Integration by substitution is the chain rule in reverse.

1. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
Sec on 5.5
Integra on by Subs tu on
V63.0121.001: Calculus I
. Professor Ma hew Leingang
New York University
May 4, 2011
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Notes
Announcements
Today: 5.5
Monday 5/9: Review in class
Tuesday 5/10: Review Sessions by TAs
Wednesday 5/11: TA office hours:
Adam 10–noon (WWH 906)
Jerome 3:30–5:30 (WWH 501)
Soohoon 6–8 (WWH 511)
Thursday 5/12: Final Exam,
2:00–3:50pm
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Notes
Resurrection Policy
If your final score beats your midterm score, we will add 10% to its
weight, and subtract 10% from the midterm weight.
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. Image credit: Sco Beale / Laughing Squid
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2. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
Objectives
Given an integral and a
subs tu on, transform the
integral into an equivalent one
using a subs tu on
Evaluate indefinite integrals
using the method of
subs tu on.
Evaluate definite integrals using
the method of subs tu on.
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Notes
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples
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Differentiation and Integration as Notes
reverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be con nuous on [a, b]. Then
∫
d x
f(t) dt = f(x)
dx a
2. Let f be con nuous on [a, b] and f = F′ for some other func on
F. Then ∫ b
f(x) dx = F(b) − F(a).
a
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3. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
Techniques of antidifferentiation?
So far we know only a few rules for an differen a on. Some are
general, like
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
Some are pre y par cular, like
∫
1
√ dx = arcsec x + C.
x x2 − 1
What are we supposed to do with that?
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No straightforward system of Notes
antidifferentiation
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 “an ” version of one of the
most important rules of differen a on: the chain rule.
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Notes
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples
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4. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Substitution for Indefinite Notes
Integrals
Example
Find ∫
x
√ dx.
x2+1
Solu on
Stare at this long enough and you no ce the the integrand is the
√
deriva ve of the expression 1 + x2 .
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Notes
Say what?
Solu on (More slowly, now)
Let g(x) = x2 + 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.
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Notes
Leibnizian notation FTW
Solu on (Same technique, new nota on)
√ √
Let u = x2 + 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.
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5. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
Useful but unsavory variation
Solu on (Same technique, new nota on, more idiot-proof)
√ √
Let u = x2 + 1. Then du = 2x dx and 1 + x2 = 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.
.
Mathema cians have serious issues with mixing the x and u like this. .
However, I can’t deny that it works.
Notes
Theorem of the Day
Theorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval I
and f is con nuous on I, then
∫ ∫
f(g(x))g′ (x) dx = f(u) du
That is, if F is an an deriva ve for f, then
∫
f(g(x))g′ (x) dx = F(g(x))
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Notes
Theorem of the Day
Theorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval I
and f is con nuous on I, then
∫ ∫
f(g(x))g′ (x) dx = f(u) du
In Leibniz nota on:
∫ ∫
du
f(u) dx = f(u) du
dx
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6. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
A polynomial example
Example
∫
Use the subs tu on u = x2 + 3 to find (x2 + 3)3 4x dx.
Solu on
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 1
= u4 = (x2 + 3)4
2 2
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Notes
A polynomial example (brute force)
Solu on
∫ ∫
( )
(x2 + 3)3 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx
1
= x8 + 6x6 + 27x4 + 54x2
2
Which would you rather do?
It’s a wash for low powers
But for higher powers, it’s much easier to do subs tu on.
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Notes
Compare
We have the subs tu on method, which, when mul plied out, gives
∫
1
(x2 + 3)3 4x dx = (x2 + 3)4
2
1( 8 )
= x + 12x6 + 54x4 + 108x2 + 81
2
1 81
= x8 + 6x6 + 27x4 + 54x2 +
2 2
and the brute force method
∫
1
(x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2
2
. Is there a difference? Is this a problem?
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7. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
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Notes
Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples
.
.
Notes
Substitution for Definite Integrals
Theorem (The Subs tu on Rule for Definite Integrals)
If g′ is con nuous and f is con nuous 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 le 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
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8. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Example Notes
∫ π
Compute cos2 x sin x dx.
0
Solu on (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 u3 + C = − 3 cos3 x + C.
3
1
Therefore
. ∫ π
1( ) 2
π
1
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . .
0 3 0 3 3
Notes
Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. 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 = −u2 du = u2 du
0 1 −1
1( ) 2
1
1 3
= u = 1 − (−1) =
3 −1 3 3
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Notes
Compare
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.
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9. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
An exponential example
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
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Notes
Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
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Notes
A third skinned cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
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10. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
Notes
A Trigonometric Example
Example
Find ∫ ( ) ( )
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
Before we dive in, think about:
What “easy” subs tu ons might help?
Which of the trig func ons suggests a subs tu on?
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Notes
Solu on
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( ) ( ) ∫
Notes
Graphs
∫ 3π/2
θ θ π/4
cot5 sec2 dθ 6 cot5 φ sec2 φ dφ
π 6 6 π/6
y y
. θ φ
π 3π ππ
2 64
The areas of these two regions are the same.
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11. . V63.0121.001: Calculus I
. Sec on 5.5: Integra on by Subs . tu on May 4, 2011
∫ ∫ Notes
Graphs π/4
6 cot5 φ sec2 φ dφ √
1
6u−5 du
π/6 1/ 3
y y
. φ u
ππ 11
√
64 3
The areas of these two regions are the same.
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Notes
u/du pairs
When deciding on a subs tu on, look for sub-expressions where
one is (a constant mul ple of) the deriva ve of the other. Such as:
√
u xn ln x sin x cos x tan x x ex
1 1
constant × du xn−1 cos x sin x sec2 x √ ex
x x
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Notes
Summary
If F is an an deriva ve for f, then:
∫
f(g(x))g′ (x) dx = F(g(x))
If F is an an deriva ve for f, which is con nuous 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)
An differen a on in general and subs tu on in par cular is a
“nonlinear” problem that needs prac ce, intui on, and
perserverance.
The whole an differen a on story is in Chapter 6.
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