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hw-3-sec-5.5-solns

# hw-3-sec-5.5-solns - 402 D CHAPTERS I 10.r 10 2e 28 26 1.10...

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Unformatted text preview: 402 D CHAPTERS INTEGRALS 10 10 I 10 .r 10 2e 28 26 1.10 68. L10 sinhx+c ha: / x , dx’f —-—dr=/ 2d1=g2\$3~10=20—(—-20)=40 2 2 __ 3 0 ‘ - _ 70.] (a: 1) ‘ ;‘ 1 _)dr=[%rz—3x+3ln1rI+-1-] 1 H II: 1 5.5 The Substitution Rule 1. Letu : 3:6. Then du = 3d:r, so dac = ﬁdu. Thus, cos 3x d2: = cos u 1 du 2 l cos u du = i sin 11 + C = 1 sin 31‘ + C. Don’t forget that it is often very easy to check 3 3 3 3 an mdeﬁmte Integratlon by dlfferentxatmg your answer. In thls case, a; (g 5111 31 + C) 2 § '6 desired result. @Letu :2 2 + :54. Then d-u = 49:3 d1: and 13 d9: = ﬁdu, “'6 1 46. dt _ —§du_ 1 -4 # lu _ ‘ f(1~6t)4"/ *“3/1‘ d”"’6~ +C”18u3+c_18(1—5t)3+0 “4 Sum = c039. Thendu = —sin9d9andsin6d9 = —du, so 4 /cos395in6d6:/u3 (—du)=—%~+C=—icos‘16+0. SECTION 5.5 THE SUBSTITUTION RULE D 403 5. Let u = 1/25. Then du = ~1/x2 d2: and 1/952 dz: = —du, so \$2 2 [de=/sec2u(—du)=—tanu+02—-tan(1/a:)+C‘. Q Letu = 252. Then du = 2\$dm and asdx = édu, so facsin(:z:2)d:c = [smug— du) : —-;-cosu+C: ——-;- cos(a:2) +0. @Letu = :63 + 5. Then du = 3m2 due and 2:2 dac = ﬁdu, so 2.4 g 2-4; _l_“_- ,-_1_ 34; /(3t+2) dt—fu (adu)-33' +Cw10_2(3t+2) .C. 11. Letu : 2x+x2. Thendu = (2+2r)da: = 2(1+;r)d:c and (x+1)d1 = §d~u, so f(:c+1)\/2:1:-é—ac2 dmzfu-uduzfu2duz §u3+C= §('21+12)3*12+C. 12.Letu=x2+1.Thendu=2xdmand\$dr=§dwso __._1‘_._._ —- ”2 l ~i..:}._., —_:£_;. -—__:_1_._;_ /(z2+1)2dm'/u (2%)“? u C—2u10-2{12+1‘) C @Letu = 771’. Then du = 7rdt anddt = éduso fsinirtdz‘ = fsinu {-i—du) = -:—(——cosu) +C = —%cosvrt+C. 14. Letu = 5t+4. Then du = Sdt anddt = édu. so —2 1 _.,... 1 1 -1- -1 _. - ____._.__. t: —>‘ gd. :_. «C:~_~— i~'__C-=___ 3+ [(5t+4)2-'d f” (a u) 5 —1.7u 25.5“ 17(5t+4)1-7 C 15. Let u = 3am + br3‘ Then du = (30 + 3br2)d.r = 3m + (made: so who +1 ldu 1 . , . .,—————- [0.“ (ix: 3 :—/u—1’2du=-§‘2u2+C= v3ax+br3+0 \/3ax + I713 711’? 3 16. Letu = 26. Then du = “2618 anddB : éduﬁo j‘secQO tan'ZGdG = fsecutanu (ﬁdu) - lsecu +C = ésec29+CC ’2 17. Let-u = x/t'. Thendu 2 3‘2? and i; dt = 2du. 50/ 005%” dt = fCosu (2cm) = 251111: + C = 9sinx/I—H- C", v v v 18. Let u. = 1 + 13":2. Then d-u 2 gr“? d: and x; air 2 gain. so fx/E sin(1 +352) dz = fsinu. (g du) : g - ‘7 C0511} 4 C = mg: €05in + \$312) 4 C. \$45 55 ® Letu = 1+tan0. Thendu =sec29d6, so f(1+tan9)5 sec26d9 = fusdu = %u6 +C =(1+’can€)6 +0. 