Scalars-4.3-ADV-1.2 ANTIDERIVATIVES - Variable (u) Substitution 2017.04.10
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2.Variable ( u ) substitution(possiblyprecededbyatrigidentity).
Therearecertainfunctionrules F ′(x) whoseantiderivatives { F(x) + C } canbe
discoveredbyrecognizingthat F ′(x) istheresultofusingthechainruleon F(x). Spottingpatternsliketheonesbelowandusingthemtodiscoverantiderivativesis
thebasisofaprocedurecalledvariablesubstitution
( u -substitution).
Pleaserereadthesectionondifferentiationusingthechainruleandpleasereview
df
themeaningof
attheendofScalars4.1.
dx
EXAMPLES:
Thefollowingexamplesalldirectlyfollowfromdoingthechainrulebackwards.
It’senlighteningtousethechainruletoverifyeachoftheresultsbelow.
F(x)
+ C
∫ F ′(x)dx
∫ sin( U(x) )⋅U ′(x) dx
∫ sec ( U(x) )⋅U ′(x)dx
2
∫e
U (x)
⋅ U ′(x)dx
U ′(x)
∫ U(x) dx
= − cos( U(x) )
+ C Use chain rule to double check these.
=
+ C
tan( U(x) )
=
e( U ( x ) )
+ C
=
ln( U(x) )
+ C
f ′(x)
∫
1− f (x)
∫
f ′(x)
dx
f (x)2 + 1
2
dx
=
=
arcsin ( f ( x ) ) + C
arctan ( f ( x ) ) + C
Itisfrequentlydifficulttodeterminethedetailsofhowtousethechainrule
backwardsbutusingthemechanicalprocedurecalledvariablesubstitution( u substitution)allowsustohandlemorecomplicatedexamplesofthistype.
TomKMadisonWI
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Hereisanoteonnotationsothatyouwillunderstandwhatyouaredoingand
whatyouarenotdoing.Rememberthatthedefinitionoftheindefiniteintegral
treats ∫ dx asasinglesymbolwhere ∫ hasnomeaningofitsownandinthis
context dx hasnomeaningofit’sown. dx isnotanindependentalgebraicobjectin
thiscontext.
THEOREM:Thevariablesubstitutionrule( u -substitution)1
Givenaninterval I If U(x) is differentiable on I and
f
is continuous
on the range of U(x) for x in I and
u = U(x)
then
∫ f ( U(x) )⋅U ′(x)dx
=
∫ f (u)du
It is instructive to think about what du actually means in this context.
This is a theorem that needs to be proved. This result is not at all obvious just
from the definition of the symbol
∫
def
f (x)dx =
{F(x)
F ′(x) = f (x)}.
Seetheendofthechapterforsomeinformalreasoningofwhythisworks.
1TakenfromStewart,6thEd,pp,333-334
TomKMadisonWI
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Amorepragmaticrestatementofthetheorem.
If ∫ G(x)dx isdifficulttointegratechecktoseeifthereexistfunctions U(x) and
f (u) suchthatyoucanrewrite ∫ G(x)dx as
Ifso,set u = U(x) .
Thenyoumaywrite:
∫ G(x)dx = ∫ f ( U(x) )⋅U ′(x) dx =
∫
f ( U(x) )⋅ U ′(x) dx .
∫ f (u)du .
Ifyoucanevaluate ∫ f (u)du thenyoucanbacksubstitute U(x) for u andthefinal
resultequals ∫ G(x)dx (intermsof x ).
Mechanically,thislookslikeyousubstituted du for U ′(x)dx butinthiscontext
dx and du arenotdefinedasindependentobjectssoconceptuallythat’snotreally
what’sgoingon.However,thismechanicalproceduredoesleadtocorrectanswers.
Stepbystepproceduretohelpyoufind U(x), U ′(x) f (u) and du withoutmaking
mechanicalerrors.
. Guess U(x) such that U(x) and U ′(x) are subexpressions of G(x) or equivalent expression. .Trytotransform ∫ G(x)dx intotheexactform ∫ f ( U(x) )⋅ U ′(x)dx forsome f (u). .Write u = U(x) .Treat dx asifitwereanindependentalgebraicobjectandwritedownthe
followingformulastohelpyoukeeptrackofthedetails.2:
du
du
= U ′(x),
du = U ′(x)dx,
dx =
U ′(x) dx
du
for dx in the expression G(x) Substitute
U ′(x)
.Thisresultsin ∫ G(x)dx = ∫ f ( U(x) ) ⋅ U ′(x)dx =
∫ f (u)du .Ifyoucanevaluate
∫ f (u)du thenyoucanbacksubstitute U(x) for u toobtainanexpressionintermsof
x .Thisexpressionrepresentsthesetofantiderivativesof G(x) ( ∫ G(x)dx ) .
