lesson 27: integration by substitution (slides)
DESCRIPTION
Integration by substitution is the chain rule in reverse.NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200TRANSCRIPT
![Page 1: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/1.jpg)
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Sec on 5.5Integra on by Subs tu on
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
May 4, 2011
![Page 2: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/2.jpg)
AnnouncementsI Today: 5.5I Monday 5/9: Review in classI Tuesday 5/10: Review Sessions by TAsI Wednesday 5/11: TA office hours:
I Adam 10–noon (WWH 906)I Jerome 3:30–5:30 (WWH 501)I Soohoon 6–8 (WWH 511)
I Thursday 5/12: Final Exam,2:00–3:50pm, CANT 200
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Resurrection PolicyIf your final score beats your midterm score, we will add 10% to itsweight, and subtract 10% from the midterm weight.
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Image credit: Sco Beale / Laughing Squid
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ObjectivesI Given an integral and asubs tu on, transform theintegral into an equivalent oneusing a subs tu on
I Evaluate indefinite integralsusing the method ofsubs tu on.
I Evaluate definite integrals usingthe method of subs tu on.
![Page 5: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/5.jpg)
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
![Page 6: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/6.jpg)
Differentiation and Integration asreverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be con nuous on [a, b]. Then
ddx
∫ x
af(t) dt = f(x)
2. Let f be con nuous on [a, b] and f = F′ for some other func onF. Then ∫ b
af(x) dx = F(b)− F(a).
![Page 7: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/7.jpg)
Techniques of antidifferentiation?So far we know only a few rules for an differen a on. Some aregeneral, like∫
[f(x) + g(x)] dx =∫
f(x) dx+∫
g(x) dx
Some are pre y par cular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
![Page 8: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/8.jpg)
Techniques of antidifferentiation?So far we know only a few rules for an differen a on. Some aregeneral, like∫
[f(x) + g(x)] dx =∫
f(x) dx+∫
g(x) dx
Some are pre y par cular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
![Page 9: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/9.jpg)
Techniques of antidifferentiation?So far we know only a few rules for an differen a on. Some aregeneral, like∫
[f(x) + g(x)] dx =∫
f(x) dx+∫
g(x) dx
Some are pre y par cular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
![Page 10: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/10.jpg)
No straightforward system ofantidifferentiation
So far we don’t have any way to find∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, we can be smart and use the “an ” version of one of themost important rules of differen a on: the chain rule.
![Page 11: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/11.jpg)
No straightforward system ofantidifferentiation
So far we don’t have any way to find∫2x√x2 + 1
dx
or ∫tan x dx.
Luckily, we can be smart and use the “an ” version of one of themost important rules of differen a on: the chain rule.
![Page 12: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/12.jpg)
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
![Page 13: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/13.jpg)
Substitution for IndefiniteIntegrals
Example
Find ∫x√
x2 + 1dx.
Solu onStare at this long enough and you no ce the the integrand is thederiva ve of the expression
√1+ x2.
![Page 14: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/14.jpg)
Substitution for IndefiniteIntegrals
Example
Find ∫x√
x2 + 1dx.
Solu onStare at this long enough and you no ce the the integrand is thederiva ve of the expression
√1+ x2.
![Page 15: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/15.jpg)
Say what?Solu on (More slowly, now)
Let g(x) = x2 + 1.
Then g′(x) = 2x and so
ddx
√g(x) =
12√
g(x)g′(x) =
x√x2 + 1
Thus ∫x√
x2 + 1dx =
∫ (ddx
√g(x)
)dx
=√
g(x) + C =√1+ x2 + C.
![Page 16: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/16.jpg)
Say what?Solu on (More slowly, now)
Let g(x) = x2 + 1. Then g′(x) = 2x and so
ddx
√g(x) =
12√
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 (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/17.jpg)
Say what?Solu on (More slowly, now)
Let g(x) = x2 + 1. Then g′(x) = 2x and so
ddx
√g(x) =
12√
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 (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/18.jpg)
Leibnizian notation FTWSolu on (Same technique, new nota on)
Let u = x2 + 1.
Then du = 2x dx and√1+ x2 =
√u. So the
integrand becomes completely transformed into∫x dx√x2 + 1
=
∫ 12du√u=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
![Page 19: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/19.jpg)
Leibnizian notation FTWSolu on (Same technique, new nota on)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u.
