hw-6-sec-8.1-solns

hw-6-sec-8.1-solns - i r 8 D TECHNIQUES OF INTEGRATION ( 1....

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Unformatted text preview: i r 8 D TECHNIQUES OF INTEGRATION ( 1. iLCt u =1I193, d1) = $2 (133 :> du = idm, v = §a33. Then by Equation 2, y/ (1115")(é133) * f (%m3)( éxs lnx —%;1:3 +0 [or l3173(1nw — +0] m3)+C bill—I || )d;,;:lm3lnx—%fx2dx:§$3lncc—%( 2 fm lnxdzv 3 le-I ll CA) 2. Let u = 0, do = cosfidfi => do 2 d6, 1; : sin 6. Then by Equation 2, f9c030d0 = Hsinfi — fsin6d0 = 65in6+cos6+C Note: A mnemonic device which is helpful for selecting u when using integration by parts is the LIATE principle of precedence for u: Logarithmic inverse trigonometric Algebraic Irigonometric Exponential If the integrand has several factors, then we try to choose among them a u which appears as high as possible on the list. For example, in f $625” do: the integrand is 1362”, which is the product of an algebraic function (at) and an exponential function (62m). Since Algebraic appears before Exponential, we choose it = 9:. Sometimes the integration turns out to be Similar regardless of the selection of u and do, but it is advisable to refer to LIATE when in .doubt. ski/Let u 2 ac, do 2 cos 5m dm :> du 2 day, 1) : ésin 5m. Then by Equation 2, f$005537d33= éxsinSx—fésin5mdx = émsin5m+ %COS5.’IZ+C. Whit“ = 56, d?) = e_z d2: => du 2 dat, u = ~64. Then fate” dm 2 —me‘$ + fe’I doc : —xe_z — (3’35 + C. 5- LEtu=T,dv=eT/2dr 2 du=dr,v=2eT/2.Thenfrer/2dr22reT/2"IQGT/szZQTBT/z—46T/2+C. 6' Let“ = 15, til) 2 sin 2t dt => du = dt, v = 4% cos 215. Then ftSiDtht = —%tcos2t+ % foostht : —%tcos2t+ isinZt—l—C. , x _ 2 . \7'}Letu—$adv=slnnxdm :> du=2mdmandv:-%cos7ra:.Then _ 2 . I—fx stnnmdac: —la:2cos7r$+%fmcos7rmdm (*).NextietU::r,dV:Cosrracdx 2 dUzdw, ’rr ,r SanflS, SO fmcosmrdx = imsmvrm — ifsmrrxdz : 3:131an + '73—2 C057” + Cl- Substltutlng for f as cos 7m dog in H), we get IZ—l 2 2 1 . . «33 COSM: -l- ;(;msm7rx + f; cosrrac + Cl) 2 ~%$2 cos 7m + fixsmnx -l— 7% cosvm: —l— C, where C = £01. 611 .WMWWWV 1 “Wu,- “Enemy.” ‘vfi’hl‘wj‘waqmmv I, w, 1 612 CHAPTER 8 TECHNIQUES OF INTEGRATION 1 8. Letu : m2, dv : cosmmdm :> du = 2mdx, v = FLSinmx. Then 1 2 I = [3:2 cosmmdm = —a32 sinmac — — msinmmdm (*). NextletU = :c,dV : sinmmdw => dU = dw, m m 1 . 