Lesson 29: Integration by Substitution

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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

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. . . . . .

Announcements

◮ MidtermIII isWednesday4/30inclass◮ Friday5/2isMovieDay!◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ Final(tentative)5/239:15am

. . . . . .

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—HenryWadsworthLongfellow

. . . . . .

RevereandDawes

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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