So theintegrand becomes completely transformed into∫
x dx√x2 + 1
=
∫ 12du√u=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
![Page 20: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/20.jpg)
Leibnizian notation FTWSolu on (Same technique, new nota on)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. So the
integrand becomes completely transformed into∫x dx√x2 + 1
=
∫ 12du√u=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
![Page 21: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/21.jpg)
Leibnizian notation FTWSolu on (Same technique, new nota on)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. So the
integrand becomes completely transformed into∫x dx√x2 + 1
=
∫ 12du√u=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
![Page 22: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/22.jpg)
Leibnizian notation FTWSolu on (Same technique, new nota on)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. So the
integrand becomes completely transformed into∫x dx√x2 + 1
=
∫ 12du√u=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
![Page 23: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/23.jpg)
Useful but unsavory variationSolu on (Same technique, new nota on, more idiot-proof)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. “Solve for dx:”
dx =du2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du2x
=
∫1
2√udu
=
∫12u
−1/2 du =√u+ C =
√1+ x2 + C.
Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.
![Page 24: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/24.jpg)
Useful but unsavory variationSolu on (Same technique, new nota on, more idiot-proof)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. “Solve for dx:”
dx =du2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du2x
=
∫1
2√udu
=
∫12u
−1/2 du =√u+ C =
√1+ x2 + C.
Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.
![Page 25: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/25.jpg)
Useful but unsavory variationSolu on (Same technique, new nota on, more idiot-proof)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. “Solve for dx:”
dx =du2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du2x
=
∫1
2√udu
=
∫12u
−1/2 du =√u+ C =
√1+ x2 + C.
Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.
![Page 26: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/26.jpg)
Useful but unsavory variationSolu on (Same technique, new nota on, more idiot-proof)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. “Solve for dx:”
dx =du2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du2x
=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.
![Page 27: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/27.jpg)
Useful but unsavory variationSolu on (Same technique, new nota on, more idiot-proof)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u. “Solve for dx:”
dx =du2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du2x
=
∫1
2√udu
=
∫12u
−1/2 du =√u+ C =
√1+ x2 + C.
Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.
![Page 28: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/28.jpg)
Theorem of the DayTheorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval Iand f is con nuous on I, then∫
f(g(x))g′(x) dx =∫
f(u) du
![Page 29: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/29.jpg)
Theorem of the DayTheorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval Iand f is con nuous on I, then∫
f(g(x))g′(x) dx =∫
f(u) du
That is, if F is an an deriva ve for f, then∫f(g(x))g′(x) dx = F(g(x))
![Page 30: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/30.jpg)
Theorem of the DayTheorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval Iand f is con nuous on I, then∫
f(g(x))g′(x) dx =∫
f(u) du
In Leibniz nota on: ∫f(u)
dudx
dx =∫
f(u) du
![Page 31: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/31.jpg)
A polynomial exampleExample
Use the subs tu on u = x2 + 3 to find∫
(x2 + 3)34x dx.
Solu onIf u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫
(x2 + 3)34x dx
=
∫u3 2du = 2
∫u3 du
=12u4 =
12(x2 + 3)4
![Page 32: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/32.jpg)
A polynomial exampleExample
Use the subs tu on u = x2 + 3 to find∫
(x2 + 3)34x dx.
Solu onIf u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫
(x2 + 3)34x dx
=
∫u3 2du = 2
∫u3 du
=12u4 =
12(x2 + 3)4
![Page 33: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/33.jpg)
A polynomial exampleExample
Use the subs tu on u = x2 + 3 to find∫
(x2 + 3)34x dx.
Solu onIf u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫
(x2 + 3)34x dx =∫
u3 2du = 2∫
u3 du
=12u4 =
12(x2 + 3)4
![Page 34: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/34.jpg)
A polynomial exampleExample
Use the subs tu on u = x2 + 3 to find∫
(x2 + 3)34x dx.
Solu onIf u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫
(x2 + 3)34x dx =∫
u3 2du = 2∫
u3 du
=12u4
=12(x2 + 3)4
![Page 35: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/35.jpg)
A polynomial exampleExample
Use the subs tu on u = x2 + 3 to find∫
(x2 + 3)34x dx.
