8.3. Integration of Rational Functions by Partial Fractions. 570 Chapter 8: Techniques of Integration

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1 570 Chapter 8: Techniques of Integration 8.3 Integration of Rational Functions b Partial Fractions This section shows how to epress a rational function (a quotient of polnomials) as a sum of simpler fractions, called partial fractions, which are easil integrated. For instance, the rational function (5-3)>( ) can be rewritten as = ,

2 8.3 Integration of Rational Functions b Partial Fractions 57 which can be verified algebraicall b placing the fractions on the right side over a common denominator s + ds - 3d. The skill acquired in writing rational functions as such a sum is useful in other settings as well (for instance, when using certain transform methods to solve differential equations). To integrate the rational function (5-3)>( + ) ( - 3) on the left side of our previous epression, we simpl sum the integrals of the fractions on the right side: 5-3 s + ds - 3d = = 2 ln ƒ + ƒ + 3 ln ƒ - 3 ƒ + C. The method for rewriting rational functions as a sum of simpler fractions is called the method of partial fractions. In the case of the above eample, it consists of finding constants A and B such that = A + + B - 3. () (Pretend for a moment that we do not know that A = 2 and B = 3 will work.) We call the fractions A>s + d and B>s - 3d partial fractions because their denominators are onl part of the original denominator We call A and B undetermined coefficients until proper values for them have been found. To find A and B, we first clear Equation () of fractions, obtaining 5-3 = As - 3d + Bs + d = sa + B - 3A + B. This will be an identit in if and onl if the coefficients of like powers of on the two sides are equal: A + B = 5, -3A + B = -3. Solving these equations simultaneousl gives A = 2 and B = 3. General Description of the Method Success in writing a rational function ƒ() > g() as a sum of partial fractions depends on two things: The degree of ƒ() must be less than the degree of g(). That is, the fraction must be proper. If it isn t, divide ƒ() b g() and work with the remainder term. See Eample 3 of this section. We must know the factors of g(). In theor, an polnomial with real coefficients can be written as a product of real linear factors and real quadratic factors. In practice, the factors ma be hard to find. Here is how we find the partial fractions of a proper fraction ƒ() > g() when the factors of g are known.

3 572 Chapter 8: Techniques of Integration Method of Partial Fractions (ƒ() > g()proper). et - r be a linear factor of g(). Suppose that s - rd m is the highest power of - r that divides g(). Then, to this factor, assign the sum of the m partial fractions: A - r + A 2 s - rd 2 + Á + Do this for each distinct linear factor of g(). 2. et 2 + p + q be a quadratic factor of g(). Suppose that s 2 + p + qd n is the highest power of this factor that divides g(). Then, to this factor, assign the sum of the n partial fractions: B + C 2 + p + q + B 2 + C 2 s 2 + p + qd 2 + Á + A m s - rd m. B n + C n s 2 + p + qd n. Do this for each distinct quadratic factor of g() that cannot be factored into linear factors with real coefficients. 3. Set the original fraction ƒ() > g() equal to the sum of all these partial fractions. Clear the resulting equation of fractions and arrange the terms in decreasing powers of. 4. Equate the coefficients of corresponding powers of and solve the resulting equations for the undetermined coefficients. EXAMPE Evaluate Distinct inear Factors using partial fractions s - ds + ds + 3d Solution The partial fraction decomposition has the form s - ds + ds + 3d = A - + B + + To find the values of the undetermined coefficients A, B, and C we clear fractions and get = As + ds + 3d + Bs - ds + 3d + Cs - ds + d = sa + B + C 2 + s4a + 2B + s3a - 3B - Cd. The polnomials on both sides of the above equation are identical, so we equate coefficients of like powers of obtaining Coefficient of 2 : A + B + C = Coefficient of : 4A + 2B = 4 Coefficient of 0 : 3A - 3B - C = C + 3.