Thehardpartisdetermining U(x) suchthat U ′(x) matchesauseful
subexpression(towithinaconstant)of G(x) .Youmayneedtodosomejudicious
factoring(orothertransformations)tospotthis.Itfrequentlyrequiressome
guessworktofindaworkable U(x) .Youneedtodoalotofexamplestoget
comfortablewiththis.
2Someauthorsdirectlyreplace U ′ ( x )dx with du which is algebraically equivalent. The dx isstillreplacedwhichdoesn'tdirectly
followfromthedefinitionof
TomKMadisonWI
∫ G(x) dx .
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Someheuristicsforfinding u = U(x) .
1.If G(x) isafraction
Try u = denominatororpartofthedenominatoror
try u = numeratororpartofthenumerator.
2.The“insidetheparentheses“heuristic.
Ifhasasubexpression U(x) insideparenthesestry u = U(x) forexample
=
U(x)
( U(x) )n
( U(x) )1/2
But if U(x) = ax + b then u = ax + b frequently works
n
m
( U(x) ) = ( U(x) )
sin ( U(x) ) , cos ( U(x) ) ,
m
n
tan ( U(x) )
e( U ( x ) )
2A.Specialcase.Ifthere’san e x or enx andnotmuchelsetry u = e x or u = enx .
3.Trigidentities.
Ifanyblueexpressionbelowisasubexpressionof G(x) substitutethefollowing
identities.
Double angle identities
1− cos 2x
2
1+ cos 2x
cos 2 x =
2
sin(2x) = 2sin x cos x
sin 2 x
=
cos(2x) = cos 2 x − sin 2 x
Combine with sin 2 x + cos 2 x = 1 to obtain to more useful identities
Other identities
sin ( A + B ) = sin A cos B + cos Asin B
cos ( A + B ) = cos A cos B − sin Asin B
Here are more general forms of some of the identities above.
2sin Asin B
= cos ( A − B ) − cos ( A + B )
2 cos A cos B = cos ( A − B ) + cos ( A + B )
2sin A cos B =
TomKMadisonWI
sin ( A − B ) + sin ( A + B )
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Note:Therearesomanytrigidentitiesthatit’seasytomissonethatwouldwork.
https://www.desmos.com/calculator/njuaughzln
4.Factortheintegrandasmuchaspossible.
Thenlookforusefulpatterns.
5.DivineInspiration:SeeSrinivasaRamanujanandMahalakshmi.
http://en.wikipedia.org/wiki/Srinivasa_Ramanujan#Personality_and_spiritual_life
EXAMPLES:
Thefirstthingtodoforanyoftheexamplesbelowistopullconstantfactorsoutside
af (x)
a
dx =
f (x)dx .
oftheintegral ∫
b
b∫
EXAMPLE1:Variablesubstitutionforadenominator
x
∫ x 2 + 1 dx
du
du
Try u = x 2 + 1,
= 2x, dx =
dx
2x
x du
= ∫ 2
x + 1 2x
1
1
=
du
2 ∫ x2 + 1
1 1
1
=
du =
lnu + C
from the tables.
∫
2 u
2
1
=
ln ( x 2 + 1) + C √
2
EXAMPLE2:Variablesubstitution(inside)
Solve for ∫ cos(7x + 5)dx =
Go “inside” the parentheses and pick u = (7x + 5),
TomKMadisonWI
=
du
= 7,
dx
dx =
du
7
du
∫ cos(7x + 5) 7
1
= ∫ cos(7x + 5)du
7
1
sin(u)
=
cosu du =
+C
∫
7
7
1
=
sin(7x + 5) + C √
7
from the tables
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EXAMPLE3:Variablesubstitution(inside)
sin 2x
sin 2x
∫ 1+ cos 2x dx = ∫ (1+ cos 2x )1/2 dx
Go "inside" the parentheses and try
u = (1+ cos 2x),
= −∫
= −
sin 2x
(1+ cos 2x )1/2
du
du
= − 2sin 2x, dx =
dx
−2sin 2x
du
1
1
= − ∫
du
2sin 2x
2 (1+ cos 2x )1/2
1 −1/2
1 u1/2
1 1
du
=
−
u
du
=
−
+ C = − u1/2 + C
2∫
2 (1 / 2)
2 ∫ u1/2
= − (1+ cos 2x ) + C √
EXAMPLE4:Variablesubstitution(inside)
1/2
∫x
2
x − 1 dx =
∫ x ( x − 1)
1/2
2
dx
Kind of tricky.
