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7535_Ch07_pp324-353.qxd 336 Chapter 7 1/10/11 10:28 AM Page 336 Differential Equations and Mathematical Modeling 7.2 Antidifferentiation by Substitution What you will learn about . . . Indefinite Integrals Indefinite Integrals • Leibniz Notation and Antiderivatives • Substitution in Indefinite Integrals • Substitution in Definite Integrals and why . . . If y = f(x) we can denote the derivative of f by either dy>dx or f¿(x). What can we use to denote the antiderivative of f ? We have seen that the general solution to the differential equation dy>dx = f(x) actually consists of an infinite family of functions of the form F(x) + C, where F¿(x) = f(x). Both the name for this family of functions and the symbol we use to denote it are closely related to the definite integral because of the Fundamental Theorem of Calculus. • Antidifferentiation techniques were historically crucial for applying the results of calculus. DEFINITION Indefinite Integral The family of all antiderivatives of a function f(x) is the indefinite integral of f with respect to x and is denoted by 1 f(x)dx. If F is any function such that F¿(x) = f(x), then 1 f(x)dx = F(x) + C, where C is an arbitrary constant, called the constant of integration. As in Chapter 6, the symbol 1 is an integral sign, the function f is the integrand of the integral, and x is the variable of integration. Notice that an indefinite integral is not at all like a definite integral, despite the similarities in notation and name. A definite integral is a number, the limit of a sequence of Riemann sums. An indefinite integral is a family of functions having a common derivative. If the Fundamental Theorem of Calculus had not provided such a dramatic link between antiderivatives and integration, we would surely be using a different name and symbol for the general antiderivative today. EXAMPLE 1 Evaluating an Indefinite Integral Evaluate 1 (x2 - sin x) dx. SOLUTION Evaluating this indefinite integral is just like solving the differential equation dy>dx = x2 - sin x. Our past experience with derivatives leads us to conclude that L (x2 - sin x) dx = x3 + cos x + C 3 (as you can check by differentiating). Now Try Exercise 3. You have actually been finding antiderivatives since Section 6.3, so Example 1 should hardly have seemed new. Indeed, each derivative formula in Chapters 3 and 4 could be turned around to yield a corresponding indefinite integral formula. We list some of the most useful such indefinite integral formulas on the next page. Be sure to familiarize yourself with these before moving on to the next section, in which function composition becomes an issue. (Incidentally, it is in anticipation of the next section that we give some of these formulas in terms of the variable u rather than x.) 7535_Ch07_pp324-353.qxd 1/10/11 10:28 AM Page 337 Section 7.2 Antidifferentiation by Substitution 337 Properties of Indefinite Integrals L L k f(x)dx = k L f(x) dx (f(x) g(x)) dx = L for any constant k f(x) dx L g(x) dx Power Formulas L un du = un+1 + C when n Z - 1 n + 1 1 du = ln ƒ u ƒ + C Lu (see Example 2) L u-1 du = L sin u du = - cos u + C Trigonometric Formulas L L L