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Integration
by substitution
Thereareoccasionswhenitispossibletoperformanapparentlydifficultpieceofintegration
byfirstmakingasubstitution.Thishastheeffectofchangingthevariableandtheintegrand.
Whendealingwithdefiniteintegrals,thelimitsofintegrationcanalsochange. Inthisunitwe
willmeetseveralexamplesofintegralswhereitisappropriatetomakeasubstitution.
Inordertomasterthetechniquesexplainedhereitisvitalthatyouundertakeplentyofpractice
exercisessothattheybecomesecondnature.
Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto:
• carryoutintegrationbymakingasubstitution
• identifyappropriatesubstitutionstomakeinordertoevaluateanintegral
Contents
1. Introduction
2
2. Integrationbysubstitutingu=ax+b
2
3. Finding
f(g(x))g
(x)dxbysubstitutingu=g(x)
6
1
c
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1.Introduction
Thereareoccasionswhenitispossibletoperformanapparentlydifficultpieceofintegration
byfirstmakingasubstitution.Thishastheeffectofchangingthevariableandtheintegrand.
Whendealingwithdefiniteintegrals,thelimitsofintegrationcanalsochange. Inthisunitwe
willmeetseveralexamplesofthistype. Theabilitytocarryoutintegrationbysubstitutionisa
skillthatdevelopswithpracticeandexperience. Forthisreasonyoushouldcarryoutallofthe
practiceexercises. Beawarethatsometimesanapparentlysensiblesubstitutiondoesnotlead
toanintegralyouwillbeabletoevaluate. Youmustthenbepreparedtotryoutalternative
substitutions.
2.Integrationbysubstitutingu=ax+b
Weintroduce thetechnique throughsome simple examples for whichalinear substitutionis
appropriate.
Example
Supposewewanttofindtheintegral
(x+4)
5
dx
(1)
Youwillbefamiliaralreadywithfindingasimilarintegral
u
5
duandknowthatthisintegral
isequalto
u
6
6
+c,wherecisaconstantofintegration.Thisisbecauseyouknowthattherule
forintegratingpowersofavariabletellsyoutoincreasethepowerby1andthendividebythe
newpower.
IntheintegralgivenbyEquation(1)thereisstillapower5,buttheintegrandismorecompli-
catedduetothepresenceofthetermx+4. Totacklethisproblemwemakeasubstitution.
Weletu=x+4. Thepointofdoingthisistochangetheintegrandintothemuchsimpleru
5
.
However,wemusttakecaretosubstituteappropriatelyforthetermdxtoo.
Intermsofdifferentialswehave
du=
du
dx
dx
Now,inthisexample,becauseu=x+4itfollowsimmediatelythat
du
dx
=1andsodu=dx.
So,substitutingbothforx+4andfordxinEquation(1)wehave
(x+4)
5
dx=
u
5
du
The resulting g integral can n be evaluated immediately y to give
u
6
6
+c. We e can revert t to an
expressioninvolvingtheoriginalvariablexbyrecallingthatu=x+4,giving
(x+4)
5
dx=
(x+4)
6
6
+c
Wehavecompletedtheintegrationbysubstitution.
c
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Example
Supposenowwewishtofindtheintegral
cos(3x+4)dx
(2)
Observethatifwemakeasubstitutionu=3x+4,theintegrandwillthencontainthemuch
simplerformcosuwhichwewillbeabletointegrate.
Asbefore,
du=
du
dx
dx
andso
withu=3x+4
and
du
dx
=3
Itfollowsthat
du=
du
dx
dx=3dx
So,substitutingufor3x+4,andwithdx=
1
3
duinEquation(2)wehave
cos(3x+4)dx =
1
3
cosudu
=
1
3
sinu+c
Wecanrevert toanexpressioninvolvingtheoriginalvariablexbyrecallingthatu=3x+4,
giving
cos(3x+4)dx=
1
3
sin(3x+4)+c
Wehavecompletedtheintegrationbysubstitution.
Itisveryeasytogeneralisetheresultofthepreviousexample.Ifwewanttofind
cos(ax+b)dx,
thesubstitutionu=ax+bleadsto
1
a
cosuduwhichequals
1
a
sinu+c,thatis
1
a
sin(ax+b)+c.
Asimilarargument,whichyoushouldtry,showsthat
sin(ax+b)dx=−
1
a
cos(ax+b)+c.
KeyPoint
sin(ax+b)dx=−
1
a
cos(ax+b)+c
cos(ax+b)dx=
1
a
sin(ax+b)+c
3
c
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Example
Supposewewishtofind
1
1−2x
dx.