du
= 1, dx = du
dx
Try u = (x − 1),
=
=
=
∫ x (u )
2
1/2
du
We want everything in terms of u so substitute
u +1 = x
∫ (u + 1) (u ) du
∫ u + 2u + u
2
5/2
1/2
3/2
=
1/2
∫ (u
du =
2
+ 2u + 1)u1/2 du
∫u
5/2
du +
∫ 2u
3/2
∫u
du +
1/2
du
2 7/2
2
2
u + 2 ⋅ u 5/2 + u 3/2 + C
7
5
3
Substitute ( x − 1) for u and you're done.
=
EXAMPLE5:Variablesubstitution(inside)
5
3
∫ x 1+ x dx =
5
3
∫ x (1+ x ) dx
1/2
u = 1+ x 3
du
3x 2
=
∫x u
=
1
(u − 1)u1/2 du
∫
3
TomKMadisonWI
5 1/2
=
du
= 3x 2
dx
dx =
du
3x 2
1 3 1/2
x u du Everything in terms of u so subsitute 3∫
u − 1 = x3
Multiply out and split into a sum.
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EXAMPLE6:Variablesubstitution(inside)
2− x
2− x
∫ 2x 2 − 8x + 1 dx = ∫ 2x 2 − 8x + 1 1/2 dx
(
)
u = 2x 2 − 8x + 1,
du
= 4x − 8 = − 4(2 − x),
dx
du
−4(2 − x)
2− x
du
= ∫
⋅
1/2
( 2x 2 − 8x + 1) −4(2 − x)
dx =
1
1
= − ∫
⋅ du
4 ( 2x 2 − 8x + 1)1/2
1 −1/2
1 u1/2
1 1
⋅
du
=
−
u
⋅
du
=
−
⋅
+C
4∫
4 (1 / 2)
4 ∫ u1/2
1
= − u1/2 + C
2
1/2
1
= − ( 2x 2 − 8x + 1) + C √
2
= −
EXAMPLE7:Trigidentity,variablesubstitutionfordenominator
sin 2x
∫ 1+ cos2 x dx =
There are so many trig identities to confound you that this may take a while.
First, let's try to get all the trig expressions in terms of just (x) or just (2x).
2sin x cos x
= ∫
dx
1+ cos 2 x
du
du
2
u = 1+ ( cos x ) ,
= 2 cos x(− sin x), dx =
dx
−2sin x cos x du
2sin x cos x
= ∫
2
1+ cos x −2sin x cos x
1
du
1+ cos 2 x
1
= − ∫ du
= − lnu + C from the tables.
u
= − ln(1+ cos 2 x) + C √
= −∫
TomKMadisonWI
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EXAMPLE8:Trigidentity,variablesubstitutionfordenominator
sin 2x
∫ 1+ sin x dx
State the trig expressions in terms of just (x) or just (2x).