cos u du = sin u + C sec2 u du = tan u + C L sec u tan u du = sec u + C L csc2 u du = - cot u + C csc u cot u du = - csc u + C Exponential and Logarithmic Formulas L A Note on Absolute Value Since the indefinite integral does not specify a domain, you should always use the absolute value when finding 1 1>u du. The function ln u + C is only defined on positive u-intervals, while the function ln ƒ u ƒ + C is defined on both the positive and negative intervals in the domain of 1>u (see Example 2). eu du = eu + C L ln u du = u ln u - u + C L loga u du = L au du = au + C ln a (See Example 2) ln u u ln u - u du = + C ln a L ln a EXAMPLE 2 Verifying Antiderivative Formulas Verify the antiderivative formulas: 1 (a) u-1 du = du = ln ƒ u ƒ + C u L L (b) L ln u du = u ln u - u + C SOLUTION We can verify antiderivative formulas by differentiating. d 1 d 1 (a) For u 7 0, we have (ln ƒ u ƒ + C) = (ln u + C) = + 0 = . u u du du d d 1 1 (ln ƒ u ƒ + C) = (ln (-u) + C) = (- 1) + 0 = . For u 6 0, we have -u u du du 1 d (ln ƒ u ƒ + C) = in either case, ln ƒ u ƒ + C is the general antiderivative of Since u du 1 the function on its entire domain. u (b) d 1 (u ln u - u + C) = 1 # ln u + u ¢ ≤ - 1 + 0 = ln u + 1 - 1 = ln u. u du Now Try Exercise 11. 7535_Ch07_pp324-353.qxd 338 Chapter 7 1/10/11 10:28 AM Page 338 Differential Equations and Mathematical Modeling Leibniz Notation and Antiderivatives b The appearance of the differential “dx” in the definite integral 1a f(x) dx is easily explained n by the fact that it is the limit of a Riemann sum of the form a f(xk) # ¢x (see Section 6.2). k=1 The same “dx” almost seems unnecessary when we use the indefinite integral 1 f(x)dx to represent the general antiderivative of f, but in fact it is quite useful for dealing with the effects of the Chain Rule when function composition is involved. Exploration 1 will show you why this is an important consideration. EXPLORATION 1 Are 1 f (u) du and 1 f (u) dx the Same Thing? Let u = x2 and let f(u) = u3. 1. Find 1 f(u) du as a function of u. 2. Use your answer to question 1 to write 1 f(u) du as a function of x. 3. Show that f(u) = x6 and find 1 f(u) dx as a function of x. 4. Are the answers to questions 2 and 3 the same? Exploration 1 shows that the notation 1 f(u) is not sufficient to describe an antiderivative when u is a function of another variable. Just as du>du is different from du>dx when differentiating, 1 f(u) du is different from 1 f(u) dx when antidifferentiating. We will use this fact to our advantage in the next section, where the importance of “dx” or “du” in the integral expression will become even more apparent. EXAMPLE 3 Paying Attention to the Differential Let f(x) = x3 + 1 and let u = x2. Find each of the following antiderivatives in terms of x: (a) L f(x) dx (b) L f(u) du (c) L f(u) dx SOLUTION (a) (b) (c) L f(x) dx = L f(u) du = L f(u) dx = L L L (x3 + 1) dx = (u3 + 1) du = (u3 + 1) dx = x4 + x + C 4 (x2)4 u4 x8 + u + C = + x2 + C = + x2 + C 4 4 4 L ((x2)3 + 1) dx = L (x6 + 1) dx = x7 + x + C 7 Now Try Exercise 15. Substitution in Indefinite Integrals A change of variables can often turn an unfamiliar integral into one that we can evaluate. The important point to