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8.3 TrigonometricSubstitutions
171
Sofarwehave seenthat it sometimeshelpstoreplaceasubexpression ofafunctionby
a singlevariable. . Occasionallyitcanhelpto o replace theoriginalvariablebysomething
morecomplicated. Thisseemslikea\reverse"substitution,butitisreallynodierentin
principlethanordinarysubstitution.
EXAMPLE8.3.1
Evaluate
Z
p
1 x2dx. . Letx=sinusodx=cosudu.Then
Z
p
1 x2dx=
Z
p
1 sin
2
ucosudu=
Z
p
cos2ucosudu:
Wewouldliketoreplace
p
cos2ubycosu,butthisisvalidonlyifcosuispositive,since
p
cos2uispositive.Consideragainthesubstitutionx=sinu.Wecouldjustaswellthink
ofthisasu=arcsinx.Ifwedo,thenbythedenitionofthearcsine, =2u=2,so
cosu0.Thenwecontinue:
Z
p
cos2ucosudu=
Z
cos
2
udu=
Z
1+cos2u
2
du=
u
2
+
sin2u
4
+C
=
arcsinx
2
+
sin(2arcsinx)
4
+C:
Thisisaperfectlygoodanswer,thoughthetermsin(2arcsinx)isabitunpleasant. Itis
possibleto simplifythis. Usingtheidentitysin2x=2sinxcosx,wecan n writesin2u=
2sinucosu=2sin(arcsinx)
p
1 sin
2
u=2x
q
1 sin
2
(arcsinx)=2x
p
1 x2:Thenthe
fullantiderivativeis
arcsinx
2
+
2x
p
1 x2
4
=
arcsinx
2
+
x
p
1 x2
2
+C:
Thistypeofsubstitutionisusuallyindicatedwhenthefunctionyouwishtointegrate
containsa polynomialexpression that might allow youtouse the fundamentalidentity
sin
2
x+cos
2
x=1inoneofthreeforms:
cos
2
x=1 sin
2
x
sec
2
x=1+tan
2
x
tan
2
x=sec
2
x 1:
Ifyourfunctioncontains1 x
2
,asintheexampleabove,tryx=sinu;ifitcontains1+x
2
tryx =tanu; ; andif it t containsx
2
1, tryx=secu. Sometimesyouwillneed d totry
somethingabitdierenttohandleconstantsotherthanone.
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172
Chapter 8 8 Techniques s ofIntegration
EXAMPLE8.3.2 Evaluate
Z
p
4 9x2dx.Westartbyrewritingthissothatitlooks
morelikethepreviousexample:
Z
p
4 9x2dx=
Z
p
4(1 (3x=2)2)dx=
Z
2
p
1 (3x=2)2dx:
Nowlet3x=2=sinuso(3=2)dx=cosuduordx=(2=3)cosudu. Then
Z
2
p
1 (3x=2)2dx=
Z
2
p
1 sin
2
u(2=3)cosudu=
4
3
Z
cos
2
udu
=
4u
6
+
4sin2u
12
+C
=
2arcsin(3x=2)
3
+
2sinucosu
3
+C
=
2arcsin(3x=2)
3
+
2sin(arcsin(3x=2))cos(arcsin(3x=2))
3
+C
=
2arcsin(3x=2)
3
+
2(3x=2)
p
1 (3x=2)2
3
+C
=
2arcsin(3x=2)
3
+
x
p
4 9x2
2
+C;
usingsomeoftheworkfromexample8.3.1.
EXAMPLE8.3.3
Evaluate
Z
p
1+x2dx. Letx=tanu,dx=sec
2
udu,so
Z
p
1+x2dx=
Z
p
1+tan
2
usec
2
udu=
Z
p
sec2usec
2
udu:
Sinceu=arctan(x), =2u=2andsecu0,so
p
sec2u=secu.Then
Z
p
sec2usec
2
udu=
Z
sec
3
udu:
Inproblemsof thistype,two integralscome e up p frequently:
Z
sec
3
udu and
R
secudu.
Bothhaverelativelyniceexpressionsbuttheyareabittrickytodiscover.
