Lesson 29: Integration by Substition (worksheet solutions)
Post on 12-May-2015
959 Views
Preview:
Transcript
. . . . . .
Section5.5IntegrationbySubstitution
V63.0121.034, CalculusI
December9, 2009
Announcements
I FinalExam: Friday12/18, 2:00-3:50pm, TischUC50I Practicefinalsonthewebsite. SolutionsFriday
. . . . . .
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
. . . . . .
ResurrectionPolicyIfyourfinalscorebeatsyourmidtermscore, wewilladd10%toitsweight, andsubtract10%fromthemidtermweight.
..Imagecredit: ScottBeale/LaughingSquid
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegralsTheoryExamples
SubstitutionforDefiniteIntegralsTheoryExamples
. . . . . .
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).
. . . . . .
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?
. . . . . .
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?
. . . . . .
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?
. . . . . .
Sofarwedon’thaveanywaytofind∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.
. . . . . .
Sofarwedon’thaveanywaytofind∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegralsTheoryExamples
SubstitutionforDefiniteIntegralsTheoryExamples
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1+ x2.
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1+ x2.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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!
. . . . . .
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!
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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).
. . . . . .
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).
. . . . . .
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).
. . . . . .
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
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegralsTheoryExamples
SubstitutionforDefiniteIntegralsTheoryExamples
. . . . . .
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
. . . . . .
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
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
Aboutthoselimits
e2(ln√3) = eln
√32
= eln 3 = 3
. . . . . .
Aboutthosefractionalpowers
93/2 = (91/2)3 = 33 = 27
43/2 = (41/2)3 = 23 = 8
. . . . . .
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
=
∫ 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
=
∫ 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
=
∫ 3
2u · udu =
13u3
∣∣∣∣32=
193
. . . . . .
ExampleFind ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
Beforewedivein, thinkabout:I What“easy”substitutionsmighthelp?I Whichofthetrigfunctionssuggestsasubstitution?
. . . . . .
ExampleFind ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
Beforewedivein, thinkabout:I What“easy”substitutionsmighthelp?I Whichofthetrigfunctionssuggestsasubstitution?
. . . . . .
SolutionLet φ =
θ
6. Then dφ =
16dθ.
∫ 3π/2
πcot5
(θ
6
)sec2
(θ
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.
. . . . . .
SolutionLet φ =
θ
6. Then dφ =
16dθ.
∫ 3π/2
πcot5
(θ
6
)sec2
(θ
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.
. . . . . .
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
. . . . . .
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
. . . . . .
Graphs
. .θ
.y
.
.π
.
.3π2
.∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ
.φ
.y
.
.π
6
.
.π
4
.∫ π/4
π/66 cot5 φ sec2 φdφ
Theareasofthesetworegionsarethesame.
. . . . . .
Graphs
. .φ
.y
.
.π
6
.
.π
4
.∫ π/4
π/66 cot5 φ sec2 φdφ
.u
.y
.∫ 1
1/√36u−5 du
.
.1√3
.
.1
Theareasofthesetworegionsarethesame.
. . . . . .
FinalThoughts
I Antidifferentiationisa“nonlinear”problemthatneedspractice, intuition, andperserverance
I Worksheetinrecitation(alsotobeposted)I ThewholeantidifferentiationstoryisinChapter 6
top related