remember is that it is not sufficient to change an integral of the form 1 f(x) dx into an integral of the form 1 g(u) dx. The differential matters. A complete substitution changes the integral 1 f(x) dx into an integral of the form 1 g(u) du. EXAMPLE 4 Using Substitution Evaluate 1 sin x ecos x dx. continued 7535_Ch07_pp324-353.qxd 1/10/11 10:28 AM Page 339 Section 7.2 Antidifferentiation by Substitution 339 SOLUTION Let u = cos x. Then du>dx = - sin x, from which we conclude that du = - sin x dx. We rewrite the integral and proceed as follows: L sin x ecos x dx = = - L L (-sin x)ecos x dx ecos x # (-sin x) dx eu du L = - eu + C = -ecos x + C = - Substitute u for cos x and du for sin x dx. Re-substitute cos x for u after antidifferentiating. Now Try Exercise 19. If you differentiate - ecos x + C, you will find that a factor of - sin x appears when you apply the Chain Rule. The technique of antidifferentiation by substitution reverses that effect by absorbing the -sin x into the differential du when you change 1 sin x ecos x dx into - 1 eu du. That is why a “u-substitution” always involves a “du-substitution” to convert the integral into a form ready for antidifferentiation. EXAMPLE 5 Using Substitution Evaluate 1 x2 25 + 2x3 dx. SOLUTION This integral invites the substitution u = 5 + 2x3, du = 6x2 dx. L x2 25 + 2x3 dx. = L (5 + 2x3)1>2 # x2 dx = 1 (5 + 2x3)1>2 # 6x2 dx 6L Set up the substitution with a factor of 6. = 1 u1>2 du 6L Substitute u for 5 + 2x3 and du for 6x2 dx. = 1 2 3>2 ¢ ≤u + C 6 3 = 1 (5 + 2x3)3>2 + C 9 Re-substitute after antidifferentiating. Now Try Exercise 27. EXAMPLE 6 Using Substitution Evaluate 1 cot 7x dx. SOLUTION We do not recall a function whose derivative is cot 7x, but a basic trigonometric identity changes the integrand into a form that invites the substitution u = sin 7x, du = 7 cos 7x dx. We rewrite the integrand as shown below. cos 7x cot 7x dx = dx Trigonometric identity L L sin 7x 1 7 cos 7x dx Note that du = 7 cos 7x dx when u = sin 7x. = 1 We multiply by # 7, or 1. 7 L sin 7x continued 7 7535_Ch07_pp324-353.qxd 340 Chapter 7 1/10/11 10:28 AM Page 340 Differential Equations and Mathematical Modeling 1 du 7L u 1 = ln ƒ u ƒ + C 7 1 = ln ƒ sin 7x ƒ + C 7 = Substitute u for 7 sin x and du for 7 cos 7x dx. Notice the absolute value! Resubstitute sin 7x for u after antidifferentiating. Now Try Exercise 29. EXAMPLE 7 Setting Up a Substitution with a Trigonometric Identity Find the indefinite integrals. In each case you can use a trigonometric identity to set up a substitution. dx cot2 3x dx (c) cos3 x dx (a) (b) 2 L cos 2x L L SOLUTION dx 1 = sec2 2x dx = sec2 2x # 2dx (a) 2 2 cos 2x L L L 1 = sec2 u du 2L 1 = tan u + C 2 1 = tan 2x + C 2 (b) L cot2 3x dx = = = = = = (c) L cos3 x dx = (csc2 3x - 1) dx L 1 (csc2 3x - 1) # 3 dx 3L 1 (csc2 u - 1) # du 3L 1 (-cot u - u) + C 3 1 (-cot 3x - 3x) + C 3 1 - cot 3x - x + C 3 L = L = L Let u = 2x and du = 2 dx. Resubstitute after antidifferentiating. Let u = 3x and du = 3 dx. Resubstitute after antidifferentiation. (cos2 x) cos x dx (1 - sin2 x) cos x dx (1 - u2) du u3 + C 3 sin3 x = sin x + C 3 Let u = sin x and du = cos x dx. = u - Resubstitute after antidifferentiating. Now Try Exercise 47. 