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Chapter 5 41

# Chapter 5 41 - SECTION 5.5 THE SUBSTITUTION RULE 473 52...

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Unformatted text preview: SECTION 5.5 THE SUBSTITUTION RULE 473 52. Letu = 3:“). so du = Zmdw. When 90 = 0.11 = 0; when w = ﬁ. 11 = 7r. Thus. foﬁ :ccos(x2) d3: 2 f0" eosu G du) : %[sin 11E : %(sin7r , sin 0) = a0 — 0) : 0. 53. Let u : t/4. so du : idt. Whent = 0,11 = 0; whent : 7r, u = 7r/4. Thus. 7r/4 f0 sec2 (t/4) )=dt [OW/4mm2 11(4du) =4[tanu]0 = 4(tang — tanO) : 4(1# 0) = 4. 54. Letu = 7rt.sodu = 7rdt. Whent = 5*“ = %;whent 2%11 2 .Thus. 7r/2 w/6:—%(1—2): — [11/62 csc 7rt cot 7rt dt 2 ﬂy: cscu cot u (% du) = i} cscu] 55. 1:11-36 tan3 6 d6 = 0 by Theorem 7(b), since f(6) = tam3 6 is an odd function. 1 . . . 56. /02 W does not exist since f ( )2 W has an inﬁnite discontinuity at a: : MN» 57. Letu = 1/:c. so du = 11/532 dw. When a: = 1.11 : 1; when a: = 2. u : %. Thus, 2 61/9: 1/2 1 2 1 2 / 2 dm:/ eu(—du)=—[eu]1/ =—(e/ —e):e—\/e. 1 1 \$ 58. Let u : A362, so du : —2;1:d:1:. When In : 0,11 = 0; when a: : 1. u = —1. Thus. 1; d. : 15161—1 du) = £16113: —%<e-l — e0>:%<1—1/e>- 59. Letu — cos6 sodu — —sin6d6. When 6 — 0. u — 1; when6 —§u : %. Thus, 1r/3 . 0 1/2 _ 1 1 / 8”; d6:/ % :/ ue2du: [—1] :—1—(—2):1. 0 cos 6 1 u 1/2 11 1/2 50 /7r/2 \$251nm6d\$ 0b Theorem 7(b) since f(a: ): 332 since is an odd function ' _./2 1+1: y 1+w6 ' 61. Letu: 1+2x sodu:2d:£.Whenm:0.u: 1;whena:: 13.11227. Thus. / f f. [1 1 62. Letu : sings. so du : cosmdm. When a: = 0.11 : 0; when ac = g u : 1. Thus. MICE (3~1):3. fox/2 cosa: sin(sinm) dm : fol sinudu : [— c0511]:J : —(cos 1 i 1): 1 — cos 1. 63.Letu::ci1.sou+1=wanddu:dm.Whenx:1.u:0;whenm:2.u:1.Thus, 1 fmem—1dm:f01(u+1)\/Hdu=f01( (113/2+u1/2)du:[gu11—5/2+%u3/2]0:§+ :i—g. (JOIN 64. Letu:1+2ac.som= é-(u— 1) anddu:2dm. Whenm:0.u: 1;whena::4.u:9.Thus, xdm 9 alt—1) du 1 9 1/2 _1/2 9 _, ____ d _1[2 3/2 21/2] 0 1 21: 1 \/_ 2 4/1(u u ) u 4 3“ u 1 9 [us/2 — 3u1/2]1 :1127— 9) — <1 — 3)] = — = — HBIH who ...
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