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8.3 TrigonometricSubstitutions
173
Firstwedo
R
secudu,whichwewillneedtocompute
Z
sec
3
udu:
Z
secudu=
Z
secu
secu+tanu
secu+tanu
du
=
Z
sec
2
u+secutanu
secu+tanu
du:
Nowlet w w = = secu+tanu, , dw w = = secutanu+sec
2
udu, exactly the e numerator r of f the
functionweareintegrating.Thus
Z
secudu=
Z
sec
2
u+secutanu
secu+tanu
du=
Z
1
w
dw=lnjwj+C
=lnjsecu+tanuj+C:
Nowfor
Z
sec
3
udu:
sec
3
u=
sec
3
u
2
+
sec
3
u
2
=
sec
3
u
2
+
(tan
2
u+1)secu
2
=
sec
3
u
2
+
secutan
2
u
2
+
secu
2
=
sec
3
u+secutan
2
u
2
+
secu
2
:
Wealreadyknowhowtointegratesecu,sowejustneedtherstquotient. Thisis\simply"
amatterofrecognizingtheproductruleinaction:
Z
sec
3
u+secutan
2
udu=secutanu:
Soputtingthesetogetherweget
Z
sec
3
udu=
secutanu
2
+
lnjsecu+tanuj
2
+C;
andrevertingtotheoriginalvariablex:
Z
p
1+x2dx=
secutanu
2
+
lnjsecu+tanuj
2
+C
=
sec(arctanx)tan(arctanx)
2
+
lnjsec(arctanx)+tan(arctanx)j
2
+C
=
x
p
1+x2
2
+
lnj
p
1+x2+xj
2
+C;
usingtan(arctanx)=xandsec(arctanx)=
q
1+tan
2
(arctanx)=
p
1+x2.
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174
Chapter 8 8 Techniques s ofIntegration
Exercises 8.3.
Findtheantiderivatives.
1.
Z
cscxdx)
2.
Z
csc
3
xdx)
3.
Z
p
x 1dx)
4.
Z
p
9+4xdx)
5.
Z
x
p
1 xdx)
6.
Z
x
2
p
1 x2dx)
7.
Z
1
p
1+x2
dx)
8.
Z
p
x+2xdx)
9.
Z
1
x2(1+x2)
dx)
10.
Z
x
2
p
4 x2
dx)
11.
Z
p
x
p
1 x
dx)
12.
Z
x
3
p
4x2 1
dx)
Wehavealreadyseenthat recognizingthe product rule canbe useful,whenwenoticed
that
Z
sec
3
u+secutan
2
udu=secutanu:
Aswith substitution, we do not t have e to relyon n insight orcleverness to o discoversuch
antiderivatives;thereisatechniquethatwilloftenhelptouncovertheproductrule.
Startwiththeproductrule:
d
dx
f(x)g(x)=f
0
(x)g(x)+f(x)g
0
(x):
Wecanrewritethisas
f(x)g(x)=
Z
f
0
(x)g(x)dx+
Z
f(x)g
0
(x)dx;
andthen
Z
f(x)g
0
(x)dx=f(x)g(x) 
Z
f
0
(x)g(x)dx:
Thismaynotseemparticularlyusefulatrstglance,butitturnsoutthatinmanycases
wehaveanintegraloftheform
Z
f(x)g
0
(x)dx
butthat
Z
f
0
(x)g(x)dx
iseasier. Thistechnique e forturningoneintegralintoanotheriscalledintegration by
parts,andisusuallywritteninmorecompactform. Ifweletu=f(x)andv=g(x)then
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8.4 Integrationby y Parts
175
du=f
0
(x)dxanddv=g
0
(x)dxand
Z
udv=uv 
Z
vdu:
Tousethistechniqueweneedtoidentifylikelycandidatesforu=f(x)anddv=g
0
(x)dx.
EXAMPLE 8.4.1
Evaluate
Z
xlnxdx. Letu=lnxsodu=1=xdx. Thenwemust
letdv=xdxsov=x
2
=2and
Z
xlnxdx=
x
2
lnx
2
Z
x
2
2
1
x
dx=
x
2
lnx
2
Z
x
2
dx=
x
2
lnx
2
x
2
4
+C:
EXAMPLE 8.4.2
Evaluate
Z
xsinxdx. Let t u=x so du =dx. Thenwe e mustlet
dv=sinxdxsov= cosxand
Z
xsinxdx= xcosx 
Z
cosxdx= xcosx+
Z
cosxdx= xcosx+sinx+C:
EXAMPLE8.4.3 Evaluate
Z
sec
3
xdx.Ofcoursewealreadyknowtheanswertothis,
butweneededtobeclevertodiscoverit.Herewe’llusethenewtechniquetodiscoverthe
antiderivative. Letu=secxanddv=sec
2
xdx. Thendu=secxtanxdxandv=tanx
and
Z
sec
3
xdx=secxtanx 
Z
tan
2
xsecxdx
=secxtanx 
Z
(sec
2
x 1)secxdx
=secxtanx 
Z
sec
3
xdx+
Z
secxdx:
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176
Chapter 8 8 Techniques s ofIntegration
Atrstthislooksuseless|we’rerightbackto
Z
sec
3
xdx. Butlookingmoreclosely:
Z
sec
3
xdx=secxtanx 
Z
sec
3
xdx+
Z
secxdx
Z
sec
3
xdx+
Z
sec
3
xdx=secxtanx+
Z
secxdx
2
Z
sec
3
xdx=secxtanx+
Z
secxdx
Z
sec
3
xdx=
secxtanx
2
+
1
2
Z
secxdx
=
secxtanx
2
+
lnjsecx+tanxj
2
+C:
EXAMPLE8.4.4
Evaluate
Z
x
2
sinxdx. Letu=x2,dv=sinxdx;thendu=2xdx
andv= cosx. . Now
Z
x
2
sinxdx= x
2
cosx+
Z
2xcosxdx. Thisisbetterthanthe
originalintegral,butweneedtodointegrationbypartsagain. Letu=2x,dv=cosxdx;
thendu=2andv=sinx,and
Z
x
2
sinxdx= x
2
cosx+
Z
2xcosxdx
= x
2
cosx+2xsinx 
Z
2sinxdx
= x
2
cosx+2xsinx+2cosx+C:
Suchrepeateduseofintegrationbypartsisfairlycommon,butitcanbeabittediousto
accomplish,anditiseasytomakeerrors,especiallysignerrorsinvolvingthesubtractionin
theformula. Thereisanicetabularmethodtoaccomplishthecalculationthatminimizes
the chance forerrorand speedsup the whole process. We e illustrate with the previous
example.Hereisthetable:
sign
u
dv
x
2
sinx
2x
cosx
2
sinx
0
cosx
or
u
dv
x
2
sinx
2x
cosx
2
sinx
0
cosx
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8.4 IntegrationbyParts
177
Toformthersttable,westartwithuatthetopofthesecondcolumnandrepeatedly
compute the derivative;startingwith dv at the topof the third column, , we e repeatedly
computetheantiderivative. Intherstcolumn,weplacea\ "ineverysecondrow. . To
formthesecondtablewecombinetherstandsecondcolumnsbyignoringtheboundary;
ifyoudothisbyhand,youmaysimplystartwithtwocolumnsandadda\ "toevery
secondrow.
To computewith thissecond table webeginat the top. Multiplytherst t entryin
columnubythesecondentryincolumndvtoget x
2
cosx,andaddthistotheintegral
oftheproductofthesecondentryincolumnuandsecondentryincolumndv.Thisgives:
x
2
cosx+
Z
2xcosxdx;
orexactlytheresultoftherstapplicationofintegrationbyparts. Sincethisintegralis
notyeteasy,wereturntothetable.Nowwemultiplytwiceonthediagonal,(x
2
)( cosx)
and( 2x)( sinx)andthenoncestraightacross,(2)( sinx),andcombinetheseas
x
2
cosx+2xsinx 
Z
2sinxdx;
givingthesameresultasthesecondapplicationofintegrationbyparts.Whilethisintegral
iseasy,wemayreturnyetoncemoretothetable.Nowmultiplythreetimesonthediagonal
to get (x
2
)( cosx),( 2x)( sinx),and d (2)(cosx), and oncestraight across,(0)(cosx).
Wecombinetheseasbeforetoget
x
2
cosx+2xsinx+2cosx+
Z
0dx= x
2
cosx+2xsinx+2cosx+C:
Typicallywewouldllinthetableonelineatatime,untilthe\straightacross"multipli-
cationgivesaneasyintegral. Ifwecanseethattheucolumnwilleventuallybecomezero,
wecaninsteadllinthewholetable;computingtheproductsasindicatedwillthengive
theentireintegral,includingthe\+C",asabove.
Exercises8.4.
Findtheantiderivatives.
1.
Z
xcosxdx)
2.
Z
x
2
cosxdx)
3.
Z
xe
x
dx)
4.
Z
xe
x
2
dx)
5.
Z
sin
2
xdx)
6.
Z
lnxdx)
178
Chapter 8 8 Techniques s ofIntegration
7.
Z
xarctanxdx)
8.
Z
x
3
sinxdx)
9.
Z
x
3
cosxdx)
10.
Z
xsin
2
xdx)
11.
Z
xsinxcosxdx)
12.
Z
arctan(
p
x)dx)
13.
Z
sin(
p
x)dx)
14.
Z
sec
2
xcsc
2
xdx)
Arational functionisafractionwithpolynomialsinthenumeratoranddenominator.
Forexample,
x
3
x2+x 6
;
1
(x 3)2
;
x
2
+1
x2 1
;
are all rational functions of f x. There e is a general technique called d \partial fractions"
that,inprinciple,allowsustointegrateanyrationalfunction. Thealgebraicstepsinthe
techniquearerathercumbersomeifthe polynomialinthe denominatorhasdegreemore
than2,andthetechniquerequiresthatwefactorthedenominator,somethingthatisnot
alwayspossible.However,inpracticeonedoesnotoftenrunacrossrationalfunctionswith
highdegreepolynomialsinthedenominatorforwhichonehastondtheantiderivative
function. So o we shallexplainhowto ndthe antiderivativeof a rationalfunctiononly
whenthedenominatorisaquadraticpolynomialax
2
+bx+c.
