Chapter 5 50

# Chapter 5 50 - 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34...

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Unformatted text preview: 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. CHAPTER 5 iNTEGRALS 1 . . 5111 1' 51nd? /_1 1 + \$2 m 0 by Theorem 5 5 7(b). smce f(:c) 1 + m2 is an odd function. fOIe-zrtdt 2 Eat]; : ;r1_(e7r _ 1) Letu 2 2 — 3m. so du 2 —3dw. When m 2 1. u 2 —1; when m 2 2, u 2 —4. Thus. 2 —4 1 1 _4 /1 2—32: dm:/.1 ;(——§du) =—§[1n|u|]_1 :2§(1n4—1n1)=—§1n4. 4 2 4 2 4 4 / +322 90 /<2 3; \$2>dmv/<\$22+1—1>dx2[ 1+ln|ati—:v] 2 m 2 2: cc as 2 z 1 2 =(—%1+1n424)2(—§+1n2—2) 21112—7: .1: \$224 10 / dm does not exnst because the functlon f (ac) 2 has an 1nﬁn1te dlscontinuity at as 2 2; 1 \$224 that is, f is discontinuous on the interval [1, 10]. Let u 2 \$2 + 41:. Then du 2 (2m + 4) da: 2 2(33 + 2) dm. so ————%xdm2/uil/2(édu)2%~2u1/2+C2\/E+C2\/\$2+4\$+C. Letu: 3t. Then du 2 3dt,sofcsc23tdt 2 fcsc2u(%du) 2 %(— cotu) +02 —%cot3t+C. Letu 2 sinmﬁ. Then du 2 7r cosntdt. so fsin7rt cos7rtdt 2 fag du) 2 % ~ §u2 + C 2 %(sin7rt)2 + C. Let u 2 (3053:. Then du 2 2sinmdav, so fsina: cos(cos:c) dcc 2 ~fcosudu 2 —sinu+C 2 —sin(cosm) +C. J5 Letu2ﬁ.Thendu2 2%,50 i/E dx22/eudu22e”+C22e‘/E+C. Letu:lna:.Thendu2ﬂaw/Wake2/c0sudu=sinu+C2sin(lna:)+0. x —sin\$ cosm ftanmln(COS\$) dzc = — fudu : —%u2 + C : —%[ln(c0s:c)]2 + 0' Let u 2 1n(cos ac). Then du 2 dm 2 2 tanxdz, so 1 sinilu+ C 2 1 sin’1(x2) +C. 1 du LetU2x2.Thendu22mdm,so/—x—dm:—/-——:_ _ x/l—zr4 2 x/l—u2 2 2 \$3 1 1 4 1+\$4 dm2Z/Edu2iln|u|+02iln(1+m )+C. Let u 21+ :34. Then du 2 4:33 dm, so/ Letu 2 1+4m. Thendu 2 4dm,sofsinh(1+4a:)dm 2711fsinhudu 2 ﬁcoshu+C 2 ﬁcosh(1+4a¢)+C. Let u 2 1 + sec0. Then du 2 sect9 tanQdO. so secﬁtanO 1 1 Z — = 2 2 =1 1 , / 1+sec0 d0 /1+sec€ (seCOtanﬁdﬂ) /udu ln|u|+C n| +sec€i+C ...
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