Techniques of Integration

1 CHAPTER7 Techniquesof Integration ,unlike differentiation,is moreofanart-formthana problemsinappliedmathematicsinvolve theintegrationoffunctionsgivenbycomplica tedformulae, beableto usetheTableseffectively. Theseare:substitution, , ( ).To integratea differen-tialf x dxwhichis notinthetable,wefirstseeka functionu u x sothatthegivendifferentialcanberewritten asa differentialg u duwhichdoesappearin ,if g u du G u C, weknowthat f x dx G u x C. x 2x 1dx ?Letu 2x 1, sothatdu 2dxandx u 1 x 2x 1dx u 12u1 2du2 14 u3 2 u1 2 du 14 25u5 2 23u3 2 C( ) 130u3 2 3u 5 C 130 2x 1 3 2 6x 2 C 115 2x 1 3 2 3x 1 C ( )whereat theendwehave replaceduby2x tanxdx ?

2 107 Chapter7 TechniquesofIntegration108 Sincethisisn t onourtables,wereverttothedefinitionofthe tangent:tanx sinx cosx. Then,lettingu cosx du sinxdxweobtain( ) tanxdx sinxcosxdx duu lnu C ln cosx C ln secx C secxdx ?.Thisistricky, , if weareguidedbytheprincipleofrewritinginte rmsofsinesandcosines,weareledtothefollow ing:( )secx 1cosx cosxcos2x cosx1 sin2x Now wecantrythesubstitutionu sinx du cosxdx. Then( ) secxdx du1 u2 Thislookslike a deadend,buta ( )11 u2 12 11 u 11 u leadsto( ) du1 u2dx 12 11 u 11 u du 12 ln 1 u ln 1 u C Usingu sinx, wefinallyendupwith( ) secxdx 12 ln 1 sinx ln 1 sinx C 12ln 1 sinx1 sinx C circlerollsalonga horizontalline,a pointonthecircletraversesa curve calledthecycloid.

3 Aloopofthecycloidis thetrajectoryofa ,inradians,andstart(att 0)withthepointofintersectionPofthecircle andthelineonwhichit is ,thepositionofthepointis nowtunitstotherightoftheoriginalpointofc ontact(assumingnoslippage),so( )x t t sint y t 1 cost To findarclength,weuseds2 dx2 dy2, wheredx 1 cost dt dy sintdt. Thus( )ds2 1 cost 2 sin2t 2dt2 2 2 cost 2dt2sods 2 1 costdt, andthearclengthis givenbytheintegral( )L 2 2 0 1 costdt costt sintPTo evaluatethisintegralbysubstitution,wenee da factorofsint. We cangetthisbymultiplyinganddividingby 1 cost:( ) 1 cost 1 cos2t 1 cost sint 1 cost Bysymmetryaroundthelinet , theintegralwillbetwicetheintegralfrom0 to.

4 Inthatinterval,sintis positive, , thesubstitutionu cost du 0 u 1, andwhent u ( )L 2 2 11u 1 2du 2 2 1 1u 1 2du 2 2 2u1 2 1 1 8 2 PartsSometimeswecanrecognizethedifferent ialtobeintegratedasa productofa functionwhichis easilydifferentiatedanda differentialwhichis ,if theproblemis tofind( ) xcosxdxthenwecaneasilydifferentiatef x x, andintegratecosxdxseparately. Whenthishappens,theintegralversionofthep roductrule,calledintegrationbyparts, maybeuseful,becauseit interchangestherolesofthetwo :d uv udv vdu, andrewriteit as( )udv d uv ,takingu x dv cosxdx, wehavedu dx v sinx.

5 :( )xcosxdx d xsinx sinxdx Chapter7 TechniquesofIntegration110andwecaneasily integratetherighthandsidetoobtain( ) xcosxdx xsinx sinxdx xsinx cosx C (IntegrationbyParts)Foranytwodifferentia blefunctionsu andv:( ) udv uv vdu To ,anda dvanddu x dv exdx. Thendu dx v ex. ( ) xexdx xex exdx xex ex C x2 dv exdx du 2xdx v exdoesn t immediatelysolve theproblem,butreducesustoexample3:( ) x2exdx x2ex 2 xexdx x2ex 2 xex ex C x2ex 2xex 2ex C find lnxdx, weletu lnx dv dx, sothatdu 1 x dx v x, and( ) lnxdx xlnx x1xdx xlnx dx xlnx x C Thissameideaworksforarctanx: Let( )u arctanx dv dxdu dx1 x2 v x andthus( ) arctanx xarctanx x1 x2dx xarctanx 12ln 1 x2 C wherethelastintegrationis accomplishedbythenew substitutionu 1 x2 du 2xdx.

6 Tointegrateexcosxdx. Weseethatanintegrationbypartsleadsustoin tegrateexsinxdx, whichis :Lettingu ex dv cosxdx du exdx v sinx, weget( ) excosxdx exsinx exsinxdx Now integratebypartsagain:lettingu ex dv sinxdx du exdx v cosx, weget( ) exsinxdx excosx excosxdx ( ) excosxdx exsinx excosx excosxdx Bringingthelasttermovertothelefthandside anddividingby2 givesustheanswer:( ) excosxdx 12 exsinx excosx C a calculationofa definiteintegralinvolvesintegrationbypar ts,it is a goodideatoevaluateassoonasintegratedterm sappear. We illustratewiththecalculationof( ) 41lnxdxLetu lnxdx dv dxsothatdu dx x v x, and( ) 41lnxdx xlnx 41 41dx 4 ln 4 x 41 4 ln 4 3 ( ) 1 20arcsinxdx ?

