Lesson 29: Integration by Substition (worksheet solutions)

Section 5.5
Integration by Substitution

V63.0121.034, Calculus I

December 9, 2009

Announcements
Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50
Practice ﬁnals on the website. Solutions Friday

. . . . . .

Schedule for next week

Monday, class: review, evaluations, movie!
Tuesday, 8:00am: Review session for all students with 8:00
recitations (Tuesday or Thursday) in CIWW 109
Tuesday, 9:30am: Review session for all students with 9:30
recitations (Tuesday or Thursday) in CIWW 109
Ofﬁce Hours continue
Friday, 2:00pm: ﬁnal in Tisch UC50

. . . . . .

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
. . . . . .

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?

. . . . . .

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

. . . . . .

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.

. . . . . .


Example
Find ∫
x
√ dx.
x 2+1

. . . . . .


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

. . . . . .

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
∫ ∫ ∫
(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:
∫ ∫
2 3
( 6 )
(x + 3) 4x dx = x + 9x4 + 27x2 + 27 4x dx
∫
( 7 )
= 4x + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2

. . . . . .


∫ ∫
2 3
( 6 )
(x + 3) 4x dx = x + 9x4 + 27x2 + 27 4x dx
∫
( 7 )
= 4x + 36x5 + 108x3 + 108x dx
1 8
= x + 6x6 + 27x4 + 54x2
2
Which would you rather do?

. . . . . .


∫ ∫
2 3
( 6 )
(x + 3) 4x dx = x + 9x4 + 27x2 + 27 4x dx
∫
( 7 )
= 4x + 36x5 + 108x3 + 108x dx
1 8
= x + 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 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 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 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.

. . . . . .

A slick example

Example
∫
sin x
cos x
Solution
Let u = cos x. Then du = − sin x dx. So
∫ ∫ ∫
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
= = 2
=
cos2 x 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)

∫ ∫ ∫ ( )
u du 1 1 1
tan x dx = = − du
1 − u2 2 1−u 1+u
1 1
= − ln |1 − u| − ln |1 + u| + C
2 2
1 1
= ln √ + C = ln √ +C
(1 − u)(1 + u) 1 − u2
1
= ln + C = ln |sec x| + C
|cos x|

. . . . . .

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.

. . . . . .

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
∫ ∫
cos x sin x dx = − u2 du
2

= − 1 u3 + C = − 1 cos3 x + C.
3 3

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

. . . . . .

Solution (Fast Way)
Do both the substitution and the evaluation at the same time.

. . . . . .

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.

. . . . . .

Solution (Fast Way)
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So
∫ π ∫ −1
cos2 x sin x dx = −u2 du
0 1
∫ 1
= u2 du
−1
1
1 1( ) 2
= u3 = 1 − (−1) =
3 −1 3 3

. . . . . .

Solution (Fast Way)
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So
∫ π ∫ −1
cos2 x sin x dx = −u2 du
0 1
∫ 1
= u2 du
−1
1
1 1( ) 2
= u3 = 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 x variable.
But the slow way is just as reliable.

. . . . . .

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

. . . . . .

Example√
∫ ln 8 √
ln 3

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

. . . . . .

Example√
∫ ln 8 √
ln 3

Solution
Let u = e2x , so du = 2e2x dx. We have
∫ √ ∫
ln 8 √ 1 8√
√ e2x 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 limits

√ √ 2
e2(ln 3)
= eln 3
= eln 3 = 3

. . . . . .

About those fractional powers

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

. . . . . .

Another way to skin that cat

Example√
∫ ln 8 √
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 √
ln 3

Solution
√
Let u = e2x + 1, so that

u2 = e2x + 1

. . . . . .

A third skinned cat

Example√
∫ ln 8 √
ln 3

Solution
√

u2 = e2x + 1 =⇒ 2u du = 2e2x dx

. . . . . .

A third skinned cat

Example√
∫ ln 8 √
ln 3

Solution
√

u2 = e2x + 1 =⇒ 2u du = 2e2x dx

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

. . . . . .

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

. . . . . .

Example
Find ∫ ( ) ( )
3π/2
5 θ 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
1 3
=6 − u−4 √
= [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 √
1 −4 3 [ −4 ]1 √ 3 [ −4 ]1/ 3
6 − u √
= −u 1 / 3 = u 1
4 1/ 3 2 2
3 [ −1/2 −4 ]
= (3 ) − (1−1/2 )−4
2
3 3
= [32 − 12 ] = (9 − 1) = 12
2 2

. . . . . .

Graphs

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

Graphs

∫ π/4 ∫ 1
. 6 cot5 φ sec2 φ dφ . √ 6u−5 du
π/6 1/ 3

y
. y
.

. . . .
φ . ..
u
ππ 1 .1
. . .√
64 3 The areas
of these two regions are the same.
. . . . . .

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

. . . . . .

Lesson 29: Integration by Substition (worksheet solutions)

More Related Content

What's hot (19)

Viewers also liked (20)

Similar to Lesson 29: Integration by Substition (worksheet solutions) (20)

More from Matthew Leingang (20)

Recently uploaded (20)

Lesson 29: Integration by Substition (worksheet solutions)