Lesson 30: Integration by Parts

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Section5.6IntegrationbyParts

Math1aIntroductiontoCalculus

April23, 2008

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Announcements

◮ MidtermIII isWednesday4/30inclass(covers§5.1–5.6)◮ Friday5/2isMovieDay!◮ Final(tentative)5/239:15am◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323

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Outline

LastTime: IntegrationbySubstitution

IntegrationbyParts

Whattomake u andwhattomake dv?

OtherdirtytricksDoubleIBP inthecaseofexpsandsines/cosinesLetting dv = dxSubstitute, thenIBPReductionformulas: “simplifying”withIBP

. . . . . .

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

f(g(x))g′(x)dx =

∫f(u)du

or ∫f(u)

∫f(u)du

Fordefiniteintegrals,∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

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Outline

IntegrationbyParts

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Introduction

Wecanverifythat ∫ln x dx = x ln x− x

bydifferentiatingtheright-handside:

(x ln x− x) = x · 1x

+ ln x− 1 = ln x

Howcouldwedothatwithoutknowingitaheadoftime?

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Introduction

Wecanverifythat ∫ln x dx = x ln x− x

bydifferentiatingtheright-handside:

(x ln x− x) = x · 1x

+ ln x− 1 = ln x

Howcouldwedothatwithoutknowingitaheadoftime?

. . . . . .

Everyruleofantidifferentiationisaruleofdifferentiationinreverse

Taketheproductrule:

(uv)′(x) = u′(x)v(x) + u(x)v′(x)

Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)

Nowintegrate!

. . . . . .

Taketheproductrule:

(uv)′(x) = u′(x)v(x) + u(x)v′(x)

Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)

Nowintegrate!

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Taketheproductrule:

(uv)′(x) = u′(x)v(x) + u(x)v′(x)

Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)

Nowintegrate!

. . . . . .

Taketheproductrule:

(uv)′(x) = u′(x)v(x) + u(x)v′(x)

Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)

Nowintegrate!

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Theorem(IntegrationbyParts)Let u and v bedifferentiablefunctions. Then∫

u(x)v′(x)dx = u(x)v(x) −∫

v(x)u′(x)dx.

Succinctly, ∫udv = uv−

∫v du.

Theorem(IntegrationbyParts, definiteform)

audv = uv

∣∣∣∣ba−

av du.

. . . . . .

Theorem(IntegrationbyParts)Let u and v bedifferentiablefunctions. Then∫

u(x)v′(x)dx = u(x)v(x) −∫

v(x)u′(x)dx.

Succinctly, ∫udv = uv−

∫v du.

Theorem(IntegrationbyParts, definiteform)

audv = uv

∣∣∣∣ba−

av du.

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Outline

IntegrationbyParts

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Example∫xex dx.

SolutionA possiblechoicewouldbe u = ex, dv = x dx. Then du = ex dxand v = 1

2x2. Thus∫

xex dx = 12x

2ex − 12

∫x2ex dx.

Thisdoesn’tmaketheintegralanysimpler, though.

Instead, tryu = x, dv = ex dx. Then du = dx, and v = ex. Thus∫

xex dx = xex −∫

ex dx = xex − ex + C

Themoralofthestoryistochoose u and dv whichmake v dueasierthan udv. Polynomialsareagoodchoicefor u.

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Example∫xex dx.

2x2. Thus∫

xex dx = 12x

2ex − 12

∫x2ex dx.

Thisdoesn’tmaketheintegralanysimpler, though. Instead, tryu = x, dv = ex dx. Then du = dx, and v = ex. Thus∫

xex dx = xex −∫

. . . . . .

Example∫xex dx.

2x2. Thus∫

xex dx = 12x

2ex − 12

∫x2ex dx.

Thisdoesn’tmaketheintegralanysimpler, though. Instead, tryu = x, dv = ex dx. Then du = dx, and v = ex. Thus∫

xex dx = xex −∫

. . . . . .

Example

Find∫

x2ex dx.

SolutionLet u = x2, du = 2x dx. Then∫

x2ex dx = u2ex − 2∫

xex dx

andwedidthelastpartalready! So∫x2ex dx = x2ex − 2xex + 2ex + C

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Example

Find∫

x2ex dx.

SolutionLet u = x2, du = 2x dx. Then∫

x2ex dx = u2ex − 2∫

xex dx

andwedidthelastpartalready! So∫x2ex dx = x2ex − 2xex + 2ex + C

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Example

Find∫

x ln x dx.

SolutionLet u = ln x, dv = x dx. Then∫

x ln x dx =12x2 ln x− 1

∫x2 · 1

=12x2 ln x− 1

∫x dx =

12x2 ln x− 1

4x2 + C

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Example

Find∫

x ln x dx.

