4.1 - 236 CHAPTER 4 INTEaaarroN TECHNIQUES The integral is...

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Unformatted text preview: 236 CHAPTER 4 * INTEaaarroN TECHNIQUES The integral is evaluated as follows: a»: 1 1 - cosr _ m : w*fldx Multrplv by l. 1-!- coax 1+ rose: 1— cosx l — cos): . . : winds Simpitfy. 1 .. cos x /'1 -" cos}; _ 3' . a : Tip; 1— cos a: = state: am a: 1 cost 2 f _ 2 d]: — f . 2 air Split up the fraction. _ sin a: sin a ll sin a sin a: 1 case: [cscfixdx _. foscrcotartir csca 2 ,cota: *2 “cot a: + csc .r. -l- C. Integrate using Table 4.1. Reiared Exercises 37—40 *( The techniques illustrated in this section are designed to transform or simplify an integrand before you apply a specific method. In fact, these ideas may help you recognize the best method to use. Keep them in mind as you learn new integration methods and improve your integration skills. SECTION 4.1 EXERCISES Review Questions 15—22. Subtle substitutions Evaiaare the foiiawiag integrate- 1. What change of variables would you use for the integral Ex 6 .322: d ".6 o — I for —- 7.3;) an 15 f3. .. genre 1 £2: .. 4.9—: r 2. Before integrating, how would you rewrite the integrand of 1 . 2 2 e _|_ a ? 9 ln s: . sin—ix fat 2) fix 17. f ( )ar 18. f 5 ex 3. What trigonometric identity is useful in evaluating fainzacix‘? - 1 it: cos .x s a . . . . . a: — 2.3: + 4 4 . + 4. Describe a first step in integrating /—--x-:-1-m ex. 19. /Ci1: .1: ex 2“ M ex 51-11-15 o Ve3+a2+4 . . . . . 10 air - till! 5. Describe a first step in 1ntegratu1g W air. 21. _ ' - 22‘ / . . .r —4.r+5 .r'l+1 y—1+},—s .rm— 23cl1+liltr2 +1 1 23—23. Splitting fractions Evaiaare Ihefoiiowing iategrais. 3r 6. Describe a first step in integrating / 9. 5/2 # iii, 23. f “I: 2 ab: 24. / “1—31,: ax Basic Skills it + ‘4 4 '3‘ 7—14. Substitution Review Evaiaare rite faiiawiag iategrais. _ _2 . star + taut 4 + e I six . 25. —-—-——2 a: 26. Tit 7- fme 3. / (9a: - 2)'"3 ex G05 i e (3 '- 5r.) 2 — 3a: 3.1: + 1 27. -—-——~—ax . fiafii 2 2 . V — Vs — a. / sin (2a m E) ex 10. fame: I x “1' CI 29—32. Division with rational functions Evaiaare the following fl air iategrais. 1 2x 11./‘11 (ix 12. w 3+2 4.1224-2 .1? -—5 4 "- a: 29. six 30. ii": a: + 4 2 it —~ 1 gr ext/EH 13/ ex 14 f3 riy iii-.2 6—14 E. + 1 ' e 32 ex V3; 31 f r + 1 [I2 + 4 . 4.1 Basic Approaches 237 33-66. Completing the square Evaiaate the following integrate. _ 56. Different methods air 2 I a. Evaluate feet a csc2 a: tit using the substitution at = cot x. 33' /m 34* A mi“ in. Evaluate j cotecsc2 rat using the substitution a = csc .r. c. Reconcile the resuits in parts (a) and (b). air 57. Different methods 2 at a. Evaluate / tixusing the substitution a = at + 1. d6 iii/WM 36. fey-e3?— w27_69—92 x+2x+1 37—40.. Multiply by 1 Evaluate thefette wing integrals. .1: + 1 as 1 - i: xi . _ .. . _ _ _ 37. m 33. We]: _ h. Evaluate a: + 1 {it after f1rstperfortn1nglcng division sin ._ a: on the integrand. 39 '51" 40 {:53 e. Reconcile the results in parts (a) and (b). 1' secs- 1. i 1 "cacti 53. Different substitutions ' eh: Further Explorations a. Show that fmg = sin—1&3: — l) + Cusing either 41. Enplain why or why not Determine whether the following state- V X —‘ I ' ments are true and give an explanation or counterexample. ' H _ _ _ l aw2.t- leta—x——. 3 3 a 2 a. a tit = e-gttt + —d.r. tix _ _ a + 4 it“ 4 b. Show that —----—2- = 2 sin_-1 a: + C usinga = Vi. 11. Long division simplifies the evaluation of the integral “v” x ‘“ x a . we; c. Prove the identity 2 sin‘1 .1: -- sin-[(235 —- i) = 3. 