**Chapter 10**

April 29 2017

**Chapter 10 Methods of Integration**
**10.1 Introduction to the Basic Formulas**
- if we start with the constants and the seven familiar functions
- x, ex, ln x, sin x, cos x, sin -1 x, cos -1 x
- and go on to build all possible finite combinations of these by
- applying the algebraic operations and th process of forming a function of a function
- then we generate the class of elementary functions
- these functions are often said to have closed form
- because they can be written down in explicit formulas involving only a finite number of familiar functions
- it is clear that the problem of calculating the derivative of an elementary function
- can always be solved by a systematic application of the rules developed in the precding chapters
- and this derivative is always an elementary function
- however, the inverse problem of integration
- which in general is much more important
- is very different and has no such clear-cut solution
- as we know, the problem of calculating the indefinite integral of a function f(x)
- S f(x) dx = F(x)
- is equivalent to finding a function F(x) such that
- d/dx F(x) = f(x)
- it is true that we have succeeded in integrating a good many elementary functions by inverting differentiation formulas
- but that doesn't carry us very far
- because it amounts to little more than calculating the integral by knowing the answer in advance
- the fact of the matter is that
- there does not exist any systematic procedure that can always be applied to any elementary function and leads step by step to a guaranteed answer in closed form
- indeed, there may not even be an answer
- for example, the function
- f(x) = e -x2 looks simple enough
- but its integral S e -x2 dx
- cannot be calculated within a class of elementary functions
- it is a statement of a deep theorem
- to the effect that no elementary function exists whose derivative is e -x2
- since integration is differentiation read backwards
- our starting point must be a short table of standard types of integrals obtained by inverting differentiation formulas as we have done in the previous chapters
- we urge students to concentrate their efforts on gaining a clear understanding of the various methods of integration and learning how to apply them
- in addition to the method of substitution
- which is already familiar to the reader
- there are three principal methods of integration to be studied in this chapter
- reduction to trigonometric integrals
- decomposition into partial fractions
- integration by parts
- these methods enable us to transform a given integral in many ways
- the object of these transformations is always to break up the given integral into a sum of simpler parts
- that can be integrated at once by means of familiar formulas
- S un du = u n+1 / n+1 + c (n != -1)
- S du/u = ln u + c
- S eu du = eu + c
- S cos u du = sin u + c
- S sin u du = - cos u + c
- S sec 2 u du = tan u + c
- S csc 2 u du = - cot u + c
- S sec u tan u du = sec u + c
- S csc u cot u du = - csc u + c
- S du / sqrt (a2 - u2) = sin -1 u/a + c
- S du / sqrt (a2 + u2) = 1/a tan -1 u/a + c
- S tan u du = - ln (cos u) + c
- S cot u du = ln (sin u) + c
- S sec u du = ln (sec u + tan u) + c
- S csc u du = - ln (csc u + cot u) + c
- the last formulas are new
- and complete our list of integrals of the six trigonometric functions
- S tan u du = S sin u du / cos u = - S d (cos u) / cos u = - ln (cos u) + c
**10.2 The Method of Substitution**
- in the method of substitution we introduce the auxiliary variable
- u as a new symbol for part of the integrand in the hope
- that its differential du will account for some other part
- and thereby reduce the complete integral to an easily recognizable form
- success in the use of this method depends on choosing a fruitful substitution
- and this in turn depends on the ability to see at a glance that part of the integrand is the derivative of some other part
- S x e -x2 = 1/2 e -x2 + c
- we point out that this integral is easy to calculate even though the similar integral
- S e -x2 dx is impossible
- the reason for this is clearly the presence of the factor x
- which is essentially the derivative of the exponent -x2
- in any particular integration problem
- the choice of the substitution is a matter of trial and error guided by experience
- if our first substitution doesn't work, we should feel no hesitation about discarding it and trying another
- we can establish the validity of the method of substitution as follows
- by showing that it is really the chain rule for derivatives read backwards
- the essence of the method is this
- we start with a complicated integral of the form
- S f[ g(x) ] g'(x) dx
- if we put u = g(x)
- then du = g'(x) dx
- and the integral takes the new form
- S f(u) du
- if we can integrate this, so that
- S f(u) du = F(u) + c
- then since u = g(x) we ought to be able to integrate by writing
- S f[ g(x) ] g'(x) dx = F[ g(x) ] + c
- all that is needed to justify our procedure is to notice that this is the right result, because
- d/dx F[ g(x) ] = F'[ g(x) ] g'(x) = f[ g(x) ] g'(x)
- by the chain rule
- the method of substitution applies to definite integrals as well as indefinite integrals
- the crucial requirement is that the limits of integration
- must be suitably changed when substitution is made
**10.3 Certain Trigonometric Integrals**
- in the next two sections we discuss several methods for reducing a given integral to one involving trigonometric functions
- it will therefore be useful to increase our ability to calculate such trigonometric integrals
- a power of a trigonometric function multiplied by its differential is easy to integrate
- S sin 3 x cos x dx = S sin 3 x d (sin x) = 1/4 sin 4 x + c
- S tan 2 x sec 2 x dx = S tan 2 x d (tan x) = 1/3 tan 3 x + c
