Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 26: Integration by Substitution (handout)
1. 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
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 Squid
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 3 / 38
Notes
Notes
Notes
1
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
2. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 4 / 38
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 5 / 38
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).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 6 / 38
Notes
Notes
Notes
2
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
3. 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?
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 most
important 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 Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 9 / 38
Notes
Notes
Notes
3
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
4. Substitution for Indefinite Integrals
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 10 / 38
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 11 / 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
=
1
2du
√
u
=
1
2
√
u
du
= 1
2u−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
Notes
Notes
Notes
4
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
5. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 13 / 38
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 14 / 38
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 15 / 38
Notes
Notes
Notes
5
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
6. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 16 / 38
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!
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 17 / 38
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 18 / 38
Notes
Notes
Notes
6
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
7. 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).
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 19 / 38
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.
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 Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 21 / 38
Notes
Notes
Notes
7
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
8. 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
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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 22 / 38
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
.
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. 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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 24 / 38
Notes
Notes
Notes
8
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
9. 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
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
√
3
2
= eln 3
= 3
we have
ln
√
8
ln
√
3
e2x
e2x + 1 dx =
1
2
8
3
√
u + 1 du
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 + 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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 27 / 38
Notes
Notes
Notes
9
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
10. 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
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 + 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
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 + 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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 30 / 38
Notes
Notes
Notes
10
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
11. 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
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
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?
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 =
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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 33 / 38
Notes
Notes
Notes
11
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
12. 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
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 34 / 38
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.
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/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.
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 36 / 38
Notes
Notes
Notes
12
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010
13. 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
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:
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))
V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 38 / 38
Notes
Notes
Notes
13
Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010