Solu onIf u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫
(x2 + 3)34x dx =∫
u3 2du = 2∫
u3 du
=12u4 =
12(x2 + 3)4
![Page 36: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/36.jpg)
A polynomial example (brute force)Solu on
∫(x2 + 3)34x dx =
∫ (x6 + 9x4 + 27x2 + 27
)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 37: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/37.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx
=
∫ (x6 + 9x4 + 27x2 + 27
)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 38: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/38.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 39: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/39.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 40: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/40.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 41: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/41.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?
I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 42: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/42.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powers
I But for higher powers, it’s much easier to do subs tu on.
![Page 43: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/43.jpg)
A polynomial example (brute force)Solu on∫
(x2 + 3)34x dx =∫ (
x6 + 9x4 + 27x2 + 27)4x dx
=
∫ (4x7 + 36x5 + 108x3 + 108x
)dx
=12x8 + 6x6 + 27x4 + 54x2
Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do subs tu on.
![Page 44: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/44.jpg)
CompareWe have the subs tu on method, which, when mul plied out, gives∫
(x2 + 3)34x dx =12(x2 + 3)4
+ C
=12(x8 + 12x6 + 54x4 + 108x2 + 81
)
+ C
=12x8 + 6x6 + 27x4 + 54x2 +
812
+ C
and the brute force method∫(x2 + 3)34x dx =
12x8 + 6x6 + 27x4 + 54x2
+ C
Is there a difference? Is this a problem?
No, that’s what+Cmeans!
![Page 45: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/45.jpg)
CompareWe have the subs tu on method, which, when mul plied out, gives∫
(x2 + 3)34x dx =12(x2 + 3)4 + C
=12(x8 + 12x6 + 54x4 + 108x2 + 81
)+ C
=12x8 + 6x6 + 27x4 + 54x2 +
812
+ C
and the brute force method∫(x2 + 3)34x dx =
12x8 + 6x6 + 27x4 + 54x2 + C
Is there a difference? Is this a problem? No, that’s what+Cmeans!
![Page 46: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/46.jpg)
A slick exampleExample
Find∫
tan x dx.
(Hint: tan x =sin xcos x
)
Solu onLet 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 47: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/47.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 48: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/48.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 49: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/49.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 50: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/50.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 51: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/51.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 52: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/52.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 53: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/53.jpg)
A slick exampleExample
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu onLet 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 54: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/54.jpg)
Can you do it another way?Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu on
Let u = sin x. Then du = cos x dx and so dx =ducos x
.∫tan x dx =
∫sin xcos x
dx =∫
ucos x
ducos x
=
∫u ducos2 x
=
∫u du
1− sin2 x=
∫u du1− u2
![Page 55: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/55.jpg)
Can you do it another way?Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu on
Let u = sin x. Then du = cos x dx and so dx =ducos x
.
∫tan x dx =
∫sin xcos x
dx =∫
ucos x
ducos x
=
∫u ducos2 x
=
∫u du
1− sin2 x=
∫u du1− u2
![Page 56: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/56.jpg)
Can you do it another way?Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solu on
Let u = sin x. Then du = cos x dx and so dx =ducos x
.∫tan x dx =
∫sin xcos x
dx =∫
ucos x
ducos x
=
∫u ducos2 x
=
∫u du
1− sin2 x=
∫u du1− u2
![Page 57: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/57.jpg)
For those who really must know allSolu on (Con nued, with algebra help)
Let y = 1− u2, so dy = −2u du. Then∫tan x dx =
∫u du1− u2
=
∫uy
dy−2u
= −12
∫dyy
= −12ln |y|+ C = −1
2ln∣∣1− u2
∣∣+ C
= ln1√
1− u2+ C = ln
1√1− sin2 x
+ C
= ln1
|cos x|+ C = ln |sec x|+ C
There are other ways to do it, too.
![Page 58: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/58.jpg)
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
![Page 59: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/59.jpg)
Substitution for Definite IntegralsTheorem (The Subs tu on Rule for Definite Integrals)
If g′ is con nuous and f is con nuous on the range of u = g(x), then∫ b
af(g(x))g′(x) dx =
∫ g(b)
g(a)f(u) du.
