Lesson 28: Integration by Subsitution

Section 5.5
Integration by Substitution

V63.0121.002.2010Su, Calculus I

New York University

June 22, 2010

Announcements
Tomorrow: Review, Evaluations, Movie
Thursday: Final Exam

. . . . . .

Announcements

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Evaluations, Movie
Thursday: Final Exam
roughly half-and-half
MC/FR
FR is all post-midterm
MC might have some
pre-midterm stuff on it

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 5.5 Integration by Substitution June 22, 2010 2 / 37

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. . . . . .
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Image credit: Scott Beale / Laughing Squid

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
∫ 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?

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

. . . . . .



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

. . . . . .



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?

. . . . . .


No straightforward system of antidifferentiation

So far we don’t have any way to find
∫
2x
√ dx
x2 + 1
or ∫
tan x dx.

. . . . . .


No straightforward system of 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 “anti” version of one of the most
important rules of differentiation: the chain rule.

. . . . . .


Outline


Theory
Examples

Theory
Examples

. . . . . .



Example
Find ∫
x
√ dx.
x2 + 1

. . . . . .



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 + x2 .

. . . . . .


Say what?

Solution (More slowly, now)
Let g(x) = x2 + 1.

. . . . . .


Say what?

Let g(x) = x2 + 1. Then g′ (x) = 2x and so

d√ 1 x
g(x) = √ g′ (x) = √
dx 2 g(x) x2 + 1

. . . . . .


Say what?

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.

. . . . . .


Leibnizian notation FTW

Solution (Same technique, new notation)
Let u = x2 + 1.

. . . . . .



√ √
Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u.

. . . . . .



√ √
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 ∫
√
x dx 2 du
√
1
√ du
= =
x2 +1 u 2 u
∫
1 −1/2
= 2u du

. . . . . .



√ √
∫ ∫ 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
Solution (Same technique, new notation, 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.

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:
∫ ∫
du
f(u) dx = f(u) du
dx

. . . . . .


A polynomial example

Example
∫
2
Use the substitution u = x + 3 to find (x2 + 3)3 4x dx.

. . . . . .



Example
∫
2

Solution
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So
∫
(x2 + 3)3 4x dx

. . . . . .



Example
∫
2

Solution
∫ ∫ ∫
(x + 3) 4x dx = u 2du = 2 u3 du
2 3 3

. . . . . .



Example
∫
2

Solution
∫ ∫ ∫
(x + 3) 4x dx = u 2du = 2 u3 du
2 3 3

1 4
= u
2

. . . . . .



Example
∫
2

Solution
∫ ∫ ∫
(x + 3) 4x dx = u 2du = 2 u3 du
2 3 3

1 4 1 2
= u = (x + 3)4
2 2

. . . . . .


A polynomial example, by brute force

Compare this to multiplying it out:
∫
(x2 + 3)3 4x dx

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫ ( )
= 4x7 + 36x5 + 108x3 + 108x dx

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫ ( )
= 4x7 + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫ ( )
= 4x7 + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2
Which would you rather do?

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫ ( )
= 4x7 + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2
It’s a wash for low powers

. . . . . .



∫ ∫ ( )
2 3
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫ ( )
= 4x7 + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2
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
∫
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?

. . . . . .


Compare

We have the substitution method, which, when multiplied out, gives
∫
1
(x2 + 3)3 4x dx = (x2 + 3)4 + C
2
1( 8 )
= x + 12x6 + 54x4 + 108x2 + 81 + C
2
1 81
= x8 + 6x6 + 27x4 + 54x2 + +C
2 2

and the brute force method
∫
1
(x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C
2
Is there a difference? Is this a problem? No, that’s what +C means!

. . . . . .


A slick example

Example
∫
Find tan x dx.

. . . . . .


A slick example

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

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
Let u = . Then du = . So
∫ ∫
sin x
tan x dx = dx
cos x

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
Let u = cos x. Then du = . So
∫ ∫
sin x
tan x dx = dx
cos x

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
Let u = cos x. Then du = − sin x dx. So
∫ ∫
sin x
tan x dx = dx
cos x

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
∫ ∫ ∫
sin x 1
tan x dx = dx = − du
cos x u

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
∫ ∫ ∫
sin x 1
cos x u
= − ln |u| + C

. . . . . .


A slick example

Example
∫
sin x
cos x

Solution
∫ ∫ ∫
sin x 1
cos x u
= − ln |u| + C
= − ln | cos x| + C = ln | sec x| + C

. . . . . .


Can you do it another way?