4 8.3 Integration of Rational Functions b Partial Fractions 573 There are several was for solving such a sstem of linear equations for the unknowns A, B, and C, including elimination of variables, or the use of a calculator or computer. Whatever method is used, the solution is A = 3>4, B = >2, and C = ->4. Hence we have where K is the arbitrar constant of integration (to avoid confusion with the undetermined coefficient we labeled as C). EXAMPE 2 Evaluate A Repeated inear Factor Solution First we epress the integrand as a sum of partial fractions with undetermined coefficients. Equating coefficients of corresponding powers of gives Therefore, EXAMPE 3 Evaluate s - ds + ds + 3d = c s + 2d 2 = = As + 2d + B A = 6 and 2A + B = 2 + B = 7, or A = 6 and B = s + 2d 2 = a s + 2d 2 b Integrating an Improper Fraction = 3 4 ln ƒ - ƒ + 2 ln ƒ + ƒ - 4 ln ƒ + 3 ƒ + K, s + 2d 2. A B s + 2d 2 = A + s2a + Bd = s + 2d-2 = 6 ln ƒ + 2 ƒ + 5s + 2d - + C Solution First we divide the denominator into the numerator to get a polnomial plus a proper fraction d Multipl both sides b s + 2d 2.

5 574 Chapter 8: Techniques of Integration Then we write the improper fraction as a polnomial plus a proper fraction. We found the partial fraction decomposition of the fraction on the right in the opening eample, so = = = = ln ƒ + ƒ + 3 ln ƒ - 3 ƒ + C. A quadratic polnomial is irreducible if it cannot be written as the product of two linear factors with real coefficients. EXAMPE 4 Evaluate using partial fractions. Integrating with an Irreducible Quadratic Factor in the Denominator s 2 + ds - d 2 Solution The denominator has an irreducible quadratic factor as well as a repeated linear factor, so we write s 2 + ds - d 2 = A + B C - + D s - d 2. (2) Clearing the equation of fractions gives = sa + Bds - d 2 + Cs - ds 2 + d + Ds 2 + d = sa + C 3 + s -2A + B - C + D 2 + sa - 2B + C + sb - C + Dd. Equating coefficients of like terms gives Coefficients of 3 : 0 = A + C Coefficients of 2 : 0 = -2A + B - C + D Coefficients of : -2 = A - 2B + C Coefficients of 0 : 4 = B - C + D We solve these equations simultaneousl to find the values of A, B, C, and D: -4 = -2A, A = 2 Subtract fourth equation from second. C = -A = -2 From the first equation B = A = 2 and C = -2 in third equation. D = 4 - B + C =. From the fourth equation

6 8.3 Integration of Rational Functions b Partial Fractions 575 We substitute these values into Equation (2), obtaining Finall, using the epansion above we can integrate: s 2 + ds - d 2 = s - d s 2 + ds - d 2 = a s - d 2 b = a s - d 2 b = ln s 2 + d + tan ln ƒ - ƒ C. EXAMPE 5 Evaluate A Repeated Irreducible Quadratic Factor Solution The form of the partial fraction decomposition is s 2 + d 2. Multipling b s 2 + d 2, we have s 2 + d 2 = A + B + C If we equate coefficients, we get the sstem D + E s 2 + d 2 = As 2 + d 2 + sb + Cs 2 + d + sd + E = As d + Bs d + Cs 3 + d + D 2 + E = sa + B 4 + C 3 + s2a + B + D 2 + sc + E + A A + B = 0, C = 0, 2A + B + D = 0, C + E = 0, A =. Solving this sstem gives A =, B = -, C = 0, D = -, and E = 0. Thus, s 2 + d 2 = c s 2 + d 2 d = = - 2 du u - 2 du u 2 s 2 + d 2 u = 2 +, du = 2 = ln ƒ ƒ - 2 ln ƒ u ƒ + 2u + K = ln ƒ ƒ - 2 ln s2 + d + = ln ƒ ƒ s 2 + d + K. 2s 2 + d + K

$g() and gsd = s - r ds - r 2 d Á s - r n d is a product of n distinct linear factors, each raised to the first power, there is a quick wa to epand ƒ() > g() b partial fractions.$