1 1 1 1 . V=——cosmm,s0 xs1nm$dx=———xcosmm+— cosmmdm:——mcosmm+—§smmm+01. m m m m m Substituting for f a: sin mm dav in (3%), we get 1 2 . 2 1 1 . 1 2 I = —-J: smma: — -— ———a:c0sma:+ jslnnm—I— Cl = —m smma: + ——xcosmm — ——smmm + C, m m m m m 2 whereC: ——C1. m ‘r9.‘Letu :1n(2ac + 1), d1) = dx => du = dx, 2) = 3:. Then \) 2x+1 2a: (2m+1) — 1 I = - d : 2 —— d /n(2m+1)dm :cln(2m+1) /2$+1 a: 331n(o:+1) 2m+1 ac 1 1 - _ _ _ _ -1 mln(2x + 1) [(1 2 1) dx mln(2x + 1) m + 2 n(2x + 1) + C =é(2m+1)ln(2x+1)#m+0 10112116111. = sin'1 w, d1; : do: => du 2 d2: ,1} = m, Then fsin‘1 sodas = msin‘lm —/ x dm. Setting \w/ \/1 —a:2 v1 —m2 mdm _ t:1-m2,wegetdt:—2xdm,so—/fiz— t1/2(—%dt)=%(2t1/2)+C:t1/2+C=\/1—:c2+C'. Hence, fsinT1 mda: = attsin’1 :6 + V1 —- 1:2 + C. 4 4 11. Letu:arctan4t,dv=dt :> duzmdtzmdt,v=t.Then t 1 t /arctan 4t dt : tarctan 4t - / 1 +416]? dt : tarctan 4t # —8—/ 1321619 dt = tarctan 4t — %1n(1 + 16t2) + C. ,. ii 1 \\12\.‘;Letu:1np,dv:p5dp 2,» du=gdp,v: %p6.Thenfp5lnpdp=%p61np—%fp5dp:%p61np— 316p6+CI / x... 13. Let u = t, dv : sec2 2t dt :> du : dt, v : % tan2t. Then ftsec2 2tdt : §ttan2t — % ftantht : ét tan2t — ilnlsec2t| +0. 14. Letu : s,dv : 23115 :> du : ds,v : $251 Then 1 1 1 1 28 2s : —— S — —— 2s : -— 23 —— 2s 1 2 "‘ 1. C . [5 ds 1n2 52 mg d3 1112 s (111?? + C [or 0an (s n ) + I 15. First let u : (1n m)2, d1} 2 dm => du = 2111:13- %dm, 1; : ac. Then by Equation 2, I : f(1n:L‘)2 d3: : $(ln$)2 —2fm1nx~ idm : $(lna:)2 -2f1nmd:c. NextletU =lnm,dV 2 d3: :> dU = 1/mdx,V : mtogetflnacdac = I1Il$—-f$-(1/$)d33 : :vlnm—fdm =xlnm—x+C1. Thus, I = 316(lnac)2 — 2(aclna: — m + Cl) 2 16(1n27)2 — ZmInx + 2x + C, where C : —2Cl. SECTION 8.1 INTEGRATION BY PARTS C 613 t m,“ _ .. WW... 1 15_ Let u = t, dz) = sinh mt dt => du = dt, v = E cosh mt. Then 1 1 1 1 . /tsinhmt dt = —tcoshmt — — coshmt dt = —tcoshmt — —2 Sinhmt + C [m 7E 0]. m m m m Q7, first let u = sin 30, d1) = 629 d0 i du : 300s 39 d0, 1) = éeze. Then ./ 1 = f 629 sin 36 d0 2 £629 sin 30 —— %f 629 cos 36 d0. Next let U = cos 30, dV 2 629 d0 => dU = —3 sin 30 d0, V : %629 to get f 620 cos 30 d0 = %e29 cos 30 + g f e26 sin 36 (19. Substituting in the previous formula gives 1 = $229 sin30 — gaze c0s36 ~ fife” sin30d€ = éezg sin30 — gegg c0530 — %1 => 1 1 = %e29 sin3t9 ~ gew cos 30 + 01. Hence, I: liseza(2sin3t9 — 3cos 30) + C, Where C : 14—301. First let u : 6‘9, dv = cos 26? d0 => du = —e_9 d6, 11 = % sin 26. Then r' f‘""’f“‘,‘,‘?