=
∫
2sin x cos x
dx
1+ sin x
u = 1+ sin x,
du
du
= cos x, dx =
dx
cos x
2sin x cos x du
1+ sin x cos x
sin x
= 2∫
du
1+ sin x
sin x
= 2∫
du
We want everything in terms of u so substitute
u
u − 1 = sin x
=
∫
= 2∫
u −1
1 ⎞
⎛
du = 2 ⎜ ∫ du − ∫ du ⎟
⎝
u
u ⎠
= 2 ( u − lnu ) + C
= 2 (1+ sin x − ln (1+ sin x ) ) + C
= 2sin x − 2 ln(1+ sin x) + (C + 2)
= 2sin x − 2 ln(1+ sin x) + C1 √
EXAMPLE9:Trigidentity,breakintosum,variablesubstitution(inside)
1
1
x
1
⎛ 1+ cos 2x ⎞
2
∫ cos x dx = ∫ ⎜⎝ 2 ⎟⎠ dx = 2 ∫ 1dx + 2 ∫ cos(2x)dx = 2 + 2 ∫ cos(2x)dx
du
du
u = 2x,
= 2, dx =
dx
2
x
1
du
=
+
cosu
∫
2
2
2
x
1
=
+
sinu
2
4
x sin(2x)
x + sin x cos x
=
+
or
+C√
2
4
2
TomKMadisonWI
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EXAMPLE10:Trigidentity,variablesubstitution(inside)
x
1
⎛ 1− cos 2x ⎞
2
∫ sin x dx = ∫ ⎜⎝ 2 ⎟⎠ dx = 2 − 2 ∫ cos(2x)dx
du
du
u = 2x,
= 2, dx =
dx
2
x
1
du
=
−
cosu
∫
2
2
2
x
1
=
−
sinu
2
4
x sin(2x)
x − sin x cos x
=
−
+ C or
+C√
2
4
2
EXAMPLE11:Factor,variablesubstitution(guesssomethingfor U ′(x) and
deducecorresponding u ),trigidentity
∫ sin
3
x dx =
Try the factoring heuristic ∫ sin 3 x dx =
∫ sin
2
du
= − sin x,
dx
u = cos x,
try
=
∫ sin x ⋅sin x ⋅sin x dx
dx = −
du
sin x
⎛ du ⎞
x ⋅sin x ⋅ ⎜ −
⎝ sin x ⎟⎠
= − ∫ sin 2 x du
We want to express sin 2 x in terms of u = cos x :
Try trig identities.
sin 2 + cos 2 x = 1,
1− cos 2 x
= − ∫ (1− cos 2 x)du
u = cos x
= − ∫ (1 − u 2 )du = − u
+
= − cos x +
TomKMadisonWI
u3
+
3
= sin 2 x
C
Make this substitution
from the tables.
cos 3 x
+ C √
3
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EXAMPLE12:Factor,variablesubstitution(guesssomethingfor U ′(x) and
deducecorresponding u ),trigidentity
∫ cos
3
x dx =
Try the factoring heuristic ∫ cos 3 x dx =
u = sin x,
=
∫ cos
=
∫ cos
2
x ⋅ cos x
2
x du
∫ cos x ⋅ cos x ⋅ cos x dx
du
du
= cos x, dx =
dx
cos x
du
cos x
We want to express cos 2 x in terms of u = sin x :
Try trig identities.
sin 2 + cos 2 x = 1,
1− sin 2 x = cos 2 x Make this substitution
=
=
∫ (1− sin
2
x)du
We want everything in terms of u so make the substitution
u = sin x
2
∫ (1 − u )du = u
= sin x −
TomKMadisonWI
−
u3
+
3
C
from the tables.
sin 3 x
+ C √
3
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EXAMPLE14:Factor,trigidentity,Variablesubstitution(inside)
2
2
1
⎛ 1+ cos(2x) ⎞
4
2
2
∫ cos x dx = ∫ cos x ⋅ cos x dx = ∫ ⎜⎝ 2 ⎟⎠ dx = 4 ∫ ( 1 + cos ( 2x ) ) dx
1
=
(1 + 2 cos(2x) + cos2 (2x))dx
4∫
1
=
dx + ∫ 2 cos(2x) + ∫ cos 2 (2x) dx
4 ∫
Solve each piece and put them all together.
(
)
∫ dx = x + C
−−−
∫ 2 cos(
2x )dx
du
du
= 2, dx =
dx
2
du
= ∫ 2 cos(u)
= ∫ cosu du = sinu = sin(2x) + C
2
u = 2x,
−−−
∫ cos (
2
2x )dx
du
du
= 2, dx =
dx
2
du
1
u + sinu cosu
= ∫ cos 2 u
=
cos 2 u du =
+ C (from Example 9)
∫
2
2
2
x + sin(2x)cos(2x)
=
2
1⎛
x + sin(2x)cos(2x) ⎞
Final answer : ⎜ x + sin(2x) +
⎟⎠ + C √
4⎝
2
u = 2x,
or
1
(12x + 8sin(2x) + sin(4x)) + C √
32
TomKMadisonWI
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EXAMPLE15:Factor,trigsubstitution,variablesubstitution(inside)
2
1
2
⎛ 1− cos(2x) ⎞
4
2
2
sin
x
dx
=
sin
x
⋅sin
x
dx
=
∫
∫
∫ ⎜⎝ 2 ⎟⎠ dx = 4 ∫ ( 1 − cos(2x) ) dx
1
=
(1 − 2 cos(2x) + cos2 (2x))dx
4∫
1
=
dx − ∫ 2 cos(2x) + ∫ cos 2 (2x) dx
4 ∫
Use the same pattern as above
(
Final answer :
or
)
1⎛
x + sin(2x)cos(2x) ⎞
⎜⎝ x − sin(2x) +
⎟⎠ + C √
4
2
3x sin 2x sin 4x
−
+
+C√
4
32
8
EXAMPLE15A:
5
∫ sin ( x ) dx EXAMPLE15B:
5
∫ cos ( x ) dx TomKMadisonWI
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EXAMPLE16:Variablesubstitution(inside)
ex
ex
∫ e2 x + 1 dx = ∫ (ex )2 + 1 dx
du
du
= e x , dx = x
Try u = e x ,
dx
e
x
e
du
= ∫ x 2
⋅ x
(e ) + 1 e
1
= ∫ x 2 du
(e ) + 1
1
= ∫ 2 du = tan −1 u + C Matches the table.
