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. . . . . .
Section5.5IntegrationbySubstitution, PartDeux
V63.0121, CalculusI
April29, 2009
Announcements
I ClassonMondaywillbereview
. . . . . .
Yes, thereisclassonMonday
I NonewmaterialI WewillreviewthecourseI Wewillanswerquestions, sobringsome
. . . . . .
Finalstuff
I Oldfinalsonline, includingFall2008I Reviewsessions: May5and6, 6:00–8:00pm, SILV 703I FinalisMay8, 2:00–3:50pminCANT 101/200
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. . . . . .
Outline
Recall: Themethodofsubstitution
Multiplesubstitutions
OddandevenfunctionsExamples
Moreexamplesandadvice
CourseEvaluations
. . . . . .
LastTime: TheSubstitutionRule
TheoremIf u = g(x) isadifferentiablefunctionwhoserangeisaninterval Iand f iscontinuouson I, then∫
f(g(x))g′(x)dx =
∫f(u)du
or ∫f(u)
dudx
dx =
∫f(u)du
. . . . . .
LastTime: TheSubstitutionRuleforDefiniteIntegrals
TheoremIf g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b
af(g(x))g′(x)dx =
∫ g(b)
g(a)f(u)du.
I Theintegralonthelefthappensin“x-land”, soitslimitsarevaluesof x
I Theintegralontherighthappensin“u-land”, soitslimitsneedtobevaluesof u
I Toconvert x to u, simplyapplythesubstitution u = g(x).
. . . . . .
LastTime: TheSubstitutionRuleforDefiniteIntegrals
TheoremIf g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b
af(g(x))g′(x)dx =
∫ g(b)
g(a)f(u)du.
I Theintegralonthelefthappensin“x-land”, soitslimitsarevaluesof x
I Theintegralontherighthappensin“u-land”, soitslimitsneedtobevaluesof u
I Toconvert x to u, simplyapplythesubstitution u = g(x).
. . . . . .
Outline
Recall: Themethodofsubstitution
Multiplesubstitutions
OddandevenfunctionsExamples
Moreexamplesandadvice
CourseEvaluations
. . . . . .
Anexponentialexample
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. Wehave∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u + 1du
Nowlet y = u + 1, dy = du. So
12
∫ 8
3
√u + 1du =
12
∫ 9
4
√y dy =
12
∫ 9
4y1/2 dy
=12· 23y3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anexponentialexample
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. Wehave∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u + 1du
Nowlet y = u + 1, dy = du. So
12
∫ 8
3
√u + 1du =
12
∫ 9
4
√y dy =
12
∫ 9
4y1/2 dy
=12· 23y3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anexponentialexample
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. Wehave∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u + 1du
Nowlet y = u + 1, dy = du. So
12
∫ 8
3
√u + 1du =
12
∫ 9
4
√y dy =
12
∫ 9
4y1/2 dy
=12· 23y3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anotherwaytoskinthatcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1
, sothat du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√udu
=13u3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anotherwaytoskinthatcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1, sothat du = 2e2x dx.
Then∫ ln√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√udu
=13u3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anotherwaytoskinthatcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1, sothat du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√udu
=13u3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anotherwaytoskinthatcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1, sothat du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√udu
=13u3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
Anotherwaytoskinthatcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1, sothat du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√udu
=13u3/2
∣∣∣∣94
=13
(27− 8) =193
. . . . . .
A thirdskinnedcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, sothat
u2 = e2x + 1
=⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
e2x√
e2x + 1dx =
∫ 3
2u · udu =
13u3
∣∣∣∣32
=193
. . . . . .
A thirdskinnedcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, sothat
u2 = e2x + 1 =⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
e2x√
e2x + 1dx =
∫ 3
2u · udu =
13u3
∣∣∣∣32
=193
. . . . . .
A thirdskinnedcat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, sothat
u2 = e2x + 1 =⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
e2x√
e2x + 1dx =
∫ 3
2u · udu =
13u3
∣∣∣∣32
=193
. . . . . .
Outline
Recall: Themethodofsubstitution
Multiplesubstitutions
OddandevenfunctionsExamples
Moreexamplesandadvice
CourseEvaluations
. . . . . .
Example
Find∫ π
−πsin(x)dx
Solution
∫ π
−πsin(x) = − cos(x)|π−π = cos(x)|−π
π = cos(−π) − cos(π) = 0
Thisisobviousfromthegraph:
. .x
.y
. . . . . .
