Lesson 26: Integration by Substitution (slides)

Section 5.5Integration by Substitution

V63.0121.006/016, Calculus I

New York University

April 27, 2010

AnnouncementsI April 29: Movie DayI April 30: Quiz 5 on §§5.1–5.4I Monday, May 10, 12:00noon Final Exam:

I SILV 703 (Section 16)I MEYR 121/122 (Section 6)

. . . . . .

. . . . . .

Announcements

I April 29: Movie DayI April 30: Quiz 5 on

§§5.1–5.4I Monday, May 10,

12:00noon Final Exam:I SILV 703 (Section 16)I MEYR 121/122 (Section

6)

V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 2 / 38

. . . . . .

Resurrection Policy

If your final score beats your midterm score, we will add 10% to itsweight, and subtract 10% from the midterm weight.

..Image credit: Scott Beale / Laughing SquidV63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 3 / 38

http://laughingsquid.com/

. . . . . .

Objectives

I Given an integral and asubstitution, transform theintegral into an equivalentone using a substitution

I Evaluate indefiniteintegrals using the methodof substitution.

I Evaluate definite integralsusing the method ofsubstitution.


. . . . . .

Outline

Last Time: The Fundamental Theorem(s) of Calculus

Substitution for Indefinite IntegralsTheoryExamples

Substitution for Definite IntegralsTheoryExamples


. . . . . .

Differentiation and Integration as reverse processes

Theorem (The Fundamental Theorem of Calculus)

1. Let f be continuous on [a,b]. Then

ddx

∫ x

af(t)dt = f(x)

2. Let f be continuous on [a,b] and f = F′ for some other function F.Then ∫ b

af(x)dx = F(b)− F(a).


. . . . . .

Techniques of antidifferentiation?

So far we know only a few rules for antidifferentiation. Some aregeneral, like ∫

[f(x) + g(x)] dx =

∫f(x)dx+

∫g(x)dx

Some are pretty particular, like∫1

x√x2 − 1

dx = arcsec x+ C.

What are we supposed to do with that?


. . . . . .



[f(x) + g(x)] dx =

∫f(x)dx+

∫g(x)dx


x√x2 − 1

dx = arcsec x+ C.



. . . . . .



[f(x) + g(x)] dx =

∫f(x)dx+

∫g(x)dx


x√x2 − 1

dx = arcsec x+ C.



. . . . . .

No straightforward system of antidifferentiation

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

dx

or ∫tan x dx.

Luckily, we can be smart and use the “anti” version of one of the mostimportant rules of differentiation: the chain rule.


. . . . . .

No straightforward system of antidifferentiation

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

dx

or ∫tan x dx.

Luckily, we can be smart and use the “anti” version of one of the mostimportant rules of differentiation: the chain rule.


. . . . . .

Outline





. . . . . .

Substitution for Indefinite Integrals

Example

Find ∫x√

x2 + 1dx.

SolutionStare at this long enough and you notice the the integrand is thederivative of the expression

√1+ x2.


. . . . . .

Substitution for Indefinite Integrals

Example

Find ∫x√

x2 + 1dx.

SolutionStare at this long enough and you notice the the integrand is thederivative of the expression

√1+ x2.


. . . . . .

Say what?

Solution (More slowly, now)

Let g(x) = x2 + 1.

Then g′(x) = 2x and so

ddx

√g(x) =

12√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.


. . . . . .

Say what?


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

ddx

√g(x) =

12√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.


. . . . . .

Say what?


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

ddx

√g(x) =

12√

g(x)g′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (ddx

√g(x)

)dx

=√

g(x) + C =√1+ x2 + C.


. . . . . .

Leibnizian notation FTW

Solution (Same technique, new notation)

Let u = x2 + 1.

Then du = 2x dx and√1+ x2 =

√u. So the integrand

becomes completely transformed into∫x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.


. . . . . .



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

√u.

So the integrandbecomes completely transformed into∫

x dx√x2 + 1

=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.


. . . . . .






=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.


. . . . . .