Why the change in the limits?I The integral on the le happens in “x-land”I The integral on the right happens in “u-land”, so the limits needto be u-values
I To get from x to u, apply g
![Page 60: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/60.jpg)
Substitution for Definite IntegralsTheorem (The Subs tu on Rule for Definite Integrals)
If g′ is con nuous and f is con nuous on the range of u = g(x), then∫ b
af(g(x))g′(x) dx =
∫ g(b)
g(a)f(u) du.
Why the change in the limits?I The integral on the le happens in “x-land”I The integral on the right happens in “u-land”, so the limits needto be u-values
I To get from x to u, apply g
![Page 61: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/61.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)
I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 62: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/62.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)
I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 63: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/63.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 64: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/64.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 65: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/65.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 66: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/66.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 67: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/67.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I First compute the indefinite integral
∫cos2 x sin x dx and then
evaluate.I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 68: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/68.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −13((−1)3 − 13
)=
23.
![Page 69: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/69.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0= −1
3((−1)3 − 13
)
=23.
![Page 70: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/70.jpg)
Example
Compute∫ π
0cos2 x sin x dx.
Solu on (Slow Way)I 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.
I Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0= −1
3((−1)3 − 13
)=
23.
![Page 71: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/71.jpg)
Definite-ly QuickerSolu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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
=13u3∣∣∣∣1−1
=13(1− (−1)
)=
23
![Page 72: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/72.jpg)
Definite-ly QuickerSolu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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
=13u3∣∣∣∣1−1
=13(1− (−1)
)=
23
![Page 73: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/73.jpg)
Definite-ly QuickerSolu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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
=13u3∣∣∣∣1−1
=13(1− (−1)
)=
23
![Page 74: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/74.jpg)
Definite-ly QuickerSolu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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
=13u3∣∣∣∣1−1
=13(1− (−1)
)=
23
![Page 75: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/75.jpg)
Definite-ly QuickerSolu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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
=13u3∣∣∣∣1−1
=13(1− (−1)
)=
23
![Page 76: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/76.jpg)
Compare
I The advantage to the “fast way” is that you completelytransform the integral into something simpler and don’t haveto go back to the original variable (x).
I But the slow way is just as reliable.
![Page 77: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/77.jpg)
An exponential exampleExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu on
![Page 78: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/78.jpg)
An exponential exampleExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x, so du = 2e2x dx. We have∫ ln
√8
ln√3
e2x√e2x + 1 dx =
12
∫ 8
3
√u+ 1 du
![Page 79: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/79.jpg)
About those limits
Since
e2(ln√3) = eln
√32
= eln 3 = 3
we have ∫ ln√8
ln√3
e2x√e2x + 1 dx =
12
∫ 8
3
√u+ 1 du
![Page 80: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/80.jpg)
An exponential exampleExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onNow let y = u+ 1, dy = du. So
12
∫ 8
3
√u+ 1 du =
12
∫ 9
4
√y dy =
12· 23y3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 81: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/81.jpg)
About those fractional powers
We have
93/2 = (91/2)3 = 33 = 2743/2 = (41/2)3 = 23 = 8
so12
∫ 9
4y1/2 dy =
12· 23y3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 82: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/82.jpg)
An exponential exampleExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onNow let y = u+ 1, dy = du. So
12
∫ 8
3
√u+ 1 du =
12
∫ 9
4
√y dy =
12· 23y3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 83: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/83.jpg)
Another way to skin that catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x + 1,
so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√e2x + 1 dx =
12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 84: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/84.jpg)
Another way to skin that catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x + 1,so that du = 2e2x dx.
Then∫ ln√8
ln√3
e2x√e2x + 1 dx =
12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 85: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/85.jpg)
Another way to skin that catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√e2x + 1 dx =
12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 86: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/86.jpg)
Another way to skin that catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√e2x + 1 dx =
12
∫ 9
4
√u du =
13u3/2
∣∣∣∣94
=13(27− 8) =
193
![Page 87: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/87.jpg)
Another way to skin that catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√e2x + 1 dx =
12
∫ 9
4
√u du =
13u3/2
∣∣∣∣94=
13(27− 8) =
193
![Page 88: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/88.jpg)
A third skinned catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u =
√e2x + 1, so that u2 = e2x + 1
=⇒ 2u du = 2e2x dx Thus∫ ln√8
ln√3
e2x√e2x + 1 dx =
∫ 3
2u · u du =
13u3∣∣∣∣32=
193
![Page 89: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/89.jpg)
A third skinned catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u =
√e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx
Thus∫ ln√8
ln√3
e2x√e2x + 1 dx =
∫ 3
2u · u du =
13u3∣∣∣∣32=
193
![Page 90: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/90.jpg)
A third skinned catExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu onLet u =
√e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus∫ ln
√8
ln√3
e2x√e2x + 1 dx =
∫ 3
2u · u du =
13u3∣∣∣∣32=
193
![Page 91: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/91.jpg)
A Trigonometric Example
Example
Find ∫ 3π/2
π
cot5(θ
6
)sec2
(θ
6
)dθ.
