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
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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
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
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
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.
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
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).
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?
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?
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?
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.
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.
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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))
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
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
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
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
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
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
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.
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.
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.
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.
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.
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.
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.
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.
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!
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!
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
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
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
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
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
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
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
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
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
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
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
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.
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite IntegralsTheoryExamples
Subs tu on for Definite IntegralsTheoryExamples
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
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
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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
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
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
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
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.
An exponential exampleExample
Find∫ ln
√8
ln√3
e2x√e2x + 1 dx
Solu on
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
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
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
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
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
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
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
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
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
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
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
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
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
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?
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?
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
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
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.
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
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
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.
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.
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
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.