4.22), we get:

I =

sin4 x cos2 x cosxdx =

sin4 x(1 sin2 x)d(sin x) =

=

v4(1 v2)dv = v

5

5 v

7

7=

sin5 x

5 sin

7 x

7.

R 8. Suppose now that cosx and sin x appear to even powersin

R(cosx, sin x). Then always one can write R(cosx, sinx)dx as

S(tanx)1

cos2 xdx = S(tanx)d(tan x) = S(u)du,

where u = tan x and S is a rational function of u (why all of these?).The substitution u = tan x can also be done even in other cases. Forinstance,

sinxdx

sin3 x+cos3 xdx can be "rationalized" by puting tan x = u.

E 21. Let us compute I =

cos2 x1+sin2 x

dx by this last method

(the general substitution t = tan x2is not good at all!-why?). Since

1 + tan2 x = 1cos2 x

, and since u = tan x implies that x = tan1 u anddx = 1

1+u2du, we get

cos2 x

1 + sin2 xdx =

1

1cos2 x

+ tan2 xdx =

1

(1 + 2u2) (1 + u2)du.

We must decompose into simple fractions the rational expression

1

(1 + 2u2) (1 + u2).


Do not you hurry to search for a decomposition of a general type:

1

(1 + 2u2) (1 + u2)=Au+B

1 + 2u2+Cu+D

1 + u2,

because we can think of this rational expression as a rational functionof t = u2, so we simply search for a decomposition of the following type:

1

(1 + 2u2) (1 + u2)=

A

1 + 2u2+

B

1 + u2.

So A = 2 and B = 1. Hence,cos2 x

1 + sin2 xdx = 2

1

1 + 2u2du

1

1 + u2du =

=2

1

1 +(

2u)2d(2u) arctan u =

=2 arctan

(2u) arctan u =

2 arctan

(2 tan x

) x.

Sometimes it is useful to extend the computations with primitivesto functions of real variables but with values in the complex numberfield C. Let f(x) = f1(x) + if2(x), where i =

1 and f1, f2 : I R,are two continuous functions of real variable x. They are usually calledthe (real) components of f. So f : I C is continuous (see Analysis I,[Po], sequences of complex numbers, etc.). Moreover, f is differentiableon I if and only if f1 and f2 are differentiable on I and then f

= f 1+if2

(see for instance Analysis I, [Po]). Thusf(x)dx =

f1(x)dx+ i

f2(x)dx.

(why?). Let us compute for instanceexp(ix)dx =

cosxdx+ i

sin xdx = sinx i cosx =

=1

i(cosx+ i sin x) =

1

iexp(ix).

We see from here that the usual rules for computing primitives of thereal valued functions extends naturally to complex valued functions ofreal variables. This is true because such extension works in the case ofthe differential calculus (see Analysis I, [Po]). So we can directly write

exp(ix)dx =1

i

exp(ix)d(ix) =

1

i

exp udu =

1

iexpu =

=1

iexp(ix) = i exp(ix).


(i is a constant complex number). These simple ideas are very helpfulin computing usual primitives of real valued functions.

E 22. Let a, b be two real numbers and letI(a, b) =

exp(ax) sin(bx)dx be a primitive of exp(ax) sin(bx). As-

sume that a, b = 0 (otherwise the integral is trivial!). We can computeI(a, b) by integrating by parts two times:

I(a, b) =

exp(ax)

[1bcos(bx)

]dx =

by parts= 1

bexp(ax) cos(bx) +

a

b

exp(ax) cos(bx)dx =

1bexp(ax) cos(bx) +

a

b2

exp(ax) [sin(bx)] dx =

by parts= 1

bexp(ax) cos(bx) +

a

b2{exp(ax) sin(bx) aI(a, b)} =

=a

bexp(ax)

[sin(bx)

b cos(bx)

a

] a

2

b2I(a, b).

So (1 +

a2

b2

)I(a, b) =

a

bexp(ax)

[sin(bx)

b cos(bx)

a

],

or

I(a, b) =ab

a2 + b2exp(ax)

[sin(bx)

b cos(bx)

a

]=

=1

a2 + b2exp(ax) [a sin(bx) b cos(bx)] .

Another smarter way to compute this integral is based on the aboveideas. Let J(a, b) =

exp(ax) cos(bx)dx and let the complex number

J(a, b) + iI(a, b) =

exp(ax) exp(ibx)dx =

exp [(a+ bi)x] dx =1

a+ biexp [(a+ bi)x] =

a bia2 + b2

exp(ax) exp(ibx) =exp(ax)

a2 + b2(a ib) [cos(bx) + i sin(bx)] =

exp(ax)

a2 + b2[a cos(bx) + b sin(bx)] + i

exp(ax)

a2 + b2[a sin(bx) b cos(bx)] .

Thus

J(a, b) =exp(ax)

a2 + b2[a cos(bx) + b sin(bx)]


and

I(a, b) =exp(ax)

a2 + b2[a sin(bx) b cos(bx)] .

R 9. A way to compute primitives of the formR(cosmx, sinnx)dx,

where m,n are natural nonzero numbers is to use Eulers formulas:

(4.23) cosmx =exp(imx) + exp(imx)

2,

(4.24) sinnx =exp(inx) exp(inx)

2i

and to reduce the computation to a primitive of a complex valued func-tion (of a real variable).

E 23. Let us compute a primitive for the trigonometricdifferential form sin 3x cos4 xdx.