7 576 Chapter 8: Techniques of Integration HISTORICA BIOGRAPHY Oliver Heaviside ( ) The Heaviside Cover-up Method for inear Factors When the degree of the polnomial ƒ() is less than the degree of g() and gsd = s - r ds - r 2 d Á s - r n d is a product of n distinct linear factors, each raised to the first power, there is a quick wa to epand ƒ() > g() b partial fractions. EXAMPE 6 Using the Heaviside Method Find A, B, and C in the partial-fraction epansion 2 + s - ds - 2ds - 3d = A - + B C - 3. (3) Solution If we multipl both sides of Equation (3) b s - d to get 2 + s - 2ds - 3d = A + Bs - d - 2 and set =, the resulting equation gives the value of A: sd 2 + s - 2ds - 3d = A , A =. Thus, the value of A is the number we would have obtained if we had covered the factor s - d in the denominator of the original fraction + Cs - d s - ds - 2ds - 3d (4) and evaluated the rest at = : A = Cover Similarl, we find the value of B in Equation (3) b covering the factor s - 2d in Equation (4) and evaluating the rest at = 2: B = sd 2 + s - d s - 2ds - 3d = 2 s -ds -2d =. s2d 2 + s2 - d s - 2d s2-3d = 5 sds -d = -5. Cover Finall, C is found b covering the s - 3d in Equation (4) and evaluating the rest at = 3: C = s3d 2 + s3 - ds3-2d s - 3d Cover = 0 s2dsd = 5.

8 8.3 Integration of Rational Functions b Partial Fractions 577 Heaviside Method. Write the quotient with g() factored: ƒsd gsd = 2. Cover the factors s - r i d of gsd one at a time, each time replacing all the uncovered s b the number r i. This gives a number for each root r i : A = A 2 = o A n = ƒsd s - r ds - r 2 d Á s - r n d. ƒsr d sr - r 2 d Á sr - r n d ƒsr 2 d sr 2 - r dsr 2 - r 3 d Á sr 2 - r n d 3. Write the partial-fraction epansion of ƒsd>gsd as ƒsd gsd = A s - r d + A 2 s - r 2 d + Á + A i ƒsr n d sr n - r dsr n - r 2 d Á sr n - r n - d. A n s - r n d. EXAMPE 7 Evaluate Integrating with the Heaviside Method Solution The degree of ƒsd = + 4 is less than the degree of gsd = , and, with g() factored, The roots of g() are r = 0, r 2 = 2, and r 3 = -5. We find A = A 2 = A 3 = s0-2ds0 + 5d = 4 s -2ds5d =-2 5 Cover = + 4 s - 2ds + 5d s - 2d s2 + 5d = 6 s2ds7d = 3 7 Cover s -5ds -5-2d s + 5d Cover = - s -5ds -7d =- 35.

9 578 Chapter 8: Techniques of Integration Therefore, + 4 s - 2ds + 5d = s - 2d - 35s + 5d, and + 4 s - 2ds + 5d =-2 5 ln ƒ ƒ ln ƒ - 2 ƒ - 35 ln ƒ + 5 ƒ + C. Other Was to Determine the Coefficients Another wa to determine the constants that appear in partial fractions is to differentiate, as in the net eample. Still another is to assign selected numerical values to. EXAMPE 8 Using Differentiation Find A, B, and C in the equation - s + d 3 = A + + B s + d 2 + C s + d 3. Solution We first clear fractions: - = As + d 2 + Bs + d + C. Substituting = - shows C = -2. We then differentiate both sides with respect to, obtaining = 2As + d + B. Substituting = - shows B =. We differentiate again to get 0 = 2A, which shows A = 0. Hence, - s + d 3 = s + d 2-2 s + d 3. In some problems, assigning small values to such as = 0, ;, ;2, equations in A, B, and C provides a fast alternative to other methods. to get EXAMPE 9 Find A, B, and C in Assigning Numerical Values to 2 + s - ds - 2ds - 3d = A - + B C - 3. Solution Clear fractions to get 2 + = As - 2ds - 3d + Bs - ds - 3d + Cs - ds - 2d.

10 8.3 Integration of Rational Functions b Partial Fractions 579 Then let =, 2, 3 successivel to find A, B, and C: Conclusion: = : sd 2 + = As -ds -2d + Bs0d + Cs0d 2 = 2A A = = 2: s2d 2 + = As0d + Bsds -d + Cs0d 5 = -B B = -5 = 3: s3d 2 + = As0d + Bs0d + Cs2dsd 0 = 2C C = s - ds - 2ds - 3d =

