Calc 2 half-341-976-1-320.pdf PDF

314 CHAPTER 5 INVERSE FUNCTIONS EXERCISES 1. The graph of f is shown. Is f one-to-one? Explain....

314 CHAPTER 5 INVERSE FUNCTIONS EXERCISES 1. The graph of f is shown. Is f one-to-one? Explain. 15. (a) ln共 x ⫹ 1兲 ⫹ ln共 x ⫺ 1兲 苷 1 (b) log 5 共c x 兲 苷 d y 16. (a) ln共1 ⫹ e⫺x 兲 苷 3 (b) sin x 苷 0.3 17–43 Differentiate. 17. y 苷 ln共x ln x兲 18. y 苷 e mx cos nx 0 x e 1兾x 19. y 苷 20. y 苷 ln sec x x2 2. The graph of t is given. 21. y 苷 sarctan x 22. y 苷 x cos⫺1 x (a) Why is t one-to-one? et (b) Estimate the value of t⫺1共2兲. 23. f 共t兲 苷 t 2 ln t 24. t共t兲 苷 1 ⫹ et (c) Estimate the domain of t⫺1. (d) Sketch the graph of t⫺1. 25. y 苷 3 x ln x 26. y 苷 共cos x兲 x 27. y 苷 x sinh共x 2 兲 28. xe y 苷 y sin x y g 29. h共␪ 兲 苷 e tan 2␪ 30. y 苷 共arcsin 2x兲 2 31. y 苷 ln sin x ⫺ 2 sin2x 32. y 苷 10 tan ␲ ␪ 1 1 33. y 苷 log 5共1 ⫹ 2x兲 34. y 苷 e cos x ⫹ cos共e x 兲 sx ⫹ 1 共2 ⫺ x兲5 0 1 x 35. y 苷 36. y 苷 sin⫺1共e x 兲 共x ⫹ 3兲7 共x 2 ⫹ 1兲 4 37. y 苷 x tan⫺1共4x兲 38. y 苷 共2x ⫹ 1兲 3共3x ⫺ 1兲 5 3. Suppose f is one-to-one, f 共7兲 苷 3, and f ⬘共7兲 苷 8. Find 39. y 苷 ln共cosh 3x兲 40. y 苷 arctan(arcsin sx ) (a) f ⫺1共3兲 and (b) 共 f ⫺1 兲⬘共3兲. ⫺1 41. y 苷 cosh 共sinh x兲 42. y 苷 x tanh⫺1sx x⫹1 4. Find the inverse function of f 共x兲 苷. 2x ⫹ 1 43. y 苷 cos(e stan 3x ) 5–9 Sketch a rough graph of the function without using a calculator. 44. Show that 冉 冊 ⫺x 5. y 苷 5 ⫺ 1 x 6. y 苷 ⫺e d 共x ⫹ 1兲 2 1 tan⫺1 x ⫹ 14 ln 苷 1 x2 ⫹ 1 共1 ⫹ x兲共1 ⫹ x 2 兲 2 7. y 苷 ⫺ln x 8. y 苷 ln共x ⫺ 1兲 dx 9. y 苷 2 arctan x 45–48 Find f ⬘ in terms of t⬘. 45. f 共x兲 苷 e t共x兲 46. f 共x兲 苷 t共e x 兲 10. Let a ⬎ 1. For large values of x, which of the functions y 苷 x a, y 苷 a x, and y 苷 log a x has the largest values and which has the smallest values? 47. f 共x兲 苷 ln t共x兲 ⱍ ⱍ 48. f 共x兲 苷 t共ln x兲 11–12 Find the exact value of each expression. 49–50 Find f 共n兲共x兲. 11. (a) e 2 ln 3 (b) log 10 25 ⫹ log 10 4 49. f 共x兲 苷 2 x 50. f 共x兲 苷 ln共2x兲 12. (a) ln e ␲ (b) tan(arcsin 12 ) 51. Use mathematical induction to show that if f 共x兲 苷 xe x, 13–16 Solve the equation for x. then f 共n兲共x兲 苷 共x ⫹ n兲e x. 13. (a) e 苷 5 x (b) ln x 苷 2 52. Find an equation of the tangent to the curve y 苷 x ln x at ex 14. (a) e 苷2 (b) tan⫺1x 苷 1 the point 共e, e兲. Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  CHAPTER 5 REVIEW 315 53. At what point on the curve y 苷 关ln共x ⫹ 4兲兴 2 is the tangent 65. lim⫹ ln共sinh x兲 66. lim e⫺x sin x xl0 xl⬁ horizontal? ; 54. If f 共x兲 苷 xe , find f ⬘共x兲. Graph f and f ⬘ on the same sin x screen and comment. 67. lim xl⬁ 1 ⫹ 2x 1 ⫺ 2x 68. lim xl⬁ 冉 冊 1⫹ 4 x x 55. (a) Find an equation of the tangent to the curve y 苷 e that x ex ⫺ 1 tan 4x is parallel to the line x ⫺ 4y 苷 1. 