EserciseEserises1.un"y12.y0ycostdrX'sinxdI,u = tan"ly, du = -qisdy,=yCostint[n-ly = anly- tanly-(a+)+Cyy-n V++2+ 008 (+)T12. u = 8in~ y, du idv=dy,vy,/--2sint+C.rs+2tJyyyly- y+sin x_(+)COn X2x2 0m x sin xdxxco6X+2x sin x+2.csx+C0A国E3ex144s2dbxercisesEaercises15./eaJnes15.x dx,ytan x(+)13. #=x, da = dx dvx[I sex dx =x tan x-tan x dx = x tan x + In icoe x1+ C款0.tan ydy=y tan yln[mc y]+C14.2sd b2]Jyydyytanyxdx=+-3x*+6xe-6*+C=(23x+6x6)*+C= 2x tom 2xla lae 2x|+C16.上1a-24p-24(0)[pt-dp=pt-9-4pl-P-12pe-P-2pe-P-2eP+C=(p*4p--12p-24p-24)e-P+C山广 JreExenciesI. [o - SedExercises19e41s.c+r+d19.17. 5s4)e4a0)x-21g(c)2~y[(3_5x)e dx=(x3-5n)(2x5)e+2e*+Cm.gg-70*+7e+C121x10)120 -=(x2-7x+7)*+Cx=xx+20+120*120e+C18.e(25z*+20802+120120)+C2+++1-(+)+e20.2+1-)e222*0[(P+#+1)e dr =(P+t+1) (2r+1)+2/+C24+)te=[G2+r+1)(2r+1)+2/+C=(2-++2)e +C"d-f-n+a+C-fa-j+++c6白人E(3-++).+C
2016/11/15 6 Exercises Exercises 11. 12. Exercises 13. 14. Exercises 15. 16. Exercises 17. 18. Exercises 19. 20
naSubstitutionEvaluate the integrals in Exercises 25-30 by using ai substitution prior+rldtan'xdsEsenisto integration by partsExertisesTevrd26V1-xd[25.27. u = x, du dx, dv'= tas*x dx, v [+9=2]2s[.Vmdtandx=[s(tanx)]g()()[ []- d3 [dd tan x(V5-5)+2+--m2-% x dx=3(- J)-(-)+C=g(V+V_V)+c8 u=(+);d=d,, n(+2)h(+2)-[ddxdv=V1xdx,v=$/(x),26.u=dX(+2)-[()x(++2)-[2±)dx=(++)2+[+1+Cl-x dx-[-VαxP-] + Vα-xp dx- [-8(x)/],-1目国29.sin (nt) dt30. (ln)4Exercisesu=r 21, [(ain u)e* du me(sin.=s)+C2. J a(ln x) drd =↓ ddx=edu8.3[x ee (In x)+x sin(ln x)]+Cumlna38. Fa(m ) dd1 Je-- do- fu duTrigonometric Integralsdeme"duu2u-122+[daf++"+C[2u2-2u+]+c[2[n a)-~2 n +1]+CSlide 4- 40sin"rcoxdnerol, We can divide the appropoddorcity sm.x1 .ooCase-2k +JanduO1.Products of Powersof Sines(sin’x)sinm (1 0osa)'sin)andCosinesCase 21F.d in fsin"anfse theideutity oocy =1 sintoobtainco'x co*+x (o0 x)00s.. ( sin*s).0o%)ebinethesinglecosxwithds ldfregual to dsisxCase3If bethandnareeveninsn"soo"xdr,wesnx L=m2tcrLg2s(2)lower powers of cos.2rto reduoe Che inbegrand to onSlide 4- 41
2016/11/15 7 Exercises Exercises Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 40 8.3 Trigonometric Integrals Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 41 1. Products of Powers of Sines and Cosines Slide 4 - 42
EXAMPLE1EvaluateEXAMPLE2Esaluatecosxdsin'rcosrdof Case 2, where m = 0 is even and n = 5 is oddSolutionThis is an example of Case 1.out[cs xdr=/costroos-xdk[(1 - sirsP d(sinx)sinxcoxd=sin xcosxsinxdmisadawyd=[(1-0)0o2(d(cs x)*i-Ja-22+=[(1-2)(do)1-Ou-+r+c-sini-号sx++snx+c- Ja- ia) dStaliply mrau号-5+0-11+0lide 4- 43Slide 4- 44EXAMPLE3EvaluatePomersof SinesandCosinmsinrco'adr.ExercisesEvaluate the integnals inExercises1-22SolistiarThis is anxarmiple of Case113m1.cosZtidarowsd- [(m)(gm)afxinib4sin2rcos2d/0821+2 2+ 20/ +22p2s[ard64rd+ 2 e2+ mpma7.sinxd[sintdoteling2ef+cosd(+)sin2x+C21-9005-3. -1s.x+CForthecogerm/og2=1sir2)os212ASL10n2x+C00s10081+sin'4x+C04S12=/ )=(sin222)Ma7. -05 +0s'x-1008 *+CCombining everything and simplitying we get[s-(-14r++s2)+slide 4- 45Slide 4- 48EXAMPLE4EvaluatV1+ cos4tdrSolutionToelintaroot, we use.the identityd8 = /+ cos 20 + 00s 20 = 2 co* 0.With e..2x-thisbecomes1 + cos 4r 2 cos 2r2.Eliminating Square RootsTherefore,i+rr-V2ozdViVezd V2 os 2]4t = e / xt0.Slide 4- 47Slide 4.48
2016/11/15 8 Slide 4 - 43 Slide 4 - 44 Slide 4 - 45 Slide 4 - 46 Exercises 1 1. sin 2 2 x C 0 9 2. [ 9cos ] 3 2 x 1 4 3. cos 4 x C 1 5 4. sin 2 10 x C 1 3 5. cos cos 3 x x C 1 1 3 6. sin 4 sin 4 4 12 x x C 2 1 3 5 7. cos cos cos 3 5 x x x C / 2 5 3 5 /2 0 0 2 1 16 8. 2 sin 2[ cos cos cos ] 3 5 15 u du u u u Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 47 2. Eliminating Square Roots Slide 4 - 48