Lesson 27: Integration by Substitution (Section 041 handout)

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.


1


Outline
Notes

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
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


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?


2


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.


Outline
Notes


Theory
Examples

Theory
Examples


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 .


3


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 .


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 .


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.

4


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


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


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.


5


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?


A slick example
Notes

Example
sin x
Find tan x dx. (Hint: tan x = )
cos x

Solution


Outline
Notes


Theory
Examples

Theory
Examples


6


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


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


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.


7


An exponential example
Notes
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3

Solution


Another way to skin that cat
Notes
Example
√
ln 8
ln 3

Solution


A third skinned cat
Notes

Example
√
ln 8
ln 3

Solution


8


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?


Solution
Notes


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.


9


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.


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


Notes

10

Lesson 27: Integration by Substitution (Section 041 handout)

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Lesson 27: Integration by Substitution (Section 041 handout)