Lesson 26: Integration by Substitution (handout)

Section 5.5
Integration by Substitution
V63.0121.006/016, Calculus I
New York University
April 27, 2010
Announcements
April 29: Movie Day
April 30: Quiz 5 on §§5.1–5.4
Monday, May 10, 12:00noon Final Exam:
SILV 703 (Section 16)
MEYR 121/122 (Section 6)
Announcements
April 29: Movie Day
April 30: Quiz 5 on
§§5.1–5.4
Monday, May 10, 12:00noon
Final Exam:
SILV 703 (Section 16)
MEYR 121/122 (Section
6)
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 2 / 38
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Notes
Notes
Notes
1
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010

Objectives
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.
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
Differentiation and Integration as reverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be continuous on [a, b]. Then
d
dx
x
a
f (t) dt = f (x)
2. Let f be continuous on [a, b] and f = F for some other function F.
Then
b
a
f (x) dx = F(b) − F(a).
Notes
Notes
Notes
2

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 − 1
dx = arcsec x + C.
What are we supposed to do with that?
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 most
important rules of differentiation: the chain rule.
Outline
Theory
Examples
Theory
Examples
Notes
Notes
Notes
3

Example
Find
x
√
x2 + 1
dx.
Solution
Stare at this long enough and you notice the the integrand is the
derivative of the expression 1 + x2.
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 + 1
dx =
d
dx
g(x) dx
= g(x) + C = 1 + x2 + C.
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
=
1
2du
√
u
=
1
2
√
u
du
= 1
2u−1/2
du
=
√
u + C = 1 + x2 + C.
Notes
Notes
Notes
4

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 + 1
dx =
x
√
u
·
du
2x
=
1
2
√
u
du
= 1
2u−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.
Theorem of the Day
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:
f (u)
du
dx
dx = f (u) du
A polynomial example
Example
Use the substitution u = x2
+ 3 to ﬁnd (x2
+ 3)3
4x dx.
Solution
If u = x2
+ 3, then du = 2x dx, and 4x dx = 2 du. So
(x2
+ 3)3
4x dx = u3
2du = 2 u3
du
=
1
2
u4
=
1
2
(x2
+ 3)4
Notes
Notes
Notes
5

A polynomial example, by brute force
Compare this to multiplying it out:
(x2
+ 3)3
4x dx = x6
+ 9x4
+ 27x2
+ 27 4x dx
= 4x7
+ 36x5
+ 108x3
+ 108x dx
=
1
2
x8
+ 6x6
+ 27x4
+ 54x2
Which would you rather do?
It’s a wash for low powers
But for higher powers, it’s much easier to do substitution.
Compare
We have the substitution method, which, when multiplied out, gives
(x2
+ 3)3
4x dx =
1
2
(x2
+ 3)4
+ C
=
1
2
x8
+ 12x6
+ 54x4
+ 108x2
+ 81 + C
=
1
2
x8
+ 6x6
+ 27x4
+ 54x2
+
81
2
+ C
and the brute force method
(x2
+ 3)3
4x dx =
1
2
x8
+ 6x6
+ 27x4
+ 54x2
+ C
Is this a problem? No, that’s what +C means!
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 x
dx = −
1
u
du
= − ln |u| + C
= − ln | cos x| + C = ln | sec x| + C
Notes
Notes
Notes
6

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 x
dx =
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 back
up and see if the other way works quicker (it does).
For those who really must know all
Solution (Continued, with algebra help)
tan x dx =
u du
1 − u2
=
1
2
1
1 − u
−
1
1 + u
du
= −
1
2
ln |1 − u| −
1
2
ln |1 + u| + C
= ln
1
(1 − u)(1 + u)
+ C = ln
1
√
1 − u2
+ C
= ln
1
|cos x|
+ C = ln |sec x| + C
There are other ways to do it, too.
Outline
Theory
Examples
Theory
Examples
Notes
Notes
Notes
7

