lesson 27: integration by substitution (section 10 version)
DESCRIPTION
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.TRANSCRIPT
![Page 1: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/1.jpg)
. . . . . .
Section5.5IntegrationbySubstitution
V63.0121, CalculusI
April27, 2009
Announcements
I Quiz6thisweekcovering5.1–5.2I Practicefinalsonthewebsite. SolutionsFriday
..Imagecredit: kchbrown
![Page 2: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/2.jpg)
. . . . . .
Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
![Page 3: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/3.jpg)
. . . . . .
OfficeHoursandotherhelpthisweekInadditiontorecitation
Day Time Who/What WhereinWWHM 1:00–2:00 LeingangOH 624
3:30–4:30 KatarinaOH 6075:00–7:00 CurtoPS 517
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R 9:00–10:00am LeingangOH 6245:00–7:00pm MariaOH 807
F 2:00–4:00 CurtoOH 1310
![Page 4: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/4.jpg)
. . . . . .
Finalstuff
I FinalisMay8, 2:00–3:50pminCANT 101/200I Oldfinalsonline, includingFall2008I Reviewsessions: May5and6, 6:00–8:00pm, SILV 703
.
.Imagecredit: Pragmagraphr
![Page 5: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/5.jpg)
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ResurrectionPolicyIfyourfinalscorebeatsyourmidtermscore, wewilladd10%toitsweight, andsubtract10%fromthemidtermweight.
..Imagecredit: ScottBeale/LaughingSquid
![Page 6: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/6.jpg)
. . . . . .
Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
![Page 7: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/7.jpg)
. . . . . .
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).
![Page 8: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/8.jpg)
. . . . . .
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?
![Page 9: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/9.jpg)
. . . . . .
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?
![Page 10: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/10.jpg)
. . . . . .
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?
![Page 11: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/11.jpg)
. . . . . .
Sofarwedon’thaveanywaytofind∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.
![Page 12: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/12.jpg)
. . . . . .
Sofarwedon’thaveanywaytofind∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, wecanbesmartandusethe“anti”versionofoneofthemostimportantrulesofdifferentiation: thechainrule.
![Page 13: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/13.jpg)
. . . . . .
Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
![Page 14: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/14.jpg)
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1 + x2.
![Page 15: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/15.jpg)
. . . . . .
SubstitutionforIndefiniteIntegrals
ExampleFind ∫
x√x2 + 1
dx.
SolutionStareatthislongenoughandyounoticethetheintegrandisthederivativeoftheexpression
√1 + x2.
![Page 16: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/16.jpg)
. . . . . .
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.
![Page 17: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/17.jpg)
. . . . . .
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.
![Page 18: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/18.jpg)
. . . . . .
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.
![Page 19: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/19.jpg)
. . . . . .
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.
![Page 20: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/20.jpg)
. . . . . .
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.
![Page 21: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/21.jpg)
. . . . . .
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.
![Page 22: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/22.jpg)
. . . . . .
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.
![Page 23: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/23.jpg)
. . . . . .
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.
![Page 24: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/24.jpg)
. . . . . .
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
![Page 25: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/25.jpg)
. . . . . .
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
![Page 26: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/26.jpg)
. . . . . .
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
![Page 27: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/27.jpg)
. . . . . .
A polynomialexample, thehardway
Comparethistomultiplyingitout:∫(x2 + 3)34x dx =
∫ (x6 + 9x4 + 27x2 + 27
)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
![Page 28: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/28.jpg)
. . . . . .
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!
![Page 29: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/29.jpg)
. . . . . .
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!
![Page 30: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/30.jpg)
. . . . . .
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
![Page 31: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/31.jpg)
. . . . . .
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
![Page 32: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/32.jpg)
. . . . . .
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
![Page 33: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/33.jpg)
. . . . . .
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
![Page 34: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/34.jpg)
. . . . . .
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
![Page 35: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/35.jpg)
. . . . . .
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
![Page 36: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/36.jpg)
. . . . . .
Outline
Announcements
LastTime: TheFundamentalTheorem(s)ofCalculus
SubstitutionforIndefiniteIntegrals
SubstitutionforDefiniteIntegralsTheoryExamples
![Page 37: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/37.jpg)
. . . . . .
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.
![Page 38: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/38.jpg)
. . . . . .
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 .
![Page 39: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/39.jpg)
. . . . . .
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 .
![Page 40: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/40.jpg)
. . . . . .
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 .
![Page 41: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/41.jpg)
. . . . . .
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
.
![Page 42: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/42.jpg)
. . . . . .
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
.
![Page 43: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/43.jpg)
. . . . . .
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
.
![Page 44: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/44.jpg)
. . . . . .
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
![Page 45: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/45.jpg)
. . . . . .
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
![Page 46: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/46.jpg)
. . . . . .
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
![Page 47: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/47.jpg)
. . . . . .
ExampleFind ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
![Page 48: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/48.jpg)
. . . . . .
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.
![Page 49: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/49.jpg)
. . . . . .
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.
![Page 50: Lesson 27: Integration by Substitution (Section 10 version)](https://reader033.vdocument.in/reader033/viewer/2022052823/5550fe98b4c9057b478b4c62/html5/thumbnails/50.jpg)
. . . . . .
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