‘”1“"'§"‘”"“’.‘f?f"‘1““" I = fe‘e cos 20 d9 = %e_9 sin 26' v f % sin 26 (—e'9 d9) 2 %e‘9 sin 20 + é—fe'e sin20 d9. Next let U = 6—0, dV 2 sin 20 d0 :5 dU : —e—"d6, V = g cos2o, so [6-9 sin20d6 : 4%6‘9 c0s29 — cos26(—e_0 d0) = — 1(3—19COS2Q — %fe_9 c0s20d0. 5 Sol 2 %e_9 sin29 + % [(—%e_0 c0520) — $1] = %e_9 sin26 — ie'e cos 20 — i1 : —9- 1 ~19 _4 EI=%6 s1n2€~ze cos26+Cl => 1—3 A %e‘9 sin 26 — ie‘e cos 20 + Cl) : %€_9 sin20 — gee—9 cos 29+C. {19.}Letu = 25, d1) = sin 3tdt => du = dt, v = —§ cos3t. Then \s. ,‘ f0”): sin3tdt 2 [—gt cos3tlgr +§foflcos3tdt =(~31‘7r—0)+%[sin3t]g: 20. First let u = x2 + 1, d1) = e-I dm :> du = 2a: dm, 1) = ~e_””. By (6), f01(z2 + De”: (1:0 2 [—(m2 + 1)e‘z]; + fol 2326—1 dx = ~2e—1 + 1+ 2]; :L‘e'x dx. Next let U : ac, dV = 6'1 d3: :> dU = dx, V : —e_z. By (6) again, 1 0 = —e_1 —e_1 +1 2 —2e‘1 +1. So fol me‘z d3: = [—1364]; + fol efiI d3: = —e‘1 + [—€_z] [01w + 1)e—I dx : —2e-1 + 1 + 2(—2e-1 + 1) = —2e*1 + 1 i 46‘1 + 2 = —6e—1 + 3. 21. Let u = t, dz; = coshtdt => du = dt, v = sinh t. Then fol tcosht dt : [t sinh tn — fol sinhtdt = (sinh 1 - sinh 0) — [cosh t]; 2 sinh 1 — (coshl — cosh O) : sinh 1 — coshl + 14 We can use the definitions of sinh and cosh to write the answer in terms of e: sinhl—coshll—lwae1 6—1) gel : 6T1) : 1: e_1+1=1~1/e. 1 22. Letu=lny7dv : idyzngl/Zdy :> du: Edyw : 2y1/2. Then x/z? /49h17%dy=[2\/731ny] 9 9 9 —/ 2yrl/2dy:(61n9—41n4)—[4w] =61n9—41n4—(12-8) r 4 4 4 ' =61n9—4ln4—4 614 (i 23. Letu : Inst, (11; z 06—2 d3: => du = — dm, 1) = —33_1. By (6), /21nw [ Inns]2 —2—dac= —— + 1 I 33 1 24. 25. 26. 27. 28. CHAPTER 8 TECHNIQUES OF INTEGRATION W 2 2 , 1 /$2d$:-%ln2+ln1+[—;] =—%ln2+0—%+1:%—%ln2. 1 1 1 1: First letu : m3, d1) = cosccdm => du = 3:62 (136,1) 2 sinx. Then 11 = [m3 cosxdw = m3 sinm — Sfmz sinmdm. Next letul = $2,dv1 = sinzvdw => (1111 = 213(15):,”01 = —-cosm. Then I2 2 fm2 sinxdm = —a:2 00820 + mecosmdz. Finally, let U2 2 at, dvg = cosmdx :> duz : dm,v2 = sin m. Then f :6 cos x dw = ac sin 96 — f sin a: dcn = as sin 36 + cos ac + C. Substituting in the expression for I 2, we get I2 2 —ac2 cos (E + 2(1) sin 36 + cos ac + C) = —:v2 cos a: + 23: sin as + 2 cos x + 20. Substituting the last expression for I2 into 3 [1 gives 11 = x3 sinx — 3(—:c2 cosm+2xsinm+2cosaj +2C) = m sinac+ 32:2 cosw —6a:sinm — 600520—60. Thus, . . 