Section 5.5Integration by Substitution
V63.0121.006/016, Calculus I
New York University
April 27, 2010
AnnouncementsI April 29: Movie DayI April 30: Quiz 5 on §§5.1–5.4I Monday, May 10, 12:00noon Final Exam:
I SILV 703 (Section 16)I MEYR 121/122 (Section 6)
. . . . . .
. . . . . .
Announcements
I April 29: Movie DayI April 30: Quiz 5 on
§§5.1–5.4I Monday, May 10,
12:00noon Final Exam:I SILV 703 (Section 16)I MEYR 121/122 (Section
6)
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 2 / 38
. . . . . .
Resurrection Policy
If your final score beats your midterm score, we will add 10% to itsweight, and subtract 10% from the midterm weight.
..Image credit: Scott Beale / Laughing SquidV63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 3 / 38
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Objectives
I Given an integral and asubstitution, transform theintegral into an equivalentone using a substitution
I Evaluate indefiniteintegrals using the methodof substitution.
I Evaluate definite integralsusing the method ofsubstitution.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 4 / 38
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Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 5 / 38
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Differentiation and Integration as reverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be continuous on [a,b]. Then
ddx
∫ x
af(t)dt = f(x)
2. Let f be continuous on [a,b] and f = F′ for some other function F.Then ∫ b
af(x)dx = F(b)− F(a).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 6 / 38
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Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some aregeneral, like ∫
[f(x) + g(x)] dx =
∫f(x)dx+
∫g(x)dx
Some are pretty particular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 7 / 38
. . . . . .
Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some aregeneral, like ∫
[f(x) + g(x)] dx =
∫f(x)dx+
∫g(x)dx
Some are pretty particular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 7 / 38
. . . . . .
Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some aregeneral, like ∫
[f(x) + g(x)] dx =
∫f(x)dx+
∫g(x)dx
Some are pretty particular, like∫1
x√x2 − 1
dx = arcsec x+ C.
What are we supposed to do with that?
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 7 / 38
. . . . . .
No straightforward system of antidifferentiation
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 “anti” version of one of the mostimportant rules of differentiation: the chain rule.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 8 / 38
. . . . . .
No straightforward system of antidifferentiation
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 “anti” version of one of the mostimportant rules of differentiation: the chain rule.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 8 / 38
. . . . . .
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 9 / 38
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Substitution for Indefinite Integrals
Example
Find ∫x√
x2 + 1dx.
SolutionStare at this long enough and you notice the the integrand is thederivative of the expression
√1+ x2.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 10 / 38
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Substitution for Indefinite Integrals
Example
Find ∫x√
x2 + 1dx.
SolutionStare at this long enough and you notice the the integrand is thederivative of the expression
√1+ x2.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 10 / 38
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Say what?
Solution (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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 11 / 38
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Say what?
Solution (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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 11 / 38
. . . . . .
Say what?
Solution (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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 11 / 38
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Leibnizian notation FTW
Solution (Same technique, new notation)
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38
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Leibnizian notation FTW
Solution (Same technique, new notation)
Let u = x2 + 1. Then du = 2x dx and√1+ x2 =
√u.
So the integrandbecomes completely transformed into∫
x dx√x2 + 1
=
∫ 12du√u
=
∫1
2√udu
=
∫12u
−1/2 du
=√u+ C =
√1+ x2 + C.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38
. . . . . .
Leibnizian notation FTW
Solution (Same technique, new notation)
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38
. . . . . .
Leibnizian notation FTW
Solution (Same technique, new notation)
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38
. . . . . .
Leibnizian notation FTW
Solution (Same technique, new notation)
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 12 / 38
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Useful but unsavory variation
Solution (Same technique, new notation, 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.
Mathematicians have serious issues with mixing the x and u like this.However, I can’t deny that it works.V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 13 / 38
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Theorem of the Day
Theorem (The Substitution Rule)
If u = g(x) is a differentiable function whose range is an interval I and fis continuous on I, then∫
f(g(x))g′(x)dx =
∫f(u)du
That is, if F is an antiderivative for f, then∫f(g(x))g′(x)dx = F(g(x))
In Leibniz notation: ∫f(u)
dudx
dx =
∫f(u)du
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 14 / 38
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A polynomial example
Example
Use the substitution u = x2 + 3 to find∫
(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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 15 / 38
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A polynomial example
Example
Use the substitution u = x2 + 3 to find∫
(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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 15 / 38
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A polynomial example, by brute force
Compare this to multiplying it out:∫(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 substitution.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 16 / 38
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A polynomial example, by brute force
Compare this to multiplying it out:∫(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 substitution.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 16 / 38
. . . . . .
