4.3 - HITI'EHH SECTION 4.3 EXERCISES Review Questiens 1...

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Unformatted text preview: HITI'EHH SECTION 4.3 EXERCISES Review Questiens 1. State the half-angle identities used te integrate sin2 .1: and ces2 1r. 2 State the three Pvthagerean identities. 3. Describe the methed used te integrate sin3 x. 4 Describe the methed used te integrate si 4.3 Trigenemetric Integrals 251 The apparent difference 1n2 the twe selutiens given here is receneiled by using the identity 1 + tanE .r :- sec2 .1: te transterm the seeend result inte the first the enly difference being an additive constant which is part ef C. b. In this case, we write the even pewer ef tan .1: in terms ef sec 1:: ftaflflxsecxir : /(sec2x "r ”3136de tan2 3;... "' sec:i _:1: -l = fsec3r1ir— jaecrdx reductien fermttia 4 WM 1 l.~ L“ Esecxtans: + E/secs‘dx — fsccxrix l 1 . .. - 1.. = —secxtan:r — —ln lseca: + tans] + C. Add imammtflviflli“ 2 2 use Theerern 4.1. Refleree’ Exercises 3 3—44 *1 Table 4. 3 summarizes the metheds used te integrate f tan .1: sec ”adv. Analegeus techniques are used fer fcet’”:r csc” .1: nix. Table 4. 3“ f tan’” .1: sec .1: din Strategy it WEI} Split eff sec2 .11., rewrite the remaining even pewer ef sec :1: in terms ef tan .11. and use 1.1 = tan Jr. 111 edd Split eff sec .1: tan 1:, rewrite the remaining even pewer et‘ tan x in terms ef sec 1:, and use 1.1. = see I. as even and 11 edd Rewrite the even pewer ef tan .1: in terms ef sec .1: te preduce a peivnemial in sec 1:; apply redueti en fermula 4 te each term. .-.-\_-.u- -._-1. — —---\—--- —-------- --. -. -.-- — .1. . .n. n...'\_-. -. ___.. ___. ... ..,...._..-.,.___,..,_._. _..____ ___ . _ __ _ --.-- ,_,,\_,,|_ _.__._______. «.n- __._____ 15—24. Integrals ef sins: and ces .1: Eveinete the feilewieg integrals. 15. / 51112111111111]: 16.18/srn31cn151ax sin3 .r cesE .1: air sin2 ti ces5 6 d6 Hi 11 :rces” .1: ferns even and 11 edd. 11/13. xv sin :1: air 2!}. 18/ sin3 ti ces‘2 ti d9 5. What is a reductien fermuia‘? —2 —3£2 3 6. Hew weuld yen evaluate fceszx sin3 .1: air? 21 [51135 .t‘ces ad): 22 fflfl .rces .1: if 7. Hew weuld e11 evaluate tanms: sec2 .1: ctr? )1 f 23. / sin2 .1: ces‘q‘ .1: r111: 24. [sin 3i'.1:ces3'*’3xrbr 8. Hew weuld yeu evaluate f sec12 .1: tans: ctr? _ _ 25—30. Integrals at tan :1: er cetx Eveiuete thefetiewing integrate Basrc Skills 2 _ 4 9-14. Integrals ef sin .1: er cesa: Emitters rhsfeliewing integreis. 25' / tan .1: dx 26' f 6 SEC I six 9. fsingxrix 12. I/ces':L 25' d9 10. fsin3 .1: six 13. f 31115 111.1 s . 11. fees mix 21 / cetfl’xdx 2g / tan3ddfi' 3 14- fCDS 20.11133 29' [2013116 I (ix. 30' fcfltfi 3xflix 252 Clear-tea 4 - INTEGRATIDN TECHNIQUES 31—44. Integrals invelving tans: and sees: Evaluate thefeilewt‘rtg integrals. 31. /10tan9.tsee2xdx 32. / tangsrsee‘ixeix 34. [Vtanesee‘iseir 3a fails}; tan a 33. ftans: seeiettt 35. /ten3 are}: 3?. fees2 .t tanUri .‘t: tit 39. / “Edie: 38. fees—2 .t tan3 a: tit 40. fesemeeetrttx set2 .1: errf4 41. / see‘1 6 d6 42. ftan5 6' see'i 6 tit? e “37.63 affai- 43. / eeti e as 44. / amt e see: a as “trfifi U Further Expleratiens 45. Explain why er why net Determine whether the fellewing state- ments are true and give an explanatien er eeunterettatnple. . . . . “FT ' a. if at 1s a pestttve Integer. then f0 eesim+1 .1: dx i t}. h. if at is a pesitive integer? then farsinmr air 2 0. 