7535_Ch07_pp324-353.qxd 1/10/11 10:28 AM Page 341 Section 7.2 Antidifferentiation by Substitution 341 Substitution in Definite Integrals Antiderivatives play an important role when we evaluate a definite integral by the Fundamental Theorem of Calculus, and so, consequently, does substitution. In fact, if we make full use of our substitution of variables and change the interval of integration to match the u-substitution in the integrand, we can avoid the “resubstitution” step in the previous four examples. EXAMPLE 8 Evaluating a Definite Integral by Substitution p>3 Evaluate 2 tan x sec x dx. L0 SOLUTION Let u = tan x and du = sec2 x dx. Note also that u(0) = tan 0 = 0 and u(p>3) = tan(p>3) = 23. So p>3 L0 13 tan x sec2 x dx = = L0 u du Substitute u-interval for x-interval. u2 13 3 3 ` = -0= . 2 0 2 2 Now Try Exercise 55. EXAMPLE 9 That Absolute Value Again 1 Evaluate x dx. L0 x - 4 2 SOLUTION Let u = x2 - 4 and du = 2x dx. Then u(0) = 02 - 4 = - 4 and u(1) = 12 - 4 = -3. So 1 1 x 2x dx 1 dx = 2 2 L0 x2 - 4 L0 x - 4 -3 = 1 du 2 L-4 u = -3 1 ln ƒ u ƒ ` 2 -4 = 1 3 1 (ln 3 - ln 4) = ln ¢ ≤ . 2 2 4 Substitute u-interval for x-interval. Notice that ln u would not have existed over the interval of integration [-4, -3]. The absolute value in the antiderivative is important. Now Try Exercise 63. Finally, consider this historical note. The technique of u-substitution derived its importance from the fact that it was a powerful tool for antidifferentiation. Antidifferentiation derived its importance from the Fundamental Theorem, which established it as the way to evaluate definite integrals. Definite integrals derived their importance from real-world applications. While the applications are no less important today, the fact that the definite 7535_Ch07_pp324-353.qxd 342 Chapter 7 1/10/11 10:28 AM Page 342 Differential Equations and Mathematical Modeling integrals can be easily evaluated by technology has made the world less reliant on antidifferentiation, and hence less reliant on u-substitution. Consequently, you have seen in this book only a sampling of the substitution tricks calculus students would have routinely studied in the past. You may see more of them in a differential equations course. Quick Review 7.2 (For help, go to Sections 4.1 and 4.4.) Exercise numbers with a gray background indicate problems that the authors have designed to be solved without a calculator. In Exercises 1 and 2, evaluate the definite integral. 2 2. L1 L0 In Exercises 3–10, find dy>dx. 2x - 1 dx L2 8. y = ln sin x 9. y = ln (sec x + tan x) x 3. y = x 3t dt 6. y = sin2 (4x - 5) 7. y = ln cos x 5 x4 dx 1. 5. y = (x3 - 2x2 + 3)4 4. y = L0 10. y = ln (csc x + cot x) 3t dt Section 7.2 Exercises In Exercises 1–6, find the indefinite integral. 1. 3. 5. L L L 2 2. (cos x - 3x ) dx 1 ¢t 2 t2 ≤ dt 4. L x 20. -2 dx 21. dt (2ex + sec x tan x - 2x) dx L In Exercises 7–12, use differentiation to verify the antiderivative formula. L csc2 u du = - cot u + C 1 9. e2x dx = e2x + C 2 L 11. 1 L 1 + u2 du = tan -1 8. L L 24. L 1 L 1 - u2 2 17. 18. 19. L sin 3x dx, L L sec 2x tan 2x dx, L 21 - r3 u = 1 - cos , u = 1 - r3 t 2 u = y4 + 4y2 + 1 26. L sec2 (x + 2) dx 2tan x sec2 x dx sec ¢ u + p p ≤ tan ¢ u + ≤ du 2 2 L 29. L tan(4x + 2) dx 30. L 31. L cos (3z + 4) dz 32. L 34. L 36. L sin2 3x 33. 