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The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples.
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. . . . . .
Section5.5IntegrationbySubstitution
V63.0121, CalculusI
April27, 2009
Announcements
I Quiz6thisweekcovering5.1–5.2I Practicefinalsonthewebsite. SolutionsFriday
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Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
. . . . . .
OfficeHoursandotherhelpthisweekInadditiontorecitation
Day Time Who/What WhereinWWHM 1:00–2:00 LeingangOH 624
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W 1:00–2:00 KatarinaOH 6072:00–3:00 LeingangOH 624
R 9:00–10:00am LeingangOH 6245:00–7:00pm MariaOH 807
F 2:00–4:00 CurtoOH 1310
. . . . . .
Finalstuff
I FinalisMay8, 2:00–3:50pminCANT 101/200I Oldfinalsonline, includingFall2008I Reviewsessions: May5and6, 6:00–8:00pm, SILV 703
.
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Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
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
SubstitutionforIndefiniteIntegrals
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 f′(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 f′(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 f′(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
SubstitutionforIndefiniteIntegrals
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) =198
. . . . . .
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) =198
. . . . . .
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) =198
. . . . . .
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.
. . . . . .
Whatdowesubstitute?
I Linearfactors (ax + b) areeasysubstitutions: u = ax + b,du = a dx
I Lookforfunction/derivativepairsintheintegrand: Onetomake u andonetomake du:
I xn and xn−1 (fudgethecoefficient)I sineandcosineI ex and exI ax and ax (fudgethecoefficient)
I√x and
1√x
I ln x and1x