Theorem (The Substitution Rule for Definite Integrals)
If g is continuous and f is continuous on the range of u = g(x), then
b
a
f (g(x))g (x) dx =
g(b)
g(a)
f (u) du.
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
Example
Compute
π
0
cos2
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
= −1
3u3
+ C = −1
3 cos3
x + C.
Therefore
π
0
cos2
x sin x dx = −
1
3
cos3
x
π
0
= −
1
3
(−1)3
− 13
=
2
3
.
Definite-ly Quicker
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
0
cos2
x sin x dx =
−1
1
−u2
du =
1
−1
u2
du
=
1
3
u3
1
−1
=
1
3
1 − (−1) =
2
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.
Notes
Notes
Notes
8

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
4
y1/2
dy
=
1
2
·
2
3
y3/2
9
4
=
1
3
(27 − 8) =
19
3
About those limits
Since
e2(ln
√
3) = eln
√
3
2
= eln 3
= 3
we have
ln
√
8
ln
√
3
e2x
e2x + 1 dx =
1
2
8
3
√
u + 1 du
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
1
2
8
3
√
u + 1 du =
1
2
9
4
√
y dy =
1
2
9
4
y1/2
dy
=
1
2
·
2
3
y3/2
9
4
=
1
3
(27 − 8) =
19
3
Notes
Notes
Notes
9

About those fractional powers
We have
93/2
= (91/2
)3
= 33
= 27
43/2
= (41/2
)3
= 23
= 8
so
1
2
9
4
y1/2
dy =
1
2
·
2
3
y3/2
9
4
=
1
3
(27 − 8) =
19
3
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
1
2
8
3
√
u + 1 du =
1
2
9
4
√
y dy =
1
2
9
4
y1/2
dy
=
1
2
·
2
3
y3/2
9
4
=
1
3
(27 − 8) =
19
3
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
3
u3/2
9
4
=
1
3
(27 − 8) =
19
3
Notes
Notes
Notes
10

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
2
u · u du =
1
3
u3
3
2
=
19
3
A Trigonometric Example
Example
Find
3π/2
π
cot5 θ
6
sec2 θ
6
dθ.
Before we dive in, think about:
What “easy” substitutions might help?
Which of the trig functions suggests a substitution?
Solution
Let ϕ =
θ
6
. Then dϕ =
1
6
dθ.
3π/2
π
cot5 θ
6
sec2 θ
6
dθ = 6
π/4
π/6
cot5
ϕ sec2
ϕ dϕ
= 6
π/4
π/6
sec2 ϕ dϕ
tan5 ϕ
Now let u = tan ϕ. 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
4
u−4
1
1/
√
3
=
3
2
[9 − 1] = 12.
Notes
Notes
Notes
11

The limits explained
tan
π
4
=
sin π/4
cos π/4
=
√
2/2
√
2/2
= 1
tan
π
6
=
sin π/6
cos π/6
=
1/2
√
3/2
=
1
√
3
6 −
1
4
u−4
1
1/
√
3
=
3
2
−u−4 1
1/
√
3
=
3
2
u−4 1/
√
3
1
=
3
2
(3−1/2
)−4
− (1−1/2
)−4
=
3
2
[32
− 12
] =
3
2
(9 − 1) = 12
Graphs
θ
y
π 3π
2
3π/2
π
cot5 θ
6
sec2 θ
6
dθ
ϕ
y
π
6
π
4
π/4
π/6
6 cot5
ϕ sec2
ϕ dϕ
The areas of these two regions are the same.
Graphs
ϕ
y
π
6
π
4
π/4
π/6
6 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.
Notes
Notes
Notes
12

Final Thoughts
Antidiﬀerentiation is a “nonlinear” problem that needs practice,
intuition, and perserverance
Worksheet in recitation (also to be posted)
The whole antidiﬀerentiation story is in Chapter 6
Summary
If F is an antiderivative for f , then:
f (g(x))g (x) dx = F(g(x))
b
a
f (g(x))g (x) dx =
g(b)
g(a)
f (u) du = F(g(b)) − F(g(a))
Notes
Notes
Notes
13

Lesson 26: Integration by Substitution (handout)

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