69. lim 70. lim xl0 tan x xl0 x ⫹ sin 2x (b) Find an equation of the tangent to the curve y 苷 e x that passes through the origin. e 4x ⫺ 1 ⫺ 4x e 4x ⫺ 1 ⫺ 4x 71. lim 72. lim xl0 x2 xl⬁ x2 56. The function C共t兲 苷 K共e⫺at ⫺ e⫺bt 兲, where a, b, and K are positive constants and b ⬎ a, is used to model the concen- 73. lim 共x 2 ⫺ x 3 兲e 2x 74. lim⫺ 共x ⫺ ␲兲 csc x x l ⫺⬁ x l␲ tration at time t of a drug injected into the bloodstream. (a) Show that lim t l ⬁ C共t兲 苷 0. (b) Find C⬘共t兲, the rate at which the drug is cleared from circulation. 75. lim⫹ xl1 冉 x x⫺1 ⫺ 1 ln x 冊 76. lim 共tan x兲cos x x l 共␲兾2兲 ⫺ (c) When is this rate equal to 0? 57. A bacteria culture contains 200 cells initially and grows at 77–88 Evaluate the integral. a rate proportional to its size. After half an hour the popula- tion has increased to 360 cells. (a) Find the number of bacteria after t hours. 77. y 冉 冊 1⫺x x 2 dx 78. y 0 4 1 16 ⫹ t 2 dt (b) Find the number of bacteria after 4 hours. 1 5 dr 2 (c) Find the rate of growth after 4 hours. 79. y ye ⫺2y dy 80. y 0 2 1 ⫹ 2r (d) When will the population reach 10,000? 1 ex ␲兾2 cos x 58. Cobalt-60 has a half-life of 5.24 years. 81. y 0 1 ⫹ e 2x dx 82. y 0 1 ⫹ sin 2x dx (a) Find the mass that remains from a 100-mg sample after 20 years. e sx cos共ln x兲 (b) How long would it take for the mass to decay to 1 mg? 83. y sx dx 84. y x dx 59. Let C共t兲 be the concentration of a drug in the bloodstream. x⫹1 csc 2 x As the body eliminates the drug, C共t兲 decreases at a rate 85. y x 2 ⫹ 2x dx 86. y 1 ⫹ cot x dx that is proportional to the amount of the drug that is pres- ent at the time. Thus C⬘共t兲 苷 ⫺kC共t兲, where k is a positive x number called the elimination constant of the drug. 87. y tan x ln共cos x兲 dx 88. y s1 ⫺ x 4 dx (a) If C0 is the concentration at time t 苷 0, find the concentration at time t. (b) If the body eliminates half the drug in 30 hours, how 89. If f 共x兲 苷 ln x ⫹ tan⫺1 x , find 共 f ⫺1兲⬘共␲兾4兲. long does it take to eliminate 90% of the drug? 60. A cup of hot chocolate has temperature 80⬚C in a room 90. What is the area of the largest rectangle in the first kept at 20⬚C. After half an hour the hot chocolate cools quadrant with two sides on the axes and one vertex on the to 60⬚C. curve y 苷 e⫺x? (a) What is the temperature of the chocolate after another half hour? 91. What is the area of the largest triangle in the first quadrant (b) When will the chocolate have cooled to 40⬚C ? with two sides on the axes and the third side tangent to the curve y 苷 e⫺x? 61–77 Evaluate the limit. 92. Show that 61. lim⫹ tan⫺1共1兾x兲 2 62. lim e x⫺x 冑 x l0 xl⬁ x2 ⫹ 1 63. lim⫺ e 2兾共x⫺3兲 64. lim arctan共x 3 ⫺ x兲 cos 兵arctan关sin共arccot x兲兴其 苷 xl3 xl⬁ x2 ⫹ 2 Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  6 TECHNIQUES OF INTEGRATION Because of the Fundamental Theorem of Calculus, we can integrate a function if we know an anti- derivative, that is, an indefinite integral. We summarize here the most important integrals that we have learned so far. x n1 1 yx n dx 苷 n1  C, n 苷 1 y x dx 苷 ln x  C ⱍ ⱍ ax ye x dx 苷 e x  C ya x dx 苷 C ln a y sin x dx 苷 cos x  C y cos x dx 苷 sin x  C y sec x dx 苷 tan x  C 2 y csc x dx 苷 cot x  C 2 y sec x tan x