{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw-6-sec-8.2-solns

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

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

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

View Full Document

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

View Full Document
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

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern