hw-6-sec-8.2-solns

hw-6-sec-8.2-solns - 622 CHAPTER 8 TECHNIQUES OF...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 622 CHAPTER 8 TECHNIQUES OF INTEGRATION I2n 12,1 Now from part (b), the left term is equal to i i, so the expression becomes i i S 121:: < 1. Now 1 n "lingo : "131201 2 1,50 by the Squeeze Theorem, 512:1 = 1. (d) We substitute the results from Exercises 45 and 46 into the result from part (c): 2.4.6....4(2n) . 12"+1 _ 3.5‘7....,(2n+1) _ 2.4.6.....(2n) 2.4.6.....(2n) 2 1: 1 z 1 ————————= 1 ____— _ "in; 12" 711330 1-3~5---~(2n—1): 71L"; 3~5~7~--~(2n+1) 1-3-5--~~(2n—1)<7r>] 2.4.6.....(2n) 2 _hm2.2 34.9.9 _2n_.2 [reananems] _n—>ool 3 3 5 5 7 2n—12n-I-1 77 g (e) The area of the lath rectangle is k4 At the 271th step, the area is increased from 2n — 1 to 271 by multiplying the width by 2:” 1, and at the (2n + 1)th step, the area is increased from 2n to 271 + 1 by multiplying the height by 272+ 1 — n . These 2 n and 1 2n 2n— 1 (2n+1)/(2n) : 2n-i-1 two steps multiply the ratio of width to height by respectively. So, by part ((1), the limitin ratioisg-E‘é 9 ~—£ g 1 3 3 5 5 7 ‘2' .yr .\ 8.2 f Trigonometric Integrals The symbols ; and é indicate the use of the substitutions {u = sin as, du : cos a: date} and {u 2 cos m, du = — sinm dm}, respectively. 1. fsin3 av cos2 mda: : fsin2 :1: cos2 w sinxdm = f(1— cos2 x) cos2 ac sinccdx é f(1 — u2)u2(—du) =f(u2—1)u2du:f(u4—u2)du=%u5—%u3+C:écos5x—écosgm—I—C fijsinew cossmdzt : fsinsa: c052cc cosacdx : fsin6m(1 — sin2 x) cosmdm ; fu6(1 — u2)du :f(u6—u8)du:%u7—%u9+0:%sin7m—%sin9x+0 I3./f3"/4 sin5sc cossmdcc : f37r/451n5 a: cos2x coszvdx 2 [MM sin5$(1 — sin2 m)cosxdx ; IVE/21150 — u2)du v 7r/2 7r/2 7r 2 — mes—MMM[éuG—AuBJfi/2—(% ya”) (3 e): 11 fl 1 8 1 384 7 4.\If7r/2 cos5 mdm = f07r/2(cos2 3:)2 cosx dm = far/20 w sin2 1:)2 cosx dw ; f01(1 — u2)2 du :f01(1—2u2+u4)du= [u—gu3+§u5];: (1—§+§)—0=38E SECTION 8.2 TRlGONOMETRIC INTEGRALS 5. Let y = 7m, so dy : 7rdac and fsin2(7rw) c055(7ra:) dm 2 ifsin2 y cos5 ydy : ifsin2 y cos4 y cosydy = % fsin2y(1 — sin2 y)2 cosydy é ifual —u2)2 du = —71;I(u2 — 2u4 +126) du __1132517 _1'-3 2-5 1-7 —;(§u —3u +7u)+C—§;sm y—s—Wsm y+fism y+C 11 El; sin3(7ra:) — g2; sin5 (7m) + 7—17; sin7(7rx) + C' 1 2x5 diL‘ and d2: : 2y dy. Then 