21. Letu:1+z3.Thendu=332dzanddez= ﬁduso f—ji—dz-fu m /______cos(7r/\$) dx = /cosu(—ldu) = —isinu+C = -lsin-7: +C. 7r 1r 7? 1: 1-2 23. Let u = cot :3. Then du r: — csc2 :5 d9: and csc2 2: dx = —du, so 3/2 [Veot\$csc2\$dr=/\/ql(—du) = —:/2 +C: ——§(cota:)3'/2+C. 24. Letu = 1 + tant. Then du == sec2tdt, so sec 2ttdt du _1,12l11‘/2 _ --= d =-——+C==2\/1 t C. foos2t¢1+tant =\/'—_/ 1+tan J; [u u 1/2 + ant + 25. Letu=secm. Then du 2 seem tanzdsc, so fsecaxtanxd\$=fseczx(secwtan\$)dz=f'u2du=§u3+C—= ﬁsec‘r’x-i—C. 26. Letu = cost. Then du = ——sintdt and sintdt = ——du», so fsint sec2(cost) dt = fsec2u. (—du) : —ta.nu + C = —tan(cost) + C. ' cosa: —2 1 1 Letu=smzc.Thendu=cosxdx,SO/2 dx=/—1—du=/u du:%+c:v—+C:#sinm+c 51:12 J: [or —csc:1: + C]. 28. Letu:1—m.Thenm=1~uanddm=—du,so (D SECTION 5.5 THE SUBSTITUTlON RULE C] 405 @Letuzm2+1 [502:2 :u—l].Thendu=2\$da:anda:dz= %du,so fx3Vm2+1das= foVw2+1xdz=f(uwl)ﬂ @du) : %f(u3/2 —u1/2)du = ﬁguw - gum) +0 = gen? + W2 - gas? + 1)“? +0. 0r:Letuzi/x2+1.Thenu2=a:2+l 2) 2110311221419: => udu:mda:,so fscz‘x/a‘?+1dcc=fr2vaz2+1xdac=ﬂu2—1)u-udu=f(u4—u2)du = gm" - %u3 + =51? + 1)“? — —§-(a:2 + 1)“2 + 0. Note: This answer can be written as 133 «x2 + 1(31?4 + x2 e 2) + C. In Exercises 31—34, let f (ac) denote the integrand and F (as) its antlderivatlve (with .C = 0). 31. f(:1;)=:c(\$2—1)3. u=x2—1 2 du=2xdac.so fzr(:c2 ——1)3dm= fu3(-]§du) = %u4+C= 38-(172 — 1)4+C Where f is positive (negative), F is increasing (decreasing). Where f changes from negative to positive (positive to negative), F has a local minimum (maximum). sun/53 u=\/§ :> duzéx—L’dezédaxﬁo f(a:)= \/5 . 2V3: sin \5‘ . \/_ d3: = [Sinu(2du) : ~2coqu+ C m Where f is positive (negative), F is increasing( ive (positive to negative). F has a local decreasing). Where f changes from negative to posit minimum (maximum). 3x cosr. u = sinx => du : cosrdr, so f(z)=sin fsinszccosscdmzfquuz §u4+C= ésin4I+C Note that at a: = g , f changes from positive to negative and F has a local maximum. Also, both f and F are periodic with period a, so at a: = 0 and e and F has local minima. at :2: = 7r, f changes from negative to positiv f(9) = tan2 9 sec2 9. u = tan9 => (in = sec2 6d6, so ftanzﬁsec26d0=fu2du=éu3+C=§tan39+C Note that f is positive and F is