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Lesson 27: Integration by Substitution (Section 041 slides)
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Transcript of Lesson 27: Integration by Substitution (Section 041 slides)
Section 5.5Integration by Substitution
V63.0121.041, Calculus I
New York University
December 13, 2010
Announcements
I ”Wednesday”, December 15: Review, Movie
I Monday, December 20, 12:00pm–1:50pm: Final Exam
Announcements
I ”Wednesday”, December 15:Review, Movie
I Monday, December 20,12:00pm–1:50pm: FinalExam
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37
Resurrection Policy
If your final score beats your midterm score, we will add 10% to its weight,and subtract 10% from the midterm weight.
Image credit: Scott Beale / Laughing SquidV63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 3 / 37
Objectives
I Given an integral and asubstitution, transform theintegral into an equivalentone using a substitution
I Evaluate indefinite integralsusing the method ofsubstitution.
I Evaluate definite integralsusing the method ofsubstitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37
Differentiation and Integration as reverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be continuous on [a, b]. Then
d
dx
∫ 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).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37
Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some are general,like ∫
[f (x) + g(x)] dx =
∫f (x) dx +
∫g(x) dx
Some are pretty particular, like∫1
x√
x2 − 1dx = arcsec x + C .
What are we supposed to do with that?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some are general,like ∫
[f (x) + g(x)] dx =
∫f (x) dx +
∫g(x) dx
Some are pretty particular, like∫1
x√
x2 − 1dx = arcsec x + C .
What are we supposed to do with that?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
Techniques of antidifferentiation?
So far we know only a few rules for antidifferentiation. Some are general,like ∫
[f (x) + g(x)] dx =
∫f (x) dx +
∫g(x) dx
Some are pretty particular, like∫1
x√
x2 − 1dx = arcsec x + C .
What are we supposed to do with that?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
No straightforward system of antidifferentiation
So far we don’t have any way to find∫2x√
x2 + 1dx
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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
No straightforward system of antidifferentiation
So far we don’t have any way to find∫2x√
x2 + 1dx
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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37
Substitution for Indefinite Integrals
Example
Find ∫x√
x2 + 1dx .
Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression√
1 + x2.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
Substitution for Indefinite Integrals
Example
Find ∫x√
x2 + 1dx .
Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression√
1 + x2.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
Say what?
Solution (More slowly, now)
Let g(x) = x2 + 1.
Then g ′(x) = 2x and so
d
dx
√g(x) =
1
2√
g(x)g ′(x) =
x√x2 + 1
Thus ∫x√
x2 + 1dx =
∫ (d
dx
√g(x)
)dx
=√
g(x) + C =√
1 + x2 + C .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
Say what?
Solution (More slowly, now)
Let g(x) = x2 + 1. Then g ′(x) = 2x and so
d
dx
√g(x) =
1
2√
g(x)g ′(x) =
x√x2 + 1
Thus ∫x√
x2 + 1dx =
∫ (d
dx
√g(x)
)dx
=√
g(x) + C =√
1 + x2 + C .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
Say what?
Solution (More slowly, now)
Let g(x) = x2 + 1. Then g ′(x) = 2x and so
d
dx
√g(x) =
1
2√
g(x)g ′(x) =
x√x2 + 1
Thus ∫x√
x2 + 1dx =
∫ (d
dx
√g(x)
)dx
=√
g(x) + C =√
1 + x2 + C .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x=
∫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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x
=
∫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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x=
∫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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x=
∫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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x=
∫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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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 =du
2x
So the integrand becomes completely transformed into∫x√
x2 + 1dx =
∫x√u· du
2x=
∫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.V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
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)
du
dxdx =
∫f (u) du
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37
A polynomial example
Example
Use the substitution u = x2 + 3 to find
∫(x2 + 3)34x dx .
Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx
=
∫u3 2du = 2
∫u3 du
=1
2u4 =
1
2(x2 + 3)4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
A polynomial example
Example
Use the substitution u = x2 + 3 to find
∫(x2 + 3)34x dx .
Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx
=
∫u3 2du = 2
∫u3 du
=1
2u4 =
1
2(x2 + 3)4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
A polynomial example
Example
Use the substitution u = x2 + 3 to find
∫(x2 + 3)34x dx .
Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx =
∫u3 2du = 2
∫u3 du
=1
2u4 =
1
2(x2 + 3)4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
A polynomial example
Example
Use the substitution u = x2 + 3 to find
∫(x2 + 3)34x dx .
Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx =
∫u3 2du = 2
∫u3 du
=1
2u4
=1
2(x2 + 3)4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
A polynomial example
Example
Use the substitution u = x2 + 3 to find
∫(x2 + 3)34x dx .
Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx =
∫u3 2du = 2
∫u3 du
=1
2u4 =
1
2(x2 + 3)4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
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
=1
2x8 + 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 substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
Compare
We have the substitution method, which, when multiplied out, gives∫(x2 + 3)34x dx =
1
2(x2 + 3)4
+ C
=1
2
(x8 + 12x6 + 54x4 + 108x2 + 81
)
+ C
=1
2x8 + 6x6 + 27x4 + 54x2 +
81
2
+ C
and the brute force method∫(x2 + 3)34x dx =
1
2x8 + 6x6 + 27x4 + 54x2
+ C
Is there a difference? Is this a problem?
No, that’s what +C means!
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
Compare
We have the substitution method, which, when multiplied out, gives∫(x2 + 3)34x dx =
1
2(x2 + 3)4 + C
=1
2
(x8 + 12x6 + 54x4 + 108x2 + 81
)+ C
=1
2x8 + 6x6 + 27x4 + 54x2 +
81
2+ C
and the brute force method∫(x2 + 3)34x dx =
1
2x8 + 6x6 + 27x4 + 54x2 + C
Is there a difference? Is this a problem? No, that’s what +C means!
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
A slick example
Example
Find
∫tan x dx .
(Hint: tan x =sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx
= −∫
1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx
= −∫
1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx
= −∫
1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx
= −∫
1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx
= −∫
1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx = −
∫1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx = −
∫1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
A slick example
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = cos x . Then du = − sin x dx . So∫tan x dx =
∫sin x
cos xdx = −
∫1
udu
= − ln |u|+ C
= − ln | cos x |+ C = ln | sec x |+ C
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
Can you do it another way?
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.∫
tan x dx =
∫sin x
cos xdx =
∫u
cos x
du
cos x
=
∫u du
cos2 x=
∫u du
1− sin2 x=
∫u du
1− u2
At this point, although it’s possible to proceed, we should probably backup and see if the other way works quicker (it does).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
Can you do it another way?
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.
∫tan x dx =
∫sin x
cos xdx =
∫u
cos x
du
cos x
=
∫u du
cos2 x=
∫u du
1− sin2 x=
∫u du
1− u2
At this point, although it’s possible to proceed, we should probably backup and see if the other way works quicker (it does).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
Can you do it another way?
Example
Find
∫tan x dx . (Hint: tan x =
sin x
cos x)
Solution
Let u = sin x. Then du = cos x dx and so dx =du
cos x.∫
tan x dx =
∫sin x
cos xdx =
∫u
cos x
du
cos x
=
∫u du
cos2 x=
∫u du
1− sin2 x=
∫u du
1− u2
At this point, although it’s possible to proceed, we should probably backup and see if the other way works quicker (it does).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
For those who really must know all
Solution (Continued, with algebra help)
Let y = 1− u2, so dy = −2u du. Then∫tan x dx =
∫u du
1− u2=
∫u
y
dy
−2u
= −1
2
∫dy
y= −1
2ln |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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 20 / 37
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite IntegralsTheoryExamples
Substitution for Definite IntegralsTheoryExamples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37
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 tobe u-values
I To get from x to u, apply g
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
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 tobe u-values
I To get from x to u, apply g
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)
=2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
= −13u3 + C = −1
3 cos3 x + C .
Therefore∫ π
0cos2 x sin x dx = −1
3cos3 x
∣∣∣∣π0
= −1
3
((−1)3 − 13
)=
2
3.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
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
=
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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
=
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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
=
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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 =
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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 =
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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 =
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
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 =
∫ −11−u2 du =
∫ 1
−1u2 du
=1
3u3
∣∣∣∣1−1
=1
3
(1− (−1)
)=
2
3
I The advantage to the “fast way” is that you completely transform theintegral into something simpler and don’t have to go back to theoriginal variable (x).