3x4 + x 2 I: air = 1n lsina: + 1 | + C I (Source: The Coiiege Mathemties Jettraat 32, 5, Nov 2001) ‘ sins: + l. ‘ _ * iii an i Applications d' at h in E + C“ 59. Area of a region between curves Find the area of the region . a .t 1 42—54. Miscellaneous integrals Use the appreaches discussed in this PDUfldfld by the curves )2 _ x3 _ 3.x: and 3'" __ .113 _ 3.x D” the seetien to evaluate thefeiiewing integrals. mmfifll [2r 4} a dx {:1 x {it}. Area of a region between curves Find the area of the entire 42. m 43. / Tw tit . a3 8x 4 1 # Vii _1x + 2x + 2 region bounded by the curves y = 2 and y = a x + i .1: + l 44 ft m at}; 45 fsiu x sin 22: [ix 61. Volumes of solids Consider the region R bounded by the graph of ' .3. ‘ f{..t) = we? + 1 and the it—Eixli} on the interval [a a]. a. Find the volume of the solid formed when R is revolved about we . . 46. / V1 + casaxax 47. L thee-em- i} xii? + 1-329 1'). Find the volume of the solid formed when R is revolved about the yards. 1 43_ / d_;; 49‘ / 4:13"... ix 62. Volumes of solids Consider the region R bounded by the graph of D 4 -— v3 3:2 + 6.131? + 13 1 flat) = + 2 and the x—aais on the interval [0, 3]. .1: 11'er .1: 50. / 3V1 + sin 2): (1x 51. fax—EM (ix a. Find the volume of the solid fanned when R is revolved about {3' 3 + 253: + 1 the .t-aJ-tis. 11. Find the volume of the solid formed when R is revolved about 1:13 3 2 h . 52. V l 11 cos dear 53. / e—ZH—H—gx t e yearns. II] 1 .1? + 2x + 1 63. Are length Find the length of the curve )2 = am flfl the interval [07. 1]. (Hint: Write the arc length integral and let a 2 54. fmtfl “2 "T" 1 + (3)3 fl.) gs+3s+33+l ' tit-l. Surface area Find the area of the surface generated when the re- 55. Different substitutions gion bounded by the graph of y = at + if" on the interval a. Evaluate ftan a: sec,3 :1: air using the substitution a w tan .t. [0; III 2] is revolved about the #398115. b. Evaluate f tan a secirait using the substitution at w sec 3:. c. Reconciie the results in parts (a) and (b). 238 Cnanraa 4 - Inraeaarren Tecnmeens as. Surface area Let fix) = Va + 1. Find the area ef the surface QU‘C" CHECK ANSWERS generated when the regien beunded by the graph eff en the 1. Let it == 6 + 5x. 2. Write the integrand its 133’; 2 + 2.7—1. interval [0, l] is revelved abeut the r—aaisi * _ . _ . 2 3. Use leng divisien te write the integrand as l + , 66. Skydiving A skydiver in free fall subject in gravitatienal I *" I acceleratien and air resistance has a velecity given by 4. (r + 3)2 + 7 all a“ - 1 , . . v(r) : v7( m. + 1 ) where vT is the terminal velecity and e 3e 0 E is a physical censtant. Find the distance that the skydiver falls 1‘ aftertsecends,which is d(r) = / v(y)dy. e 4-2~'&”t,egra“°”_byliafirt5W The Substitutien Rule (Sectien 3.5) arises when we reverse the Chain Rule fer derivatives. In this sectien, we empley a similar strategy and reverse the Prcduct Rule fer derivatives. The result is an integratien technique called integrerien by parts. Te illustrate the imper- tance ef integratien by parts, censider the indefinite integrals [exdx=ex+C and [nexdx=? The first integral is an elementary integral that we have alreadyr enceuntered, The secend integral is enly slightly different—and yet: the appearance ef the preduct are“: in the inte- grand makes this integral (at the mement) impessible te evaluate. Integratien by parts is ideally suited fer evaluating integrals ef precincts ef functiens. Integration by Parts fer Indefinite Integrals Given twe differentiable functiens a and v, the Preduct Rule states that granite) -—— mean) + were By integrating beth sides, we can write this rule in terms ef an indefinite integral: a(x)v(x) = f(tt’(r)v(;r] + a(r)v’(:t))rix. Rearranging this eapressien in the ferm /n('r)w£r = n(x)v(x) — /v(r)n’(r) ab: (iv cits leads te the basic relatienship fer integrnrien by parts. It is expressed cempactly by eating that dis = a ’ (r) ab: and rip = v ’ (x) doc. Suppressing the independent variable x, we have fadv=ave~ [vein The integral f a dv is viewed as the given integral, and we use integratien by parts te ex- press it in terms ef a new integral fr rain. The technique is successful if the new integral can be evaluated. Snppese that a and v are differentiable functiens. Then [adv = av - fvdn. Integratien by Parts ' | | | l ...
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