u +1
= tan −1 (e x ) + C √
TomKMadisonWI
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EXAMPLE17:Factor,Variablesubstitution(inside)
( x+e ) dx =
∫e
x
(e )
x
x
∫ e ⋅ e dx
u = ex
du
= ex ,
dx
dx =
du
ex
(e ) du
x
∫ e ⋅e
=
x
ex
(e )
x
∫ e du
∫ e du =
=
=
u
eu + C
x
√
= e(e ) + C
EXAMPLE18:Variablesubstitution(inside)
2
3
∫ x sin(x ) dx =
u = x 3,
=
∫x
2
sin(x 3 )
du
du
= 3x 2 , dx = 2
dx
3x
du
3x 2
1
= ∫ sin(x 3 )du
3
1
− cos(u)
=
sinu du =
+ C
∫
3
3
− cos(x 3 )
+ C √
=
3
TomKMadisonWI
from the tables.
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EXAMPLE19:Variablesubstitution(inside)
1
∫ cos2 (2x)dx =
du
du
= 2, dx =
u = U(x) = 2x,
2
dx
1 du
= ∫
cos 2 (u) 2
1
1
tanu
=
du =
+C√
from the tables
2
∫
2 cos u
2
EXAMPLE20:Variablesubstitution(inside)
∫
2z
3
z2 + 1
dz =
∫ (z
2z
2
+ 1)
1/3
dz
du
du
= 2z, dz =
dz
2z
u = z 2 + 1,
2z
=
∫ (z
2
=
∫ (z
2
+ 1)
1/3
1
+ 1)
1/3
du
2z
du
1
u 2/3
3
−1/3
du
=
u
du
=
+ C = u 2/3 + C
∫ u1/3
∫
2
2/3
2/3
3 2
=
( z + 1) + C √
2
=
from the tables.
EXAMPLE21:Variablesubstitution(inside)
∫ x (1+ x ) dx =
1/2
u = 1+ x,
=
TomKMadisonWI
∫ x ⋅u
1/2
du
du
= 1,
dx
du = dx
We want everything in terms of u so make substitution
u −1 = x
=
∫ (u − 1) ⋅u
=
u 5/2
u 3/2
−
substitute back for u
(5 / 2) ( 3 / 2)
1/2
du =
∫u
3/2
du − ∫ u1/2 du
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EXAMPLE22:Variablesubstitution,denominatororsubexpressionthereof.
1
∫ r ln r dr
=
1
∫ ( r )( ln r )dr
Try u = a subexpression of the denominator because u = r ln r
just looks too complicated (pure intuition).