Example
Find∫ π
−πsin(x)dx
Solution
∫ π
−πsin(x) = − cos(x)|π−π = cos(x)|−π
π = cos(−π) − cos(π) = 0
Thisisobviousfromthegraph:
. .x
.y
. . . . . .
Example
Find∫ π
−πsin(x)dx
Solution
∫ π
−πsin(x) = − cos(x)|π−π = cos(x)|−π
π = cos(−π) − cos(π) = 0
Thisisobviousfromthegraph:
. .x
.y
. . . . . .
EvenandOddFunctions
DefinitionA function f is even ifforall x,
f(−x) = f(x)
A function f is odd ifforall x,
f(−x) = −f(x).
Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis
. . . . . .
EvenandOddFunctions
DefinitionA function f is even ifforall x,
f(−x) = f(x)
A function f is odd ifforall x,
f(−x) = −f(x).
Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis
. . . . . .
EvenandOddFunctions
DefinitionA function f is even ifforall x,
f(−x) = f(x)
A function f is odd ifforall x,
f(−x) = −f(x).
Thesepropertiesarerevealedinthegraph.
I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis
. . . . . .
EvenandOddFunctions
DefinitionA function f is even ifforall x,
f(−x) = f(x)
A function f is odd ifforall x,
f(−x) = −f(x).
Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.
I Anevenfunctionhasreflectivesymmetryinthe y-axis
. . . . . .
EvenandOddFunctions
DefinitionA function f is even ifforall x,
f(−x) = f(x)
A function f is odd ifforall x,
f(−x) = −f(x).
Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis
. . . . . .
EvenandOddfunctionspictured
. .x
.y.odd
.x
.y.even
. . . . . .
Examplesofsymmetricfunctions
Evenandoddfunctionsabound.I x 7→ xn isoddwhen n isoddandevenwhen n iseven.Funny, that!
I sin isoddand cos iseven.
. . . . . .
Combiningsymmetricfunctions
Theorem
(a) Thesumofevenfunctionsiseven. Thesumofoddfunctionsisodd.
(b) Theproductofevenfunctionsiseven. Theproductofoddfunctionsiseven. Theproductofanoddfunctionandanevenfunctionisanoddfunction.
(c) If g iseven, then f ◦ g iseven. Thecompositionoftwooddfunctionsisodd. Thecompositionofanevenfunctionandanoddfunctioniseven.
. . . . . .
Integratingsymmetricfunctions
TheoremLet a beanynumber.
(a) If f isodd, then ∫ a
−af(x)dx = 0.
(b) If f iseven, then ∫ a
−af(x)dx = 2
∫ a
0f(x)dx.
. . . . . .
Integratingsymmetricfunctions
TheoremLet a beanynumber.
(a) If f isodd, then ∫ a
−af(x)dx = 0.
(b) If f iseven, then ∫ a
−af(x)dx = 2
∫ a
0f(x)dx.
. . . . . .
Proof(odd f).
Tocompute∫ a
−af(x)dx, let u = −x. Then du = −dx andwehave
∫ a
−af(x)dx = −
∫ −a
af(−u)du
=
∫ −a
af(u)du
= −∫ a
−af(u)du.
Theonlynumberwhichisequaltoitsownnegativeiszero.
. . . . . .
Proof(even f).Withthesamesubstitutionwehave∫ 0
−af(x)dx = −
∫ 0
af(−u)du
= −∫ 0
af(u)du
=
∫ a
0f(u)du.
So ∫ a
−af(x)dx =
∫ 0
−af(x)dx +
∫ a
0f(x)dx = 2
∫ a
0f(x)dx.
. . . . . .
ExampleCompute ∫ e
√π+1
−e√
π−1sin(x)
√1 + cos3(x)dx.
SolutionTheintegrandisodd! Sotheansweriszero.
. . . . . .
ExampleCompute ∫ e
√π+1
−e√
π−1sin(x)
√1 + cos3(x)dx.
SolutionTheintegrandisodd! Sotheansweriszero.
. . . . . .
ExampleCompute ∫ 2
−2
(x4 + x2 + 3
)dx.
. . . . . .
SolutionBecausetheintegrandisevenwecansimplifyourarithmetic. It’sespeciallynicetopluginzerosincetheresultisoftenzero.∫ 2
−2
(x4 + x2 + 3
)dx = 2
∫ 2
0
(x4 + x2 + 3
)dx
= 2[x5
5+
x3
3+ 3x
]20
= 2[25
5+
23
3+ 3(2)
]= 2
[325
+83
+ 6]
=215
(32 · 3 + 8 · 5 + 6 · 15)
=2 · 22615
. . . . . .