dx 苷 sec x  C y csc x cot x dx 苷 csc x  C y sinh x dx 苷 cosh x  C y cosh x dx 苷 sinh x  C y tan x dx 苷 ln ⱍ sec x ⱍ  C y cot x dx 苷 ln ⱍ sin x ⱍ  C y 1 x2  a2 1 dx 苷 tan1 a x a 冉冊 C y 1 sa 2  x 2 dx 苷 sin1 x a 冉冊  C, a  0 In this chapter we develop techniques for using these basic integration formulas to obtain indefinite integrals of more complicated functions. We learned the most important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 6.1. Then we learn methods that are special to particular classes of functions such as trigonometric functions and rational functions. 6.1 INTEGRATION BY PARTS Every differentiation rule has a corresponding integration rule. For instance, the Sub- stitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for inte- gration by parts. The Product Rule states that if f and t are differentiable functions, then d 关 f 共x兲t共x兲兴 苷 f 共x兲t共x兲  t共x兲 f 共x兲 dx In the notation for indefinite integrals this equation becomes y 关 f 共x兲t共x兲  t共x兲 f 共x兲兴 dx 苷 f 共x兲t共x兲 or y f 共x兲t共x兲 dx  y t共x兲 f 共x兲 dx 苷 f 共x兲t共x兲 Unless otherwise noted, all content on this page is © Cengage Learning. 317 www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 318 CHAPTER 6 TECHNIQUES OF INTEGRATION We can rearrange this equation as 1 y f 共x兲t共x兲 dx 苷 f 共x兲t共x兲  y t共x兲 f 共x兲 dx Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u 苷 f 共x兲 and v 苷 t共x兲. Then the differentials are du 苷 f 共x兲 dx and dv 苷 t共x兲 dx, so, by the Substitution Rule, the formula for inte- gration by parts becomes 2 y u dv 苷 uv  y v du EXAMPLE 1 Find y x sin x dx. SOLUTION USING FORMULA 1 Suppose we choose f 共x兲 苷 x and t共x兲 苷 sin x. Then f 共x兲 苷 1 and t共x兲 苷 cos x. (For t we can choose any antiderivative of t.) Thus, using Formula 1, we have y x sin x dx 苷 f 共x兲t共x兲  y t共x兲 f 共x兲 dx 苷 x 共cos x兲  y 共cos x兲 dx 苷 x cos x  y cos x dx 苷 x cos x  sin x  C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected. SOLUTION USING FORMULA 2 Let It is helpful to use the pattern: u苷x dv 苷 sin x dx u苷䊐 dv 苷 䊐 du 苷 䊐 v苷䊐 Then du 苷 dx v 苷 cos x and so u d√ u √ √ du y x sin x dx 苷 y x sin x dx 苷 x 共cos x兲  y 共cos x兲 dx 苷 x cos x  y cos x dx 苷 x cos x  sin x  C NOTE Our aim in using integration by parts is to obtain a simpler integral than the one we started with. Thus in Example 1 we started with x x sin x dx and expressed it in terms of the simpler integral x cos x dx. If we had chosen u 苷 sin x and dv 苷 x dx, Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  SECTION 6.1 INTEGRATION BY PARTS 319 then du 苷 cos x dx and v 苷 x 2兾2, so integration by parts gives x2 1 y x sin x dx 苷 共sin x兲  yx 2 cos x dx 2 2 Although this is true, x x 2 cos x dx is a more difficult integral than the one we started with. In general, when deciding on a choice for u and dv, we usually try to choose u 苷 f 共x兲 to be a function that becomes simpler when differentiated (or at least not more complicated) as long as dv 苷 t共x兲 dx can be readily integrated to give v. V EXAMPLE 2 Evaluate y ln x dx. SOLUTION Here we don’t have much choice for u and dv. Let u 苷 ln x dv 苷 dx 1 Then du 苷 dx v苷x x Integrating by parts, we get dx y ln x dx 苷 x ln x  y x x It’s customary to write x 1 dx as x dx. 