Chapter 5 39

Chapter 5 39 - SECTION 5.5 THE SUBSTiTUTION RULE 33 Let u =...

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Unformatted text preview: SECTION 5.5 THE SUBSTiTUTION RULE 33. Let u = cot cc. Then du : — csc2 a: due and csc2 at d3: = —du. so 3/2 /\/cot:csc2mda::/\/EI(—du)= ﬁzzy—2+0: —§(c0tm)3/2+C. 1 1 34. Letu— — — .Then du — ——2dac and —2dw— — ———du so m m an 7r Law/”dz: cosu —ldu :—lsinu+C=—lsian-+C. . \$2 71' a: W W 35. /c0tmdm=/Coszdm. Letuzsincc. Then du zcosmdac so smm footxdajz fiduzlnlm +C:ln|sina:| +C. 36. Let u : cos x. Then du = — sinmda: and sinmda: : —du, so sina: —du _1 71 /1+0052mdw:/1+u2=—tan u+C:!tan (cosm)+C. 37. Let u 2 sec 3:. Then du 2 seem tang: data, so fsec3w tanzcdcc : fsec2w(secm tanm)dm : fu2 du = 1713 +0 : lsec332+CC 3 3 38. Letu = \$3 + 1. Then 9:3 = u! 1 and du : 3SE2 d313, so fx3/\$3+1m5dm:f\3/x3+1-x3 -m2dm:fu1/3(u—1)(%du) : = get.“ 7 gum) + o = %(\$3 + 1)7/3 - W3 + 1)4/3 + 0 39. Let u : b + 033““. Then du : (a + Decca dm. so 1 1 3/2 a. b “+1d : 1/2 d : (g 3/2) C: b n+1 C. /" V +05" ”5 /“ (a+1)c “ (a+1)c 3“ + 3c(a+1)( +63” ) + 40. Let u : cos t. Then du : ~ sintdt and sintdt = —du. so fsint sec2(cost) dt = fsec2u - (—du) 2 Atanu+ C = —tan(cost) + C. 41. Letu : 1+ 252. Then du = 2m clan, so 1+r 1 ac 71%du /1+\$2 a: /1+m2d\$+/1+\$2 1: tan m+ u tan m+2ln|u|+C %f(u4/3 - u1/3)du :tan_1m+%ln|1+m2|+C':tatn_1att+%1n(1+:152)+0[sincel—i—a:2 > 0]. 1 -d 42.Letu2372.Thendu=2atd33,so/1aC dm:/ 2 u :étan"1u+C: +224 1—t—u2 43. Letu : x+ 2. Then du : dac. so 5” dmz u4_2du— u3/4 211‘1/4 du: m f _:;(m+2)7/4 §(x+2)3/4+ C 44. Letu:1Am.Thenx:1—uanddw:~du.so «mu. % tan’1(m2) + C. 117/4—2-gu3/4+C “32 (lolly 1—2u+u2 d 7 d : 1 _ —1/2 _ 1/2 3/2 __1_ x a: / ﬂ ( u) ﬂ du /(u 2n + u )du :_.2(2u1/2_2 wag u5/2)+C:—2\/1———a:+§(17x )3/2 — go 4 as)“2 +0 471 ...
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