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Integration by parts "undoes" the product rule
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. . . . . .
Section5.6IntegrationbyParts
Math1aIntroductiontoCalculus
April23, 2008
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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.Image: FlickruserPowi...(ponanwi)
. . . . . .
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
. . . . . .
Outline
LastTime: IntegrationbySubstitution
IntegrationbyParts
Whattomake u andwhattomake dv?
OtherdirtytricksDoubleIBP inthecaseofexpsandsines/cosinesLetting dv = dxSubstitute, thenIBPReductionformulas: “simplifying”withIBP
. . . . . .
LastTime: IntegrationbySubstitution
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
Fordefiniteintegrals,∫ b
af(g(x))g′(x)dx =
∫ g(b)
g(a)f(u)du.
. . . . . .
Outline
LastTime: IntegrationbySubstitution
IntegrationbyParts
Whattomake u andwhattomake dv?
OtherdirtytricksDoubleIBP inthecaseofexpsandsines/cosinesLetting dv = dxSubstitute, thenIBPReductionformulas: “simplifying”withIBP
. . . . . .
Introduction
Wecanverifythat ∫ln x dx = x ln x− x
bydifferentiatingtheright-handside:
ddx
(x ln x− x) = x · 1x
+ ln x− 1 = ln x
Howcouldwedothatwithoutknowingitaheadoftime?
. . . . . .
Introduction
Wecanverifythat ∫ln x dx = x ln x− x
bydifferentiatingtheright-handside:
ddx
(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!
. . . . . .
Everyruleofantidifferentiationisaruleofdifferentiationinreverse
Taketheproductrule:
(uv)′(x) = u′(x)v(x) + u(x)v′(x)
Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)
Nowintegrate!
. . . . . .
Everyruleofantidifferentiationisaruleofdifferentiationinreverse
Taketheproductrule:
(uv)′(x) = u′(x)v(x) + u(x)v′(x)
Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)
Nowintegrate!
. . . . . .
Everyruleofantidifferentiationisaruleofdifferentiationinreverse
Taketheproductrule:
(uv)′(x) = u′(x)v(x) + u(x)v′(x)
Sou(x)v′(x) = (uv)′(x) − v(x)u′(x)
Nowintegrate!
. . . . . .
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)
∫ b
audv = uv
∣∣∣∣ba−
∫ b
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)
∫ b
audv = uv
∣∣∣∣ba−
∫ b
av du.
. . . . . .
Outline
LastTime: IntegrationbySubstitution
IntegrationbyParts
Whattomake u andwhattomake dv?
OtherdirtytricksDoubleIBP inthecaseofexpsandsines/cosinesLetting dv = dxSubstitute, thenIBPReductionformulas: “simplifying”withIBP
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
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
. . . . . .
Example
Find∫
x ln x dx.
SolutionLet u = ln x, dv = x dx. Then∫
x ln x dx =12x2 ln x− 1
2
∫x2 · 1
xdx
=12x2 ln x− 1
2
∫x dx =
12x2 ln x− 1
4x2 + C
. . . . . .
Example
Find∫
x ln x dx.
SolutionLet u = ln x, dv = x dx. Then∫
x ln x dx =12x2 ln x− 1
2
∫x2 · 1
xdx
=12x2 ln x− 1
2
∫x dx =
12x2 ln x− 1
4x2 + C
. . . . . .
Worksheet1–7
. . . . . .
Outline
LastTime: IntegrationbySubstitution
IntegrationbyParts
Whattomake u andwhattomake dv?
OtherdirtytricksDoubleIBP inthecaseofexpsandsines/cosinesLetting dv = dxSubstitute, thenIBPReductionformulas: “simplifying”withIBP
. . . . . .
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
. . . . . .
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
. . . . . .
Todothesecondintegral, let u = cos(3x), dv = e−2x dx. Then
I = −12e
−2x sin(3x) + 32
{−1
2e−2x cos(3x) − 3
2
∫e−2x sin(3x)dx
}= −1
2e−2x sin(3x) − 3
4e−2x cos(3x) − 9
4 I
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.
. . . . . .
Todothesecondintegral, let u = cos(3x), dv = e−2x dx. Then
I = −12e
−2x sin(3x) + 32
{−1
2e−2x cos(3x) − 3
2
∫e−2x sin(3x)dx
}= −1
2e−2x sin(3x) − 3
4e−2x cos(3x) − 9
4 I
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.
. . . . . .
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−
∫x
1 + x2dx
Nowwecanintegratethelastoneby substituting u = 1 + x2,du = 2x dx. Weget∫
arctan x dx = arctan x− 12 ln(1 + x2) + C
. . . . . .
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−
∫x
1 + x2dx
Nowwecanintegratethelastoneby substituting u = 1 + x2,du = 2x dx. Weget∫
arctan x dx = arctan x− 12 ln(1 + x2) + C
. . . . . .
Integralofthelogarithm
Example
Find∫
ln x dx.
SolutionLet u = ln x, dv = dx. Then∫
ln x dx = x ln x− x + C
aspredicted.
. . . . . .
Integralofthelogarithm
Example
Find∫
ln x dx.
SolutionLet u = ln x, dv = dx. Then∫
ln x dx = x ln x− x + C
aspredicted.
. . . . . .
Onemore
Example ∫arcsin x dx =
√1− x2 + x arcsin x + C
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
Worksheet#11–13
. . . . . .
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
. . . . . .
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