SolutionLet u = ln x, dv = x dx. Then∫

x ln x dx =12x2 ln x− 1

∫x2 · 1

=12x2 ln x− 1

∫x dx =

12x2 ln x− 1

4x2 + C

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Worksheet1–7

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Outline

IntegrationbyParts

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DoubleIBP inthecaseofexpsandsines/cosinesWorksheet#8

Example

Find∫

e−2x sin(3x)dx

SolutionLet u = sin(3x), dv = e−2x dx. Then du = 3 cos(3x)dx and

v = −12e−2x. So

I = −12e

−2x sin(3x) + 32

∫e−2x cos(3x)dx

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DoubleIBP inthecaseofexpsandsines/cosinesWorksheet#8

Example

Find∫

e−2x sin(3x)dx

SolutionLet u = sin(3x), dv = e−2x dx. Then du = 3 cos(3x)dx and

v = −12e−2x. So

I = −12e

−2x sin(3x) + 32

∫e−2x cos(3x)dx

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Todothesecondintegral, let u = cos(3x), dv = e−2x dx. Then

I = −12e

−2x sin(3x) + 32

2e−2x cos(3x) − 3

∫e−2x sin(3x)dx

}= −1

2e−2x sin(3x) − 3

4e−2x cos(3x) − 9

Wearebackwherewestarted, butinagoodway.

134 I = −1

2e−2x sin(3x) − 3

4e−2x cos(3x)

=⇒ I = − 213e

−2x sin(3x) − 313e

−2x cos(3x)

Danger! Don’tundoyourwork. Thechoice u = e−2x,dv = cos(3x)dx willcanceleverythingontheright-handsidebutI.

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Todothesecondintegral, let u = cos(3x), dv = e−2x dx. Then

I = −12e

−2x sin(3x) + 32

2e−2x cos(3x) − 3

∫e−2x sin(3x)dx

}= −1

2e−2x sin(3x) − 3

4e−2x cos(3x) − 9

Wearebackwherewestarted, butinagoodway.

134 I = −1

2e−2x sin(3x) − 3

4e−2x cos(3x)

=⇒ I = − 213e

−2x sin(3x) − 313e

−2x cos(3x)

Danger! Don’tundoyourwork. Thechoice u = e−2x,dv = cos(3x)dx willcanceleverythingontheright-handsidebutI.

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Letting dv = dxWorksheet#9–10

Example

Find∫

arctan x dx

SolutionLet u = arctan x, dv = dx. So du =

11 + x2

dx and v = x. Thus∫arctan x dx = x arctan x−

1 + x2dx

Nowwecanintegratethelastoneby substituting u = 1 + x2,du = 2x dx. Weget∫

arctan x dx = arctan x− 12 ln(1 + x2) + C

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Letting dv = dxWorksheet#9–10

Example

Find∫

arctan x dx

SolutionLet u = arctan x, dv = dx. So du =

11 + x2

dx and v = x. Thus∫arctan x dx = x arctan x−

1 + x2dx

Nowwecanintegratethelastoneby substituting u = 1 + x2,du = 2x dx. Weget∫

arctan x dx = arctan x− 12 ln(1 + x2) + C

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Integralofthelogarithm

Example

Find∫

ln x dx.

SolutionLet u = ln x, dv = dx. Then∫

ln x dx = x ln x− x + C

aspredicted.

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Integralofthelogarithm

Example

Find∫

ln x dx.

SolutionLet u = ln x, dv = dx. Then∫

ln x dx = x ln x− x + C

aspredicted.

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Onemore

Example ∫arcsin x dx =

√1− x2 + x arcsin x + C

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Substitute, thenIBP

Example

Find∫

z(ln z)2 dz

SolutionLet u = ln z, so z = eu and dz = eu du. Thus∫

z(ln z)2 dz =

∫e2uu2 du

Nowintegratebyparts.

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Substitute, thenIBP

Example

Find∫

z(ln z)2 dz

z(ln z)2 dz =

∫e2uu2 du

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Substitute, thenIBP

Example

Find∫

z(ln z)2 dz

z(ln z)2 dz =

∫e2uu2 du

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Worksheet#11–13

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Reductionformulas: “simplifying”withIBP

Example

Find∫

cosn x dx.

SolutionLet u = cosn−1 x, dv = cos x dx. Then v = sin x anddu = (n− 1) cosn−2(x)(− sin x)dx Weget∫

cosn x dx = cosn−1 x sin x +

∫cosn−1 x sin2 x dx

Nowwrite sin2 x = 1− cos2 x, expand, andcollectthe∫cosn x dx ontheleft. Youget

∫cosn x dx =

1ncosn−1 x sin x +

n− 1n

∫cosn−2 x dx

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Reductionformulas: “simplifying”withIBP

Example

Find∫

cosn x dx.

SolutionLet u = cosn−1 x, dv = cos x dx. Then v = sin x anddu = (n− 1) cosn−2(x)(− sin x)dx Weget∫

cosn x dx = cosn−1 x sin x +

∫cosn−1 x sin2 x dx

Nowwrite sin2 x = 1− cos2 x, expand, andcollectthe∫cosn x dx ontheleft. Youget

∫cosn x dx =

1ncosn−1 x sin x +

n− 1n

∫cosn−2 x dx

Lesson 30: Integration by Parts

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Transcript of Lesson 30: Integration by Parts