Example
∫
sin x
cos x

. . . . . .



Example
∫
sin x
cos x

Solution
du
Let u = sin x. Then du = cos x dx and so dx = .
cos x

. . . . . .



Example
∫
sin x
cos x

Solution
du
Let u = sin x. Then du = cos x dx and so dx = .
cos x
∫ ∫ ∫
sin x u du
tan x dx = dx =
cos x cos x cos x
∫ ∫ ∫
u du u du u du
= = =
cos2 x 2
1 − sin x 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)
Let y = 1 − u2 , so dy = −2u du. Then
∫ ∫ ∫
u du u dy
tan x dx = =
1 − u2 y −2u
∫
1 dy 1 1
=− = − ln |y| + C = − ln 1 − u2 + C
2 y 2 2
1 1
= ln √ + C = ln √ +C
1 − u2 1 − sin2 x
1
= ln + C = ln |sec x| + C
|cos x|

There are other ways to do it, too.
. . . . . .


Outline


Theory
Examples

Theory
Examples

. . . . . .



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)

. . . . . .



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
Compute cos2 x sin x dx.
0

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
First compute the indefinite integral cos2 x sin x dx and then
evaluate. Let u = . Then du = and
∫
cos2 x sin x dx

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
evaluate. Let u = cos x. Then du = and
∫
cos2 x sin x dx

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
evaluate. Let u = cos x. Then du = − sin x dx and
∫
cos2 x sin x dx

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
∫ ∫
cos2 x sin x dx = − u2 du

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
∫ ∫

= − 1 u3 + C = − 3 cos3 x + C.
3
1

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
∫ ∫

= − 1 u3 + C = − 3 cos3 x + C.
3
1

Therefore
∫ π π
1
cos2 x sin x dx = − cos3 x
0 3 0

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
∫ ∫

= − 1 u3 + C = − 3 cos3 x + C.
3
1

Therefore
∫ π
1 π
1( )
cos2 x sin x dx = − cos3 x =− (−1)3 − 13
0 3 0 3

. . . . . .


Example
∫ π
0

Solution (Slow Way)
∫
∫ ∫

= − 1 u3 + C = − 3 cos3 x + C.
3
1

Therefore
∫ π
1 π
1( ) 2
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = .
0 3 0 3 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) = and u(π) = . So
∫ π
cos2 x sin x dx
0

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = . So
∫ π
cos2 x sin x dx
0

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = − 1. So
∫ π
cos2 x sin x dx
0

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
∫ π ∫ −1 ∫ 1
2
cos x sin x dx = −u du =
2
u2 du
0 1 −1

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
∫ π ∫ −1 ∫ 1
2
2
u2 du
0 1 −1
1 3 1
1( ) 2
= u = 1 − (−1) =
3 −1 3 3

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
∫ π ∫ −1 ∫ 1
2
2
u2 du
0 1 −1
1 3 1
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).

. . . . . .


Definite-ly Quicker

Solution (Fast Way)
∫ π ∫ −1 ∫ 1
2
2
u2 du
0 1 −1
1 3 1
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.
. . . . . .


An exponential example
Example
∫ ln √8 √
2x
Find √ e e2x + 1 dx
ln 3

. . . . . .


Example
∫ ln √8 √
2x
ln 3

Solution
Let u = e2x , so du = 2e2x dx. We have
∫ √ ∫
ln 8 √ 1 8√
2x
√ e e2x + 1 dx = u + 1 du
ln 3 2 3

. . . . . .


About those limits

Since
√ √ 2
e2(ln 3)
= eln 3
= eln 3 = 3

we have √
∫ ln 8 √ ∫ 8√
2x 1
√ e e2x + 1 dx = u + 1 du
ln 3 2 3

. . . . . .


Example
∫ ln √8 √
2x
ln 3

Solution
∫ √ ∫
ln 8 √ 1 8√
2x
√ e e2x + 1 dx = u + 1 du
ln 3 2 3

Now let y = u + 1, dy = du. So
∫ 8√ ∫ 9√ ∫ 9
1 1 1
u + 1 du = y dy = y1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y3/2 = (27 − 8) =
2 3 4 3 . 3 . . . . .


About those fractional powers

We have

93/2 = (91/2 )3 = 33 = 27
43/2 = (41/2 )3 = 23 = 8

so ∫ 9 9
1 1 2 3/2 1 19
y1/2 dy = · y = (27 − 8) =
2 4 2 3 4 3 3

. . . . . .