Wemakethesubstitutionu=1−2xinordertosimplify theintegrandto
1
u
. Recallthatthe
integralof
1
u
withrespecttouisthenaturallogarithmofu,ln|u|.Asbefore,
du=
du
dx
dx
andso
with
u=1−2x and
du
dx
=−2
Itfollowsthat
du=
du
dx
dx=−2dx
Theintegralbecomes
1
u
1
2
du
= −
1
2
1
u
du
= −
1
2
ln|u|+c
= −
1
2
ln|1−2x|+c
The result of the e previous s example can be generalised: if f we want to find
1
ax+b
dx, the
substitutionu=ax+bleadsto
1
a
1
u
duwhichequals
1
a
ln|ax+b|+c.
Thismeans,forexample,thatwhenfacedwithanintegralsuchas
1
3x+7
dxwecanimme-
diatelywritedowntheansweras
1
3
ln|3x+7|+c.
KeyPoint
1
ax+b
dx=
1
a
ln|ax+b|+c
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A little e more e care e must be e taken with the e limits of integration n when n dealing with definite
integrals.Considerthefollowingexample.
Example
Supposewewishtofind
3
1
(9+x)
2
dx
Wemakethesubstitutionu=9+x.Asbefore,
du=
du
dx
dx
andso
with
u=9+x and
du
dx
=1
Itfollowsthat
du=
du
dx
dx=dx
Theintegralbecomes
x=3
x=1
u
2
du
wherewehaveexplicitlywrittenthevariableinthelimitsofintegrationtoemphasisethatthose
limitswereonthevariablexandnotu.Wecanwritetheseaslimitsonuusingthesubstitution
u=9+x.Clearly,whenx=1,u=10,andwhenx=3,u=12.Sowerequire
u=12
u=10
u
2
du =
1
3
u
3
12
10
=
1
3
12
3
−10
3
=
728
3
Notethatinthisexamplethereisnoneedtoconverttheanswergivenintermsofubackinto
oneintermsofxbecausewehadalreadyconvertedthelimitsonxintolimitsonu.
Exercises1.
1.Ineachcaseuseasubstitutiontofindtheintegral:
(a)
(x−2)
3
dx (b)
1
0
(x+5)
4
dx (c)
(2x−1)
7
dx (d)
1
−1
(1−x)
3
dx.
2.Ineachcaseuseasubstitutiontofindtheintegral:
(a)
sin(7x−3)dx (b)
e
3x−2
dx (c)
π/2
0
cos(1−x)dx (d)
1
7x+5
dx.
5
c
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3.Finding
f(g(x))g
(x)dxbysubstitutingu=g(x)
Example
Supposenowwewishtofindtheintegral
2x
1+x
2
dx
(3)
Inthisexamplewemakethesubstitutionu=1+x
2
,inordertosimplifythesquare-rootterm.
Weshallseethattherest oftheintegrand, , 2xdx,willbe e takencareofautomatically inthe
substitutionprocess,andthatthisisbecause2xisthederivativeofthatpartoftheintegrand
usedinthesubstitution,i.e.1+x
2
.
Asbefore,
du=
du
dx
dx
andso
withu=1+x
2
and
du
dx
=2x
Itfollowsthat
du=
du
dx
dx=2xdx
So,substitutingufor1+x
2
,andwith2xdx=duinEquation(3)wehave
2x
1+x
2
dx =
udu
=
u
1/2
du
=
2
3
u
3/2
+c
Wecanreverttoanexpressioninvolvingtheoriginalvariablexbyrecallingthatu=1+x
2
,
giving
2x
1+x
2
dx=
2
3
(1+x
2
)
3/2
+c
Wehavecompletedtheintegrationbysubstitution.
Letus analysethis examplealittlefurtherbycomparingtheintegrandwiththegeneralcase
f(g(x))g
(x).
Supposewewrite
g(x)=1+x
2
and
f(u)=
u
Thenwenotethatthecomposition
1
ofthefunctionsfandgisf(g(x))=
1+x
2
.
1
whenfinding the compositionof functions s f f andg g it is the output from g whichis usedas input tof,
resultinginf(g(x))
c
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Further,wenotethatifg(x)=1+x
2
theng
(x)=2x.Sotheintegral
2x
1+x
2
dx
isoftheform
f(g(x))g
(x)dx
Toperformtheintegrationweusedthesubstitutionu=1+x
2
. Inthegeneralcaseitwillbe
appropriatetotrysubstitutingu=g(x).Thendu=
du
dx
dx=g
(x)dx.
Oncethesubstitutionwasmadetheresultingintegralbecame
udu. Inthegeneralcaseit
willbecome
f(u)du.Providedthatthisfinalintegralcanbefoundtheproblemissolved.
Forpurposesofcomparisonthespecificexampleandthegeneralcasearepresentedside-by-side:
2x
1+x
2
dx
f(g(x))g
(x)dx
letu=1+x
2
letu=g(x)
du=
du
dx
dx=2xdx
du=
du
dx
dx=g
(x)dx
2x
1+x
2
dx=
udu
f(g(x))g
(x)dx=
f(u)du
2
3
u
3/2
+c
2
3
(1+x
2
)
3/2
+c
KeyPoint
Toevaluate
f(g(x))g
(x)dx
substituteu=g(x),anddu=g
(x)dxtogive
f(u)du
Integrationisthencarriedoutwithrespecttou,beforerevertingtotheoriginalvariablex.