Since sin 3x = exp(3ix)exp(3ix)2i

and since

cos2 x =1 + cos 2x

2=

1

2+

exp(2ix) + exp(2ix)4

=

=1

4[2 + exp(2ix) + exp(2ix)] ,

cos4 x =1

16[6 + exp(4ix) + exp(4ix) + 4 exp(2ix) + 4 exp(2ix)] ,

one has sin 3x cos4 xdx =

1

32i

[exp(3ix) exp(3ix)]

[6 + exp(4ix) + exp(4ix) + 4 exp(2ix) + 4 exp(2ix)] dx =

=1

32i

[6 exp(3ix) + exp(7ix) + exp(ix) + 4 exp(5ix) ++4 exp(ix)

6 exp(3ix) exp(ix) exp(7ix) 4 exp(ix) 4 exp(5ix)]dx =

=1

16

[6 sin(3x) + sin(7x) + 3 sin x+ 4 sin 5x]dx =

18cos 3x 1

112cos 7x 3

16cosx 1

20cos 5x.


E 2. Let In =sinn xdx, where n > 2, and n = 2k is even

(k > 1). If n is odd it is easier to write

sinn xdx = sinn1 x d(cosx) = (1 u2)n12 du,etc. Let us compute In by firstly finding a recurrence formula:

I2k =

sin2k xdx =

(sin2k2 x

)(1 cos2 x)dx =

I2k2

sin2k2 x cosxd(sin x)by parts=

= I2k2 [sin2k1 x cosx

sinx{(2k 2) sin2k3 x cos2 x sin2k1 x} dx] =

I2k2 sin2k1 x cosx+ (2k 2)

sin2k2 x(1 sin2 x)dx I2k =

= (2k 1)I2k2 (2k 1)I2k sin2k1 x cosx,or

(4.25) I2k =2k 12k

I2k2 12k

sin2k1 x cosx, k 1,because it is easy to see that this last formula works even for k = 1.

E 24. For instance, since

I2 =

sin2 xdx =

1

2

(1 cos 2x)dx =

1

2

(x sin 2x

2

)=

1

2(x sin x cosx) ,

one has from (4.25) that

(4.26)

sin4 xdx =

3

4

[1

2(x sinx cosx)

] 1

4sin3 x cosx.

Another way to compute such an integral is to express sinn x as apolynomial of different powers of exp(ix) or exp(ix) :

sinn x =

[exp(ix) exp(ix)

2i

]n=

=1

2nin[exp(inx)

(n

1

)exp i(n 2)x+

+

(n

2

)exp i(n 4)x ...].


So sinn xdx =

=1

2nin[

exp(inx)dx

(n

1

)exp i(n 2)xdx+

+

(n

2

)exp i(n 4)xdx ...].

For instance,sin4 xdx =

1

16

[exp(4ix)dx 4

exp(2ix)dx

]+

+6

16

[dx 4

exp(2ix)dx+

exp(4ix)dx

]=

1

16

[exp(4ix)

4i 4 exp(2ix)2i +6x 4 exp(2ix)2i +

exp(4ix)4i

]=

1

16

[1

2sin 4x 4 sin 2x+ 6x

].

An elementary trigonometric computation tells us that this expressionis exactly that one from (4.26). Moreover, it is more beautiful, becauseit has no powers of trigonometric functions! These last ones are "notdesirable" during the integration computation.

R 10. It is very useful to know that the following primitivesare not elementary functions:

expxxdx;

sinxxdx;

cos xxdx;

sinhxx

dx;coshxx

dx;sin x2dx;

cos x2dx;

exp(x2)dx; 1

lnxdx;

d

1k2 sin2 ;

1 k2 sin2 d; d(1+h sin2 )

1k2 sin2

, where k, h (1, 1) \{0} (elliptic integrals). Proofs for such statements belong to very highMathematics. It implies deep knowledge of Algebraic Geometry and Al-gebraic Functions Theory. We note that not all the above statementsare "independent" one to each other". For instance, if

expxxdx is not

an elementary function, then

1lnx

dx is also not an elementary func-tion. Indeed, let F (x) =

1

lnxdx, a primitive of the differential form

1lnx

dx. Let us make the substitution x = exp(t) in this last differentialform:

1

ln xdx =

exp t

tdt.

So a primitive of exp ttdt is F (exp(t)) on any interval I which does

not contain 0 (prove it!). But, if F were elementary, then F (exp(t)

5. PROBLEMS AND EXERCISES 47

would also be elementary (as a composition between two elementaryfunctions), a contradiction (explain everything slowly!). Try to findother pairs of "dependent" primitives in the above row of primitives!

Remark 10 is very useful in practice. For instance, if in the state-ment of an exercise there is a mistake: instead of

1

x lnxdx, x > 0, one

omitted "an x and it appears

1lnx

dx, you may spend a lot of time "tocompute" this last primitive!! All this effort is for nothing! Because"to compute" usually means to work only with elementary functions!Indeed, the primitive

1

lnxdx is not an elementary function (see remark

10), while1

x ln xdx =

1

ln xd(lnx) =

1

udu = ln |u| = ln |ln x| ,

is an elementary one!

E 25. We know that a canonical parametrization of the

ellipse x2

a2+ y

2

b2= 1, a, b > 0, and a < b, is the following

x = x(t) = a cos t, y = y(t) = b sin t, t [0, 2).To compute the length of an arc of this ellipse, one reaches the followingprimitive (see Chapter 2, Section 8):

a2 sin2 t+ b2 cos2 tdt =

a2 sin2 t+ b2(1 sin2 t)dt =

=

b2 (b2 a2) sin2 tdt = b

1 k2 sin2 tdt,

where k =b2a2b

. This last primitive is not an elementary function(see remark 10).

curs analiza matematica

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