A polynomial example, by brute force
Compare this to multiplying it out:∫(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 substitution.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 16 / 38
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Compare
We have the substitution method, which, when multiplied 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 this a problem?
No, that’s what +C means!
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 17 / 38
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Compare
We have the substitution method, which, when multiplied 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 this a problem? No, that’s what +C means!
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 17 / 38
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A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
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A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
. . . . . .
A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
. . . . . .
A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
. . . . . .
A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
. . . . . .
A slick example
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
. . . . . .
Can you do it another way?
Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.∫
tan x dx =
∫sin xcos x
dx =
∫u
cos xdu
cos x
=
∫uducos2 x
=
∫udu
1− sin2 x=
∫u du1− u2
At this point, although it’s possible to proceed, we should probablyback up and see if the other way works quicker (it does).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 19 / 38
. . . . . .
Can you do it another way?
Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.
∫tan x dx =
∫sin xcos x
dx =
∫u
cos xdu
cos x
=
∫uducos2 x
=
∫udu
1− sin2 x=
∫u du1− u2
At this point, although it’s possible to proceed, we should probablyback up and see if the other way works quicker (it does).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 19 / 38
. . . . . .
Can you do it another way?
Example
Find∫
tan x dx. (Hint: tan x =sin xcos x
)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.∫
tan x dx =
∫sin xcos x
dx =
∫u
cos xdu
cos x
=
∫uducos2 x
=
∫udu
1− sin2 x=
∫u du1− u2
At this point, although it’s possible to proceed, we should probablyback up and see if the other way works quicker (it does).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 19 / 38
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For those who really must know all
Solution (Continued, with algebra help)
∫tan x dx =
∫udu1− u2
=
∫12
(1
1− u− 1
1+ u
)du
= −12ln |1− u| − 1
2ln |1+ u|+ C
= ln1√
(1− u)(1+ u)+ C = ln
1√1− u2
+ C
= ln1
|cos x|+ C = ln |sec x|+ C
There are other ways to do it, too.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 20 / 38
. . . . . .
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 21 / 38
. . . . . .
Substitution for Definite Integrals
Theorem (The Substitution Rule for Definite Integrals)
If g′ is continuous and f is continuous 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 left happens in “x-land”I The integral on the right happens in “u-land”, so the limits need to
be u-valuesI To get from x to u, apply g
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 22 / 38
. . . . . .
Substitution for Definite Integrals
Theorem (The Substitution Rule for Definite Integrals)
If g′ is continuous and f is continuous 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 left happens in “x-land”I The integral on the right happens in “u-land”, so the limits need to
be u-valuesI To get from x to u, apply g
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 22 / 38
. . . . . .
Example
Compute∫ π
0cos2 x sin x dx.
Solution (Slow Way)
First compute the indefinite integral∫
cos2 x sin x dx and then
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
3cos3 x
∣∣∣∣π0= −1
3((−1)3 − 13
)=
23.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 23 / 38
. . . . . .
Example
Compute∫ π
0cos2 x sin x dx.
Solution (Slow Way)
First compute the indefinite integral∫
cos2 x sin x dx and then
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
3cos3 x
∣∣∣∣π0= −1
3((−1)3 − 13
)=
23.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 23 / 38
. . . . . .
Example
Compute∫ π
0cos2 x sin x dx.
Solution (Slow Way)
First compute the indefinite integral∫
cos2 x sin x dx and then
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
3cos3 x
∣∣∣∣π0= −1
3((−1)3 − 13
)=
23.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 23 / 38
. . . . . .
Definite-ly Quicker
Solution (Fast Way)
Do both the substitution and the evaluation at the same time.
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
I The advantage to the “fast way” is that you completely transformthe integral into something simpler and don’t have to go back tothe original variable (x).
I But the slow way is just as reliable.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38
. . . . . .
Definite-ly Quicker
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. 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
I The advantage to the “fast way” is that you completely transformthe integral into something simpler and don’t have to go back tothe original variable (x).
I But the slow way is just as reliable.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38
. . . . . .
Definite-ly Quicker
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. 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
I The advantage to the “fast way” is that you completely transformthe integral into something simpler and don’t have to go back tothe original variable (x).
I But the slow way is just as reliable.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38
. . . . . .
Definite-ly Quicker
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. 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
I The advantage to the “fast way” is that you completely transformthe integral into something simpler and don’t have to go back tothe original variable (x).
I But the slow way is just as reliable.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38
. . . . . .
An exponential example
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. We have∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u+ 1du
Now let 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) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 25 / 38
. . . . . .
An exponential example
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. We have∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u+ 1du
Now let 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) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 25 / 38
. . . . . .
About those limits
Since
e2(ln√3) = eln
√32 = eln 3 = 3
we have ∫ ln√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u+ 1du
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 26 / 38
. . . . . .