Itisworthpointingoutthatintegrationbysubstitutionissomethingofanart-andyourskill
atdoingitwillimprovewithpractice. Furthermore,asubstitutionwhichatfirstsight t might
seemsensible,canleadnowhere. Forexample,ifyouweretrytofind
1+x
2
dxbyletting
u=1+x
2
youwouldfindyourselfupablindalley. Bepreparedtopersevereandtrydifferent
approaches.
7
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mathcentreDecember1,2008
Example
Supposewewishtoevaluate
4x
2x
2
+1
dx
By writing the integrand d as
1
2x
2
+1
·4x we note that it takes the form
f(g(x))g
(x)dx
wheref(u)=
1
u
,g(x)=2x
2
+1andg
(x)=4x.
Thesubstitutionu=g(x)=2x
2
+1transformstheintegralto
f(u)du=
1
u
du
Thisisevaluatedtogive
1
u
du =
u
−1/2
du
= 2u
1/2
+c
Finally,usingu=2x
2
+1toreverttotheoriginalvariablegives
4x
2x
2
+1
dx=2(2x
2
+1)
1/2
+c
orequivalently
2
2x
2
+1+c
Example
Supposewewishtofind
sin
x
x
dx.
Considerthesubstitutionu=
x.Then
du =
du
dx
dx
=
1
2
x
−1/2
dx
=
1
2x
1/2
dx
=
1
2
x
dx
sothat
sin
x
x
dx=2
sinudu
fromwhich
2
sinudu = −2cosu+c
= −2cos
x+c
Wecanalsomakethefollowingobservations:
theintegrandcanbewrittenintheformsin
1
x
.
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mathcentreDecember1,2008
8
Writingf(u)=sinuandg(x)=
xtheng
(x)=
1
2
x
−1/2
=
1
2x
1/2
=
1
2
x
.
Further,f(g(x))=sin
x.
Hencewewritethegivenintegralas
2
sin
x
1
2
x
dx
whichisoftheform
2
f(g(x))g
(x)dx
withfandgasgivenabove.
Asbeforethesubstitutionu=g(x)=
xproducestheintegral
2
f(u)du=2
sinudu
fromwhich
2
sinudu = −2cosu+c
= −2cos
x+c
Exercises2
1. Ineachcasetheintegrandcanbewrittenasf(g(x))g
(x). Identifythefunctionsf f andg
andusethegeneralresultonpage7tocompletetheintegration.
(a)
2xe
x
2
−5
dx
(b)
−2xsin(1−x
2
)dx
(c)
cosx
1+sinx
dx.
2.Ineachcaseusethegivensubstitutiontofindtheintegral:
(a)
−2xe
−x
2
dx,
u=−x
2
.
(b)
xsin(2x
2
)dx,
u=2x
2
.
(c)
5
0
x
3
x
4
+1dx,
u=x
4
+1.
3.Ineachcaseuseasuitablesubstitutiontofindtheintegral.
(a)
5x
1−x
2
dx
(b)
dx
x(1+
x)
2
(c)
x
4
(1+x
5
)
3
dx
(d)
x
3
x
4
+16
dx
(e)
cosx
(5+sinx)
2
(f)
1
0
x
3
x
4
+12
dx
(g)
5x
2
1−x
3
dx (h)
e
cosx
sinxdx
(i)
e
sinx
cosxdx.
9
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mathcentreDecember1,2008
AnswerstoExercises
Exercises1
1. (a)
(x−2)
4
4
+c
(b)
4651
5
=930
1
5
(c)
(2x−1)
8
16
+c
(d)
4.
2.(a)−
cos(7x−3)
7
+c
(b)
e
3x−2
3
+c
(c)1.382(3dp)
(d)
1
7
ln|7x+5|+c
Exercises2
1. (a) f(u)=e
u
,g(x)=x
2
−5,e
x
2
−5
+c,
(b) f(u)=sinu,g(x)=1−x
2
,−cos(1−x
2
)+c
(c) f(u)=
1
u
,g(x)=1+sinx,ln|1+sinx|+c.
2. (a) e
−x
2
+c (b) −
cos(2x
2
)
4
+c (c) 2610(4sf).
3. (a) −
5
3
(1−x
2
)
3/2
+c
(b) −
2
1+
x
+c
(c)
1
20
(1+x
5
)
4
+c
(d)
1
2
(x
4
+16)
1/2
+c
(e) −
1
5+sinx
+c (f) 0.0707
(g) −
10
9
(1−x
3
)
3/2
+c (h) −e
cosx
+c
(i) e
sinx
+c.
c
mathcentreDecember1,2008
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