lesson 29: integration by substition (worksheet solutions)

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. . . . . . Section 5.5 Integration by Substitution V63.0121.034, Calculus I December 9, 2009 Announcements I Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50 I Practice finals on the website. Solutions Friday

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Page 1: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Section5.5IntegrationbySubstitution

V63.0121.034, CalculusI

December9, 2009

Announcements

I FinalExam: Friday12/18, 2:00-3:50pm, TischUC50I Practicefinalsonthewebsite. SolutionsFriday

Page 2: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Schedulefornextweek

I Monday, class: review, evaluations, movie!I Tuesday, 8:00am: Reviewsessionforallstudentswith8:00recitations(TuesdayorThursday)inCIWW 109

I Tuesday, 9:30am: Reviewsessionforallstudentswith9:30recitations(TuesdayorThursday)inCIWW 109

I OfficeHourscontinueI Friday, 2:00pm: finalinTischUC50

Page 3: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

ResurrectionPolicyIfyourfinalscorebeatsyourmidtermscore, wewilladd10%toitsweight, andsubtract10%fromthemidtermweight.

..Imagecredit: ScottBeale/LaughingSquid

Page 4: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Outline

LastTime: TheFundamentalTheorem(s)ofCalculus

SubstitutionforIndefiniteIntegralsTheoryExamples

SubstitutionforDefiniteIntegralsTheoryExamples

Page 5: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

DifferentiationandIntegrationasreverseprocesses

Theorem(TheFundamentalTheoremofCalculus)

1. Let f becontinuouson [a,b]. Then

ddx

∫ x

af(t)dt = f(x)

2. Let f becontinuouson [a,b] and f = F′ forsomeotherfunction F. Then ∫ b

af(x)dx = F(b)− F(a).

Page 6: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Techniquesofantidifferentiation?

Sofarweknowonlyafewrulesforantidifferentiation. Somearegeneral, like∫

[f(x) + g(x)] dx =∫

f(x)dx+∫

g(x)dx

Someareprettyparticular, like∫1

x√x2 − 1

dx = arcsec x+ C.

Whatarewesupposedtodowiththat?

Page 7: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Techniquesofantidifferentiation?

Sofarweknowonlyafewrulesforantidifferentiation. Somearegeneral, like∫

[f(x) + g(x)] dx =∫

f(x)dx+∫

g(x)dx

Someareprettyparticular, like∫1

x√x2 − 1

dx = arcsec x+ C.

Whatarewesupposedtodowiththat?

Page 8: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Techniquesofantidifferentiation?

Sofarweknowonlyafewrulesforantidifferentiation. Somearegeneral, like∫

[f(x) + g(x)] dx =∫

f(x)dx+∫

g(x)dx

Someareprettyparticular, like∫1

x√x2 − 1

dx = arcsec x+ C.

Whatarewesupposedtodowiththat?

Page 9: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Sofarwedon’thaveanywaytofind∫2x√x2 + 1

dx

or ∫tan x dx.

Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.

Page 10: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Sofarwedon’thaveanywaytofind∫2x√x2 + 1

dx

or ∫tan x dx.

Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.

Page 11: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Outline

LastTime: TheFundamentalTheorem(s)ofCalculus

SubstitutionforIndefiniteIntegralsTheoryExamples

SubstitutionforDefiniteIntegralsTheoryExamples

Page 12: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

SubstitutionforIndefiniteIntegrals

ExampleFind ∫

x√x2 + 1

dx.

SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression

√1+ x2.

Page 13: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

SubstitutionforIndefiniteIntegrals

ExampleFind ∫

x√x2 + 1

dx.

SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression

√1+ x2.

Page 14: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Saywhat?

Solution(Moreslowly, now)Let g(x) = x2 + 1.

Then g′(x) = 2x andso

ddx

√g(x) =

1

2√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.

Page 15: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Saywhat?

Solution(Moreslowly, now)Let g(x) = x2 + 1. Then g′(x) = 2x andso

ddx

√g(x) =

1

2√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.

Page 16: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Saywhat?

Solution(Moreslowly, now)Let g(x) = x2 + 1. Then g′(x) = 2x andso

ddx

√g(x) =

1

2√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.