35. ln6x dx L x L s1/3 cos (s4/3 - 8) ds sin (2t + 1) 37. L cos (2t + 1) 39. L x ln x u = 2x2 u = 2x 9r2 dr 28. u = 3x x cos (2x2) dx, t dt, 2 dx L 14. f(u) = u and u = x In Exercises 17–24, use the indicated substitution to evaluate the integral. Confirm your answer by differentiation. 2 22. 8(y4 + 4y2 + 1)2(y3 + 2y) dy, 27. du = sin u + C 16. f(u) = sin u and u = 4x t 2 L (1 - x)2 5 15. f(u) = eu and u = 7x x 3 u = ¢ 1 - cos ≤ sin 25. -1 In Exercises 13–16, verify that 1 f(u) du Z 1 f(u) dx 13. f(u) = 2u and u = x2 (x 7 0) , u = 7x - 2 In Exercises 25–46, use substitution to evaluate the integral. csc u cot u = - csc u + C 1 x 5x dx = 10. 5 + C ln 5 L u + C 12. L x2 + 9 23. (3x4 - 2x-3 + sec2 x) dx 28(7x - 2)3 dx, dx L t2 + 1 6. 7. L 2 dx dt 3(sin x)-2 dx 2cot x csc2 x dx x x tan7 ¢ ≤ sec2 ¢ ≤ dx 2 2 dx 6 cos t 38. L (2 + sin t)2 40. L dt tan2 x sec2 x dx 7535_Ch07_pp324-353.qxd 1/10/11 10:28 AM Page 343 Antidifferentiation by Substitution Section 7.2 x dx 41. Lx + 1 dx 43. L cot 3x 45. L 40 dx 42. 2 1 67. L x + 25 dx 2 44. x3 dx L0 2x4 + 9 p/3 L 25x + 8 68. (1 - cos 3x) sin 3x dx Lp/6 69. Show that sec x dx [Hint: Multiply the integrand by y = ln ` sec x + tan x sec x + tan x L csc x dx [Hint: Multiply the integrand by (See the discussion following Example 4, Section 6.4.) csc x + cot x csc x + cot x 70. Show that y = ln ` and then use a substitution to integrate the result.] In Exercises 47–52, use the given trigonometric identity to set up a u-substitution and then evaluate the indefinite integral. 3 47. L 48. L 2 2 sin 2x dx, sin 2x = 1 - cos 2x sec4 x dx, sec2 x = 1 + tan2 x 49. L 50. L 51. L 52. L p/4 71. True or False By u-substitution, 10 p/4 3 10 u du. Justify your answer. 4 cos2 x dx, cos 2x = 2 cos2 x - 1 72. True or False If f is positive and differentiable on [a, b], then tan4 x dx, b f¿(x)dx tan2 x = sec2 x - 1 (cos4 x - sin4 x)dx, La cos 2x = cos2 x - sin2 x 1 2y + 1 dy L0 54. L0 102u du L0 (1 + u3/2)2 56. p 58. 2t5 + 2t (5t4 + 2) dt 60. dx x + 2 L0 dt L1 t - 3 3 dx -3 2u sin 2u du 5 62. 2 63. cos L0 dx 2x - 3 L2 3p/4 64. Lp/4 cot x dx 2 ex dx x L-1 x2 + 1 L0 3 + e Two Routes to the Integral In Exercises 67 and 68, make a substitution u = Á (an expression in x), du = Á . Then 65. x dx b. Justify your answer. e 2 (C) ln ƒ cos x ƒ + C (E) - ln ƒ cot x ƒ + C 2 2x 10 e (B) e4 - 1 dx = (C) e4 - 2 (D) 2e4 - 2 (E) e4 - 1 2 5 p>6 7 61. cos x f(a) (D) - ln ƒ cos x ƒ + C 4 dr L-p 24 + 3 sin x 1 L0 5r f(b) 73. Multiple Choice 1 tan x dx = tan2 x + C (B) ln ƒ cot x ƒ + C (A) 2 (A) L-1 (4 + r2)2 f(x) = ln a 74. Multiple Choice r 21 - r2 dr 1 tan x sec2 x dx L-p>4 1 59. tan3 x sec2 x dx = cos 2x = 1 - 2 sin2 x 0 57. is the solution to the initial value problem dy = cot x, f(2) = 6. dx 2 sin2 x dx, 3 55. sin x ` +6 sin 2 Standardized Test Questions In Exercises 53–66, make a u-substitution and integrate from u(a) to u(b). 53. cos 3 ` +5 cos x is the solution to the initial value problem dy = tan x, f(3) = 5. dx and then use a substitution to integrate the result.] 46. 