7r fowxscosxdxz [$3511127-l-31E2COSZE - 6xsmm — 6cosx — 6C]0 _(0 371'2 0+6 60) (0+0 0 6 6C)~12—37r2 Letuzy d2): 9—y- :e_2ydy => du:dy v: —le'2y Then a 62y : 2 ~ 1 y 1 1 1 Tdy:1~%ye-Zyl +é/ e‘gydy=<—%e-2+o>—118‘2”l =—%e-2—1e-2+1=1—%e‘2~ 0 e?! 0 0 0 _1 # Letu=arctan(1/a:),dv_~dac 4 du— 1+(11/xP-CU2 d$—m2:x1,v=x. Then \/§ x/E \/§ 1 d 1 V5 /1 arctan<g> d$= {marctan<%>]l +£ 3:20:31 =f%—1O%+§[ln(x2+l)]l 7r 3 7T 1 7r 3 7r 1 4 7r 3 7r 1 =——*— — ’1 ————— —l —=——~—— —12 6 4+2(1n4 n2) 6 4 2112 6 4 2n Letu—c0s_1m dv dac ~> du dm v—x Then 1/2 1/2 dag 3/4 I: cosilacdm: xcosTlx 1/2-1— —m——=l-1+/ 13—1/2 —-ldt,wheret=1—a:2 => /0 1 1. 0 W 2 3 1 1 2 1 dt:—2mdm.Thus,I—E+%f3/4t1/2dt—g-I—[x/fl3/4—E+1——2§—%(7r+6—3\/§). Letu = (lnx)2, do 2 96—3 d3: => du = 21:$ (120,11 : figs—2. Then 2 2 2 2 2 1:/ (Inf) d$=[_(lni)] +/ fldx.NowletU=ln:c,dV=3f3dm :» dUzldx,V:—%x_2. 1 a: 2m 1 1 2:3 2: Then 2 2 2 2 lnm lnx 1 7 1 1 1 1 1 1 1 3 1 —— — ~— — d =——l 2 — —— =———l 2 fl «— — =———l 2 /1x3dm [2x2]1+2/1m m 8n+0+2l2m2ll 8n 2(8+2) 16 811 fir SECTION 8.1 INTEGRATION BY PARTS :1 615 COS ac 29_Letu 1n(s1nm),d'u cosm :3 => du Sing: dm, 1) : sin ac. Then i 1 = f cosmln(sin m) dw : sinxln(sin a?) — fcoswdm : sinm 1n(sinm) — sing: + C. , Another method: Substitute t = sin x, so dt = cos 3: dx. Then I = I hit dt = tlnt — t + C (see Example 2) and so ; I : sinxflnsinm # 1) + C. 30. Letu=r2,dv : fidr => du=2rdr,v : \/4+7"2iBy(6), 1 dr=[r2m];_2/1T 4+r2dr:\/g_%[(4+T2)3/2] 0 =¢5—§(5>3/2+%<8>=¢5(1—1—£-)+%=%?—%¢5 0 1 T3 /0 V4+T2 5 l 31. Letu = (lnw)2, do = 2:4 d3: :> du = 2%:E (1.x, 1) = By (6), 2 x5 2 2 3104 2$4 m4(lnx)2da¢= ——(1n:e)2 #2/ —-lnmdx: §52(1n2)2—0~2/ ——1n3:dm. 1 5 1 1 5 1 5 4 5 LetU=1nm,dV=—m—dm => dUzldxyzi. 5 :B 25 Then 2m41nacdac— [Eilnm]2—/2£4—dm— g21112—0— - 21n2— (iz _L) 1 5 25 1 1 25 25 125 1 5 125 125 ‘ So 274(1nm)2 d3: = 352(1n2)2 — 2(%ln2 — 13213 = %(ln2)2 v %—ln2+ 16725. 32. Let u = sin(t # 3), d1) : es ds => du = —cos(t — 5) ds, 1) : 65. Then t I = jot eS sin(t — 3) d3 2 [es sin(t — 5)] + L: as cos(t — 3) d8 : e‘ sinO — 60 sint + I1. For I1, let U = cos(t - s), 0 deesds => dU=sin(t~—s)ds,V=es.Soh=[escos(t-s)]g—fgessin(t—s)ds=e‘c0s0—eocost—I. Thus,I:—sint+et—c0st—I :> 2I=et—cost*sint :> I:%(et—cost—sint). 