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