Example
∫ ln √8 √
2x
ln 3

Solution
∫ √ ∫
ln 8 √ 1 8√
2x
√ e e2x + 1 dx = u + 1 du
ln 3 2 3

Now let y = u + 1, dy = du. So
∫ 8√ ∫ 9√ ∫ 9
1 1 1
u + 1 du = y dy = y1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y3/2 = (27 − 8) =
2 3 4 3 . 3 . . . . .


Another way to skin that cat

Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3

Solution
Let u = e2x + 1,

. . . . . .



Example
∫ ln √8 √
ln 3

Solution
Let u = e2x + 1,so that du = 2e2x dx.

. . . . . .



Example
∫ ln √8 √
ln 3

Solution
Let u = e2x + 1,so that du = 2e2x dx. Then
∫ √ ∫
ln 8 √ 1 9√
2x
√ e e2x + 1 dx = u du
ln 3 2 4

. . . . . .



Example
∫ ln √8 √
ln 3

Solution
∫ √ ∫
ln 8 √ 1 9√
2x
√ e e2x + 1 dx = u du
ln 3 2 4
9
1 3/2
= u
3 4

. . . . . .



Example
∫ ln √8 √
ln 3

Solution
∫ √ ∫
ln 8 √ 1 9√
2x
√ e e2x + 1 dx = u du
ln 3 2 4
1 3/2 9
= u
3 4
1 19
= (27 − 8) =
3 3

. . . . . .


A third skinned cat

Example
∫ ln √8 √
2x
ln 3

Solution
√
Let u = e2x + 1, so that

u2 = e2x + 1

. . . . . .


A third skinned cat

Example
∫ ln √8 √
2x
ln 3

Solution
√

u2 = e2x + 1 =⇒ 2u du = 2e2x dx

. . . . . .


A third skinned cat

Example
∫ ln √8 √
2x
ln 3

Solution
√

u2 = e2x + 1 =⇒ 2u du = 2e2x dx

Thus √
∫ ln 8 ∫ 3 3
1 19
√ = u · u du = u3 =
ln 3 2 3 2 3

. . . . . .


A Trigonometric Example

Example
Find ∫ ( ) ( )
3π/2
5θ 2 θ
cot sec dθ.
π 6 6

. . . . . .


A Trigonometric Example

Example
Find ∫ ( ) ( )
3π/2
5θ 2 θ
cot sec dθ.
π 6 6

Before we dive in, think about:
What “easy” substitutions might help?
Which of the trig functions suggests a substitution?

. . . . . .


Solution
θ 1
Let φ = . Then dφ = dθ.
6 6
∫ 3π/2 ( ) ( ) ∫ π/4
5 θ 2 θ
cot sec dθ = 6 cot5 φ sec2 φ dφ
π 6 6 π/6
∫ π/4
sec2 φ dφ
=6
π/6 tan5 φ

. . . . . .


Solution
θ 1
Let φ = . Then dφ = dθ.
6 6
∫ 3π/2 ( ) ( ) ∫ π/4
5 θ 2 θ
cot sec dθ = 6 cot5 φ sec2 φ dφ
π 6 6 π/6
∫ π/4
sec2 φ dφ
=6
π/6 tan5 φ

Now let u = tan φ. So du = sec2 φ dφ, and
∫ π/4 ∫ 1
sec2 φ dφ −5
6 =6 √ u du
π/6 tan5 φ 1/ 3
( )
1 −4 1 3
=6 − u √
= [9 − 1] = 12.
4 1/ 3 2

. . . . . .


The limits explained

√
π sin π/4 2/2
tan = =√ =1
4 cos π/4 2/2
π sin π/6 1/2 1
tan = =√ =√
6 cos π/6 3/2 3

. . . . . .


The limits explained

√
π sin π/4 2/2
tan = =√ =1
4 cos π/4 2/2
π sin π/6 1/2 1
tan = =√ =√
6 cos π/6 3/2 3

( ) √
1 −4 1
3 [ −4 ]1 3 [ −4 ]1/ 3
6 − u √
= −u √ = u
4 1/ 3 2 1/ 3 2 1

3 [ ]
= (3−1/2 )−4 − (1−1/2 )−4
2
3 3
= [32 − 12 ] = (9 − 1) = 12
2 2

. . . . . .


Graphs
∫ 3π/2 ( ) ( ) ∫ π/4
θ 2 θ
. cot 5
sec 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.

. . . . . .


Graphs ∫ ∫
π/4 1
5 2 −5
. 6 cot φ sec 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

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

. . . . . .


Lesson 28: Integration by Subsitution

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