. . . . . . Section 5.5 Integration by Substitution Math 1a Introduction to Calculus April 21, 2008 Announcements ◮ Midterm III is Wednesday 4/30 in class ◮ Friday 5/2 is Movie Day! ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 ◮ Office hours Tues, Weds, 2–4pm SC 323 ◮ Final (tentative) 5/23 9:15am . Image: Flickr user kchbrown
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. . . . . .
Section5.5IntegrationbySubstitution
Math1aIntroductiontoCalculus
April21, 2008
Announcements
◮ MidtermIII isWednesday4/30inclass◮ Friday5/2isMovieDay!◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ Final(tentative)5/239:15am
.
.Image: Flickruserkchbrown
. . . . . .
Announcements
◮ MidtermIII isWednesday4/30inclass◮ Friday5/2isMovieDay!◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ Final(tentative)5/239:15am
. . . . . .
HappyPatriot’sDay!
Listen, mychildren, andyoushallhearOfthemidnightrideofPaulRevere,OntheeighteenthofApril, inSeventy-Five;HardlyamanisnowaliveWhoremembersthatfamousdayandyear.
—HenryWadsworthLongfellow
. . . . . .
RevereandDawes
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
Worksheet
. . . . . .
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
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
Worksheet
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1 + x2.
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1 + x2.
. . . . . .
Solution(Moreslowly, now)
Let u = x2 + 1. Thendudx
= 2x andso
ddx
√u =
12√ududx
=x√
x2 + 1
Thus ∫x√
x2 + 1dx =
√1 + x2 + C.
. . . . . .
Solution(Sametechnique, newnotation)Let u = x2 + 1. Then du = 2x dx and
√1 + x2 =
√u. Sothe
integrandbecomescompletelytransformedinto∫x√
x2 + 1dx =
∫1√u
(12du
)=
∫12u
−1/2 du
=√u + C =
√1 + x2 + C.
. . . . . .
TheoremoftheDay
Theorem(TheSubstitutionRule)If 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
. . . . . .
Example
Find∫
tan x dx.
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
. . . . . .
Example
Find∫
tan x dx.
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
. . . . . .
Example
Find∫
tan x dx.
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
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
Worksheet
. . . . . .
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.
. . . . . .
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
3 cos3 x
∣∣π0 = 2
3 .
. . . . . .
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
3 cos3 x
∣∣π0 = 2
3 .
. . . . . .
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
3 cos3 x
∣∣π0 = 2
3 .
. . . . . .
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
= 13u
3∣∣1−1 =
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
= 13u
3∣∣1−1 =
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
= 13u
3∣∣1−1 =
23
.
. . . . . .
ExampleFind ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
. . . . . .
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 φdφ
tan5 φ
Nowlet u = tanφ. So du = sec2 φdφ, and
6∫ π/4
π/6
sec2 φdφ
tan5 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 φdφ
tan5 φ
Nowlet u = tanφ. So du = sec2 φdφ, and
6∫ π/4
π/6
sec2 φdφ
tan5 φ= 6
∫ 1
1/√3u−5 du
= 6(−14u−4
)∣∣∣∣11/
√3
=32
[9− 1] = 12.
. . . . . .
Outline
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
Worksheet
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