Page 17: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

LeibniziannotationFTW

Solution(Sametechnique, newnotation)Let u = x2 + 1.

Then du = 2x dx and√1+ x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Page 18: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

LeibniziannotationFTW

Solution(Sametechnique, newnotation)Let u = x2 + 1. Then du = 2x dx and

√1+ x2 =

√u.

Sotheintegrandbecomescompletelytransformedinto∫

x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Page 19: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

LeibniziannotationFTW

Solution(Sametechnique, newnotation)Let u = x2 + 1. Then du = 2x dx and

√1+ x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Page 20: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

LeibniziannotationFTW

Solution(Sametechnique, newnotation)Let u = x2 + 1. Then du = 2x dx and

√1+ x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Page 21: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

LeibniziannotationFTW

Solution(Sametechnique, newnotation)Let u = x2 + 1. Then du = 2x dx and

√1+ x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Page 22: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Usefulbutunsavoryvariation

Solution(Sametechnique, newnotation, moreidiot-proof)Let u = x2 + 1. Then du = 2x dx and

√1+ x2 =

√u. “Solvefor

dx:”

dx =du2x

Sotheintegrandbecomescompletelytransformedinto∫x√

x2 + 1dx =

∫x√u· du2x

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.

Mathematicianshaveseriousissueswithmixingthe x and u likethis. However, I can’tdenythatitworks.

Page 23: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

TheoremoftheDay

Theorem(TheSubstitutionRule)If u = g(x) isadifferentiablefunctionwhoserangeisaninterval Iand f iscontinuouson I, then∫

f(g(x))g′(x)dx =∫

f(u)du

Thatis, if F isanantiderivativefor f, then∫f(g(x))g′(x)dx = F(g(x))

InLeibniznotation: ∫f(u)

dudx

dx =∫

f(u)du

Page 24: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A polynomialexample

Example

Usethesubstitution u = x2 + 3 tofind∫

(x2 + 3)34x dx.

SolutionIf u = x2 + 3, then du = 2x dx, and 4x dx = 2du. So∫

(x2 + 3)34x dx =∫

u3 2du = 2∫

u3 du

=12u4 =

12(x2 + 3)4

Page 25: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A polynomialexample

Example

Usethesubstitution u = x2 + 3 tofind∫

(x2 + 3)34x dx.

SolutionIf u = x2 + 3, then du = 2x dx, and 4x dx = 2du. So∫

(x2 + 3)34x dx =∫

u3 2du = 2∫

u3 du

=12u4 =

12(x2 + 3)4

Page 26: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A polynomialexample, bybruteforce

Comparethistomultiplyingitout:∫(x2 + 3)34x dx =

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Whichwouldyouratherdo?I It’sawashforlowpowersI Butforhigherpowers, it’smucheasiertodosubstitution.

Page 27: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A polynomialexample, bybruteforce

Comparethistomultiplyingitout:∫(x2 + 3)34x dx =

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Whichwouldyouratherdo?

I It’sawashforlowpowersI Butforhigherpowers, it’smucheasiertodosubstitution.

Page 28: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A polynomialexample, bybruteforce

Comparethistomultiplyingitout:∫(x2 + 3)34x dx =

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Whichwouldyouratherdo?I It’sawashforlowpowersI Butforhigherpowers, it’smucheasiertodosubstitution.

Page 29: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Compare

Wehavethesubstitutionmethod, which, whenmultipliedout,gives∫

(x2 + 3)34x dx =12(x2 + 3)4

+ C

=12

(x8 + 12x6 + 54x4 + 108x2 + 81

)

+ C

=12x8 + 6x6 + 27x4 + 54x2 +

812

+ C

andthebruteforcemethod∫(x2 + 3)34x dx =

12x8 + 6x6 + 27x4 + 54x2

+ C

Isthisaproblem?

No, that’swhat +C means!