苷 x ln x  y dx Check the answer by differentiating it. 苷 x ln x  x  C Integration by parts is effective in this example because the derivative of the function f 共x兲 苷 ln x is simpler than f. V EXAMPLE 3 Find y t 2 e t dt. SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is unchanged when differentiated or integrated), so we choose u 苷 t2 dv 苷 e t dt Then du 苷 2t dt v 苷 et Integration by parts gives 3 y t e dt 苷 t e 2 t 2 t  2 y te t dt The integral that we obtained, x te t dt, is simpler than the original integral but is still not obvious. Therefore we use integration by parts a second time, this time with u 苷 t and dv 苷 e t dt. Then du 苷 dt, v 苷 e t, and y te dt 苷 te t t  y e t dt 苷 te t  e t  C Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 320 CHAPTER 6 TECHNIQUES OF INTEGRATION Putting this in Equation 3, we get yt e dt 苷 t 2 e t  2 y te t dt 2 t 苷 t 2 e t  2共te t  e t  C兲 苷 t 2 e t  2te t  2e t  C1 where C1 苷 2C V EXAMPLE 4 Evaluate y e x sin x dx. www.stewartcalculus.com SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try An easier method, using complex choosing u 苷 e x and dv 苷 sin x dx anyway. Then du 苷 e x dx and v 苷 cos x, so numbers, is given under Additional integration by parts gives Topics. Click on Complex Numbers and see Exercise 50. 4 ye x sin x dx 苷 e x cos x  y e x cos x dx The integral that we have obtained, x e x cos x dx, is no simpler than the original one, but at least it’s no more difficult. Having had success in the preceding example inte- grating by parts twice, we persevere and integrate by parts again. This time we use u 苷 e x and dv 苷 cos x dx. Then du 苷 e x dx, v 苷 sin x, and 5 ye x cos x dx 苷 e x sin x  y e x sin x dx At first glance, it appears as if we have accomplished nothing because we have Figure 1 illustrates Example 4 by arrived at x e x sin x dx, which is where we started. However, if we put the expression showing the graphs of f 共x兲 苷 e x sin x and F共x兲 苷 12 e x 共sin x  cos x兲. As a for x e x cos x dx from Equation 5 into Equation 4 we get visual check on our work, notice that f 共x兲 苷 0 when F has a maximum or ye x sin x dx 苷 e x cos x  e x sin x  y e x sin x dx minimum. 12 This can be regarded as an equation to be solved for the unknown integral. Adding F x e x sin x dx to both sides, we obtain f 2 y e x sin x dx 苷 e x cos x  e x sin x _3 6 Dividing by 2 and adding the constant of integration, we get _4 FIGURE 1 ye x sin x dx 苷 12 e x 共sin x  cos x兲  C If we combine the formula for integration by parts with the Evaluation Theorem, we can evaluate definite integrals by parts. Evaluating both sides of Formula 1 be- tween a and b, assuming f  