=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.


. . . . . .






=

∫ 12du√u

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.


. . . . . .

Useful but unsavory variation

Solution (Same technique, new notation, more idiot-proof)


√u. “Solve for dx:”

dx =du2x

So the integrand becomes completely transformed into∫x√

x2 + 1dx =

∫x√u· du2x

=

∫1

2√udu

=

∫12u

−1/2 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.V63.0121.006/016, Calculus I (NYU) Section 5.5 Integration by Substitution April 27, 2010 13 / 38

. . . . . .

Theorem of the Day

Theorem (The Substitution Rule)

If u = g(x) is a differentiable function whose range is an interval I and fis 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: ∫f(u)

dudx

dx =

∫f(u)du


. . . . . .

A polynomial example

Example

Use the substitution u = x2 + 3 to find∫

(x2 + 3)34x dx.

SolutionIf u = x2 + 3, then du = 2x dx, and 4x dx = 2du. So∫

(x2 + 3)34x dx =

∫u3 2du = 2

∫u3 du

=12u4 =

12(x2 + 3)4


. . . . . .

A polynomial example

Example

Use the substitution u = x2 + 3 to find∫

(x2 + 3)34x dx.

SolutionIf u = x2 + 3, then du = 2x dx, and 4x dx = 2du. So∫

(x2 + 3)34x dx =

∫u3 2du = 2

∫u3 du

=12u4 =

12(x2 + 3)4


. . . . . .

A polynomial example, by brute force

Compare this to multiplying it out:∫(x2 + 3)34x dx =

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do substitution.


. . . . . .



∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Which would you rather do?

I It’s a wash for low powersI But for higher powers, it’s much easier to do substitution.


. . . . . .



∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=12x8 + 6x6 + 27x4 + 54x2

Which would you rather do?I It’s a wash for low powersI But for higher powers, it’s much easier to do substitution.


. . . . . .

Compare

We have the substitution method, which, when multiplied out, gives∫(x2 + 3)34x dx =

12(x2 + 3)4

+ C

=12

(x8 + 12x6 + 54x4 + 108x2 + 81

)

+ C

=12x8 + 6x6 + 27x4 + 54x2 +

812

+ C

and the brute force method∫(x2 + 3)34x dx =

12x8 + 6x6 + 27x4 + 54x2

+ C

Is this a problem?

No, that’s what +C means!


. . . . . .

Compare

We have the substitution method, which, when multiplied out, gives∫(x2 + 3)34x dx =

12(x2 + 3)4 + C

=12

(x8 + 12x6 + 54x4 + 108x2 + 81

)+ C

=12x8 + 6x6 + 27x4 + 54x2 +

812

+ C

and the brute force method∫(x2 + 3)34x dx =

12x8 + 6x6 + 27x4 + 54x2 + C

Is this a problem? No, that’s what +C means!


. . . . . .

A slick example

Example

Find∫

tan x dx.

(Hint: tan x =sin xcos x

)

SolutionLet u = cos x. Then du = − sin x dx. So∫

tan x dx =

∫sin xcos x

dx = −∫

1udu

= − ln |u|+ C= − ln | cos x|+ C = ln | sec x|+ C


. . . . . .

A slick example

Example

Find∫

tan x dx. (Hint: tan x =sin xcos x

)


tan x dx =

∫sin xcos x

dx = −∫

1udu



. . . . . .

A slick example

Example

Find∫


)

SolutionLet u = cos x. Then du = − sin x dx.

So∫tan x dx =

∫sin xcos x

dx = −∫

1udu



. . . . . .

A slick example

Example

Find∫


)


tan x dx =

∫sin xcos x

dx = −∫

1udu



. . . . . .

A slick example

Example

Find∫


)


tan x dx =

∫sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C


. . . . . .

A slick example

Example

Find∫


)


tan x dx =

∫sin xcos x

dx = −∫

1udu



. . . . . .

Can you do it another way?

