《微积分》课程教学课件（Calculus）08. Techniques of Integration

CALCULUSTHOMAS'CALCULUSRYTRANSCEChapter 8AMESSTEWARTechniques of IntegrationAASCNAfia-htoherk2tansd-hieca+((x + 1)3tinrl +[4-njl + c4+8.1-[-+(a>aa1)Aindxdr=codr +BasicIntegration Formulasn.bs-aar+c /v"()c /-()-c[-()+(11+ /v.0-(3)-c 6>1>9)41.Making a Simplifyang Substitution2.Completing the Square2x-9EvaluatedxEraatey8x-x2x29x+1Solution Let u=x -9x +1,then du =(2x-9)kSohutiom Wecomplete thesure towrite theradicandas[=Juriedu8xx=-(8x)=(x4)+16Sodx=yx-9x+1dxSo-uzy-8x-xy16(x-4)+C= 2/"+C =2/P9r+1+C(-1/2)+1Let u=x-4,then du=dkTherefore+0= sin(=)+C=sin4-uV8x-x人产

2016/11/15 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 8 Techniques of Integration Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8.1 Basic Integration Formulas Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Making a Simplifying Substitution 2 2 9 9 1 x Evaluate dx x x     2 Solution Let u x x then du      9 1, (2 9) x dx 1/2 2 2 9 9 1 x du So dx u du x x u          ( 1/2) 1 1/2 2 2 2 9 1 ( 1/ 2) 1 u C u C x x C             Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Completing the Square 2 8 dx Evaluate x x   Solution Wecomplete the square to write the radicand as 2 2 2 8 ( 8 ) ( 4) 16 x x x x x         2 2 8 16 ( 4) dx dx So x x x       Let u x then du dx    4, 1 1 2 2 2 4 sin ( ) sin ( ) 8 4 4 4 dx du u x Therefore C C x x u            

3.Expanding a Power and Using a Tnigononetne Identity4.Elimmnating a Square RootEvaluate fl" fi+cos 4x dtEwaluate [(secx+tanx)'dxSolution cos4x dJ2-cos 2xSolution [(secx+tanx)'dx = [(sec* x+2seex tanx+ tan' x)dtx [(sec° x+2sec x tan x+sec x1)dr=[os2dx=-]"cos2x d=[(2sec x+2secxtanx-1)dx-0= 2 tan x+2sec xx+CAA一5:Reducing anImproperFraction6.SeparatingaFraction3x+2r-7dEraluateEvalwate3x+23x27x6Solurion We first separate the integrand to gerSolution-J(x-3+x+3x + 2xdx3x ++2一4d = 3V--yi-x3x+21n3x+2+CIn the first of these new integrals, we suhsitndeu=1-x,du=-2xd,and xdx=-du3 u//2=3J(-1/2)dt--jur"du=-$+G,=-3/-x +CJi-xVu21/2The second of the new integnals is astandard form, 2f=2sin-"x+CVi-xCombining these resalts and renaming C,+C, as Cghves3x + 2dx=-3/+2sin*x+Ci-x8.21.ProductRuleinIntegralFormIntegration byParts人AFe

2016/11/15 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Expanding a Power and Using a Trigonometric Identity 2 Evaluate x x dx (sec tan )   2 2 2 Solution (sec tan ) (sec 2sec tan tan ) x x dx x x x x dx       2 2     (sec 2sec tan sec 1) x x x x dx  2    (2sec 2sec tan 1) x x x dx      2tan 2sec x x x C Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 4. Eliminating a Square Root /4 0 Evaluate x dx 1 cos4    /4 /4 2 0 0 Solution 1 cos4 2 cos 2 x dx x dx        /4 /4 0 0 2 cos2 2 cos2 x dx x dx       /4 0 sin 2 1 2 2[ ] 2[ 0] 2 2 2 x      Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 5. Reducing an Improper Fraction 2 3 7 3 2 x x Evaluate dx x    2 3 7 6 ( 3 ) 3 2 3 2 x x Solutio dx x dx x n x         2 3 2ln 3 2 2 x      x x C Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 6. Separating a Fraction 2 3 2 1 x Evaluate dx x    Solution We first separate the t integrand o get 2 2 2 3 2 3 2 1 1 1 x xdx dx dx x x x          In the first of these new egrals we subsitute int , 2 1 1 , 2 , 2 u x du xdx and xdx du       1/2 1/2 2 1 1 2 ( 1/ 2) 3 3 3 3 3 1 1 2 2 1/ 2 xdx du u u du C x C x u                  The of the new is a form second integrals standard , 1 2 2 2 2sin 1 dx x C x      1 2 Combining these results and C as C gives renaming C  2 1 2 3 2 3 1 2sin 1 x dx x x C x          Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8.2 Integration by Parts Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 1. Product Rule in Integral Form