and t are continuous, and using the Evaluation Theorem, we obtain b 6 y a ] f 共x兲t共x兲 dx 苷 f 共x兲t共x兲 a  y t共x兲 f 共x兲 dx b a b Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  SECTION 6.1 INTEGRATION BY PARTS 321 1 EXAMPLE 5 Calculate y tan1x dx. 0 SOLUTION Let u 苷 tan1x dv 苷 dx dx Then du 苷 v苷x 1  x2 So Formula 6 gives 1 1 x y 0 tan1x dx 苷 x tan1x 0  y ] 1 0 1  x2 dx 1 x 苷 1 ⴢ tan1 1  0 ⴢ tan1 0  y dx 0 1  x2 Since tan 1 x  0 for x  0 , the integral  1 x in Example 5 can be interpreted as the 苷 y 2 dx 4 0 1  x area of the region shown in Figure 2. y To evaluate this integral we use the substitution t 苷 1  x 2 (since u has another y=tan–!x meaning in this example). Then dt 苷 2x dx, so x dx 苷 12 dt. When x 苷 0, t 苷 1; when x 苷 1, t 苷 2; so 0 x dt ⱍ ⱍ] 1 2 2 dx 苷 2 苷 12 ln t 1 1 x y 0 1  x2 y 1 t 1 苷 12 共ln 2  ln 1兲 苷 12 ln 2 1  1 x  ln 2 FIGURE 2 Therefore y tan1x dx 苷 y 2 dx 苷  0 4 0 1  x 4 2 EXAMPLE 6 Prove the reduction formula Equation 7 is called a reduction 1 n1 formula because the exponent n has been reduced to n  1 and n  2. 7 y sin x dx 苷  n cos x sin n n1 x n y sin n2 x dx where n  2 is an integer. SOLUTION Let u 苷 sin n1x dv 苷 sin x dx Then du 苷 共n  1兲 sin n2x cos x dx v 苷 cos x so integration by parts gives y sin x dx 苷 cos x sin n n1 x  共n  1兲 y sin n2x cos 2x dx Since cos 2x 苷 1  sin 2x, we have y sin x dx 苷 cos x sin n x  共n  1兲 y sin n2x dx  共n  1兲 y sin n x dx n1 Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 322 CHAPTER 6 TECHNIQUES OF INTEGRATION As in Example 4, we solve this equation for the desired integral by taking the last term on the right side to the left side. Thus we have n y sin n x dx 苷 cos x sin n1x  共n  1兲 y sin n2x dx 1 n1 or y sin x dx 苷  n cos x sin n n1 x n y sin n2 x dx The reduction formula 7 is useful because by using it repeatedly we could even- tually express x sin n x dx in terms of x sin x dx (if n is odd) or x 共sin x兲0 dx 苷 x dx (if n is even). 6.1 EXERCISES 1–2 Evaluate the integral using integration by parts with the 1兾2 1 r3 indicated choices of u and dv. 23. y cos 1x dx 24. y dr 0 0 s4  r 2 1. yx 2 ln x dx ; u 苷 ln x, dv 苷 x 2 dx 25. y 2 共ln x兲2 dx 26. y t e s sin共t  s兲 ds 1 0 2. y  cos  d ; u 苷 , dv 苷 cos  d 27–30 First make a substitution and then use integration by parts to evaluate the integral. 3–26 Evaluate the integral. 2 3 t 27. y cos sx dx 28. yte dt 3. y x cos 5x dx 4. y ye 0.2y dy s 4 29. y  3 cos共 2 兲 d 30. y e sx dx s 兾2 1 5. y te 3t dt 6. y 共x  1兲 sin  x dx 31. (a) Use the reduction formula in Example 6 to show that 7. y 共x 2  2x兲 cos x dx 8. yt 2 sin t dt x sin 2x y sin x dx 苷 2  C 9. y ln共2x  1兲 dx 10. yp 5 ln p dp 2 4 (b) Use part (a) and the reduction formula to evaluate 1 x sin 4x dx. 11. y arctan 4t dt 12. y sin x dx 32. (a) Prove the reduction formula 13. y e  sin 3 d 2 14. ye  cos 2 d 1 n1 xe 2x y cos x dx 苷 n cos n n1 x sin x  n y cos n2 x dx 3 t 15. y 共1  2x兲2 dx 16. y t e dt (b) Use