Outline
Recall: Themethodofsubstitution
Multiplesubstitutions
OddandevenfunctionsExamples
Moreexamplesandadvice
CourseEvaluations
. . . . . .
Example∫x3
(5x4 + 2)2dx
SolutionLet u = 5x4 + 2, so du = 20x3 dx. Then∫
x3
(5x4 + 2)2dx =
120
∫1u2
du
= − 120
· 1u
+ C
= − 120(5x4 + 2)
+ C
. . . . . .
Example∫x3
(5x4 + 2)2dx
SolutionLet u = 5x4 + 2, so du = 20x3 dx. Then∫
x3
(5x4 + 2)2dx =
120
∫1u2
du
= − 120
· 1u
+ C
= − 120(5x4 + 2)
+ C
. . . . . .
Example∫sin(sin(θ)) cos(θ)dθ
SolutionLet u = sin(θ), so du = cos(θ)dθ. Then∫
sin(sin(θ)) cos(θ)dθ =
∫sin(u)du
= − cos(u) + C
= − cos(sin(θ)) + C
. . . . . .
Example∫sin(sin(θ)) cos(θ)dθ
SolutionLet u = sin(θ), so du = cos(θ)dθ. Then∫
sin(sin(θ)) cos(θ)dθ =
∫sin(u)du
= − cos(u) + C
= − cos(sin(θ)) + C
. . . . . .
Example∫ex + e−x
ex − e−x dx
SolutionThenumeratoristhederivativeofthedenominator! Letu = ex − e−x, so du =
(ex + e−x) dx. Then∫
ex + e−x
ex − e−x dx =
∫1udu
= ln |u| + C = ln∣∣ex − e−x
∣∣ + C
. . . . . .
Example∫ex + e−x
ex − e−x dx
SolutionThenumeratoristhederivativeofthedenominator! Letu = ex − e−x, so du =
(ex + e−x) dx. Then∫
ex + e−x
ex − e−x dx =
∫1udu
= ln |u| + C = ln∣∣ex − e−x
∣∣ + C
. . . . . .
Example∫3x
1 + 9xdx
SolutionNotice 9x = (32)x = 32x = (3x)2. Solet u = 3x,du = (ln 3) · 3x dx. Then∫
3x
(1 + 9x)dx =
1ln 3
∫1
1 + u2du
=1ln 3
arctan(u) + C
=1ln 3
arctan(3x) + C
. . . . . .
Example∫3x
1 + 9xdx
SolutionNotice 9x = (32)x = 32x = (3x)2. Solet u = 3x,du = (ln 3) · 3x dx. Then∫
3x
(1 + 9x)dx =
1ln 3
∫1
1 + u2du
=1ln 3
arctan(u) + C
=1ln 3
arctan(3x) + C
. . . . . .
Example∫sec2
√x√
xdx
SolutionLet u =
√x, so du =
12√xdu. Then
∫sec2
√x√
xdx = 2
∫sec2(u)du
= 2 tan(u) + C
= 2 tan(√
x)
+ C
. . . . . .
Example∫sec2
√x√
xdx
SolutionLet u =
√x, so du =
12√xdu. Then
∫sec2
√x√
xdx = 2
∫sec2(u)du
= 2 tan(u) + C
= 2 tan(√
x)
+ C
. . . . . .
Example∫dx
x ln x
SolutionLet u = ln x, so du =
1xdx. Then∫dx
x ln x=
∫1udu
= ln |u| + C
= ln |ln x| + C
. . . . . .
Example∫dx
x ln x
SolutionLet u = ln x, so du =
1xdx. Then∫dx
x ln x=
∫1udu
= ln |u| + C
= ln |ln x| + C
. . . . . .
Whatdowesubstitute?
I Linearfactors (ax + b) areeasysubstitutions: u = ax + b,du = a dx
I Lookfor function/derivativepairs intheintegrand, onetomake u andonetomake du:
I xn and xn−1 (fudgethecoefficient)I sineandcosine(fudgetheminussign)I ex and exI ax and ax (fudgethecoefficient)
I√x and
1√x(fudgethefactorof 2)
I ln x and1x
. . . . . .
Outline
Recall: Themethodofsubstitution
Multiplesubstitutions
OddandevenfunctionsExamples
Moreexamplesandadvice
CourseEvaluations
. . . . . .
CourseEvaluations
I PleasefilloutCAS anddepartmentalevaluationsI CAS goestoSILV 909(needavolunteer)I departmentalgoestoWWH 627(needanothervolunteer)I Thankyouforyourinput!