33. Lety = x/E, so thatdy : éwTI/Z dz = fidm : Q—Igidx. Thus, fcosx/Edw = fcosy(2ydy) : 2fycosydy. Now a: i usepartswithuzy, d1; :cosydy,du=dy,v =sinyto getfycosydy:ysiny—fsinydy=ysiny+cosy+Cl, sofcosx/Edm=2ysiny+2cosy+0:2x/Esin\/a:+ZCOS\/E+C. @Letm = -t2, so that dm : —2t dt. Thus, fitseft2 dt = f(—t2)e_‘2 (%)(#2t dt) = i—fxex dsc. Now use parts with u=$,dv =em dz,du=dm,v :e”3 toget §fmexdm = %($ez ‘ fem doc) = émex — éem +C' = —-;—(1 — flex +C = —%(1 +t2)e”t2 + C. 62 cos(92) - %(20 d6) = mcosmdac. Now use 1r/2 J77 J? 03cos(62) d6 : / 35. Let x = 02, so that dm : 26 d0. Thus, / m x/1r/2 parts with u : w, dv = cos 2: dm, du = dm, 1} : sina: to get 2 r W 1 _ . 7r . . _ 2/ accosacdx— ([xsmmhm / smxdw) — m2 m2 (7rsin7r+cosrr)#%(%sin%+cos—E) =§(1r-0—1)—§(325'1+0) :# 7r ‘rr/2 NlH [m sin :5 + cos 20] {Oh- ml..- pl: l 2 616 CHAPTER 8 TECHNIQUES OF INTEGRATION @}et :1: : cost, so that dm 2 — sintdt. Thus, s i? ,. f0” emst sin2t dt 2 f0” e°°5t(2 sint cost) dt = ffl e” ~2x (—d1) = 2K1 mez dm. Now use parts with u = :5, d1) 2 ex dm, du = d1, 1) 2 em to get 2fi1mexd$=2([$ex]l_l -—file$dw) 22(61 +e'1 — [exll -—1 )2 2<e+ — [e1 -e"11> / du:%dy,v=%y2—ytoget f(y‘1)lnydy—(%y2 y)1ny f(%y 1)dy—%y(y—2)lny—iy2+y+0 =%(1+x)(x#1)ln(1+x)—i(1+$)2+1+x+C, which can be written as %($2 — 1) ln(1 + at) —— £932 + %m + g + C. 38. Let y = In as, so that dy = idm :> d1: = m dy 2 6y dy. Thus, f sin(ln cs) da: : f Sing 63’ dy = éey (sin y — cos y) + C [by Example 4] = §zv[sin(ln :v) — cos(ln + C. In Exercises 39~42, let f (1:) denote the integrand and F(x) its antiderivative (with C = 0). 39. Letu=2m+3,dv=ezdm => du=2dm,v=e”‘.Then f(2x+3)ezdzv=(2m+3)e$ ——2fexdm: (2ac—l—3)eI —2eI+C =(2a:+1)e”+0 We see from the graph that this is reasonable, since F has a minimum where f changes from negative to positive. 40. Letu=lnx,dv=w3/2dw => du:%dm,v=§x5/2.Then 3 fws/zlnmdsc= gm5/21n$~%fm3/2dm= §m5/2lnm~ (%)2m5/2+C =§x5/2lnx~%x5/2+C We see from the graph that this is reasonable, since F has a minimum where f changes from negative to positive. 