6. Let y = x/E, so that dy = - 3 - 3 dez/S—XZ—yflydy)=2/sin3ydy22/sin2ysinydyz2/(1—cos2y)sinydy a: é2f(1—u2)(—du):2f(u2#1)du=2(%u3—u)+0:ig-Cos3y—2cosy—kC = 2cos3(\/E)—2cosx/E+C 3 f0"/2cos2 6d6 : 0"” 51+ c0520)d9 [half-angle identity] =%[6+%sin29]3/2=%[(%+0)—(0+0)] :§ Kié./IJOW/2sin2(20)d6 = 5/251 — cos49)d6 : fie — isin40]3/2 2 gm — 0) — (0 — 0)] = g 9. f0" sin4(3t) dt = H [sin2(3t)]2 dt = H [a1 — cos 615)]2 dt = g f0"(1 — 2cos6t + cos26t)dt : if; [1w2cos6t+%(1+c0512t)]dt= fifow —2COSGt+%COS].2t) dt %[%t—§sin6t+2i45in12t]g*HQ} 0+0) (0 0:0)]~3§ f0"(cos2 9)3 d0 2 f0" [1(1 + cos 26)]3 d6 = g [5(1 + 3cos 29 + 3c0s2 26 + cos?’ 26) d0 10. f0" 00566d9 2 i I I I i 1 : %[6 + % sin 26];r + é f0” + cos 40)] d0 —I— § f0” [(1 — sin2 29) cos 26] d0 2 g« + 116 [0 + $11149]; + g- f0°(1 — u2)(%du) [u : sin20, du = 2cos29 d6] 7r 37r _5_7r —§+16+0’1 1i9f(1-|—cos€)2do9 =f(1+2cos€+00526)d€=0+2sin9+%f(1+cos26)d6 20+2sin0+%0+isin20+C: t39—k2sin6—l—isin29—l—C . \2 ,fi12‘Letuzm,dvzcoszzL‘da: => duzda:,'u:fcos2md$=f%(1+cos2x)dx: éx+isin2aaso J / fIECOS2IIIdCL‘ = $617+ isin2$) — f gas—t isian) dw : %$2 + ixsin2m— ix2 + §00s233+C : inc? + ixsian—I— ficos2m+C 13. fJ/2sin2mC032wdxz OW/2i(4sin2xcos2m)da:: OW/QflZSina: cosm)2d$=fif0"/2sin22xdm 2% Ofl/2%(1-cos4m)dx:§ ofl/2(1—C0s4w)dm=§[a:—isin4mH/2zéfi) :1—"6 624 D CHAPTERS TECHNIQUES OF INTEGRATION 14. I; sin2 t cos4 tdt = i [0" (4 sinzt cos2 t) cos2 tdt = i 0"(2 sint cos t)2 %(1 + 00521:) dt : % 0"(sin 2t)2(1 + cos 2t) dt : % f0” (sin2 2t + sin2 2t cos 2t) dt = éfoflsin2 2tdt+§fgr sin2 2t cos2tdt =§ c:r%(1 *cos4t)dt+%[% - lsin3 2t];r =—11—6[t—isin4t];+%(0—0)=%g[(7r—0)—0]=fi = {gal/R45 — 18a2 + 5u4) + C = émm — 18sin2 a + 5sin4 a) + C 16. Let u = sin 6. Then du = cos0d0 and fcosQ c0s5(sin9) d0 = fcos5 udu = f(cos2 u)2 cosudu = f(1 — sin2 u)2 cosudu = [(1 # 25in2u+sin4u)cosudu = I Now let a: = sin u. Then dis 2 cosudu and I=f(1~2:c2+av4)dw=x—%m3+%m5+C=sinu—§sin3u+%sin5u+0 = sin(sin 0) — gsin3 (sin 0) + % sin5 (sin 0) + C ‘ri '3 __ 2 - _ {9/c0s2xtan3xdw2/Sm xdwé/(l u du) =/[—l+u] du \ coszz u u =-1n|u|+%u2+C=%cosza:—ln|cosw|+C 5 5 4 1_ - 29 2 18. /cot563in40d0=/COS5Z Sin40d02/COS 0d02/COS 60 cosOd0=/( sm ) cos0d0 sin sin 9 sin sin 6 1_ 22 _ 2 4 2U +u du:/<l—2u+u3)du u u u =lnu —u2+lu4+C=1nsinOI—sin20+lsin49+C 4 4 .A H\ ' O “ 19' )/cos:v+sm2x d$:/cosm+251ncc cosmdm:fcosmd$+/2cosmdmé/ldu+25inx sin :8 sin :1: sin a: u /' =lnlu|+2sin$+C:ln|sinxl+2sinx+C 0r: Use the formula I cot a: due : 1n lsinccl + C. {Obfcogm sin2xdcsz2fcossa: sinmdwé —2fu3du= —%u4+C——— —%cos4a:+C 21. Letu :tanm,du: sec2mdm. Then fsec2a: tanmda: = fudu: éuZ +0 2 %tan2;v+C. 0r: Letv =secm,dv :secx tanwdz. Thenfsec2$ tanmda: = fvdv 2%122 +C = §seczw+CC 22. fJ/zsec4(t/2)dt= 5/4sec4a:(2dx) [xzt/2,dm:%dt] =2fJ/4sec2$(1+tan2m)d:c =2f01(1+u2)du [uztanagduzsecgmdm] :2[u+%u3];=2(1+—1§):§ SECTION 8.2 TRIGONOMETRIC INTEGRALS 23. ftanzxdmzf(sec2mw1)dx:tanm_w+c {’24. )f(tan2 w + tan4 10113: = [tan2 :0 (1 + tam2 21:)d1t = ftan2 m sec2 asda: = fu2 du [u = tanx, du : sec2 xdx] \J :%u3+C=%tan3x+C {'25) sec6 tdt = fSBC4 t - sec2 tdt = f(’ca.n2 t + 1)2 sec2 tdt = f(u2 + 1)2 du [u = tan t, du 2 sec2 tdt] :f(u4+2u2+1)du=éus+§u3+u+C=étan5t+gtan3t+tant+0 526:: "/4 sec4 0 tan4 Qdf) = "/4 tan2 «9 + 1 tan‘1 0 sec2 9d9 : 1 U2 +1 1L4 du u = tan 0, du 2 sec2 0d0 \J’ 0 0 0 = f01(u6 +“4)d“ =[%“7 + éuslé = % + % = i: ,. WOW/3 tan5 w sec4 a: clan = [OW/3 tan5 a: (tan2 :10 + 1) sec2 3: dm 2 fofi 11,5012 + 1) du [u = tan ac, du : $802 ac dav] :If<u7+u5>du=[§u8+éu61 =%+%=%+%=%+%=% Alternate solution: 3 3 3 3 3 f"/ tan5 a: sec4 aids: = fad tam4 a: sec 2: secz tanzcdx : fow/ (8802 :c — 1)2 sec 30 seczc tanmdm 0 = f12(u2 — 1)2u3 du [11. = secw, du = secx tanxdw] = f12(u4 — 2112 + 1)u3 du =ff(u7—2u5+u3)du=[éu8—%u6+iu4]f=(3 —%+4>—<—:~§+i>=1—;1 28. ftan3(2:c) secs(2a:) d3: = ftan2(2x) sec4(2a:) - sec(2ac) tan(2x) dx = f(u2 — 1)u4(% du) [u = sec(2m), du = 25ec(2;c) tan(2m) dLL‘] %f(u6 — u4)du : T121417 — 1-151? + C = fi sec7(2w) — 315 sec5(2$) + C H 29. ftan3zzc secmdav = ftanzx secs: tanmdac = f(sec2:c ~ 1) seem tanwdw 1 3 =f(u2#1)du [u=secac,du=seC$tanwd$] :3u3—u+C=%seC$—secx+0 30. fofl/s tan5 a: sec6 xdx = fofl/s tan5 1: sec4 :3 sec2 33 do: : [OW/3 tan5 a: (1 + tan2 ac)2 sec2 ac dz = 0‘5 u5(1 + u2)2 du [u = tanzc, du 2 sec2 xdm] : fo‘fi u5(1 + 2u2 + u4) du =ff<u5+2u7+u9>du= [éu6+iu8+%oul°]f= 261%: = Alternate solution: 1? 