An exponential example
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. We have∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u+ 1du
Now let 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) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 27 / 38
. . . . . .
About those fractional powers
We have
93/2 = (91/2)3 = 33 = 27
43/2 = (41/2)3 = 23 = 8
so12
∫ 9
4y1/2 dy =
12· 23y3/2
∣∣∣∣94=
13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 28 / 38
. . . . . .
An exponential example
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x, so du = 2e2x dx. We have∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 8
3
√u+ 1du
Now let 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) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 29 / 38
. . . . . .
Another way to skin that cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1,
so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94
=13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
. . . . . .
Another way to skin that cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1,so that du = 2e2x dx.
Then∫ ln√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94
=13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
. . . . . .
Another way to skin that cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94
=13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
. . . . . .
Another way to skin that cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94
=13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
. . . . . .
Another way to skin that cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln
√8
ln√3
e2x√
e2x + 1dx =12
∫ 9
4
√u du
=13u3/2
∣∣∣∣94
=13(27− 8) =
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
. . . . . .
A third skinned cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, so that
u2 = e2x + 1
=⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · udu =
13u3
∣∣∣∣32=
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 31 / 38
. . . . . .
A third skinned cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, so that
u2 = e2x + 1 =⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · udu =
13u3
∣∣∣∣32=
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 31 / 38
. . . . . .
A third skinned cat
Example
Find∫ ln
√8
ln√3
e2x√
e2x + 1dx
SolutionLet u =
√e2x + 1, so that
u2 = e2x + 1 =⇒ 2udu = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · udu =
13u3
∣∣∣∣32=
193
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 31 / 38
. . . . . .
A Trigonometric Example
Example
Find ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
Before we dive in, think about:I What “easy” substitutions might help?I Which of the trig functions suggests a substitution?
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 32 / 38
. . . . . .
A Trigonometric Example
Example
Find ∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ.
Before we dive in, think about:I What “easy” substitutions might help?I Which of the trig functions suggests a substitution?
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 32 / 38
. . . . . .
Solution
Let 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 φdφtan5 φ
Now let u = tanφ. So du = sec2 φdφ, and
6∫ π/4
π/6
sec2 φdφtan5 φ
= 6∫ 1
1/√3u−5 du
= 6(−14u−4
)∣∣∣∣11/
√3=
32[9− 1] = 12.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 33 / 38
. . . . . .
Solution
Let 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 φdφtan5 φ
Now let u = tanφ. So du = sec2 φdφ, and
6∫ π/4
π/6
sec2 φdφtan5 φ
= 6∫ 1
1/√3u−5 du
= 6(−14u−4
)∣∣∣∣11/
√3=
32[9− 1] = 12.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 33 / 38
. . . . . .
The limits explained
tanπ
4=
sinπ/4cosπ/4
=
√2/2√2/2
= 1
tanπ
6=
sinπ/6cosπ/6
=1/2√3/2
=1√3
6(−14u−4
)∣∣∣∣11/
√3=
32
[−u−4
]11/
√3=
32
[u−4
]1/√3
1
=32
[(3−1/2)−4 − (1−1/2)−4
]=
32[32 − 12] =
32(9− 1) = 12
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 34 / 38
. . . . . .
The limits explained
tanπ
4=
sinπ/4cosπ/4
=
√2/2√2/2
= 1
tanπ
6=
sinπ/6cosπ/6
=1/2√3/2
=1√3
6(−14u−4
)∣∣∣∣11/
√3=
32
[−u−4
]11/
√3=
32
[u−4
]1/√3
1
=32
[(3−1/2)−4 − (1−1/2)−4
]=
32[32 − 12] =
32(9− 1) = 12
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 34 / 38
. . . . . .
Graphs
. .θ
.y
..π
.
.3π2
.∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ
.φ
.y
.
.π
6
.
.π
4
.∫ π/4
π/66 cot5 φ sec2 φdφ
The areas of these two regions are the same.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 35 / 38
. . . . . .
Graphs
. .φ
.y
.
.π
6
.
.π
4
.∫ π/4
π/66 cot5 φ sec2 φdφ
.u
.y
.∫ 1
1/√36u−5 du
.
.1√3
.
.1
The areas of these two regions are the same.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 36 / 38
. . . . . .
Final Thoughts
I Antidifferentiation is a “nonlinear” problem that needs practice,intuition, and perserverance
I Worksheet in recitation (also to be posted)I The whole antidifferentiation story is in Chapter 6
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 37 / 38
. . . . . .
Summary
If F is an antiderivative for f, then:I ∫
f(g(x))g′(x)dx = F(g(x))
I ∫ b
af(g(x))g′(x)dx =
∫ g(b)
g(a)f(u) du = F(g(b))− F(g(a))
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 38 / 38