This has the e64256ect of changing the variable and the integran d When dealing with de64257nite integrals the limits of integrat ion can also change In this unit we will meet several examples of integrals where it is appropri ate to make a substitu ID: 27546
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IntegrationbysubstitutionThereareoccasionswhenitispossibletoperformanapparentlydicultpieceofintegrationby rstmakingasubstitution.Thishasthee ectofchangingthevariableandtheintegrand.Whendealingwithde niteintegrals,thelimitsofintegrationcanalsochange.Inthisunitwewillmeetseveralexamplesofintegralswhereitisappropriatetomakeasubstitution.Inordertomasterthetechniquesexplainedhereitisvitalthatyouundertakeplentyofpracticeexercisessothattheybecomesecondnature.Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto:carryoutintegrationbymakingasubstitutionidentifyappropriatesubstitutionstomakeinordertoevaluateanintegralContents1.Introduction22.Integrationbysubstitutingu=ax+b23.FindingZf(g(x))g0(x)dxbysubstitutingu=g(x)61c\rmathcentreDecember1,2008 1.IntroductionThereareoccasionswhenitispossibletoperformanapparentlydicultpieceofintegrationby rstmakingasubstitution.Thishasthee ectofchangingthevariableandtheintegrand.Whendealingwithde niteintegrals,thelimitsofintegrationcanalsochange.Inthisunitwewillmeetseveralexamplesofthistype.Theabilitytocarryoutintegrationbysubstitutionisaskillthatdevelopswithpracticeandexperience.Forthisreasonyoushouldcarryoutallofthepracticeexercises.Beawarethatsometimesanapparentlysensiblesubstitutiondoesnotleadtoanintegralyouwillbeabletoevaluate.Youmustthenbepreparedtotryoutalternativesubstitutions.2.Integrationbysubstitutingu=ax+bWeintroducethetechniquethroughsomesimpleexamplesforwhichalinearsubstitutionisappropriate.ExampleSupposewewantto ndtheintegralZ(x+4)5dx(1)Youwillbefamiliaralreadywith ndingasimilarintegralZu5duandknowthatthisintegralisequaltou6 6+c,wherecisaconstantofintegration.Thisisbecauseyouknowthattheruleforintegratingpowersofavariabletellsyoutoincreasethepowerby1andthendividebythenewpower.IntheintegralgivenbyEquation(1)thereisstillapower5,buttheintegrandismorecompli-catedduetothepresenceofthetermx+4.Totacklethisproblemwemakeasubstitution.Weletu=x+4.Thepointofdoingthisistochangetheintegrandintothemuchsimpleru5.However,wemusttakecaretosubstituteappropriatelyforthetermdxtoo.Intermsofdi erentialswehavedu=du dxdxNow,inthisexample,becauseu=x+4itfollowsimmediatelythatdu dx=1andsodu=dx.So,substitutingbothforx+4andfordxinEquation(1)wehaveZ(x+4)5dx=Zu5duTheresultingintegralcanbeevaluatedimmediatelytogiveu6 6+c.Wecanreverttoanexpressioninvolvingtheoriginalvariablexbyrecallingthatu=x+4,givingZ(x+4)5dx=(x+4)6 6+cWehavecompletedtheintegrationbysubstitution.c\rmathcentreDecember1,20082 ExampleSupposenowwewishto ndtheintegralZcos(3x+4)dx(2)Observethatifwemakeasubstitutionu=3x+4,theintegrandwillthencontainthemuchsimplerformcosuwhichwewillbeabletointegrate.Asbefore,du=du dxdxandsowithu=3x+4anddu dx=3Itfollowsthatdu=du dxdx=3dxSo,substitutingufor3x+4,andwithdx=1 