Weshouldmentionaspecialtypeofrationalfunctionthatwealreadyknowhowto
integrate: If f the denominatorhasthe form(ax+b)
n
,thesubstitutionu = = ax+b will
alwayswork. Thedenominatorbecomesu
n
,andeachxinthenumeratorisreplacedby
(u b)=a, , and dx x = = du=a. While e it maybe tedioustocomplete the integration if the
numeratorhashighdegree,itismerelyamatterofalgebra.
8.5 RationalFunctions
179
EXAMPLE8.5.1
Find
Z
x
3
(3 2x)5
dx:Usingthesubstitutionu=3 2xweget
Z
x
3
(3 2x)5
dx=
1
2
Z
u 3
2
3
u5
du=
1
16
Z
u
3
9u
2
+27u 27
u5
du
=
1
16
Z
u
2
9u
3
+27u
4
27u
5
du
=
1
16
u
1
1
9u
2
2
+
27u
3
3
27u
4
4
+C
=
1
16
(3 2x)
1
1
9(3 2x)
2
2
+
27(3 2x)
3
3
27(3 2x)
4
4
+C
1
16(3 2x)
+
9
32(3 2x)2
9
16(3 2x)3
+
27
64(3 2x)4
+C
Wenowproceedtothecaseinwhichthedenominatorisaquadraticpolynomial.We
canalwaysfactoroutthecoecientofx
2
andputitoutsidetheintegral,sowecanassume
thatthedenominatorhastheformx
2
+bx+c. Therearethreepossiblecases,depending
onhowthequadraticfactors: eitherx
2
+bx+c=(x r)(x s),x
2
+bx+c=(x r)
2
,
oritdoesn’tfactor. Wecanusethequadraticformulatodecidewhichofthesewehave,
andtofactorthequadraticifitispossible.
EXAMPLE8.5.2
Determinewhetherx
2
+x+1factors,andfactoritifpossible. The
quadraticformulatellsusthatx
2
+x+1=0when
x=
1
p
1 4
2
:
Sincethereisnosquarerootof 3,thisquadraticdoesnotfactor.
EXAMPLE8.5.3
Determinewhetherx
2
x 1factors,andfactoritifpossible. . The
quadraticformulatellsusthatx
2
x 1=0when
x=
1
p
1+4
2
=
1
p
5
2
:
Therefore
x
2
x 1=
1+
p
5
2
p
5
2
!
:
180
Chapter 8 8 Techniques s ofIntegration
Ifx
2
+bx+c=(x r)
2
thenwehavethespecialcasewehavealreadyseen,thatcan
behandledwithasubstitution.Theothertwocasesrequiredierentapproaches.
Ifx
2
+bx+c=(x r)(x s),wehaveanintegraloftheform
Z
p(x)
(x r)(x s)
dx
wherep(x)isapolynomial. Therststepistomakesurethatp(x)hasdegreelessthan
2.
EXAMPLE8.5.4 Rewrite
Z
x
3
(x 2)(x+3)
dxintermsofanintegralwithanumerator
thathasdegreelessthan2. Todothisweuselongdivisionofpolynomialstodiscoverthat
x
3
(x 2)(x+3)
=
x
3
x2+x 6
=x 1+
7x 6
x2+x 6
=x 1+
7x 6
(x 2)(x+3)
;
so
Z
x
3
(x 2)(x+3)
dx=
Z
x 1dx+
Z
7x 6
(x 2)(x+3)
dx:
Therstintegraliseasy,soonlythesecondrequiressomework.
Nowconsiderthefollowingsimplealgebraoffractions:
A
x r
+
B
x s
=
A(x s)+B(x r)
(x r)(x s)
=
(A+B)x As Br
(x r)(x s)
:
Thatis,addingtwofractionswithconstantnumeratoranddenominators(x r)and(x s)
producesafractionwithdenominator(x r)(x s)andapolynomialofdegreelessthan
2forthenumerator. Wewanttoreversethisprocess: : startingwithasinglefraction,we
wanttowriteitasasumoftwosimplerfractions. Anexampleshouldmakeitclearhow
toproceed.
EXAMPLE8.5.5
Evaluate
Z
x
3
(x 2)(x+3)
dx. Westartbywriting
7x 6
(x 2)(x+3)
asthesumoftwofractions.Wewanttoendupwith
7x 6
(x 2)(x+3)
=
A
x 2
+
B
x+3
:
Ifwegoaheadandaddthefractionsontherighthandsideweget
7x 6
(x 2)(x+3)
=
(A+B)x+3A 2B
(x 2)(x+3)
:
SoallweneedtodoisndAandB sothat7x 6=(A+B)x+3A 2B,whichisto
say,weneed7=A+Band 6=3A 2B.Thisisaproblemyou’veseenbefore:solvea
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