. . . . . .

Section5.5IntegrationbySubstitution

V63.0121, CalculusI

April28, 2009

Announcements

I Quiz6thisweekcovering5.1–5.2I Practicefinalsonthewebsite. SolutionsFriday

..Imagecredit: kchbrown

. . . . . .

Outline

Announcements

LastTime: TheFundamentalTheorem(s)ofCalculus

SubstitutionforIndefiniteIntegralsTheoryExamples

SubstitutionforDefiniteIntegralsTheoryExamples

. . . . . .

OfficeHoursandotherhelp

Day Time Who/What WhereinWWHM 1:00–2:00 LeingangOH 624

5:00–7:00 CurtoPS 517T 1:00–2:00 LeingangOH 624

4:00–5:50 CurtoPS 317W 2:00–3:00 LeingangOH 624R 9:00–10:00am LeingangOH 624F 2:00–4:00 CurtoOH 1310

. . . . . .

Finalstuff

I FinalisMay8, 2:00–3:50pmin19W4101/102I Oldfinalsonline, includingFall2008I Reviewsessions: May5and6, 6:00–8:00pm, SILV 703

.

.Imagecredit: Pragmagraphr

. . . . . .

ResurrectionPolicyIfyourfinalscorebeatsyourmidtermscore, wewilladd10%toitsweight, andsubtract10%fromthemidtermweight.

..Imagecredit: ScottBeale/LaughingSquid

. . . . . .

Outline

Announcements

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

Announcements

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.

. . . . . .

Leibniziannotationwinsagain

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

Then du = 2x dx and√1 + x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x√

x2 + 1dx =

∫1√u

(12du

)=

∫1

2√udu

=

∫12u

−1/2 du

=√u + C =

√1 + x2 + C.

. . . . . .

Leibniziannotationwinsagain

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

√1 + x2 =

√u.

Sotheintegrandbecomescompletelytransformedinto∫

x√x2 + 1

dx =

∫1√u

(12du

)=

∫1

2√udu

=

∫12u

−1/2 du

=√u + C =

√1 + x2 + C.

. . . . . .

Leibniziannotationwinsagain

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

√1 + x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x√

x2 + 1dx =

∫1√u

(12du

)=

∫1

2√udu

=

∫12u

−1/2 du

=√u + C =

√1 + x2 + C.

. . . . . .

Leibniziannotationwinsagain

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

√1 + x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x√

x2 + 1dx =

∫1√u

(12du

)=

∫1

2√udu

=

∫12u

−1/2 du

=√u + C =

√1 + x2 + C.

. . . . . .

Leibniziannotationwinsagain

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

√1 + x2 =

√u. Sothe

integrandbecomescompletelytransformedinto∫x√

x2 + 1dx =

∫1√u

(12du

)=

∫1

2√udu

=

∫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

. . . . . .

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, thehardway

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

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

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

. . . . . .

Compare

Wehave∫(x2 + 3)34x dx =

12(x2 + 3)4

+ C

∫(x2 + 3)34x dx =

12x8 + 6x6 + 27x4 + 54x2

+ C

Now

12

(x2 + 3)4 =12

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

)=

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

812

Isthisaproblem?

No, that’swhat +C means!

. . . . . .

Compare

Wehave∫(x2 + 3)34x dx =

12(x2 + 3)4 + C∫

(x2 + 3)34x dx =12x8 + 6x6 + 27x4 + 54x2 + C

Now

12

(x2 + 3)4 =12

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

)=

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

812

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

. . . . . .

Outline

Announcements

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.

. . . . . .

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

.

. . . . . .

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

=

∫ 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 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.

. . . . . .

Graphs

. .θ

.y

.

.π

.

.3π

2

.∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ

.φ

.y

.

.π

6

.

.π

4

.∫ π/4

π/66 cot5 φ sec2 φdφ

. . . . . .

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

. . . . . .

Summary: 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