3duinEquation(2)wehaveZcos(3x+4)dx=Z1 3cosudu=1 3sinu+cWecanreverttoanexpressioninvolvingtheoriginalvariablexbyrecallingthatu=3x+4,givingZcos(3x+4)dx=1 3sin(3x+4)+cWehavecompletedtheintegrationbysubstitution. Itisveryeasytogeneralisetheresultofthepreviousexample.Ifwewantto ndZcos(ax+b)dx,thesubstitutionu=ax+bleadsto1 aZcosuduwhichequals1 asinu+c,thatis1 asin(ax+b)+c.Asimilarargument,whichyoushouldtry,showsthatZsin(ax+b)dx= 1 acos(ax+b)+c. KeyPointZsin(ax+b)dx= 1 acos(ax+b)+cZcos(ax+b)dx=1 asin(ax+b)+c 3c\rmathcentreDecember1,2008 ExampleSupposewewishto ndZ1 1 2xdx.Wemakethesubstitutionu=1 2xinordertosimplifytheintegrandto1 u.Recallthattheintegralof1 uwithrespecttouisthenaturallogarithmofu,lnjuj.Asbefore,du=du dxdxandsowithu=1 2xanddu dx= 2Itfollowsthatdu=du dxdx= 2dxTheintegralbecomesZ1 u 1 2du= 1 2Z1 udu= 1 2lnjuj+c= 1 2lnj1 2xj+c Theresultofthepreviousexamplecanbegeneralised:ifwewantto ndZ1 ax+bdx,thesubstitutionu=ax+bleadsto1 aZ1 uduwhichequals1 alnjax+bj+c.Thismeans,forexample,thatwhenfacedwithanintegralsuchasZ1 3x+7dxwecanimme-diatelywritedowntheansweras1 3lnj3x+7j+c. KeyPointZ1 ax+bdx=1 alnjax+bj+c c\rmathcentreDecember1,20084 Alittlemorecaremustbetakenwiththelimitsofintegrationwhendealingwithde niteintegrals.Considerthefollowingexample.ExampleSupposewewishto ndZ31(9+x)2dxWemakethesubstitutionu=9+x.Asbefore,du=du dxdxandsowithu=9+xanddu dx=1Itfollowsthatdu=du dxdx=dxTheintegralbecomesZx=3x=1u2duwherewehaveexplicitlywrittenthevariableinthelimitsofintegrationtoemphasisethatthoselimitswereonthevariablexandnotu.Wecanwritetheseaslimitsonuusingthesubstitutionu=9+x.Clearly,whenx=1,u=10,andwhenx=3,u=12.SowerequireZu=12u=10u2du=1 3u31210=1 3 123 103=728 3Notethatinthisexamplethereisnoneedtoconverttheanswergivenintermsofubackintooneintermsofxbecausewehadalreadyconvertedthelimitsonxintolimitsonu.Exercises1.1.Ineachcaseuseasubstitutionto ndtheintegral:(a)Z(x 2)3dx(b)Z10(x+5)4dx(c)Z(2x 1)7dx(d)Z1 1(1 x)3dx.2.Ineachcaseuseasubstitutionto ndtheintegral:(a)Zsin(7x 3)dx(b)Ze3x 2dx(c)Z=20cos(1 x)dx(d)Z1 7x+5dx.5c\rmathcentreDecember1,2008 3.FindingZf(g(x))g0(x)dxbysubstitutingu=g(x)ExampleSupposenowwewishto ndtheintegralZ2xp 1+x2dx(3)Inthisexamplewemakethesubstitutionu=1+x2,inordertosimplifythesquare-rootterm.Weshallseethattherestoftheintegrand,2xdx,willbetakencareofautomaticallyinthesubstitutionprocess,andthatthisisbecause2xisthederivativeofthatpartoftheintegrandusedinthesubstitution,i.e.1+x2.Asbefore,du=du dxdxandsowithu=1+x2anddu dx=2xItfollowsthatdu=du dxdx=2xdxSo,substitutingufor1+x2,andwith2xdx=duinEquation(3)wehaveZ2xp 1+x2dx=Zp udu=Zu1=2du=2 3u3=2+cWecanreverttoanexpressioninvolvingtheoriginalvariablexbyrecallingthatu=1+x2,givingZ2xp 1+x2dx=2 3(1+x2)3=2+cWehavecompletedtheintegrationbysubstitution. Letusanalysethisexamplealittlefurtherbycomparingtheintegrandwiththegeneralcasef(g(x))g0(x).Supposewewriteg(x)=1+x2andf(u)=p uThenwenotethatthecomposition1ofthefunctionsfandgisf(g(x))=p 1+x2. 