Math116 notes Indefinite Integrals1 Antiderivatives

bFTCItellsusthattocalculateadefiniteintegralfafandx allwehavetodois 111findanantiderivativeFanoffanthen121usetheformulafabfaddx Fcb Fca Anyantiderivativewillwork Fortunately anytwoantiderivativeswillonlydifferbyaconstant

FAIT tf FCHandFix aretwoantiderivativesoffox thenFiCx Fix CforsomeconstantCeIRProof WeknowthatFiatandFix havethesamederivativebecauseFIM fan E'Cx ThusFix Fix 0 Theonlywayafunctioncanhavezeroderivative everywhereisif itisaconstantfunction thusEN Fax CforsomeconstantCcIR Blas

Therefore antiderivativesareuniquelydetermineduptoaconstant Anotherwordforantiderivativeis indefiniteintegral andweusethenotation

JfcxDX Fix C a generalantiderivative

toexpressthatF'Cx fix akaindefiniteintegral

Easyexamples f2xdx X'tC fsincxidx cosh C I edx exC ftxdx InAltC etc

So ffindxrepresentsthegeneralantiderivativeoffanReversed'owwewrigheaddduineaurity assumingnto

sincethederivativeofX is nx thismeansthattheantiderivativeoffix is x Therefore wehavethefollowing reverse versionofthepowerrule

find IT C a worksforany n1 1Nelewhenn 1 fx dx InCx CAlsosincedifferentiationislinear so isintegration

fflxitglxldx ffcxldx fg.INdx Jkfixidx Kffexidx foranyconstantKEIR

Examplet Calculatetheindefiniteintegral153 3 7dx Uselinearitytosplititup thenapplythereversepowerruletoeachterm f1dxfdx

153 3 7dx 513dx 3Xdx 7IDI ITX 7 2 7 9 ponyforget cExampleI Calculate 191Sin3 1 te Dx Againuselinearitytosplititup

1 91SinBx e dx 1 9dx Isin131d fe4xd If SHHdx FanthenIfCkxdx f Kx

tox Eason te C egig II w

2 Substitution ReverseChainRule

IfFMisanantiderivativeoffix thenbythechainRule FCgullisanantiderivativeofflgcxllgtxl.BRfCgcxDg4x1dx FCgcx CAlternativelysetu gun ThenTY gM sotheaboveformulacanbewrittenas IfTheexpressiondug4x7dx

ffcgcxhgkxidx ffcuddI dx f.fiudu Flu C niggledandifferenutiaFJ

Thisiscalledthesubstitutionmethod becausewesubstitute uforgallNOTATION peopletendtowrite du gCHdxforsubstitutions Thisiscalledadifferential

Examplet f2xSinaidx Tomakeasubstitution locateaninsidefunctionwhosedifferentialisalsoafactorintheintegrand Inthiscase u x seemstoworkbecauseitsdifferential u 2xDXappearsintheintegrand

f2xsincedx fsina.ug4zgxudx fs.inudu coscu t C costx7 t C

ExampleI I fissimnexdx Inthiscasewenotice a Sinanisagoodsubstitution becauseitsdifferentialdu DXappearsas afactor

I I x dx sindx fTudu InAtul C InCltSinai C

Example3 1 2 I1 3132dx Onceagainidentifyan insidefunction Inthiscase U seemstoworkandthedifferentialisdu DXSo

1 2111 31312DX f u312du Zzu t C Zz113,52 C

Examplett Stancxidx Writetanks yougetStanandx Y DX Nowthesubstitutionshouldbeobvious U cosCx sothatdu Sinaidx or du SinxDX So