Integration by parts is a technique for simplifying integrals of the formIf f and g are differentiable functions of r, the Product Rule says thatUcng( - (g() + fixig (c),F(x)g(x) dxRerngng thes ohslatonwItis useful whenfcan be diferentiatedrepeatedy andg[g()dr=[Uk(]r)g)a,can be integrated repeatedlywithout difficulty.In tenns of indefinite integrals, this equation becomes1. xcos.xdtrf()=x,g(x)=cosx[()r)+R(2.jredrf(x)=x,g(x)=)ledingothe laegrayprsfula[eudit)Jam目AEEXAMPLE1FindcosxdIntegration by Parts FormmlSolutianWe une the formuls/rdewith(2)Mdr :W=X.de oosxdt,dr = d,U=snYtiirefmThenWitha properchoice ofu and v,the secondintegralcosx dr =axd xsinx+ cos+ Cmay be easier to evaluate than the first. In using theformula, various choices may be available for u and dv.There are four choices available for u and dv in Example 1:Let w = X and du = cos x dx.1Letu=Iand du= xcosxdr.2.3Letu= xcosxand du =dx.4.Letw= cosxand du= xdxBEXAMPLE2Find[axd. Choice 2 was used in Eixample I.The other three choices lead to integrals we don'tknowhow to integrate. For instance, Choioe 3 leads to the integralSolutionJinxdx can be writen as Jinx-1dr,we use the fomulaSince[(r08xPsmn)dfude-ww-Jedrwithw=Inxdr=dSinplifieswhesditlermtiateEag tointegnst1The goal ofintegration by parts is to go froms an integral J uahe that we don'tsee bouto evaluatetoan integral Fduthatwe can evaluate.Generally,you choosedufirstto bedu = ↓dr,U = x.Sinplestacaiwiyas much ofthe integrand, including dr, as you can readily imegrate; is thelefloverpartWhen finding from de,any antidernvative will work and we usually pick the simpThen from Equation (2),onbityonmofimeatioseedediuewoudsimplynouof the right-hand side of Equation (2),n-xnx-1=rn-axxnx-x+c.Sometimes we have to use integration by parts more than once.EE

2016/11/15 3 It is useful when ƒ can be differentiated repeatedly and g can be integrated repeatedly without difficulty. 1. cos x xdx  2 2. x x e dx  f x x g x x ( ) , ( ) cos   2 ( ) , ( ) x f x x g x e   With a proper choice of u and v, the second integral may be easier to evaluate than the first. In using the formula, various choices may be available for u and dv

EXAMPLE3EvalunteredSolutionWitha=,du=e'd,du= 2rd,and= er,wehaved--2xed2.Repeated UseThenew imegnal lesscomplicated than the originalbecausetheexponemt onredaced by one.To evaluate the integral on the righ, we integrate by parts again withw s, de. e' dr. Then dr dr, v e', andd-d-m'-e+c.Using this last evaluation, we thenobtain[Ped=-2adrer- 2xe+2e+C.AAThe techniqueof Example 3worksfor any integnal Jredr in which a is a positiveEinteger,becausedifferentiatingwilleventuallylead tozero and integratinge'iseasy.EXAMPLESObtain a formela that expresthe inteprEXAMPLE4EvaluateJoosxde'cos.xdrin termsofan imtegralofa loerpowerofcosSolutionLetue'and dcos xdr.Then du = edr,w =sinx.andWmaythinkThwSelutionW".00xanddr = cosx.dk,Jerosxdf e*sins/ersinxdr.so that dv = (n 1)cog* x(sinx adi)= sinxandThseondinaifihsinplafsluIegraticypantsegeseatcor-m+(n-)cdwe use integration by parts withThe evekuationf=e,du = sinxd,=005X,dy e'a.+/nrequirestwoThenecsx='sinr-(ecs-(se)integrations by= oe-xsinr + ( / or-xc (a D / 1kparts=e'sinx +e'cosre'cosxdIf we add (e = 1)/ or' x dr to both sides ofthis equation, we obtainrThe unknoun integal now appears en both sides of the eguation, Adding the integral toconc0ssi+ (nboth sides and adding the constant ofimtegration giveWe thendividhoghby,dthefinal resu2ecsin+e'cx+GDiiding by 2 and renaming the constant ofintegration gieFord-+ordBn+CTategration by FartsEatc theimegasinExercises 24using mtegaeby partsExercisesJrmtt2fooBaThe formula found in Example S is called a reduction formula becauseit replaces an is-galotainingsomepoweroffunctioewithanintegnalofthesmefomhaingtepower rduced, Whnis apesitive integer, wemay apply the fomulareeatedly unti tI. u=xdudx dy=ndx, V=2 oremaining integral is easy to evaluate.For example, the result im Example S5 tells us that+/Findx=2x c [(-2 co)dx=2x co()+4sn()+Ccosxsinr+sint+C2. =0, du= de; dvcos r9 d, v=→ sin 8;[cossinsin9=sinw8+cos+EE

