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Lesson 27: Integration by Substitution (Section 041 handout)
1. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Notes
Section 5.5
Integration by Substitution
V63.0121.041, Calculus I
New York University
December 13, 2010
Announcements
”Wednesday”, December 15: Review, Movie
Monday, December 20, 12:00pm–1:50pm: Final Exam
Announcements
Notes
”Wednesday”, December 15:
Review, Movie
Monday, December 20,
12:00pm–1:50pm: Final
Exam
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37
Objectives
Notes
Given an integral and a
substitution, transform the
integral into an equivalent
one using a substitution
Evaluate indefinite integrals
using the method of
substitution.
Evaluate definite integrals
using the method of
substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37
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2. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37
Differentiation and Integration as reverse processes
Notes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be continuous on [a, b]. Then
x
d
f (t) dt = f (x)
dx a
2. Let f be continuous on [a, b] and f = F for some other function F .
Then
b
f (x) dx = F (b) − F (a).
a
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37
Techniques of antidifferentiation?
Notes
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
√ dx = arcsec x + C .
x x2 − 1
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
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3. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
No straightforward system of antidifferentiation
Notes
So far we don’t have any way to find
2x
√ dx
x2 + 1
or
tan x dx.
Luckily, we can be smart and use the “anti” version of one of the most
important rules of differentiation: the chain rule.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37
Substitution for Indefinite Integrals
Notes
Example
Find
x
√ dx.
x2 + 1
Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression 1 + x 2 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
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4. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Say what?
Notes
Solution (More slowly, now)
Let g (x) = x 2 + 1. Then g (x) = 2x and so
d 1 x
g (x) = g (x) = √
dx 2 g (x) x2 + 1
Thus
x d
√ dx = g (x) dx
x2 + 1 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
Notes
Solution (Same technique, new notation)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand
becomes completely transformed into
1
√
x dx 2 du
√
1
√ du
= =
x2 + 1 u 2 u
1 −1/2
= 2u 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
Notes
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
du
dx =
2x
So the integrand becomes completely transformed into
x x du 1
√ dx = √ · = √ du
x2 + 1 u 2x 2 u
1 −1/2
= 2u 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
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5. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Theorem of the Day
Notes
Theorem (The Substitution Rule)
If u = g (x) is a differentiable function whose range is an interval I and f
is 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:
du
f (u) dx = f (u) du
dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37
A polynomial example
Notes
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
1 1
= u 4 = (x 2 + 3)4
2 2
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
A polynomial example, by brute force
Notes
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
Which would you rather do?
It’s a wash for low powers
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
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6. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Compare
Notes
We have the substitution method, which, when multiplied out, gives
1
(x 2 + 3)3 4x dx = (x 2 + 3)4
2
1 8
= x + 12x 6 + 54x 4 + 108x 2 + 81
2
1 81
= x 8 + 6x 6 + 27x 4 + 54x 2 +
2 2
and the brute force method
1
(x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2
2
Is there a difference? Is this a problem?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
A slick example
Notes
Example
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solution
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
Outline
Notes
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37
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7. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Substitution for Definite Integrals
Notes
Theorem (The Substitution Rule for Definite Integrals)
If g is continuous and f is continuous on the range of u = g (x), then
b g (b)
f (g (x))g (x) dx = f (u) du.
a g (a)
Why the change in the limits?
The integral on the left happens in “x-land”
The integral on the right happens in “u-land”, so the limits need to
be u-values
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
π Notes
Compute cos2 x sin x dx.
0
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 = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
Therefore
π π
1 1 2
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = .
0 3 0 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
Definite-ly Quicker
Notes
Solution (Fast Way)
Do both the substitution and the evaluation at the same time. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = − 1. So
π −1 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
1
1 3 1 2
= u = 1 − (−1) =
3 −1 3 3
The advantage to the “fast way” is that you completely transform the
integral into something simpler and don’t have to go back to the
original variable (x).
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
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8. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
An exponential example
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
Another way to skin that cat
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
A third skinned cat
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
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9. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
A Trigonometric Example
Notes
Example
Find
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
Before we dive in, think about:
What “easy” substitutions might help?
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
Notes
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
Graphs
3π/2
θ θ π/4 Notes
cot5 sec2 dθ 6 cot5 ϕ sec2 ϕ dϕ
π 6 6 π/6
y y
θ ϕ
3π π ππ
2 64
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
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10. V63.0121.041, Calculus I Section 5.5 : Integration by Substitution December 13, 2010
Graphs
π/4 1
−5 Notes
6 cot5 ϕ sec2 ϕ dϕ √ 6u du
π/6 1/ 3
y y
ϕ u
ππ 1 1
64 √
3
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
Notes
If F is an antiderivative for f , then:
f (g (x))g (x) dx = F (g (x))
If F is an antiderivative for f , which is continuous on the range of g ,
then:
b g (b)
f (g (x))g (x) dx = f (u) du = F (g (b)) − F (g (a))
a g (a)
Antidifferentiation in general and substitution in particular is a
“nonlinear” problem that needs practice, intuition, and perserverance
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
Notes
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