Before we dive in, think about:I What “easy” subs tu ons might help?I Which of the trig func ons suggests a subs tu on?
![Page 92: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/92.jpg)
A Trigonometric Example
Example
Find ∫ 3π/2
π
cot5(θ
6
)sec2
(θ
6
)dθ.
Before we dive in, think about:I What “easy” subs tu ons might help?I Which of the trig func ons suggests a subs tu on?
![Page 93: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/93.jpg)
Solu on
Let φ =θ
6. Then dφ =
16dθ.
∫ 3π/2
π
cot5(θ
6
)sec2
(θ
6
)dθ = 6
∫ π/4
π/6cot5 φ sec2 φ dφ
= 6∫ π/4
π/6
sec2 φ dφtan5 φ
![Page 94: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/94.jpg)
Solu on
Let φ =θ
6. Then dφ =
16dθ.
∫ 3π/2
π
cot5(θ
6
)sec2
(θ
6
)dθ = 6
∫ π/4
π/6cot5 φ sec2 φ dφ
= 6∫ π/4
π/6
sec2 φ dφtan5 φ
Now let u = tanφ. So du = sec2 φ dφ, and
![Page 95: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/95.jpg)
Solu onNow let u = tanφ. So du = sec2 φ dφ, and
6∫ π/4
π/6
sec2 φ dφtan5 φ
= 6∫ 1
1/√3u−5 du
= 6(−14u−4
)∣∣∣∣11/
√3=
32[9− 1] = 12.
![Page 96: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/96.jpg)
The limits explained
tanπ
4=
sin(π/4)cos(π/4)
=
√2/2√2/2
= 1
tanπ
6=
sin(π/6)cos(π/6)
=1/2√3/2
=1√3
![Page 97: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/97.jpg)
The limits explained
6(−14u−4
)∣∣∣∣11/
√3=
32[−u−4]1
1/√3 =
32[u−4]1/√3
1
=32
[(3−1/2)−4 − (1−1/2)−4
]=
32[32 − 12] =
32(9− 1) = 12
![Page 98: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/98.jpg)
Graphs
.. θ.
y
..π
..3π2
.
∫ 3π/2
π
cot5(θ
6
)sec2
(θ
6
)dθ
. φ.
y
..π
6
..π
4
.
∫ π/4
π/66 cot5 φ sec2 φ dφ
The areas of these two regions are the same.
![Page 99: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/99.jpg)
Graphs
.. φ.
y
..π
6
..π
4
.
∫ π/4
π/66 cot5 φ sec2 φ dφ
. u.
y
.
∫ 1
1/√36u−5 du
..1√3
..1
The areas of these two regions are the same.
![Page 100: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/100.jpg)
u/du pairs
When deciding on a subs tu on, look for sub-expressions whereone is (a constant mul ple of) the deriva ve of the other. Such as:
u xn ln x sin x cos x tan x√x ex
constant× du xn−1 1x
cos x sin x sec2 x1√x
ex
![Page 101: Lesson 27: Integration by Substitution (slides)](https://reader034.vdocuments.us/reader034/viewer/2022042623/54813d29b37959c22b8b46e4/html5/thumbnails/101.jpg)
SummaryI If F is an an deriva ve for f, then:∫
f(g(x))g′(x) dx = F(g(x))
I If F is an an deriva ve for f, which is con nuous on the rangeof g, then:∫ b
af(g(x))g′(x) dx =
∫ g(b)
g(a)f(u) du = F(g(b))− F(g(a))
I An differen a on in general and subs tu on in par cular is a“nonlinear” problem that needs prac ce, intui on, andperserverance.
I The whole an differen a on story is in Chapter 6.