2016/11/15 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Repeated Use The evaluation requires two integrations by parts. Exercises

IategratioubnPartRtig(ndr-finig()1'Gg(t)d(3)EXAMPLE6Find the area ofthe regionbounded by the curvey = xe-and thex-axisfromr=0to=4SolutionThergonisshadediFigueisa2.Evaluating Definite Integrals4xed1-ibyPartsnd dlt dr, ThenLet w =I, t[ - *~ [()d* [-4e (0) + e*dRORE&sTherpminEmpee1 (2) = 1 5e~4 0.91AA一6.TpinxdExercsesLxinxd5umln x, du=安; dv=xdx, v=号 xdx-[m],-婴=21n2-[],=2 m2--=1n4-3.Tabular Integration6.u=nx,du=d=d,V=[mx-[学],-[孕变[],-+]We have seen that integrals of the fom J f(t)g(x) dr, in which / can be differenttaterepeatedly to become zero and g can be integrated repeatedly without difficulty,arenatural candidates for integration by parts, However, if many tepetitions are requiredthe calculations can be cumbersome; or, you choose substitutions for a repeated inite-grationbypartsthat justends up giving back theoriginal integral you weretrying tofind. In situations like these, there is a way to organize the caleulations that preventthese pitfalls and makes the work much easier. It is called tabular integration and isillustrated in the following examples.六国EXAMPLEBEvaluateEXAMPLE7EvaluateredTaWin fix)SolutioWith f(r) = and g(x) = e', we lisSolutiorand gfx)Fuay and its derivativesgixy and its integrahKty and is deriativese(t) and its integrals(+)2(+)sin.2()3r().2(+)frdo(+)06cosx()We combine the products of the fiunctions connected by the arrows accoeding to the opera-tion signs above the arrows to obtaiAgainwe combinetheprodacts oftheflunctiSonsconnccted by the arrowsaccording.tothced=Pe-2+20+C.operation signsabovethearrowsto ob* sinxdt = 'cosx + 3r° sinx + 6r cosr .6sinx+ C.白目

2016/11/15 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 2. Evaluating Definite Integrals by Parts Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley 3. Tabular Integration

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

ExercisesEralute dhe ingnas isEscrcises23-32广24V1-00z25Vi-mria2V1-r024. - -2cs x =2V223. - 2/0os_ = 43.Integrals of Powers of tanx and25. s d= "sd=[sin[sin=2secx26. "pino)do=], sinedo =-cos 0 =2Slide 4- 49Slide 4- 50wwhoonetetngendsdthiruT ingatehigh-1andsecxtanr+1,andintegraks, we use the identitiesEXAMPLE6Evaluatcbyparts whennecessary toreducethe higherpouers tolowerpowersEXAMPLESEvaluatesxdJSelutionWe integrate by partsSotatdur = see* x dx.sec x tn x dSOCTtan['id[uny-tiesdt.ux (st 1)dThen[uridrxdsecrdr(tanx)(secrtandh)sectan[o Je,-1M se.x tan.(secx-1)secxdatax[aseerdr-see'xdt.In thefirst itegal, welt scx tan.x +au-scxdrw =tans,Combiningthe osecamf-cubed integralsghand have2see' xdr = sec tanx +secxd[2a-o+eandThe emg integrals are standaed forms, so[see xdr=+secxtanx+In sex +tanx + C.unsdr+ur=tn++x+cSlide 4- 51owersof TangentsandSecantsEsereisesEvaluate the integrals inExercihses33-50ec xtan x)-sec.tantdsectanxdae xtan ).see(seexta).s -tan4rmd36.seextardtn37.xd18secx.tanx.d4.ProductsofSinesandCosines33.tanx+C34.seer+C35.e'x+C36.sec'i+37.itan's+c38.tan'1+Csecx-Ean'x+Slide 4- 53Slide 4-54