7 We make thesubstitutionu arcsinx dv dx du dx 1 x2 v x. Then( ) 1 20arcsinxdx xarcsinx 1 20 1 20xdx 1 x2 Now, tocompletethelastintegral,letu 1 x2 du 2xdx, leadingusto( ) 1 20arcsinxdx 12 6 12 3 41u 1 2du 12 u1 2 3 41 12 32 1 Chapter7 TechniquesofIntegration112 thatintegrationofa rationalfunctioncanbereducedtothefollowi ngformulae,oncewehave ) dxx a ln x a C b) duu2 b2 1barctan ub C c) uduu2 b2 12ln u2 b2 C Theseareeasilyverifiedbydifferentiatingt herighthandsides(orbyusingprevioustechni ques). ve findtheintegral( ) dx x a x b wecheckthat( )1 x a x b 1a b 1x a 1x b sothat( ) dx x a x b 1a b ln x a ln x b C 1a bln x ax b C polynomialcanbewrittenasa productoffactorsoftheformx ror x a 2 b2, whereris a realrootandthequadratictermscorrespondto theconjugatepairsofcomplex sumoftermswhosedenominatorsareofthesefor ms,andthustheintegrationis rationalfunctionR x , if thedegreeofthenumeratorisnotlessthanthed egreeofthedenominator, bylongdivision,wecanwrite( )R x Q x p x q x wherenow degp x therootsarealldistinct(therearenomultipl eroots)

8 ,writep qasa sumoftermsoftheform( )Ax r B x a 2 b2 Cx x a 2 b2 B C . therootsarenotdistinct,theexpansionis morecomplicated; ,andhow tofindthecoefficientsA B xdx x 1 x 2 Firstwewrite( )x x 1 x 2 Ax 1 Bx 2 Now multiplythisequationby x 1 x 2 , getting( )x A x 2 B x 1 If wesubstitutex 1, weget1 A 1 2 , soA 1;now lettingx 2, weget2 B 2 1, soB 2, ( )x x 1 x 2 1x 1 2x 2 Integrating,weget( ) xdx x 1 x 2 ln x 1 2 ln x 2 C ln x 2 2 x 1 C So,thisis equation; x2 3 dx x2 1 x 3 Heretherootsare 1 3, sowehave theexpansion( )x2 3 x2 1 x 3 Ax 1 Bx 1 Cx 3leadingto( )x2 3 A x 1 x 3 B x 1 x 3 C x 1 x 1 Substitutex 1 : 1 3 A 2 4 , soA 1 1 : 1 3 B 2 2 , soB 1 3 : 9 3 C 4 2 , soC 3 4, ( )x2 3 x2 1 x 3 14 1x 1 12 1x 1 34 1x 3 andtheintegralis( ) x2 3 dx x2 1 x 3 14ln x 1 12ln x 1 34ln x 3 C Chapter7 TechniquesofIntegration114 dxx2 4x 5 ?

9 Herewecanfactor:x2 4x 5 x 1 x 5 , sowecanwrite( )1x2 4x 5 Ax 1 Bx 5andsolve forAandBasabove:A 1 6 B 1 6, sowehave( )1x2 4x 5 16 1x 5 1x 1 andtheintegralis( ) dxx2 4x 5 16ln x 5x 1 C dxx2 4x 5 ?Herewecan t findrealfactors, :x2 4x 5 x 2 2 1, andnow :( ) dxx2 4x 5 dx x 2 2 1 arctan x 2 C x 3 dxx2 4x 5 ? Herewehave tobea ,wecompletethesquare,giving( )x 3x2 4x 5 x 3 x 2 2 1 If onlythatx 3 werex 2, ,withu x ,sincex 3 x 2 5,thereis noproblem: x 3 dxx2 4x 5 x 2 dx x 2 2 1 5dx x 2 2 1 12ln x 2 2 1 5 arctan x 2 C ( ) 2x 1 dxx2 6x 14 ? First,wecompletethesquareinthedenominato r:x2 6x 14 x 3 2 , writethenumeratorintermsofx 3 : 2x 1 2 x 3 :( ) 2x 1 dxx2 6x 14 7x2 6x 14 2x 3x2 6x 14 , :( ) 2x 1 dxx2 6x 14 7 dx x 3 2 5 2 x 3 dx x 3 2 5( ) 7 5arctanx 3 5 ln x 3 2 5 C x 1 dxx x2 1 ?

10 Herewehave , sowetryanexpressionoftheform( )x 1x x2 1 Ax Bx2 1 Cxx2 1 Clearingthedenominatorsontheright,wearel edtotheequation( )x 1 A x2 1 Bx Cx2 Settingx 0 gives1 A. Butwehave nomorerootstosubstitutetofindBandC, 0,andontherightisA C, soA C 0;sinceA 1,welearnthatC B. ( )x 1x x2 1 1x 1x2 1 xx2 1 andourintegralis( ) x 1 dxx x2 1 ln x arctanx 12ln x2 1 C x2 1 dxx x2 4x 5 ? Thedenominatorisx x 2 2 1 , soweexpecta partialfractionsexpansionoftheform( )x2 1x x2 4x 5 Ax B x 2 2 1 C x 2 x 2 2 1 Clearingofdenominators,weobtaintheequati on( )x2 1 A x 2 2 1 Bx C x 2 x Forx 0,weobtain1 A 5 , soA 1 A C, soC 1 4A B 2C, so0 4 5 B 2 5,soB 2 5 ( )x2 1x x2 4x 5 15 1x 25 1 x 2 2 1 15 x 2 x 2 2 1 Chapter7 TechniquesofIntegration116whichwecaninte grateto( )