343 66. (a) integrate with respect to u from u(a) to u(b). (b) find an antiderivative with respect to u, replace u by the expression in x, then evaluate from a to b. 75. Multiple Choice If 13 f(x - a) dx = 7 where a is a constant, 5-a then 13-a f(x) dx = (A) 7 + a (B) 7 (C) 7 - a (D) a - 7 (E) - 7 76. Multiple Choice If the differential equation dy>dx = f(x) leads to the slope field shown below, which of the following could be 1 f(x) dx? (A) sin x + C (D) - cos x + C (B) cos x + C (E) sin x + C 2 (C) - sin x + C 7535_Ch07_pp324-353.qxd 344 Chapter 7 1/10/11 10:28 AM Page 344 Differential Equations and Mathematical Modeling Explorations 77. Constant of Integration Consider the integral L 2x + 1 dx. 2 2x + 1 dx = (x + 1)3/2 + C. (a) Show that 3 L (b) Writing to Learn Explain why x y1 = L0 x 2t + 1 dt and y2 = L3 2t + 1 dt are antiderivatives of 2x + 1. (c) Use a table of values for y1 - y2 to find the value of C for which y1 = y2 + C. (c) Writing to Learn Explain why the different-looking answers in parts (a) and (b) are actually equivalent. 80. Different Solutions? Consider the integral 1 2 sec2 x tan x dx. (a) Evaluate the integral using the substitution u tan x. (b) Evaluate the integral using the substitution u sec x. (c) Writing to Learn Explain why the different-looking answers in parts (a) and (b) are actually equivalent. Extending the Ideas 81. Trigonometric Substitution Suppose u = sin-1 x. Then cos u 7 0. (a) Use the substitution x = sin u, dx = cos u du to show that dx 3 C = L0 2x + 1 dx. 78. Group Activity Making Connections Suppose that f(x) dx = F(x) + C. L (a) Explain how you can use the derivative of F(x) + C to confirm the integration is correct. (b) Explain how you can use a slope field of f and the graph of y = F(x) to support your evaluation of the integral. (c) Explain how you can use the graphs of y1 = F(x) and x y2 = 10 f(t) dt to support your evaluation of the integral. (d) Explain how you can use a table of values for y1 - y2, y1 and y2 defined as in part (c), to support your evaluation of the integral. (e) Explain how you can use graphs of f and NDER of F(x) to support your evaluation of the integral. x (f) Illustrate parts (a)–(e) for f(x) = . 2x2 + 1 79. Different Solutions? Consider the integral 1 2 sin x cos x dx. (a) Evaluate the integral using the substitution u sin x. (b) Evaluate the integral using the substitution u cos x. = L 21 - x2 (d) Writing to Learn Give a convincing argument that L 1 du. dx 1 du to show that = sin-1x + C. L L 21 - x2 82. Trigonometric Substitution Suppose u = tan-1 x. (b) Evaluate (a) Use the substitution x = tan u, dx = sec2 u du to show that dx = L 1 + x2 (b) Evaluate L L 1 du to show that 1 du. dx L 1 + x2 = tan-1 x + C. 83. Trigonometric Substitution Suppose 2x = sin y. (a) Use the substitution x = sin2 y, dx = 2 sin y cos y dy to show that 1/2 L0 p/4 2x dx 21 - x = L0 2 sin2 y dy. (b) Use the identity given in Exercise 49 to evaluate the definite integral without a calculator. 84. Trigonometric Substitution Suppose u = tan-1 x. (a) Use the substitution x = tan u, dx = sec2 u du to show that 23 L0 p/3 dx 21 + x2 = L0 sec u du. (b) Use the hint in Exercise 45 to evaluate the definite integral without a calculator.

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