Example

Find∫


)

Solution

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

cos x.∫

tan x dx =

∫sin xcos x

dx =

∫u

cos xdu

cos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫u du1− u2

At this point, although it’s possible to proceed, we should probablyback up and see if the other way works quicker (it does).


. . . . . .


Example

Find∫


)

Solution


cos x.

∫tan x dx =

∫sin xcos x

dx =

∫u

cos xdu

cos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫u du1− u2



. . . . . .


Example

Find∫


)

Solution


cos x.∫

tan x dx =

∫sin xcos x

dx =

∫u

cos xdu

cos x

=

∫uducos2 x

=

∫udu

1− sin2 x=

∫u du1− u2



. . . . . .

For those who really must know all

Solution (Continued, with algebra help)

∫tan x dx =

∫udu1− u2

=

∫12

(1

1− u− 1

1+ u

)du

= −12ln |1− u| − 1

2ln |1+ u|+ C

= ln1√

(1− u)(1+ u)+ C = ln

1√1− u2

+ C

= ln1

|cos x|+ C = ln |sec x|+ C

There are other ways to do it, too.


. . . . . .

Outline





. . . . . .

Substitution for Definite Integrals

Theorem (The Substitution Rule for Definite Integrals)

If g′ is continuous and f is continuous on the range of u = g(x), then∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

Why the change in the limits?I The integral on the left happens in “x-land”I The integral on the right happens in “u-land”, so the limits need to

be u-valuesI To get from x to u, apply g


. . . . . .

Substitution for Definite Integrals

Theorem (The Substitution Rule for Definite Integrals)

If g′ is continuous and f is continuous on the range of u = g(x), then∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

Why the change in the limits?I The integral on the left happens in “x-land”I The integral on the right happens in “u-land”, so the limits need to

be u-valuesI To get from x to u, apply g


. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

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 = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0= −1

3((−1)3 − 13

)=

23.


. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

Solution (Slow Way)



evaluate.

Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π


3cos3 x

∣∣∣∣π0= −1

3((−1)3 − 13

)=

23.


. . . . . .

Example

Compute∫ π

0cos2 x sin x dx.

Solution (Slow Way)



evaluate. Let u = cos x. Then du = − sin x dx and∫cos2 x sin x dx = −

∫u2 du

= −13u

3 + C = −13 cos

3 x+ C.

Therefore∫ π


3cos3 x

∣∣∣∣π0= −1

3((−1)3 − 13

)=

23.


. . . . . .

Definite-ly Quicker

Solution (Fast Way)

Do both the substitution and the evaluation at the same time.

Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du =

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13(1− (−1)

)=

23

I The advantage to the “fast way” is that you completely transformthe integral into something simpler and don’t have to go back tothe original variable (x).

I But the slow way is just as reliable.


. . . . . .

Definite-ly Quicker

Solution (Fast Way)

Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1.

So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du =

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13(1− (−1)

)=

23




. . . . . .

Definite-ly Quicker

Solution (Fast Way)

Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du =

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13(1− (−1)

)=

23




. . . . . .

Definite-ly Quicker

Solution (Fast Way)

Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So∫ π

0cos2 x sin x dx =

∫ −1

1−u2 du =

∫ 1

−1u2 du

=13u3

∣∣∣∣1−1

=13(1− (−1)

)=

23




. . . . . .

An exponential example

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u = e2x, so du = 2e2x dx. We have∫ ln

√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u+ 1du

Now let y = u+ 1, dy = du. So

12

∫ 8

3

√u+ 1du =

12

∫ 9

4

√y dy =

12

∫ 9

4y1/2 dy

=12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx


√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u+ 1du


12

∫ 8

3

√u+ 1du =

12

∫ 9

4

√y dy =

12

∫ 9

4y1/2 dy

=12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


. . . . . .

About those limits

Since

e2(ln√3) = eln

√32 = eln 3 = 3

we have ∫ ln√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u+ 1du


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx


√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u+ 1du


12

∫ 8

3

√u+ 1du =

12

∫ 9

4

√y dy =

12

∫ 9

4y1/2 dy

=12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


. . . . . .