increasing. At 1’ = 0. f = O and F has a horizontal tangent. Letu=m»1,sodu=d:c.Whena::0,u= ——1:when:c=2,u =1.Thus, Theorem 6(b), since f (u) = 11.25 is an odd function. SM 515 406 U CHAPTER5 INTEGRALS .Letu = 4+3m, so du = 3dm. Whena: = 0,21. = 4;whenx = 7,11 : 2.5. Thus, 3/2 25 1 , fg1/‘4'+'3'x dx = ffié du): 413—72] - -3053 ‘2 ~43”) = 3(125 8): 23- = 26. 37. Let u = 1 + 2233, so d1: = 63:2 d1. When 1‘ = 0, u = 1; when .r : 1, u = 3. Thus, 1 13 r 1.3 a 7 f0 \$2(1+2x35) dw: u°(1du)= éééusil ———3i6(36—16 =-3%5129—1)= \$.17. .Letuzwz, sodu:2:rdm. Whenx20. 11:0; whenxzxfrr, u::r.1‘hus, foﬁ cccos(x2) d3: = f; cosu(% du): %[sinu]3 = é-(sinm' — sinO): %(O — O) = 0. 39. Letu = t/4, so du : ﬁdt. When t = O. u : 0; when t = 7r, u —— 7r/4. Thus, f: sec2(t/4) dt = 5/4 sec2 u(4d'u) = 4itan a}?“1 = 4(tan4} — tanO) = 4(1 -— 0) = 4. 40. Letu =7rt,sodu : 7rdt, Whent = %, u = %; _l§_ 17,32___i ‘- “1 —.11. “swim/e— «(1 2F" V f1/1/62 csc7rt cot 7rt dt = f3: cscu cotu (% du 41. f:/_6 63mm 9 d6: 0 by Theorem 6(b) since f ((9) — tam3 615 an odd function 3:2 sin .1- 1 ) r6 is an odd function. 1r/2 42. l 12 sinz: dm — 0 by Theorem 6(1)) since f (x) “/2 1 + \$6 43. Letu == 1 +2cc, so do = 2613:. Whenx = 0,21 = 1; when;~ : 13.11 = ‘27. Thus, 13 _ (1' W23 1,13" / \/__1_.__ =/27u Edll)=§L-;~3u E; %=(3-1)=3 +2.1 44. Letu = sinsc, so du = cosxdx. When 1' : 0, u : 0; when a: = g u z 1. Thus, jaw/2 cosa: sin(sin 3:) d1: :- fol sinu du = [-— cos u_§3 : —(cos 1 —— 1) = 1 — cos 1. 45. Letu : x2 + (12, so du : 2xdx and xdsc = édu. When 1‘ = 0.11.: a2;when 3: 2: a, u = 2a2. Thus, a 20 2 . 2&2 r 2 2 I A \$1/\$2+a2 d3: 2/20 11% (171:): %[§U3/2] 2 = iéu3f2]a: 2%[(2a2)3/2_(a2)3/2] :%(2\/§‘1)a3 46. Assume a > O. Letu = (1.2 —— 3:2, so du = ~2rc Bx. Whenm = 0, u = a2; when 3: = a, u = 0. Thus, a m 0 1/2 1 1 a2 1/2 1 {2 3X2”2 1 3 fox 0. —a: dmzfazu (—gdu)=§f0 u du=§-L§u'JO =§a. i47’Letu:x—1,sou+1:manddu:dm.Whenz=1,u=O;whenac=2,u=1.Thus 2 1 1 1 /1:lcx/aic—ldalcz/(u—1—1)ﬁdu=/(213/2—1-ul/2)du=‘;§11.5/2+§u3/2]0 o o .Letu:1+2m,sox::(u—1)anddu=2d.~z. Whenr:0_ u=1; whencc=4, u:9. Thus, ‘1 I xdac _ ﬂu )du= 3/910 u1,2___ u—1/2)d'u,=-1~4 [221 ”3/2 —2u1/2]1=ﬁ-§[u uu3/2—3 1/2]: 0 1/1+2:c— ﬁt 3 2 2_1_6 5+3_15' ll ...
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hw-3-sec-5.5-solns - 402 D CHAPTERS I 10.r 10 2e 28 26 1.10...

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