du 1
du
u = ln r,
= , dx =
= r ⋅ du
⎛ 1⎞
dr r
⎜⎝ ⎟⎠
r
1
=
∫ r ln r rdu
=
∫ u du
1
=
1
∫ ln r du
= lnu + C
1
∫ r ln r dr = ln(ln r)) + C √
TomKMadisonWI
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EXAMPLE23:VariableSubstitution
∫
1
dx =
x +1
∫ ( x + 1)
−1/2
dx
du
= 1 dx = du
dx
u = x + 1,
=
−1/2
∫ u du =
u1/2
= 2u1/2
⎛ 1⎞
⎜⎝ ⎟⎠
2
= 2 ( x + 1) √
1/2
EXAMPLE24:VariableSubstitution
x
x
∫ x + 1 dx = ∫ ( x + 1)1/2 dx
du
=1
dx
try u = x + 1,
(no useful subexpression match)
try u = x + 1 = ( x + 1) ,
1/2
=
x
∫ ( x + 1) 2 ( x + 1)
1/2
1/2
= 2 ∫ x du
TomKMadisonWI
du
We want everything interms of u so make the substitution
= 2 ∫ ( u 2 − 1) du
=
du 1
−1/2
1/2
= ( x + 1) , 2 ( x + 1) du = dx
dx 2
u2 − 1 = x
2
( x + 1)3/2 − 2 ( x + 1)1/2 √
3
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Scalars-4.3-ADV-1.2 ANTIDERIVATIVES - Variable (u) Substitution 2017.04.10
A .docx
Page 18 of 20
EXAMPLE25:VariableSubstitution
∫
x2
x +1
dx =
x
∫ ( x + 1)
1/2
dx
try u = x + 1
du
dx
=1
(no useful subexpression match)
x + 1 = ( x + 1) ,
try u =
1/2
du
dx
=
1
2
( x + 1)−1/2 ,
du
= dx
1
( x + 1)−1/2
2
2 ( x + 1) du = dx
1/2
x2
∫ ( x + 1) 2 ( x + 1)
=
1/2
1/2
= 2 ∫ x 2 du
We want everything in terms of u so make the substitution
(u
2
2
)
2
− 1 = x2
2
4
4
= 2
=
∫ (u − 1) du = 2 ∫ (u − 2u
2 ∫ u du − 4 ∫ u du + 2 ∫ 1du
= 2
=
du
u5
2
5
TomKMadisonWI
5
2
)
+ 1 du
2
−
4u 3
+ 2u
3
4
( x + 1)5/2 − ( x + 1)3/2 + 2 ( x + 1)1/2 √
3
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Scalars-4.3-ADV-1.2 ANTIDERIVATIVES - Variable (u) Substitution 2017.04.10
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EXAMPLE27:VariableSubstitution
1
∫ x ( x + 1)1/2 dx =
u = ( x + 1)
1/2
=
1
∫ x ( x + 1)
1/2
du 1
−1/2
1/2
= ( x + 1) , 2 ( x + 1) du = dx
dx 2
= x + 1,
2 ( x + 1) du
1/2
1
2 ∫ du We want everytyhing in terms of u so make the substitution
x
u2 − 1 = x
1
1
= 2∫ 2
du = − 2 ∫
du
Matches the tables
u −1
1− u 2
=
(
= − 2 tanh −1 (u) = − 2 tanh −1 ( x + 1)
TomKMadisonWI
1/2
)√
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Scalars-4.3-ADV-1.2 ANTIDERIVATIVES - Variable (u) Substitution 2017.04.10
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Page 20 of 20
Informalexplanationofwhy u -substitutionworks
Startingwiththetwofunctions, F(u) and U(x) ,thederivativewithrespectto x of
theircomposition F( U(x) )′ equals F ′( U(x) )
⋅ U ′(x) Chainrule.
wrt x wrtvaluesof U(x) wrt x Thusifwearepresentedwithanexpressionoftheform F ′( U(x) ) ⋅ U ′(x) (wherewe
know F ′ and U and U ′ )wehopethatwecandeterminethefunction F .Theclearest
waytofocusonthenatureof F ′ and F istoexpressthemintermsofasingle
variable.For u = U(x) we have F ′(u) and F(u). Weknow F ′(u) sowecancompute
F(u) like this : F(u) + C = ∫ F ′(u)du .Thisisthekeypoint.
Wenowknow F(u) and U(x) sowecanreconstructtheoriginalcompositefunction
F( U(x) ) .Inthestatementofthetheoremwewrote f (u) to represent F ′(u) .So
puttingitalltogether F(u) + C =
∫ F ′(u)du
=
∫ f (u)du andofcourseattheend
wewanttobacksubstitute U(x) for u sowehaveanexpressionsolelyintermsof x .
Thusfrom F ′( U(x) ) ⋅ U ′(x) wehavediscovered F( U(x) ) + C .
Insolvingaproblemthetrickistopicka u = U(x) suchthatwecanexpressthe
problem’soriginalfunctionintheform F ′( U(x) ) ⋅ U ′(x) SeeStewart.6thed,p.333forsimilarreasoning.
Hereareexplanationsoftechniques1and2(inScalars4.3ADV-1)formanually
findingantiderivatives.
TomKMadisonWI
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