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Unformatted text preview: CHAPTER 7 Techniques of Integration Integration, unlike differentiation, is more of an artform than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given by complicated formulae, and practitioners consult a Table of Integrals in order to complete the integration. There are certain methods of integration which are essential to be able to use the Tables effectively. These are: substitution, integration by parts and partial fractions. In this chapter we will survey these methods as well as some of the ideas which lead to the tables. After the examination on this material, students will be free to use the Tables to integrate. The idea of substitution was introduced in section 4.1 (recall Proposition 4.4). To integrate a differential f x dx which is not in the table, we first seek a function u u x so that the given differential can be rewritten as a differential g u du which does appear in the table. Then, if g u du G u C, we know that f x dx G u x C. Finding and employing the function u often requires some experience and ingenuity as the following examples show. Example 7.2 tan xdx ? where at the end we have replaced u by 2x 1. 107 $ # (7.2) 1 3 2 3x 1 3u 5 C 1 1 3 2 30 u 1 2x 30 1 2x 15 ! (7.1) 3 2 6x 2 " x 2x 1dx 2 5 u 5 2 2 3 u 3 u 1 1 2 du 2 u 2 Let u 2x 1, so that du Example 7.1 x 2x 1dx ? 2dx and x u 1 2. Then 1 4 1 4 u3
2 u1 2 du
2 C C C 7.1. Substitution Chapter 7 Techniques of Integration 108 This is tricky, and there are several ways to find the integral. However, if we are guided by the principle of rewriting in terms of sines and cosines, we are led to the following: This looks like a dead end, but a little algebra pulls us through. The identity 1 u2 1 u leads to Example 7.4 As a circle rolls along a horizontal line, a point on the circle traverses a curve called the cycloid. A loop of the cycloid is the trajectory of a point as the circle goes through one full rotation. Let us find the length of one loop of the cycloid traversed by a circle of radius 1. Let the variable t represent the angle of rotation of the circle, in radians, and start (at t 0) with the point of intersection P of the circle and the line on which it is rolling. After the circle has rotated through t radians, the position of the point is as given as in figure 7.1. The point of contact of the circle with the line is now t units to the right of the original point of contact (assuming no slippage), so 0 (7.11) so ds 21 costdt, and the arc length is given by the integral
2 L 2 1 costdt (7.10) ds2 1 cost 2 sin2 t 2 dt 2 2 2 cost 2 dt 2 $ To find arc length, we use ds2 dx2 dy2 , where dx 1 cost dt dy $ (7.9) xt t sint yt 1 cost sintdt. Thus ! (7.8) sec xdx sinx ln 1 sinx C " Using u sin x, we finally end up with 1 ln 1 2 1 1 ln 2 1 sinx sinx C 1 u 1 u ! (7.7) " du dx 1 u2 1 2 1 1 du 1 ln 1 2 ! (7.6) 1 u " 1 1 2 1 1 1 (7.5) sec xdx $ Now we can try the substitution u sin x du cos xdx. Then du u2 (7.4) Example 7.3 sec xdx ?. sec x 1 cos x cos x cos2 x cos x 1 sin2 x u ln 1 u C (7.3) tan xdx sin x dx cos x du u ln u C lncos x C ln sec x C Since this isn't on our tables, we revert to the definition of the tangent: tanx letting u cos x du sinxdx we obtain sin x cos x. Then, $ To evaluate this integral by substitution, we need a factor of sint. We can get this by multiplying and dividing by 1 cost: By symmetry around the line t , the integral will be twice the integral from 0 to . In that interval, sint is positive, so we can drop the absolute value signs. Now, the substitution u cost du sintdt will work. When t 0 u 1, and when t u 1. Thus