11 8.3 Integration of Rational Functions b Partial Fractions 579 EXERCISES 8.3 Epanding Quotients into Partial Fractions Epand the quotients in Eercises 8 b partial fractions s - 3ds - 2d s + d z + z 6. z 2 sz - d z 3 - z 2-6z 7. t t t 2-5t + 6 t 4 + 9t 2 Nonrepeated inear Factors In Eercises 9 6, epress the integrands as a sum of partial fractions and evaluate the integrals d >2 2 + d t 3 + t 2-2t Repeated inear Factors In Eercises 7 20, epress the integrands as a sum of partial fractions and evaluate the integrals s 2 - d 2 s - ds d Irreducible Quadratic Factors In Eercises 2 28, epress the integrands as a sum of partial fractions and evaluate the integrals s + ds 2 + d s 2 + d 2 d 2s + 2 s ss 2 + dss - d 3 ds sss 2 + 9d 2 ds u 3 + 5u 2 + 8u + 4 su 2 + 2u + 2d 2 du u 4-4u 3 + 2u 2-3u + su 2 + d 3 du Improper Fractions s4 2 + d 2 In Eercises 29 34, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral d d 23 3t 2 + t + 4 t 3 + t

12 580 Chapter 8: Techniques of Integration Evaluating Integrals Evaluate the integrals in Eercises e t e e 2t - e t e 2t e 2t + 3e t cos d sin u du 38. sin 2 + sin - 6 cos 2 u + cos u Initial Value Problems Solve the initial value problems in Eercises 4 44 for as a function of t s - 2d 2 tan - s2d s4 2 + ds - 2d 2 s + d 2 tan - s3d s9 2 + ds + d 2 st 2-3t + 2d s3t 4 + 4t 2 + d st 2 + 2td = st, 7 0d, sd = st + d = 2 + st 7 -d, s0d = p>4 Applications and Eamples In Eercises 45 and 46, find the volume of the solid generated b revolving the shaded region about the indicated ais. 45. The -ais 46. The -ais 2 = st 7 2d, s3d = 0 (0.5, 2.68) (2.5, 2.68) = 223, sd = -p23> ( )(2 ) 0 T 47. Find, to two decimal places, the -coordinate of the centroid of the region in the first quadrant bounded b the -ais, the curve = tan -, and the line = 23. T T T 48. Find the -coordinate of the centroid of this region to two decimal places. (3,.83) 49. Social diffusion Sociologists sometimes use the phrase social diffusion to describe the wa information spreads through a population. The information might be a rumor, a cultural fad, or news about a technical innovation. In a sufficientl large population, the number of people who have the information is treated as a differentiable function of time t, and the rate of diffusion, >, is assumed to be proportional to the number of people who have the information times the number of people who do not. This leads to the equation = ksn - d, (5, 0.98) where N is the number of people in the population. Suppose t is in das, k = >250, and two people start a rumor at time t = 0 in a population of N = 000 people. a. Find as a function of t. b. When will half the population have heard the rumor? (This is when the rumor will be spreading the fastest.) 50. Second-order chemical reactions Man chemical reactions are the result of the interaction of two molecules that undergo a change to produce a new product. The rate of the reaction tpicall depends on the concentrations of the two kinds of molecules. If a is the amount of substance A and b is the amount of substance B at time t = 0, and if is the amount of product at time t, then the rate of formation of ma be given b the differential equation = ksa - dsb - d, or = k, sa - dsb - d where k is a constant for the reaction. Integrate both sides of this equation to obtain a relation between and t (a) if a = b, and (b) if a Z b. Assume in each case that = 0 when t = An integral connecting p to the approimation 22> 7 4 s - d 4 a. Evaluate b. How good is the approimation p 22>7? Find out b epressing a 22 as a percentage of p. 7 - pb

13 8.3 Integration of Rational Functions b Partial Fractions 58 c. Graph the function = 4 s - d 4 for 0. Eperiment with the range on the -ais set between 0 and, then 2 + between 0 and 0.5, and then decreasing the range until the graph can be seen. What do ou conclude about the area under the curve? 52. Find the second-degree polnomial P() such that P s0d = 0, and Psd 3 s - d 2 is a rational function. Ps0d =,

7.7. Inverse Trigonometric Functions. Defining the Inverses

7.7. Inverse Trigonometric Functions. Defining the Inverses 7.7 Inverse Trigonometric Functions 57 7.7 Inverse Trigonometric Functions Inverse trigonometric functions arise when we want to calculate angles from side measurements in triangles. The also provide useful