I But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
An exponential example
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x , so du = 2e2x dx. We have∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 8
3
√u + 1 du
Now let y = u + 1, dy = du. So
1
2
∫ 8
3
√u + 1 du =
1
2
∫ 9
4
√y dy =
1
2
∫ 9
4y1/2 dy
=1
2· 2
3y3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
An exponential example
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x , so du = 2e2x dx. We have∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 8
3
√u + 1 du
Now let y = u + 1, dy = du. So
1
2
∫ 8
3
√u + 1 du =
1
2
∫ 9
4
√y dy =
1
2
∫ 9
4y1/2 dy
=1
2· 2
3y3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
About those limits
Since
e2(ln√3) = e ln
√32
= e ln 3 = 3
we have ∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 8
3
√u + 1 du
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 26 / 37
An exponential example
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x , so du = 2e2x dx. We have∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 8
3
√u + 1 du
Now let y = u + 1, dy = du. So
1
2
∫ 8
3
√u + 1 du =
1
2
∫ 9
4
√y dy =
1
2
∫ 9
4y1/2 dy
=1
2· 2
3y3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 27 / 37
About those fractional powers
We have
93/2 = (91/2)3 = 33 = 27
43/2 = (41/2)3 = 23 = 8
so1
2
∫ 9
4y1/2 dy =
1
2· 2
3y3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 28 / 37
An exponential example
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x , so du = 2e2x dx. We have∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 8
3
√u + 1 du
Now let y = u + 1, dy = du. So
1
2
∫ 8
3
√u + 1 du =
1
2
∫ 9
4
√y dy =
1
2
∫ 9
4y1/2 dy
=1
2· 2
3y3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 29 / 37
Another way to skin that cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x + 1,
so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 9
4
√u du
=1
3u3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
Another way to skin that cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x + 1,so that du = 2e2x dx.
Then∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 9
4
√u du
=1
3u3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
Another way to skin that cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x + 1,so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 9
4
√u du
=1
3u3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
Another way to skin that cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x + 1,so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 9
4
√u du
=1
3u3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
Another way to skin that cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u = e2x + 1,so that du = 2e2x dx. Then∫ ln√8
ln√3
e2x√
e2x + 1 dx =1
2
∫ 9
4
√u du
=1
3u3/2
∣∣∣∣94
=1
3(27− 8) =
19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
A third skinned cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u =√
e2x + 1, so that
u2 = e2x + 1
=⇒ 2u du = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · u du =
1
3u3
∣∣∣∣32
=19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
A third skinned cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u =√
e2x + 1, so that
u2 = e2x + 1 =⇒ 2u du = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · u du =
1
3u3
∣∣∣∣32
=19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
A third skinned cat
Example
Find
∫ ln√8
ln√3
e2x√
e2x + 1 dx
Solution
Let u =√
e2x + 1, so that
u2 = e2x + 1 =⇒ 2u du = 2e2x dx
Thus ∫ ln√8
ln√3
=
∫ 3
2u · u du =
1
3u3
∣∣∣∣32
=19
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
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?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
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?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
Solution
Let ϕ =θ
6. Then dϕ =
1
6dθ.
∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ = 6
∫ π/4
π/6cot5 ϕ sec2 ϕ dϕ
= 6
∫ π/4
π/6
sec2 ϕ dϕ
tan5 ϕ
Now let u = tanϕ. So du = sec2 ϕ dϕ, and
6
∫ π/4
π/6
sec2 ϕ dϕ
tan5 ϕ= 6
∫ 1
1/√3
u−5 du
= 6
(−1
4u−4
)∣∣∣∣11/√3
=3
2[9− 1] = 12.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
Solution
Let ϕ =θ
6. Then dϕ =
1
6dθ.
∫ 3π/2
πcot5
(θ
6
)sec2
(θ
6
)dθ = 6
∫ π/4
π/6cot5 ϕ sec2 ϕ dϕ
= 6
∫ π/4
π/6
sec2 ϕ dϕ
tan5 ϕ
Now let u = tanϕ. So du = sec2 ϕ dϕ, and
6
∫ π/4
π/6
sec2 ϕ dϕ
tan5 ϕ= 6
∫ 1
1/√3
u−5 du
= 6
(−1
4u−4
)∣∣∣∣11/√3
=3
2[9− 1] = 12.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
Graphs
θ
y
π 3π
2
∫ 3π/2
πcot5
(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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 35 / 37
Graphs
ϕ
y
π
6
π
4
∫ π/4
π/66 cot5 ϕ sec2 ϕ dϕ
u
y
∫ 1
1/√3
6u−5 du
1√3
1
The areas of these two regions are the same.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 36 / 37
Summary
I If F is an antiderivative for f , then:∫f (g(x))g ′(x) dx = F (g(x))
I If F is an antiderivative for f , which is continuous on the range of g ,then:∫ b
af (g(x))g ′(x) dx =
∫ g(b)
g(a)f (u) du = F (g(b))− F (g(a))
I Antidifferentiation in general and substitution in particular is a“nonlinear” problem that needs practice, intuition, and perserverance
I The whole antidifferentiation story is in Chapter 6
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 37 / 37