1when ndingthecompositionoffunctionsfandgitistheoutputfromgwhichisusedasinputtof,resultinginf(g(x))c\rmathcentreDecember1,20086 Further,wenotethatifg(x)=1+x2theng0(x)=2x.SotheintegralZ2xp 1+x2dxisoftheformZf(g(x))g0(x)dxToperformtheintegrationweusedthesubstitutionu=1+x2.Inthegeneralcaseitwillbeappropriatetotrysubstitutingu=g(x).Thendu=du dxdx=g0(x)dx.OncethesubstitutionwasmadetheresultingintegralbecameZp udu.InthegeneralcaseitwillbecomeZf(u)du.Providedthatthis nalintegralcanbefoundtheproblemissolved.Forpurposesofcomparisonthespeci cexampleandthegeneralcasearepresentedside-by-side:Z2xp 1+x2dx Zf(g(x))g0(x)dxletu=1+x2 letu=g(x)du=du dxdx=2xdx du=du dxdx=g0(x)dxZ2xp 1+x2dx=Zp udu Zf(g(x))g0(x)dx=Zf(u)du2 3u3=2+c 2 3(1+x2)3=2+c KeyPointToevaluateZf(g(x))g0(x)dxsubstituteu=g(x),anddu=g0(x)dxtogiveZf(u)duIntegrationisthencarriedoutwithrespecttou,beforerevertingtotheoriginalvariablex. Itisworthpointingoutthatintegrationbysubstitutionissomethingofanart-andyourskillatdoingitwillimprovewithpractice.Furthermore,asubstitutionwhichat rstsightmightseemsensible,canleadnowhere.Forexample,ifyouweretryto ndRp 1+x2dxbylettingu=1+x2youwould ndyourselfupablindalley.Bepreparedtopersevereandtrydi erentapproaches.7c\rmathcentreDecember1,2008 ExampleSupposewewishtoevaluateZ4x p 2x2+1dxBywritingtheintegrandas1 p 2x2+14xwenotethatittakestheformZf(g(x))g0(x)dxwheref(u)=1 p u,g(x)=2x2+1andg0(x)=4x.Thesubstitutionu=g(x)=2x2+1transformstheintegraltoZf(u)du=Z1 p uduThisisevaluatedtogiveZ1 p udu=Zu 1=2du=2u1=2+cFinally,usingu=2x2+1toreverttotheoriginalvariablegivesZ4x p 2x2+1dx=2(2x2+1)1=2+corequivalently2p 2x2+1+cExampleSupposewewishto ndZsinp x p xdx.Considerthesubstitutionu=p x.Thendu=du dxdx=1 2x 1=2dx=1 2x1=2dx=1 2p xdxsothatZsinp x p xdx=2Zsinudufromwhich2Zsinudu= 2cosu+c= 2cosp x+cWecanalsomakethefollowingobservations:theintegrandcanbewrittenintheformsinp x1 p x.c\rmathcentreDecember1,20088 Writingf(u)=sinuandg(x)=p xtheng0(x)=1 2x 1=2=1 2x1=2=1 2p x.Further,f(g(x))=sinp x.Hencewewritethegivenintegralas2Zsinp x1 2p xdxwhichisoftheform2Zf(g(x))g0(x)dxwithfandgasgivenabove.Asbeforethesubstitutionu=g(x)=p xproducestheintegral2Zf(u)du=2Zsinudufromwhich2Zsinudu= 2cosu+c= 2cosp x+cExercises21.Ineachcasetheintegrandcanbewrittenasf(g(x))g0(x).Identifythefunctionsfandgandusethegeneralresultonpage7tocompletetheintegration.(a)Z2xex2 5dx(b)Z 2xsin(1 x2)dx(c)Zcosx 1+sinxdx.2.Ineachcaseusethegivensubstitutionto ndtheintegral:(a)Z 2xe x2dx,u= x2.(b)Zxsin(2x2)dx,u=2x2.(c)Z50x3p x4+1dx,u=x4+1.3.Ineachcaseuseasuitablesubstitutionto ndtheintegral.(a)Z5xp 1 x2dx(b)Zdx p x(1+p x)2(c)Zx4(1+x5)3dx(d)Zx3 p x4+16dx(e)Zcosx (5+sinx)2(f)Z10x3 p x4+12dx(g)Z5x2p 1 x3dx(h)Zecosxsinxdx(i)Zesinxcosxdx.9c\rmathcentreDecember1,2008 AnswerstoExercisesExercises11.(a)(x 2)4 4+c(b)4651 5=9301 5(c)(2x 1)8 16+c(d)4.2.(a) cos(7x 3) 7+c(b)e3x 2 3+c(c)1.382(3dp)(d)1 7lnj7x+5j+cExercises21.(a)f(u)=eu,g(x)=x2 5,ex2 5+c,(b)f(u)=sinu,g(x)=1 x2, cos(1 x2)+c(c)f(u)=1 u,g(x)=1+sinx,lnj1+sinxj+c.2.(a)e x2+c(b) cos(2x2) 4+c(c)2610(4sf).3.(a) 5 3(1 x2)3=2+c(b) 2 1+p x+c(c)1 20(1+x5)4+c(d)1 2(x4+16)1=2+c(e) 1 5+sinx+c(f)0.0707(g) 10 9(1 x3)3=2+c(h) ecosx+c(i)esinx+c.c\rmathcentreDecember1,200810