41. Letu: ém2,dv= 2xV1+x2dw :> du:;rdx,v : §(1+x2)3/2. Then fats \/1+a:2 din: éflvz [%(1+a:2)3/2] — 2 m 1 + x2)3/2dm 3 : __:r2)3/2 __ g . é . +$2)5/2 : %m2(1 —— m2)3/2 — T23(1+12)5/2 + C Another method: Use substitution with u 2 1 + 332 to get %(1 + m2)5/2 — —;—(1 + x2)3/2 + C. 2(2e—1): 4/e. 37. yet y = 1 + s, so that dy = dar. Thus, f a: ln(1 + m) dw = [(y — 1) lnydy. Now use parts with u = 1n y, do = (y — 1) dy, K . SECTION 8.1 INTEGRATION BY PARTS 619 52_ By repeated applications of the reduction formula in Exercise 48, ‘fav‘ttaI da: : x46? — 4fazse” dx 2 flex —— 4(cc3eI — Bfwzew dm) : $463: — 4m3e$ + 12(cczeZ — 2fatleI dm) 2 flex — 433381 + 123026” —— 24(3316" — fwoez dcc) : $462 F 436365 + 129861 _ 24”” 24635 + C [or 63%“ — 4303 + 12x2 — 24:15 + 24) + Cl (-513. Area = f: xe—0'4zdm. Let u = x, d” _—_ 6—0‘4z dm => «Av/d” ____ dm’ 0 : _2.56—0.4z' Then _ _ E 5 : ——12.56 2 +0+2.5[—2.5e 0-4 ]0 : —12.5e—2 — 625(e-2 _ 1) = 625 - 18758—2 or T _ Te 54.Thecurvesy=az1nazandy=Slnmintersectwhenxlnm:Slnm <=> mum—51113520 <=> (m—5)lnx:0; S,whena: = 1orx : 5. Forl < at < 5,wehave51nx > xlnmsincelnm > 0.Thus, thati m)dm —> du—dm/x,v:5m—§Lv2. areasz5 (51nm~a:lnx)da:=f15 [(5—m)ln$]dar. Letu—lnx,dv— (5 Then area: [(lnm)(5m — ~ 15 [(593 — dcc : (ln5)( =2§1n5-[5x—§x2]i_2—§1n5 [(25 T) (5 55. The curves y : ac sins: and y = (m — 2)2 intersect at a % 1.04748 and b % 2.87307, SO area = f: [m sinw — (a: — 2)2] dac : [—30 cosa: + sinm — — 2)3]: [by Example 1] % 2.81358 — 0.63075 : 2.18283 56. The curves y = arctan 3a: and y = gm intersect at x = in z $291379, SO area = f; larctan 3m — d3: 2 2f: (arctan 3m — also 2 2 [at arctan 3m — % ln(1 + 95") — if]: [see Example 5] z 2(1.39768) = 2.79536 57- V = [0127mcos(7m:/2)dm. Letu : ac, d1; : cos(7rm/2) dac :> du 2 dm, 1; = %sin(7r:1:/2). V=27r[2$sin(fl)]; —27r-—2—/01 sin(7—r§) dm : 27T<§ 70> —4[—%COS(%)]: :4+ 3(0— 1) :4— 7r 2 7r 58. Volume : fol 27rac(eI — e”) dm 2 27r f01(1781 — xe'x) dag : 271'“; mei dm — fol are” elm] [both integrals by parts] = 27r[(acez — ex) — (—xe‘r — 3””) L1) : 27r[2/e # 0] : 47r/e ...
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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.

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hw-6-sec-8.1-solns - i r 8 D TECHNIQUES OF INTEGRATION ( 1....

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