1 7.2. Integration by Parts Sometimes we can recognize the differential to be integrated as a product of a function which is easily differentiated and a differential which is easily integrated. For example, if the problem is to find then we can easily differentiate f x x, and integrate cos xdx separately. When this happens, the integral version of the product rule, called integration by parts, may be useful, because it interchanges the roles of the two factors. udv vdu, and rewrite it as Recall the product rule: d uv $ (7.16) x cos xdx d x sin x sinxdx $ $ In the case of 7.14, taking u x dv cos xdx, we have du (7.15) udv (7.14) x cos xdx d uv vdu dx v sin x. Putting this all in 7.15: 1 1 1 (7.13) L 2 2 u du 2 2 u du 1 2 1 1 2 2 2 2u1 2 1 8 2 $ 1 $ (7.12) 1 cost 1 cos2 t 1 cost sint cost $ 7.2 Integration by Parts 109 Figure 7.1 PSfrag replacements 1 1 cos t 1 t P t sint t Chapter 7 Techniques of Integration 110 and we can easily integrate the right hand side to obtain Proposition 7.1 (Integration by Parts) For any two differentiable functions u and v: (7.18) udv uv vdu To integrate by parts: 1. First identify the parts by reading the differential to be integrated as the product of a function u easily differentiated, and a differential dv easily integrated. 2. Write down the expressions for u dv and du v. 3. Substitute these expressions in 7.18. 4. Integrate the new differential vdu. This same idea works for arctanx: Let 1 and thus 1 x2 $ where the last integration is accomplished by the new substitution u 1 x 2 du 2xdx. $ (7.23) arctanx x arctanx x dx x arctan x 1 ln 1 2 $ $ $ (7.22) u arctanx dv dx du dx v x2 x x2 (7.21) ln xdx x ln x 1 x dx x x ln x dx x ln x x C $ $ Example 7.7 To find lnxdx, we let u ln x dv dx, so that du 1 x dx v x, and C (7.20) x2 ex dx x 2 ex 2 xex dx x 2 ex 2 xex $ $ The substitution u x2 dv reduces us to example 3: $ Example 7.6 Find x2 ex dx. ex dx du 2xdx v ex doesn't immediately solve the problem, but ex C (7.19) xex dx xex $ $ Let u x dv ex dx. Then du Example 7.5 Find xex dx. dx v ex . 7.18 gives us ex dx xex ex C x 2 ex $ $ (7.17) x cos xdx x sin x sin xdx x sin x cosx C 2xex 2ex C Example 7.8 These ideas lead to some clever strategies. Suppose we have to integrate e x cos xdx. We see that an integration by parts leads us to integrate ex sin xdx, which is just as hard. But suppose we integrate by parts again? See what happens: Letting u ex dv cosxdx du ex dx v sin x, we get Inserting this in 7.24 leads to Bringing the last term over to the left hand side and dividing by 2 gives us the answer: Example 7.9 If a calculation of a definite integral involves integration by parts, it is a good idea to evaluate as soon as integrated terms appear. We illustrate with the calculation of
4 4 4 Example 7.10
1 2 0 0 1 1 (7.32) arcsin xdx u du 1 2 u1 2 1 2 1 2 6 1 2 3 4 $ Now, to complete the last integral, let u 1 x2 du 2xdx, leading us to 12 3 4 0 0 1 x2 (7.31) arcsin xdx x arcsin x 1 2 0 1 2 1 2 xdx $ $ $ We make the substitution u arcsinx dv dx du dx (7.30) arcsinxdx ? 1 x2 v 1 1 1 1 x. Then 12 (7.29) ln xdx x ln x $ $ Let u ln xdx dv dx so that du dx x v x, and
