《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