2016/11/15 9 Slide 4 - 49 Exercises 2 0 23. 2[cos ] 4 2 x    0 24. 2[cos ] 2 2 x    / 2 / 2 0 /2 0 0 /2 25. cos cos cos [sin ] [sin ] 2 t dt tdt tdt t t                0 0 0 26. sin sin [cos ] 2 d d               Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 50 3. Integrals of Powers of tanx and secx Slide 4 - 51 Slide 4 - 52 Slide 4 - 53 Exercises 1 2 33. tan 2 x C 1 2 34. sec 2 x C 1 3 35. sec 3 x C (sec tan ) sec x x x  2 (sec tan ) sec x x x  2 2 (sec tan ) sec tan x x x x   2 2 2 sec tan sec x x x   1 1 5 3 36. sec sec 5 3 x x C   1 3 37. tan 3 x C 1 1 3 5 38. tan tan 3 5 x x C   Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 54 4. Products of Sines and Cosines

The integralsProducts of Sines and CosinesExercisesmx sin ar drin mr cos ar dt,tcosirdEvalate the integnalsin Exercses 5156arise in mamy applications ievohing periodic functions, We can enaloate these integals51.52.[sin2x cos 3rdesin 3x c0s 2x drthroagh integration by paris,but tao suchintezrationsarereguired in each casc.It is simplerto use the identities154.53.sin 3r sin 3r drsin x cos x dt(3)sin mr sin ex = [cos (m n)r cos (m + n)i],(sin(ma) + sin (e + ah)](9)sin mux. cos a55.csJrcos4tdr56.cosx.cos7rdecos mscosr -[cos (m n)s + cos(m + m)(5)51..-iocosSr+c52. ↓0os.x-1ocowsSr+cEXAMPLE7COSTEvaloatesin 3r cos 5r dt.54 -100 2r -53. 5+-1sim6x,= .TFrom Equation (4) withth m =3 and w= 5, we.getolutia55.sinx+sin7x+c5s Hgsnor+Jamaxn =0srcos5r-lsin(22)+tin8/a2)+slide 4-55Slide 4-58Trigonometric substitutions occur when we replace the variable of integration by atrigonometric function, The most common suhstitutions are x = a tan e,x = α sin e, andXsecThesesubstitutionsare eflective intransfoming integrals imvolvingV+VaandVainto itegals we canevaluatedirectly since thy omefrom threference righttriangles in Figure 8.28.4VV-Va-PTrigonometric Substitutions1etntV-edes=dsdV+-acVa=-oneV-7-maFIGUREB.2Refriangles fiur the firce bodemtifying tbesides labeledxanda forcachsubstintioWithr -asine,-aas0-(in0)-oWithratano,a+xma"+atan"=a'(l+tane)=a"sec"eWith.x aseco,Xg2"sec0αa(sec*01)=atan.0Slide 4- s7Slide 4- 5a1*= atane roguires 6= tin (with号 0.angle e for reversibility.4. Draw an appropriate reference triangle to reverse the substitution in the inte-gration result and comvert it back to the original variable x.堂sFnGuRt s.3The arctingemt aresine, andirosecant ofx/a, gnphed as functions ofx/a.0号xasecerequiresesecSlide 4-60

2016/11/15 10 Slide 4 - 55 Slide 4 - 56 Exercises 1 1 51. cos cos5 2 10    x x C 1 1 52. cos cos5 2 10 x x C   1 1 53. [ sin 6 ] 2 12 x x      /2 0 1 1 54. [cos 2 ] 4 2 x    1 1 55. sin sin 7 2 14 x x C   /2 /2 1 1 1 56. [ sin 6 sin8 ] 0 2 6 8 x x     Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 57 8.4 Trigonometric Substitutions Slide 4 - 58 Slide 4 - 59 Slide 4 - 60