Stankidx 1 542 dx futdu IntultC InKosal Cx11

Example5Hereisanexampleusingadefiniteintegral x2 2 3DX Ifyousubstituteu x2 2 3 thenthedifferential isdu 2 2dx whichyoucanwriteas Idu Cx11Dx Withdefiniteintegrals youalsohavetokeeptrackoftheBOUNDSoftheintegral thesymbol S means from I to 2 butwechangedvariablesto uSowehavetoputtheboundsintermsofutoo

x I u 112 24 3 6X 2 u 212212 3 11 Notation

Sothenewintegralaftersubstituting a x2 2 3 is gt a F

1,2 24 3dx f ut Edu If tudu Elnlull In1111 tha th tExercisesEvaluatethefollowingintegralsbysubstitution

earctanct il Ix'sin dx 2 fFjdx 3 I dt 4 foi dx5 frattantodo 6 fxedx 7 fr dt 8 fivedy

3 IntegrationByParts ReverseProductRuledu duTodifferentiateaproduct uv youusetheproductrule luv u d v d Ifyouintegratebothsides

withrespecttoX yougetuv fansdx fu vd dx fu dXtfvf DX fudv Ivdu

Nowsolveforfudvtwo fudv w fvda

Thisiscalledtheintegrationbypartsformula Here'stheidea ifyouwanttosolveanintegralofaproducti e Somethinglike Sfangkxidx youmaketwosubstitions u fix anddv g41DX Then

ffcxigixldx fudv uv fvdu flxlgcxl fgcxf.liDxThehopeisthattheintegral fg fixdxiseasierthantheoriginal fflxlg4xldxExamplet Let'suseintegrationbypartstosolve Ixcos DXYouhavetofigureoutuanddvsothattheintegralhastheformJudvWehavetwooptions

fxwc dy or feskyxdu dv U dv

Let'susethefirstone sou xanddv cos DXThenduDxandVSinai sotheIBPformulagives

Ixcoscxidx fudv uv fvdu xsinlxl fsinlxldxxsinlxltcoslxltc.eduExerciseforyoutryoption andconvinceyourselfthatitdoesn'tworkExampled fv dx TouseIBPonthisonesubin u x42xanddv p dx sothatdu dxx11 1andv ThentheIBPformulagives

x2t2x dx fudv w vdu 12121Witt 2VxtT2 2dx II 21427717 4 1 113dx 212 21VxT 415 111 C Ix22xNxtT f x11752 C

Example3 I InH1dx Letu In11anddu dx sothatdu DXandv Then

I xY3lnlxtdx fudv uv fvdu fx4l3lnlx7 fZx4l3yfdx o.oGo.o 3

443 Z143dx 4 43 ff t C C

Example4 Inkdx Thisisaweirdonebutwe'lluseIBP letu In11anddvdx There'sreallynootheroptionThendu XDXandv x so

IInlxidx w fdu XInk fdx Inx X t CExercise UseIBPtoevaluate farctancxidx

Example5 Here'satoughie feisincxDx Let'stryIBPwith aex du sincxidx thendu e dxandv cosX so

Iexsincxdx excosx f coshedx ecosCx Jexcos DXHmmWegot fexcosandx Let'stryIBPagainwithu e dv cos dx

fexcoscxidx exsincxl fexs.inCxDxWhat We'rebackto feisincxidxAGAIN Here'sthetrick let IAI fexsinandx Thenwejustshowed

I x excoscx tf excoscxidx ecos1 1 Isinx IHSowecansolveforIG

Iexcosandx I e IsinCH Cosa c2

Moreexercises UseANYmethodtoevaluatethefollowingintegralsy HARD ga 15sin3 1DX b If 7dt c 1 2 4 7Dt HintforCclcompletethesquareThinkaboutarctan

d 7wVFwTdw e 18W dw Ifl f upDX HintforCfl completethesquareagain

g fyedy 1h11YeYdy lil 131nextdx j cos4oldo HintforCjltrigidentityHARD HARD xk Jxarctankidx let fzezdx m fVmaMdm FMIV dx Hintforinisin'tcost

MoreexercisesfromthetextbookSection8l substitution p493 I 222426293134Section8.2 integrationbyparts p498 I 183638