46—47. Integrals ef cuts: and esex 46. Use a ehange ef variables te preve that feetaeit = In Isinttl + C. 47. Preve that feseedx =~= -~ln [eser + eetrl + C. (Hist: See the preef ef Theerern 4.1.) 43. Cemparing areas The regien R-l is beunded by the graph ef _v = tan .1: and the .easis en the interval [0, err/3 ]. The regien R1 is heuuded by the graph ef y I see a: and the .‘t-attis en the interval [0. er/ 6]. Whieh re gien has the greater area? 49. Regine between enrves Find the area ef the regien beuuded by the graphs efv = tans: and y = sees: en the interval [0, er/4]. 50—57. Additienal integrals Eveittrtte the felletvtttg integrals. " ”£2 + 3 2 see4(ln6‘) 50. stein (.1: )tix 51. Weft? e 9 affid / 9’ 1766 s1ny will 54. / tani .t: see2 a: tit -rrf4 will 53. / 'v’seeitiI — 1ee —rrf3 GIT 55. /(1wees2s)3iidt e 56.. fesemseetirdx 57. /e" see (e‘ + Uri-x 53—61. Square rents Evetttttte the feiiewirtg integrate. rrffil- will. 58. f V1— + ees 4x tit 59. ‘V1 — ees Zara's: 'n'j'él f} wffi . “trifli- 60. / 'Vl - eeschtx 61. f (1 + ees 4r)3”2ttr e e 64. A sine rednetien fermnla Use integratiee by parts te ebtain a 62. Sine feetball Find the velunte ef the seiid generated when the regien hennded by the graph ef y = sin a and the .t-asis en the interval [0. er] is revelved abeut the taxis. 63. Are length Find the length ef the enrve y = he (see I). fer Dixie/4. reduntien fermuia fer pesitive integers a: /sin”tteix = *---sin“"*1 .tteess: + (rt — l)/sin”_2aees2rdx. Then use an identity te ehtein the redeetien fermela . __1 . sin” reesr. a—1 _ H fsin”xttx = —w + fem” Emir. a a Use this reductien fertnula te evaluate f sin‘fi .r six. 65. A tangent rednetien fertnula Preve that fer pesitive integers rt 5e 1, Use the fermela te evaluate jg 66. A seeant rednetien fermula Preve that fer pesitive integers rt # 1. eel (Hint: Integrate by parts with tt = see .1: and dv = seega tit.) Applieatiens 67—71. Integrals ef the fern] f sin ms: ees as: six Use the feliewt'ttg three identities reevaluate the given integrate. sin tax sinrtr w (ees ((re _.. a).t) -- ees ((m + a).:t)) sinrttreesrta = (sinflrrt — rt).:t) + sin (Us + rt).t)) l. 2 l 2 l. 2 eesetaees as: = (ees ((ttt — tilt) + CUE (OH + 4%)) 67. / sin 3.x: ees 7's: tit: 63. / sin 5t: sin 7r tit 69. / sin 33: sin 2t: six 70. fees .t see 23: tit 71. Freve the feitewing erthegenality reiatiens (whieh are need he generate Fearier series). Assume m and rt are integers with m5ét’t. 'iIT E/Siflmfilflflxdxfifl e s h. feeseeteesttacix=0 e 7.. e. fsinetreeseadxwfljerlet+e even it ails-i Trigonometric Substitutions 253 72. Mercator map projection The Mercator map projection was c. Ptove that jgfirs'in2 (as) (it has the same value for all positive proposed by the Flemish geographer Gerardus Mercator _ . integers n. (1512“1594l- The “gimme fact“ “i the MEI-"33ml“ We 35 e a. Does the conclusion of part to) hold if sine is replaced with function of the latitude d is given by the function cosine? a e. Repeat parts (a), (b), and (c) with sin2 i: replaced with sin4 it. Comment on your observations. G t? = sec tit. - ( J ff, I f. Challenge problem: Show that fer at w 1, 2, 3, . . . , Graph G, for 0 5. 