4 4 (7.28) ln xdx
1 dx 4 ln 4 x (7.27) ex cos xdx 1 x e sin x 2 ex cos x (7.26) ex cos xdx ex sin x ex cos x ex cos xdx C 4 ln 4 (7.25) ex sin xdx ex cos x ex cosxdx $ $ $ Now integrate by parts again: letting u ex dv sin xdx du (7.24) ex cos xdx $ $ $ 7.2 Integration by Parts 111 ex sin x ex sin xdx ex dx v cos x, we get 3 3 2 1 Chapter 7 Techniques of Integration 112 The point of the partial fractions expansion is that integration of a rational function can be reduced to the following formulae, once we have determined the roots of the polynomial in the denominator. Proposition 7.2 dx a) ln x a C x a du 1 u arctan b) C u2 b2 b b udu 1 ln u2 b2 C c) 2 b2 u 2 These are easily verified by differentiating the right hand sides (or by using previous techniques). Example 7.11 Let us illustrate with an example we've already seen. To find the integral we check that x b a b x a x b so that The trick 7.34 can be applied to any rational function. Any polynomial can be written as a product of factors of the form x r or x a 2 b2 , where r is a real root and the quadratic terms correspond to the conjugate pairs of complex roots. The partial fraction expansion allows us to write the quotient of polynomials as a sum of terms whose denominators are of these forms, and thus the integration is reduced to Proposition 7.2. Here is the partial fractions procedure. 1. Given a rational function R x , if the degree of the numerator is not less than the degree of the denominator, by long division, we can write where now deg p deg q. 2. Find the roots of q x 0. If the roots are all distinct (there are no multiple roots), write p q as a sum of terms of the form x r x b2 x b2 3. Find the values of A B C . 4. Integrate term by term using Proposition 7.2. $ $ (7.37) A B a2 (7.36) Rx Qx px qx Cx a2 $ $ $ x b a b a b # (7.35) ln x a ln x b C ln dx a x 1 1 $ ! (7.34) " 1 a x 1 1 (7.33) $ $ 7.3. Partial Fractions x dx a x b 1 x x a b C If the roots are not distinct, the expansion is more complicated; we shall resume this discussion later. For the present let us concentrate on the case of distinct roots, and how to find the coefficients A B C in 7.37. xdx Example 7.12 Integrate x 1 x 2 First we write A B x (7.38) x 1 x 2 x 1 x 2 Integrating, we get x 2 So, this is the procedure for finding the coefficients of the partial fractions expansion when the roots are all real and distinct: 1. Write down the expansion with unknown coefficients. 2. Multiply through by the product of all the terms x r. 3. Substitute each root in the above equation; each substitution determines one of the coefficients. x2 leading to and the integral is x2 (7.45) 1 1 3 x2 3 dx 1 x 3 1 ln x 4 1 ln x 2 3 ln x 4 ! ! ! x2 x 1 x 1 x 3 C $ (7.44) " " " x2 3 1 x 3 1 4 1 Substitute x Substitute x Substitute x 1: 1 3 A 2 4 , so A 1 4. 1: 1 3 B 2 2 , so B 1 2. 