3 3 3 f0/ tam5 ac sec6 atdx : f0"/ tan4 an sec5 w secac tanmda: = fow/ (sec2 at — 1)2 sec5 ac secs: tanwdm : ffluz “ 1)2U5 du [u : secrc, du : seem tanmdm] : f12(u4 _ 2u2 + 1)u5 du : ff<u9 _ 2u7 + u5)du =[%ul°-%u8+%u“]i= %Z-64+¥)—(z%-i+%)=9-§% 31- ftansccd$ = [(seczm * 1)2 tanmdw = fsec4zc tanwdzc ~ 2fseczx tanardm + ftanmdx = fsec3a: secac tanwdw—2ftanm sec2mdx+ftanxdm = isec‘lx —tan2m+ln|secacl + C [or fisec4m — seczx+ln|secx| + C] 625 626 CHAPTER 8 TECHNIQUES OF INTEGRATION “:1 32. ftan6 ay dy 2 ftan4 ay (sec2 ay — 1) dy 2 ftan4 ay sec2 ay dy — ftan4 ay dy g 2 5+1 tan5 ay — ftam2 ay (sec2 ay — 1) dy 2 % tan5 ay — ftan2 ay sec2 ay dy + f(sec2 ay - 1) dy g zétansay~fitan3ay+étanay—y+0 L d9 = /tan30 sec46d0 = /tan30 . (tan26+ 1) . 56026d6 2 fu3(u2 + 1) du [u 2 tan 6, du 2 sec2 0d9] 2 f(u5 +u3)du2 éu6+ iu4+02 étan66+ itan40+C 34. ftanzm secccd$ 2 f(sec2x — 1) secmda: 2 fseCBmdac — fsecxda: 2 %(secm tana: +1n|secav + tan — 1n |sec33 + tanml + C [by Example 8 and (1)] 2 %(secm tanm — 1n |secm +tanx|) + C 35. Let u 2 x,dv 2 seem tanatda: 2> du 2 (136,1) 2 secx. Then fw seem tanmda: 2xsecm Afsecwdx 2 msecx ~1n|secm+tan$| +0. ‘ ‘ 1 36‘ /c::s¢:z>d¢:/:;::‘cos2¢d¢:/tan¢secz¢d¢=/udu [u=tan¢.du=sec2¢d¢] 2%u2+C2%tan2¢+C Alternate solution: Let u 2 cos <1) to get % sec2 (15 + C. 37. Ly: cotzmdm 2 f://62(CSC2.T — 1) dx 2 [— cotx ~ m] 2 (0 — — (—\/§ — g) 2 \/§ — g 77/2 77/2 7r/2 77/2 38. / cot3 mdx 2/ cotm (csc2 a: — 1) dx 2 / cotw csc2 :cdm —/ C951? d2: 7r/4 77/4 7174 7r/4 Sula? 7r/2 2[—%cot2ac~1n|sin$|] 2(0—1n1)—[—~%—1n%] =§+lnfi 2%(1—1n2) 7r/4 2 39. foot3 a CSC3 ada 2 foot2 a csc a~csca cotada 2 f(csc2a —1)csc2a-csca cotada 2 f(u2 —1)u2-(—du) [u 2 csc a, du 2 2 cscacotada] :f(u2—u4)du=éu3iéu5+03§CSCSOt—%CSCSO(+C 40. f csc4 a: cot6 atdx 2 [cot6 an (cot2 :10 + 1) CSC2 :6 dm 2 fu6(u2 + 1) - (~du) [u 2 cot m, du 2 — (:sc2 xdm] 2 fu6(u2 + 1) - (—du) [u 2 cot x, du 2 — csc2 xdm] 2f(~u8—u6)du2—%u9—%u7+02—%C0t9$—%cot7m+0 CSC$ cscm—cotx —Cscmc0tx+csc2:c 41.12 cscxdx2 —(—_)d$2 ———dm.Letu2cscm—00tx 2> csccc — cota: cscx — cota: du 2 (—cscx cote: +Csc2 3:) dz. Then I 2 fdu/u 21n|u| 2ln|cscm — cotzr| + C. ...
View Full Document

This note was uploaded on 10/10/2011 for the course M 408S taught by Professor Shirley during the Spring '11 term at UT Arlington.

Page1 / 5

hw-6-sec-8.2-solns - 622 CHAPTER 8 TECHNIQUES OF...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online