About those fractional powers

We have

93/2 = (91/2)3 = 33 = 27

43/2 = (41/2)3 = 23 = 8

so12

∫ 9

4y1/2 dy =

12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx


√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u+ 1du


12

∫ 8

3

√u+ 1du =

12

∫ 9

4

√y dy =

12

∫ 9

4y1/2 dy

=12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


. . . . . .

Another way to skin that cat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u = e2x + 1,

so that du = 2e2x dx. Then∫ ln√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94

=13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

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

Then∫ ln√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94

=13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u = e2x + 1,so that du = 2e2x dx. Then∫ ln

√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94

=13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx


√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94

=13(27− 8) =

193


. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx


√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94

=13(27− 8) =

193


. . . . . .

A third skinned cat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, so that

u2 = e2x + 1

=⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193


. . . . . .

A third skinned cat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, so that

u2 = e2x + 1 =⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193


. . . . . .

A third skinned cat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, so that

u2 = e2x + 1 =⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

=

∫ 3

2u · udu =

13u3

∣∣∣∣32=

193


. . . . . .

A Trigonometric Example

Example

Find ∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ.

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


. . . . . .

A Trigonometric Example

Example

Find ∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ.

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


. . . . . .

Solution

Let φ =θ

6. Then dφ =

16dθ.

∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φdφ

= 6∫ π/4

π/6

sec2 φdφtan5 φ

Now let u = tanφ. So du = sec2 φdφ, and

6∫ π/4

π/6

sec2 φdφtan5 φ

= 6∫ 1

1/√3u−5 du

= 6(−14u−4

)∣∣∣∣11/

√3=

32[9− 1] = 12.


. . . . . .

Solution

Let φ =θ

6. Then dφ =

16dθ.

∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φdφ

= 6∫ π/4

π/6

sec2 φdφtan5 φ

Now let u = tanφ. So du = sec2 φdφ, and

6∫ π/4

π/6

sec2 φdφtan5 φ

= 6∫ 1

1/√3u−5 du

= 6(−14u−4

)∣∣∣∣11/

√3=

32[9− 1] = 12.


. . . . . .

The limits explained

tanπ

4=

sinπ/4cosπ/4

=

√2/2√2/2

= 1

tanπ

6=

sinπ/6cosπ/6

=1/2√3/2

=1√3

6(−14u−4

)∣∣∣∣11/

√3=

32

[−u−4

]11/

√3=

32

[u−4

]1/√3

1

=32

[(3−1/2)−4 − (1−1/2)−4

]=

32[32 − 12] =

32(9− 1) = 12


. . . . . .

The limits explained

tanπ

4=

sinπ/4cosπ/4

=

√2/2√2/2

= 1

tanπ

6=

sinπ/6cosπ/6

=1/2√3/2

=1√3

6(−14u−4

)∣∣∣∣11/

√3=

32

[−u−4

]11/

√3=

32

[u−4

]1/√3

1

=32

[(3−1/2)−4 − (1−1/2)−4

]=

32[32 − 12] =

32(9− 1) = 12


. . . . . .

Graphs

. .θ

.y

..π

.

.3π2

.∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ

.φ

.y

.

.π

6

.

.π

4

.∫ π/4

π/66 cot5 φ sec2 φdφ

The areas of these two regions are the same.


. . . . . .

Graphs

. .φ

.y

.

.π

6

.

.π

4

.∫ π/4

π/66 cot5 φ sec2 φdφ

.u

.y

.∫ 1

1/√36u−5 du

.

.1√3

.

.1

The areas of these two regions are the same.


. . . . . .

Final Thoughts

I Antidifferentiation is a “nonlinear” problem that needs practice,intuition, and perserverance

I Worksheet in recitation (also to be posted)I The whole antidifferentiation story is in Chapter 6


. . . . . .

Summary

If F is an antiderivative for f, then:I ∫

f(g(x))g′(x)dx = F(g(x))

I ∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u) du = F(g(b))− F(g(a))


Lesson 26: Integration by Substitution (slides)

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