3 : 9 3 C 4 2 , so C 3 4, and 7.42 becomes 1 2 1 3 4 1 (7.43) x2 3 Ax 1 x 3 Bx 1 x 3 C x x2 x 1 x 1 (7.42) $ Here the roots are x2 3 1 x 3 Example 7.13 Integrate 3 dx 1 x 3 1 3, so we have the expansion x2 A B x C 3 1 x 1 (7.41) ln x 1 2 ln x 2 C ln xdx 1 x x 2 x x 2 x 22 x 1 (7.40) x 1 x 1 1 2 C If we substitute x 1, we get 1 and 7.38 becomes A1 2 , so A 1; now letting x (7.39) x Ax Now multiply this equation by x 1 x 2 , getting 2 Bx 1 2, we get 2 B2 1, so B $ $ 7.3 Partial Fractions 113 2, Chapter 7 Techniques of Integration 114 and the integral is x2 5 x2 14 (7.52) 2 x2 6x 2x 1 dx x2 6x 14 7 6x x 3 14 Example 7.17 2x 1 dx ? x2 6x 14 First, we complete the square in the denominator: x2 6x 14 x 3 numerator in terms of x 3 : 2x 1 2 x 3 7. This gives the expansion: (7.51) x 3 dx x 2 dx 5dx x2 4x 5 x 22 1 x 22 1 1 ln x 2 2 1 5 arctan x 2 C 2 2 5. Now, write the If only that x 3 were x there is no problem: 2, we could use Proposition 7.2c, with u x2 x 2 2 1 x 2. Well, since x 3 x 2 5, (7.50) x2 Example 7.16 3 dx ? 4x 5 Here we have to be a little more resourceful. Again, we complete the square, giving x x 3 4x 5 x2 5 x 1 x 3 (7.49) x2 Example 7.15 dx ? x2 4x 5 Here we can't find real factors, because the roots are complex. But we can complete the square: 4x 5 x 2 2 1, and now use Proposition 7.2b: dx 4x dx 22 arctan x 2 C (7.48) dx 4x 1 x ln 6 x 5 1 x2 5 x 1 C (7.47) 1 4x $ and solve for A and B as above: A 1 6 B 1 6, so we have 1 1 6 x 5 1 x2 5 x 1 (7.46) 1 4x x2 Example 7.14 7.3.1 Quadratic Factors dx ? 4x 5 Here we can factor: x2 4x 5 x 1 x 5 , so we can write A x B 5 so, using Proposition 7.2: x 5 Clearing the denominators on the right, we are led to the equation Setting x 0 gives 1 A. But we have no more roots to substitute to find B and C, so instead we equate coefficients. The coefficient of x2 on the left is 0, and on the right is A C, so A C 0; since A 1, we learn that C 1. Comparing coefficients of x we learn that 1 B. Thus 7.55 becomes x2 1 x2 1 and our integral is Clearing of denominators, we obtain the equation For x 0, we obtain 1 A 5 , so A 1 5. Comparing coefficients of x2 we obtain 1 A C, so C 1 5. Comparing coefficients of x we obtain 0 4A B 2C, so 0 4 5 B 2 5, so B 2 5 and 7.59 becomes x x2
2 2 ! x 2 1 x 2 1 $ ! (7.61) " " x2 1 4x 5 1 5 1 x 2 5 1 1 5 x 2 (7.60) x2 1 A x 2 2 1 Bx C x 2x x x2 x 1 (7.59) x2 1 4x 5 2 1 , so we expect a partial fractions expansion of the form A x B 22 C x 2 x 22 1 Example 7.19 x2 1 dx x x2 4x 5 The denominator is x x 2 ? (7.58) ln x x 1 dx x x2 1 arctanx 1 ln x2 2 $ (7.57) x 1 x x2 1 1 x 1 x 1 C (7.56) x 1 A x2 1 Bx Cx2 x2 1 x2 1 (7.55) Example 7.18 x 1 dx ? x x2 1 Here we have to expect each of the terms in Proposition 7.2 to appear, so we try an expression of the form x 1 x x2 1 A x B Cx (7.54) 7 x 3 arctan 5 5 ln x 3 2 (7.53) 7 2 5 7.3 Partial Fractions 115 2x 1 dx x2 6x 14 dx 32 x x 3 dx 32 5 C Chapter 7 Techniques of Integration 116 which we can integrate to Multiple Roots If the denominator has a multiple root, that is there is a factor x r raised to a power, then we have to allow for the possibility of terms in the partial fraction of the form 1 x r raised to the same power. But then the numerator can be (as we have seen above in the case of quadratic factors) a polynomial of degree as much as one less than the power. This is best explained through a few examples. Clearing of denominators, we obtain Substituting x 0 we obtain 1 C 1 , so C 1. Substituting x 1, we obtain 2 D. To find A and B we have to compare coefficients of powers of x. Equating coefficients of x 3 , we have 0 A D, so A 2. Equating coefficients of x2 , we have 1 A B, so B 1 A 1. Thus the expansion 7.63 is x3 which we can integrate term by term: 