- other trigonometric integrals can often be reduced to problems of this type by using appropriate trigonometric identities
- we begin by considering integrals of the form
- S sin m x cos n x dx
- where one of the exponents is an odd positive integer
- if n is odd, we factor out cos x dx
- which is d (sin x)
- and since an even power of cos x remains
- we can use the identity cos 2 x = 1 - sin 2 x
- to express the remaining part of the integrand entirely in terms of sin x
- and if m is odd, we factor out sin x dx, which is - d (cos x)
- and use the identity sin 2 x = 1 - cos 2 x in a similar way
- if one of the exponents is an odd positive integer that is quite large
- it may be necessary to use the binomial theorem
- for instance, every odd positive power of cos x, whether large or small, has the form
- cos 2n+1 x = cos 2n x cos x = (cos 2 x) n cos x = (1 - sin 2 x) n cos x
- where n is a non-negative integer
- if we put u = sin x and du = cos x dx
- then
- S cos 2n+1 x dx = S (1 - sin 2 x)n cos x dx = S (1 + u2) n du
- if necessary, the expression (1 - u2) n can now be expanded by applying the binomial theorem
- and the resulting polynomial in u is easy to integrate term by term
- if both exponents are non-negative even integers
- then it is necessary to change the form of the integral by using the half-angle formulas
- cos 2 angle = 1/2 ( 1 + cos 2 angle )
- as these examples show
- the value of the half-angle formulas for this work
- lies in the fact that they allow us to reduce the exponent by a factor of 1/2
- at the expense of multiplying the angle by 2
- which is a considerable advantage purchased at a very low cost
- we next consider integrals of the form
- S tan m x sec n x dx
- where n is an even positive integer or m is an odd positive integer
- our work is based on the fact that
- d (tan x) = sec 2 x dx
- and d (sec x) = sec x tan x
- and we exploit the identity tan 2 x + 1 = sec 2 x
- in essentially the same way we can handle integrals of the form
- S cot m x csc n x dx
- where n is an even positive integer or m is an odd positive integer
- our tools in this case are the formulas
- d (cot x) = - csc 2 x dx
- and d (csc x) = - csc x cot x dx
- and when necessary we use the identity 1 + cot 2 x = csc 2 x
- another approach to trigonometric integrals that is sometimes useful
- is to express each function occurring in the integral in terms of sines and cosines alone
**10.4 Trigonometric Substitutions**
- an integral involving one of the radical expressions
- sqrt (a2 - x2), sqrt (a2 + x2), sqrt (x2 - a2)
- where a is a positive constant
- can often be transformed into a familiar trigonometric integral by using a suitable trigonometric substitution or change of variable
- there are three cases, which depend on the trigonometric identities
- 1 - sin 2 angle = cos 2 angle
- 1 + tan 2 angle = sec 2 angle
- sec 2 angle - 1 = tan 2 angle
- x = a sin angle replaces sqrt (a2 - x2) by a cos angle
- x = a tan angle replaces sqrt (a2 + x2) by a sec angle
- x = a sec angle replaces sqrt (x2 - a2) by a tan angle
**10.5 Completing the Square**
- in section 10.4 we used trigonometric substitutions to calculate integrals
- containing sqrt (a2 - x2), sqrt (a2 + x2), and sqrt (x2 - a2)
- the case sqrt (-a2 - x2) is clearly of no interest
- by the algebraic device of completing the square
- we can extend these methods to integrals involving general quadratic polynomials and their square roots
- that is, expressions of the form ax2 + bx + c and sqrt (ax2 + bx + c)
- we remind students that the process of completing the square is based on the simple fact that
- (x + A) 2 = x2 + 2Ax + A2
- if an integral involves the square root of a third-, fourth-, or higher-degree polynomial
- then it can be proved that there does not exist any general method for carrying out the integration
**10.6 The Method of Partial Fractions**
- we recall that a rational function is a quotient of two polynomials
- by taking the denominator of such a quotient to be 1
- we see that the polynomials themselves are included among the rational functions
- we already know the integrals of simple rational functions
- 2x + 1, x2 + x
- 1/x2, -/1x
- 1/x, ln x
- x/x2 + 1, 1/2 ln (x2 + 1)
- 1/x2 + 1, tan -1 x
- our purpose in this section is to describe a systematic procedure for computing the integral of any rational function
- we shall find that this integral can always be expressed in terms of polynomials, rational functions, logarithms, and inverse tangents
- the basic idea is to break up a given rational function into a sum of simpler functions called partial fractions
- which can be integrated by methods discussed earlier
- a rational function is called proper if the degree of the numerator is less than the degree of the denominator
- otherwise, it is said to be improper
- if we have to integrate an improper rational function
- it is essential to begin by performing long division until we reach a remainder whose degree is less than that of the denominator
- by applying this process, any improper rational function can be expressed as the sum of a polynomial and a proper rational function
- where the integral can be solved by using the integrals of simple rational functions
- these remarks tell us that we can restrict our attention to proper rational functions
- this restriction is not only convenient, but also necessary
- because it is only to proper rational functions that the following discussions apply
- in elementary algebra we learned how to combine fractions over a common denominator
- we must learn how to reverse this process
- and split a given fraction in a sum of fractions having simple denominators
- this procedure is called decomposition into partial fractions
- the purpose of this decomposition is to enable us to integrate the given function by using elementary integrals