9 e1 wfl. (See the Guided Project Mercator . . . . .. ii,” “if , 1*3r5n-(2m—1) pTGJECHEZ-‘PES fer a derivation of thls integral.) sin in: (is 2 cos ”1 it is = or - w. ,, ,, 2 - s - a - - - 2m Additional Exercises 73. Exploring powers of sine and cosine QU'CK EHECK AN5WERS . . . i - _ . - _ 3. Graph the functions ffir) = 31112.1: and fight) = am2 it on the 1. geesqx — CD83: *1“ C 2.. Write fs1n3 it cos3 .1: air —-— interval [0, or]. Find the area under these curves on [0, w]. 3 .1: sin x dx : f (1 _ €032 x) cos3 x sin x d}; 1}. Graph afew more of the functions f,, (it) = Slflgffltfli‘l thciu- Th 11 b , , _ O b , b _ _ terval [0, er], where a: is a positive integer, Find the area under E“ use I e 3” “1““an ” " C95 35- 1' egm Y WHERE these curves on [0, or]. Comment on your observations. [sin3 it cos3 it air = f sin3 .75? cos2 it cos x dx. 4i f sin2 it cos 4-4Trig,Geometrissabsinthe __ We wrote the arc length integral fer the segment of the parabola y : s2 on the interval [0, 2] as 2 2 [V1+ ~412er = fl‘x/i + nice. o o At the time, we did net have the analytical methods needed to evaluate this integral. The difficulty with fa: V l + 43:2 tit is that the square root of a sum (or difference) of two squares is net easily simplified. 0n the ether hand, the square root of a product of two squares is easily simplified: ‘s/AEB2 3 [AB [. If we could somehow replace l + 4.322 with a product of squares, this integral might be easier to evaluate. The goal of this section is to introduce techniques that transform sums of squares e2 + x2 (and the difference of 3* To magi-stand how a sum flfflqums squares d2 — s2 and r2 — r12) into products of squares. is rewritten as a product of squares, Integrals similar to the arc length integral for the parabola arise in many different think of the Pythagorean Theorem: situations. For examme, electrostatic, magnetic, and gravitational forces obey an inverse e3 + e2 : 1:2,.51. rearrangement efthie square law (their strength is proportional to 1 /r2, where r is a distance). Computing these theorem leads to the standard substitution . . _ _ . six ix fflf intggfalg [mpg-lying thfi difffiffi'nflfi {3f fUI'CB flfilds 111 tWfl dImEHSIUHS lfifldfl ED Intflgrfllg SUUh as W 01' W- squares or2 — II. The term Vet: .. it: , , .- 2 + 2 2x 2'51 . is thfl ,Ength of Um, Etch: of a right It turns out that integrals containin g the terms at .... . I [or at. — rt .’ where or 1s a constant, triangle whose hypfltflmse has length can be Simpllfled usmg somewhat unexpected substitutions involving trigonometric fence 5, and whflfig 0mg, side, has langm ,,_ tions. The new integrals produced by these substitutions are often trigonometric integrals Labeling one acute angle El, we see that [if that variety Stfldlfld in the preceding SECEiDI’l. s = rs sin 3. 2 Integrals Involving e2 -- 1: Suppose you are faced with an integral whose integrand contains the term a2 - .152, where a I at is a positive constant. Observe what happens when .r is replaced with rs sin 6: I tag — x2 I d2 *— (tl sin 9)? Replace :swith ct sint}. I = e2 — :32 sin2 9 Simplify. 'v’ {12 "‘ IE 2 . 2 = or (1 - sin 8) Factor. , ___________________ a l I I I g 2 Li__5_iiiif.t at titles2 6'. l. — sinflh' Z cos‘fltil ...
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