part (a) to evaluate x cos 2x dx. 1兾2 1 (c) Use parts (a) and (b) to evaluate x cos 4x dx. 17. y x cos  x dx 18. y 共x 2  1兲ex dx 0 0 33. (a) Use the reduction formula in Example 6 to show that 3 9 ln y 19. y r 3 ln r dr 20. y dy 兾2 n1 兾2 1 4 sy y sin n x dx 苷 y sin n2x dx 0 n 0 1 s3 21. y0 t cosh t dt 22. y 1 arctan共1兾x兲 dx where n  2 is an integer. Unless otherwise noted, all content on this page is © Cengage Learning. ; Graphing calculator or computer required CAS Computer algebra system required 1. Homework Hints at stewartcalculus.com www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  SECTION 6.2 TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS 323 (b) Use part (a) to evaluate x0兾2 sin 3x dx and x0兾2 sin 5x dx. rate r, and the exhaust gases are ejected with constant veloc- (c) Use part (a) to show that, for odd powers of sine, ity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation 兾2 2ⴢ4ⴢ6ⴢ ⴢ 2n y sin 2n1x dx 苷 m  rt 0 3ⴢ5ⴢ7ⴢ ⴢ 共2n  1兲 v共t兲 苷 tt  ve ln m 34. Prove that, for even powers of sine, where t is the acceleration due to gravity and t is not too 兾2 1ⴢ3ⴢ5ⴢ ⴢ 共2n  1兲  large. If t 苷 9.8 m兾s 2, m 苷 30,000 kg, r 苷 160 kg兾s, and y sin 2nx dx 苷 ve 苷 3000 m兾s, find the height of the rocket one minute 0 2ⴢ4ⴢ6ⴢ ⴢ 2n 2 after liftoff. 35–38 Use integration by parts to prove the reduction 43. A particle that moves along a straight line has velocity formula. v共t兲 苷 t 2et meters per second after t seconds. How far will it travel during the first t seconds? 35. y 共ln x兲 n dx 苷 x 共ln x兲n  n y 共ln x兲n1 dx 44. If f 共0兲 苷 t共0兲 苷 0 and f and t are continuous, show that a a n x dx 苷 x ne x  n y x n1e x dx y f 共x兲 t 共x兲 dx 苷 f 共a兲 t共a兲  f 共a兲 t共a兲  y f 共x兲 t共x兲 dx 36. yxe 0 0 tan n1 x 45. Suppose that f 共1兲 苷 2, f 共4兲 苷 7, f 共1兲 苷 5, f 共4兲 苷 3, 37. y tan n x dx 苷  y tan n2 x dx 共n 苷 1兲 and f is continuous. Find the value of x14 x f 共x兲 dx. n1 46. (a) Use integration by parts to show that tan x sec n2x n2 38. y sec n x dx 苷  y sec n2x dx n1 n1 y f 共x兲 dx 苷 x f 共x兲  y x f 共x兲 dx 共n 苷 1兲 (b) If f and t are inverse functions and f  is continuous, prove that 39. Use Exercise 35 to find x 共ln x兲3 dx. b f 共b兲 40. Use Exercise 36 to find x x 4e x dx. y f 共x兲 dx 苷 bf 共b兲  af 共a兲  y t共 y兲 dy a f 共a兲 41. Calculate the average value of f 共x兲 苷 x sec2 x on the interval [Hint: Use part (a) and make the substitution y 苷 f 共x兲.] 关0, 兾4兴. (c) In the case where f and t are positive functions and 42. A rocket accelerates by burning its onboard fuel, so its mass b  a  0, draw a diagram to give a geometric interpre- decreases with time. Suppose the initial mass of the rocket tation of part (b). at liftoff (including its fuel) is m, the fuel is consumed at (d) Use part (b) to evaluate x1e ln x dx. 6.2 TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS In this section we look at integrals involving trigonometric functions and integrals that can be transformed into trigonometric integrals by substitution. TRIGONOMETRIC INTEGRALS Here we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. 