Page 30: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Compare

Wehavethesubstitutionmethod, which, whenmultipliedout,gives∫

(x2 + 3)34x dx =12(x2 + 3)4 + C

=12

(x8 + 12x6 + 54x4 + 108x2 + 81

)+ C

=12x8 + 6x6 + 27x4 + 54x2 +

812

+ C

andthebruteforcemethod∫(x2 + 3)34x dx =

12x8 + 6x6 + 27x4 + 54x2 + C

Isthisaproblem? No, that’swhat +C means!

Page 31: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx.

(Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 32: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 33: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx.

So∫tan x dx =

∫sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 34: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 35: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 36: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

A slickexample

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C

Page 37: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Canyoudoitanotherway?

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = sin x. Then du = cos x dx andso dx =

ducos x

.∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫udu1− u2

Atthispoint, althoughit’spossibletoproceed, weshouldprobablybackupandseeiftheotherwayworksquicker(itdoes).

Page 38: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Canyoudoitanotherway?

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = sin x. Then du = cos x dx andso dx =

ducos x

.

∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫udu1− u2

Atthispoint, althoughit’spossibletoproceed, weshouldprobablybackupandseeiftheotherwayworksquicker(itdoes).

Page 39: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Canyoudoitanotherway?

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)

SolutionLet u = sin x. Then du = cos x dx andso dx =

ducos x

.∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫udu1− u2

Atthispoint, althoughit’spossibletoproceed, weshouldprobablybackupandseeiftheotherwayworksquicker(itdoes).

Page 40: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Forthosewhoreallymustknowall

Solution(Continued, withalgebrahelp)

∫tan x dx =

∫udu1− u2

=

∫12

(1

1− u− 1

1+ u

)du

= −12ln |1− u| − 1

2ln |1+ u|+ C

= ln1√

(1− u)(1+ u)+ C = ln

1√1− u2

+ C

= ln1

|cos x|+ C = ln |sec x|+ C

Page 41: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Outline

LastTime: TheFundamentalTheorem(s)ofCalculus

SubstitutionforIndefiniteIntegralsTheoryExamples

SubstitutionforDefiniteIntegralsTheoryExamples

Page 42: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Theorem(TheSubstitutionRuleforDefiniteIntegrals)If g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

Whythechangeinthelimits?I Theintegralonthelefthappensin“x-land”I Theintegralontherighthappensin“u-land”, sothelimitsneedtobe u-values

I Togetfrom x to u, apply g

Page 43: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Theorem(TheSubstitutionRuleforDefiniteIntegrals)If g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

Whythechangeinthelimits?I Theintegralonthelefthappensin“x-land”I Theintegralontherighthappensin“u-land”, sothelimitsneedtobe u-values

I Togetfrom x to u, apply g

Page 44: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

Solution(SlowWay)

Firstcomputetheindefiniteintegral∫

cos2 x sin x dx andthen

evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0= −1

3

((−1)3 − 13

)=

23.

Page 45: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

Solution(SlowWay)

Firstcomputetheindefiniteintegral∫

cos2 x sin x dx andthen

evaluate.

Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0= −1

3

((−1)3 − 13

)=

23.

Page 46: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

Solution(SlowWay)

Firstcomputetheindefiniteintegral∫

cos2 x sin x dx andthen

evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0= −1

3

((−1)3 − 13

)=

23.

Page 47: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Solution(FastWay)Doboththesubstitutionandtheevaluationatthesametime.

Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13

(1− (−1)

)=

23

I Theadvantagetothe“fastway”isthatyoucompletelytransformtheintegralintosomethingsimpleranddon’thavetogobacktotheoriginal x variable.

I Buttheslowwayisjustasreliable.

Page 48: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Solution(FastWay)Doboththesubstitutionandtheevaluationatthesametime. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.

So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13

(1− (−1)

)=

23

I Theadvantagetothe“fastway”isthatyoucompletelytransformtheintegralintosomethingsimpleranddon’thavetogobacktotheoriginal x variable.

I Buttheslowwayisjustasreliable.

Page 49: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Solution(FastWay)Doboththesubstitutionandtheevaluationatthesametime. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13

(1− (−1)

)=

23

I Theadvantagetothe“fastway”isthatyoucompletelytransformtheintegralintosomethingsimpleranddon’thavetogobacktotheoriginal x variable.

