Lesson 27: Integration by Substitution (Section 041 slides)

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

”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

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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



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



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 + x 2 .


Say what?

Solution (More slowly, now)
Let g (x) = x 2 + 1.


Say what?

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


Say what?

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

Solution (Same technique, new notation)
Let u = x 2 + 1.



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



√
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
√
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 = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
du
dx =
2x



√
du
dx =
2x
So the integrand becomes completely transformed into
x x du
√ dx = √ ·
x2 +1 u 2x



√
du
dx =
2x
x x du 1
√ dx = √ · = √ du
x2 +1 u 2x 2 u



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



√
du
dx =
2x
x x du 1
√ dx = √ · = √ du
x2 +1 u 2x 2 u
1 −1/2
= 2u du
√
= u+C = 1 + x2 + C .



√
du
dx =
2x
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
Use the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx.



Example

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

(x 2 + 3)3 4x dx



Example

Solution

(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du



Example

Solution

(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
1
= u4
2



Example

Solution

(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

Compare this to multiplying it out:

(x 2 + 3)3 4x dx




(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx




(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx

= 4x 7 + 36x 5 + 108x 3 + 108x dx




(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




(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?




(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
It’s a wash for low powers




(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
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
(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?


Compare

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

and the brute force method

1
(x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 + 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 = 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
= = =
cos 2x 1 − sin2 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 − u 2 , so dy = −2u du. Then
u du u dy
tan x dx = 2
=
1−u y −2u
1 dy 1 1
=− = − ln |y | + C = − ln 1 − u 2 + 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 = cos x . Then du = − sin x dx and

cos2 x sin x dx


Example
π
0

Solution (Slow Way)


cos2 x sin x dx = − u 2 du


Example
π
0

Solution (Slow Way)



= − 3 u 3 + C = − 1 cos3 x + C .
1
3


Example
π
0

Solution (Slow Way)



= − 3 u 3 + C = − 1 cos3 x + C .
1
3

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


Example
π
0

Solution (Slow Way)



= − 3 u 3 + C = − 1 cos3 x + C .
1
3

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


Example
π
0

Solution (Slow Way)



= − 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

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
cos2 x sin x dx
0



Solution (Fast Way)
π −1 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1



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



Solution (Fast Way)
π −1 1
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).



Solution (Fast Way)
π −1 1
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.


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


Example
√
ln 8
ln 3

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


About those limits

Since
√ √ 2
e 2(ln 3)
= e ln 3
= e ln 3 = 3

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


Example
√
ln 8
ln 3

Solution
√
ln 8 8√
1
√ e 2x e 2x + 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 = y 1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y 3/2 = (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
y 1/2 dy = · y = (27 − 8) =
2 4 2 3 4 3 3


Example
√
ln 8
ln 3

Solution
√
ln 8 8√
1
√ e 2x e 2x + 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 = y 1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y 3/2 = (27 − 8) =
2 3 4 3 3

Another way to skin that cat

Example
√
ln 8
ln 3

Solution
Let u = e 2x + 1,



Example
√
ln 8
ln 3

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



Example
√
ln 8
ln 3

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



Example
√
ln 8
ln 3

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



Example
√
ln 8
ln 3

Solution
√
ln 8 9√
1
√ e 2x e 2x + 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
ln 3

Solution
Let u = e 2x + 1, so that

u 2 = e 2x + 1


A third skinned cat

Example
√
ln 8
ln 3

Solution

u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx


A third skinned cat

Example
√
ln 8
ln 3

Solution

u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx

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


A Trigonometric Example

Example
Find
3π/2
θ θ
cot5 sec2 dθ.
π 6 6


A Trigonometric Example

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
θ 1
Let ϕ = . Then dϕ = dθ.
6 6
3π/2 π/4
θ θ
cot5 sec2 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
θ θ
cot5 sec2 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
1 3
=6 − u −4 √
= [9 − 1] = 12.
4 1/ 3 2


Graphs
3π/2 π/4
θ θ
cot 5
sec 2
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
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

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 27: Integration by Substitution (Section 041 slides)

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