3 EXAMPLE 1 Evaluate y cos x dx. SOLUTION Simply substituting u 苷 cos x isn’t helpful, since then du 苷 sin x dx. In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor. Thus here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 324 CHAPTER 6 TECHNIQUES OF INTEGRATION using the identity sin 2x  cos 2x 苷 1: cos 3x 苷 cos 2x ⴢ cos x 苷 共1  sin 2x兲 cos x We can then evaluate the integral by substituting u 苷 sin x, so du 苷 cos x dx and y cos x dx 苷 y cos x ⴢ cos x dx 苷 y 共1  sin x兲 cos x dx 3 2 2 苷 y 共1  u 2 兲 du 苷 u  13 u 3  C 苷 sin x  13 sin 3x  C In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor (and the remainder of the expression in terms of cosine) or only one cosine factor (and the remainder of the expression in terms of sine). The identity sin 2x  cos 2x 苷 1 enables us to convert back and forth between even powers of sine and cosine. V EXAMPLE 2 Find y sin 5x cos 2x dx. SOLUTION We could convert cos 2x to 1  sin 2x, but we would be left with an expression in terms of sin x with no extra cos x factor. Instead, we separate a single Figure 1 shows the graphs of the inte- sine factor and rewrite the remaining sin 4x factor in terms of cos x : grand sin 5x cos 2x in Example 2 and its indefinite integral (with C 苷 0 ). Which sin 5x cos 2x 苷 共sin2x兲2 cos 2x sin x 苷 共1  cos 2x兲2 cos 2x sin x is which? Substituting u 苷 cos x, we have du 苷 sin x dx and so 0.2 y sin x cos x dx 苷 y 共sin x兲 5 2 2 2 cos 2x sin x dx 苷 y 共1  cos 2x兲2 cos 2x sin x dx 苷 y 共1  u 2 兲2 u 2 共du兲 苷 y 共u 2  2u 4  u 6 兲 du _π π 苷 冉 u3 3 2 u5 5  u7 7 冊 C _ 0.2 苷  13 cos 3x  25 cos 5x  17 cos 7x  C FIGURE 1 In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix A): sin 2x 苷 12 共1  cos 2x兲 and cos 2x 苷 12 共1  cos 2x兲  Example 3 shows that the area of the V EXAMPLE 3 Evaluate y sin 2x dx. region shown in Figure 2 is 兾2. 0 1.5 SOLUTION If we write sin 2x 苷 1  cos 2x, the integral is no simpler to evaluate. y=sin@ x Using the half-angle formula for sin 2x, however, we have    y 0 sin 2x dx 苷 12 y 0 共1  cos 2x兲 dx 苷 [ (x  1 2 1 2 sin 2x) ] 0 0 π 苷 12 (  12 sin 2)  12 (0  12 sin 0) 苷 12  Notice that we mentally made the substitution u 苷 2x when integrating cos 2x. _0.5 Another method for evaluating this integral was given in Exercise 31 in Section 6.1. FIGURE 2 Unless otherwise noted, all content on this page is © Cengage Learning. www.MathSchoolinternational.com Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.  SECTION 6.2 TRIGONOMETRIC INTEGRALS AND SUBSTITUTIONS 325 4 EXAMPLE 4 Find y sin x dx. SOLUTION We could evaluate this integral using the reduction formula for x sin n x dx (Equation 6.1.7) together with Example 3 (as in Exercise 31 in Section 6.1), but a better method is to write sin 4x 苷 共sin 2x兲2 and use a half-angle formula: y sin x dx 苷 y 共sin x兲 4 2 2