I Buttheslowwayisjustasreliable.

Page 50: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Solution(FastWay)Doboththesubstitutionandtheevaluationatthesametime. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du

=

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13

(1− (−1)

)=

23

I Theadvantagetothe“fastway”isthatyoucompletelytransformtheintegralintosomethingsimpleranddon’thavetogobacktotheoriginal x variable.

I Buttheslowwayisjustasreliable.

Page 51: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 52: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 53: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 54: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Aboutthoselimits

e2(ln√3) = eln

√32

= eln 3 = 3

Page 55: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Aboutthosefractionalpowers

93/2 = (91/2)3 = 33 = 27

43/2 = (41/2)3 = 23 = 8

Page 56: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 57: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 58: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 59: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 60: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

Page 61: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193

Page 62: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193

Page 63: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

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

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193

Page 64: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

ExampleFind ∫ 3π/2

πcot5

6

)sec2

6

)dθ.

Beforewedivein, thinkabout:I What“easy”substitutionsmighthelp?I Whichofthetrigfunctionssuggestsasubstitution?

Page 65: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

ExampleFind ∫ 3π/2

πcot5

6

)sec2

6

)dθ.

Beforewedivein, thinkabout:I What“easy”substitutionsmighthelp?I Whichofthetrigfunctionssuggestsasubstitution?

Page 66: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

SolutionLet φ =

θ

6. Then dφ =

16dθ.

∫ 3π/2

πcot5

6

)sec2

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φdφ

= 6∫ π/4

π/6

sec2 φdφtan5 φ

Nowlet u = tanφ. So du = sec2 φdφ, and

6∫ π/4

π/6

sec2 φdφtan5 φ

= 6∫ 1

1/√3u−5 du

= 6(−14u−4

)∣∣∣∣11/

√3=

32[9− 1] = 12.

Page 67: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

SolutionLet φ =

θ

6. Then dφ =

16dθ.

∫ 3π/2

πcot5

6

)sec2

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φdφ

= 6∫ π/4

π/6

sec2 φdφtan5 φ

Nowlet u = tanφ. So du = sec2 φdφ, and

6∫ π/4

π/6

sec2 φdφtan5 φ

= 6∫ 1

1/√3u−5 du

= 6(−14u−4

)∣∣∣∣11/

√3=

32[9− 1] = 12.

Page 68: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Thelimitsexplained

tanπ

4=

sinπ/4cosπ/4

=

√2/2√2/2

= 1

tanπ

6=

sinπ/6cosπ/6

=1/2√3/2

=1√3

6(−14u−4

)∣∣∣∣11/

√3=

32

[−u−4]1

1/√3 =

32

[u−4]1/√3

1

=32

[(3−1/2)−4 − (1−1/2)−4

]=

32[32 − 12] =

32(9− 1) = 12

Page 69: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Thelimitsexplained

tanπ

4=

sinπ/4cosπ/4

=

√2/2√2/2

= 1

tanπ

6=

sinπ/6cosπ/6

=1/2√3/2

=1√3

6(−14u−4

)∣∣∣∣11/

√3=

32

[−u−4]1

1/√3 =

32

[u−4]1/√3

1

=32

[(3−1/2)−4 − (1−1/2)−4

]=

32[32 − 12] =

32(9− 1) = 12

Page 70: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Graphs

. .θ

.y

.

.

.3π2

.∫ 3π/2

πcot5

6

)sec2

6

)dθ

.y

.

6

.

4

.∫ π/4

π/66 cot5 φ sec2 φdφ

Theareasofthesetworegionsarethesame.

Page 71: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

Graphs

. .φ

.y

.

6

.

4

.∫ π/4

π/66 cot5 φ sec2 φdφ

.u

.y

.∫ 1

1/√36u−5 du

.

.1√3

.

.1

Theareasofthesetworegionsarethesame.

Page 72: Lesson 29: Integration by Substition (worksheet solutions)

. . . . . .

FinalThoughts

I Antidifferentiationisa“nonlinear”problemthatneedspractice, intuition, andperserverance

I Worksheetinrecitation(alsotobeposted)I ThewholeantidifferentiationstoryisinChapter 6