7.4. Trigonometric Methods Now, although the above techniques are all that one needs to know in order to use a Table of Integrals, there is one form which appears so often, that it is worthwhile seeing how the integration formulae are found. Expressions involving the square root of a quadratic function occur quite frequently in practice. How do we integrate 1 x2 or 1 x2 ? When the expressions involve a square root of a quadratic, we can convert to trigonometric functions using the substitutions suggested by figure 7.2. (7.66) 2 ln x 2 ln x 1 x2 1 dx x3 x 1 1 x 1 2x2 x 1 $ (7.65) x2 1 x 1 2 x 1 x2 1 x3 2 C (7.64) x2 1 Ax2 x 1 Bx x 1 C x x 1 1 (7.63) x2 1 x3 x 1 A x B x2 C x3 D Dx3 Example 7.20 x2 1 dx ? x3 x 1 We have to allow for the possibility of a term of the form Ax2 expansion of the form Bx C x3 , or, what is the same, an (7.62) x2 1 dx x x2 4x 5 1 ln x 5 2 arctan x 5 2 1 ln x2 10 4x 5 C PSfrag replacements x u u (A) This is still a hard integral, but we can discover it by an integration by parts (see Practice Problem set 4, problem 6) to be Example 7.23 x 1 x2dx ? (7.72) 1 x2dx 1 x2 ln 1 x2 x 1 x 2 C $ Now, we return to figure 7.2B to write this in terms of x: tanu x sec u (7.71) sec3 du ln sec u tanu 1 sec u tanu 2 (7.70) 1 x2 dx sec3 udu C 1 x2 We finally obtain $ $ Example 7.22 To find sec u. Then 1 x2dx, we use the substitution of figure 7.2B: x (7.69) 1 x2dx arcsin x 2 x 1 x2 4 C tanu dx sec2 udu 1 Now, to return to the original variable x, we have to use the double angle formula: sin 2u x 1 x2, and we finally have the answer: (7.68) 1 x2dx 1 cos2u du 2 Now, we use the halfangle formula: cos2 u 1 cos2u 2: u 2 sin 2u 4 C (7.67) 1 x2dx cos2 udu 2 sin u cosu x2 $ $ Example 7.21 To find cos u. Then 1 1 x2 x2dx, we use the substitution of figure 7.2A: x 1 7.4 Trigonometric Methods 117 Figure 7.2 1 x2 x 1 (B) sin u dx cosudu 1 x2 Chapter 7 Techniques of Integration 118 This integration now follows from use of double and halfangle formulae: For the remainder of this course, we shall assume that you have a table of integrals available, and know how to use it. There are several handbooks, and every Calculus text has a table of integrals on the inside back cover. There are a few tables on the web: http://math2.org/math/integrals/tableof.htm http://www.cahs1.org/lessonIcalc/table /table of integrals.htm http://www.engineering.com/community/library/textbook /calculus/calculus table integrals content.htm http://www.maths.abdn.ac.uk/~jrp/ma1002/website/int /node51.html (7.76) x2 1 x2dx arcsinx 8 x 1 x2 1 2 2x2 C Now, sin 4u 2 sin 2u cos 2u 4 sin u cosu 1 2 sin2 u 4x 1 x2 1 2x2 Finally (7.75) " ! sin2 u cos2 udu 1 4 sin2 2u du 1 1 cos 4u du 8 1 sin 4u u C 8 4 (7.74) x2 1 x2dx sin2 u cos2 udu $ $ Here simple substitution fails, and we use the substitution of figure 7.2A: x cos u. Then Example 7.24 x2 1 x2dx ? (7.73) x 1 x2dx u1 2 du x2 1 2 2 1 3 3 2 C sinu dx cosudu 1 x2 $ Don't be misled: always try simple substitution first; in this case the substitution u 2xdx leads to the formula 1 x 2 du ...
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This note was uploaded on 07/14/2011 for the course MAC 3473 taught by Professor Block during the Spring '08 term at University of Florida.
 Spring '08
 BLOCK
 Calculus, Applied Mathematics

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