- in general, any proper rational function can be decomposed
- because every polynomial can be factored completely into linear and quadratic factors
- this discussion shows that the integral of every rational function can be expressed in terms of polynomials, rational functions, logarithms, and inverse tangents
- the detailed work can be very laborious
- but at least the path that must be followed is clearly visible
**10.7 Integration by Parts**
- when the formula for the derivative of a product (the product rule) is written in the notation of differentials
- it is d(uv) = udv + vdu
- or udv = d(uv) - vdu
- by integrating, we obtain
- S u dv = uv - S v du
- this formula provides a method of finding S u dv if the second integral S v du is easier to calculate than the first
- the method is called integration by parts
- and it often works when all other methods fail
- for example
- S x cos x dx
- u = x, dv = cos x dx
- du = dx, v = sin x
- S x cos x dx = x sin x - S sin x dx = x sin x + cos x + c
- the method of integration by parts applies particularly well to products of different tpes of functions
- such as x cos x
- a product of a polynomial and a trigonometric function
- in using the method
- the given differential must be though of as a product u dv
- the part dv must be something we can integrate
- and the part u should usually be something that is simplified by differentiation
- for example
- S ln x dx
- u = ln x, dv = dx
- du = dx/x, v = x
- S ln x dx = x ln x - S x dx / x = x ln x - x + c
- in some cases it is necessary to carry out two or more integrations by parts in succession
- it sometimes happens that the integral we start with appears a second time during the integration by parts
- in which case, it is often possible to solve for this integral by elementary algebra
- the method of the following example is often used to make an integral depend on a simpler integral of the same type
- and thus to obtain a convenient reduction formula
- by repeated use of which the given integral can be easily calculated
- for example
- reducing the exponent on sin x by 2 with each use
**10.8 Functions That Cannot Be Integrated**
- we have described all the standard methods of integration that the student is expected to be acquainted with
- for most practical purposes we have reached the end of this particular road
- in spite of the many successes achieved by the methods of this chapter
- certain integrals have always resisted every attempt to express them in terms of elementary functions
- S e -x2 dx, S ex / x dx, S cos x2 dx, S dx / ln x, S sqrt(sin x) dx, S sin x / x dx
- there are also the so-called elliptic integrals, such as
- S sqrt(1 - x3) dx and S dx / sqrt(1 - x4)
- we know from our work in section 6.7 that for any continuous integrand the definite integral
- F(x) = S0x f(t) dt
- exists and has the property that
- d/dx F(x) = f(x)
- since this is equivalent to S f(x) dx = F(x)
- we see that the indefinite integral of every continuous function exists
- however this fact has nothing to do with the issue of whether the integral can be expressed in terms of elementary functions
- when such an expression is not possible
- the expression can be thought of as providing a legitimate and sometimes useful method for creating new functions
- for example
- the non-elementary function of x defined by
- 1/sqrt(2pi) S0x e -t2/2 dt
- has important applications in the theory of probability
- and for this reason it has been studied and tabulated and has thereby acquired a certain status as a "known function"
- it is easy to see that the integral
- S ex / x dx becomes S dt / ln t
- under the substitution t = ex, for x = ln t, dx = dt/t, and therefore
- S ex / x dx = S t / ln t dt / t = dt / ln t
- since we know that the first integral is not elementary
- it is clear that the second integral is also not elementary
- this is worth noticing because the function
- S2x dt / ln t
- is of great importance in the theory of prime numbers
- and the behaviour of this function for large values of x has been studied exhaustively for more than a century
**10.9 Numerical Integration**
- from the point of view of the theorist
- the main value of calculus is intellectual
- it helps us comprehend the underlying connections among natural phenomena
- however
- anyone who uses calculus as a practical tool in science or engineering must occasionally face the question
- of how the theory can be applied to yield useful methods for performing actual numerical calculations
- in this section we consider the problem of computing the numerical value of a definite integral
- Sab f(x) dx
- in decimal form to any desired degree of accuracy
- in order to find the value of this using the formula
- Sab f(x) dx = F(b) - F(a)
- we must be able to find the indefinite integral F(x)
- and we must be able to evaluate it at both x = a and x = b
- when this is not possible, the formula is useless
- this approach fails even for such simple-looking integrals as
- S0x sqrt(sin x) dx and S15 ex / x dx
- because there are no elementary functions whose derivatives are sqrt(sin x) and ex / x
- our purpose here is to describe two methods of computing the numerical value of the integral
- as accurately as we wish by simple procedures that can be applied
- regardless of whether an indefinite integral can be found or not
- the formulas we develop use only simple arithmetic
- and the values of f(x) at a finite number of points in the interval [a, b]
- in comparison with the use of approximating sums that are used in defining the integral
- the formulas of this section are much more efficient
- in the sense that they give much better accuracy for the same amount of computational labor
- the trapezoidal rule
- simpson's rule