lesson 26: integration by substitution (handout)

Section 5.5Integration by Substitution

V63.0121.006/016, Calculus I

New York University

April 27, 2010

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6)

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

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Notes

Notes

Notes

1

Section 5.5 : Integration by SubstitutionV63.0121.006/016, Calculus I April 27, 2010

http://laughingsquid.com/

Objectives

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

I Evaluate indefinite integralsusing the method ofsubstitution.

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

d

dx

∫ 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).


Notes

Notes

Notes

2


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

Some are pretty particular, like∫1

x√

x2 − 1dx = arcsec x + C .

What are we supposed to do with that?


No straightforward system of antidifferentiation

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

x2 + 1dx

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





Notes

Notes

Notes

3


Substitution for Indefinite Integrals

Example

Find ∫x√

x2 + 1dx .

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. Then g ′(x) = 2x and so

d

dx

√g(x) =

1

2√

g(x)g ′(x) =

x√x2 + 1

Thus ∫x√

x2 + 1dx =

∫ (d

dx

√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 integrandbecomes completely transformed into∫

x dx√x2 + 1

=

∫ 12du√

u=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .


Notes

Notes

Notes

4


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:”

dx =du

2x

So the integrand becomes completely transformed into∫x√

x2 + 1dx =

∫x√u· du

2x=

∫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)

du

dxdx =

∫f (u) du


A polynomial example

Example

Use the substitution u = x2 + 3 to find

∫(x2 + 3)34x dx .

Solution

If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So∫(x2 + 3)34x dx =

∫u3 2du = 2

∫u3 du

=1

2u4 =

1

2(x2 + 3)4


Notes

Notes

Notes

5


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

=1

2x8 + 6x6 + 27x4 + 54x2

Which would you rather do?

I It’s a wash for low powers

I 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 =

1

2(x2 + 3)4 + C

=1

2

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

)+ C

=1

2x8 + 6x6 + 27x4 + 54x2 +

81

2+ C

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

1

2x8 + 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 x

cos x)

Solution

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

∫sin x

cos xdx = −

∫1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


Notes

Notes

Notes

6


Can you do it another way?

Example

Find

∫tan x dx . (Hint: tan x =

sin x

cos x)

Solution

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

cos x.∫

tan x dx =

∫sin x

cos xdx =

∫u

cos x

du

cos x

=

∫u du

cos2 x=

∫u du

1− sin2 x=

∫u du

1− u2

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


For those who really must know all

Solution (Continued, with algebra help)

∫tan x dx =

∫u du

1− u2=

∫1

2

(1

1− u− 1

1 + u

)du

= −1

2ln |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





Notes

Notes

Notes

7


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 tobe u-values

I 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

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


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 =

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3

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

I But the slow way is just as reliable.


Notes

Notes

Notes

8


An exponential example

Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution

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

ln√3

e2x√

e2x + 1 dx =1

2

∫ 8

3

√u + 1 du

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

1

2

∫ 8

3

√u + 1 du =

1

2

∫ 9

4

√y dy =

1

2

∫ 9

4y1/2 dy

=1

2· 2

3y3/2

∣∣∣∣94

=1

3(27− 8) =

19

3


About those limits

Since

e2(ln√3) = e ln

√32

= e ln 3 = 3

we have ∫ ln√8

ln√3

e2x√

e2x + 1 dx =1

2

∫ 8

3

√u + 1 du



Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution


ln√3

e2x√

e2x + 1 dx =1

2

∫ 8

3

√u + 1 du


1

2

∫ 8

3

√u + 1 du =

1

2

∫ 9

4

√y dy =

1

2

∫ 9

4y1/2 dy

=1

2· 2

3y3/2

∣∣∣∣94

=1

3(27− 8) =

19

3


Notes

Notes

Notes

9


About those fractional powers

We have

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

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

so1

2

∫ 9

4y1/2 dy =

1

2· 2

3y3/2

∣∣∣∣94

=1

3(27− 8) =

19

3



Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution


ln√3

e2x√

e2x + 1 dx =1

2

∫ 8

3

√u + 1 du


1

2

∫ 8

3

√u + 1 du =

1

2

∫ 9

4

√y dy =

1

2

∫ 9

4y1/2 dy

=1

2· 2

3y3/2

∣∣∣∣94

=1

3(27− 8) =

19

3


Another way to skin that cat

Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution

Let u = e2x + 1,so that du = 2e2x dx. Then∫ ln√8

ln√3

e2x√

e2x + 1 dx =1

2

∫ 9

4

√u du

=1

3u3/2

∣∣∣∣94

=1

3(27− 8) =

19

3


Notes

Notes

Notes

10


A third skinned cat

Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution

Let u =√

e2x + 1, so that

u2 = e2x + 1 =⇒ 2u du = 2e2x dx

Thus ∫ ln√8

ln√3

=

∫ 3

2u · u du =

1

3u3

∣∣∣∣32

=19

3


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 =

1

6dθ.

∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ = 6

∫ π/4

π/6cot5 ϕ sec2 ϕ dϕ

= 6

∫ π/4

π/6

sec2 ϕ dϕ

tan5 ϕ

Now let u = tanϕ. So du = sec2 ϕ dϕ, and

6

∫ π/4

π/6

sec2 ϕ dϕ

tan5 ϕ= 6

∫ 1

1/√3

u−5 du

= 6

(−1

4u−4

)∣∣∣∣11/√3

=3

2[9− 1] = 12.


Notes

Notes

Notes

11


The limits explained

tanπ

4=

sinπ/4

cosπ/4=

√2/2√2/2

= 1

tanπ

6=

sinπ/6

cosπ/6=

1/2√3/2

=1√3

6

(−1

4u−4

)∣∣∣∣11/√3

=3

2

[−u−4

]11/√3

=3

2

[u−4

]1/√31

=3

2

[(3−1/2)−4 − (1−1/2)−4

]=

3

2[32 − 12] =

3

2(9− 1) = 12


Graphs

θ

y

π 3π

2

∫ 3π/2

πcot5

(θ

6

)sec2

(θ

6

)dθ

ϕ

y

π

6

π

4

∫ π/4

π/66 cot5 ϕ sec2 ϕ dϕ

The areas of these two regions are the same.


Graphs

ϕ

y

π

6

π

4

∫ π/4

π/66 cot5 ϕ sec2 ϕ dϕ

u

y

∫ 1

1/√3

6u−5 du

1√3

1

The areas of these two regions are the same.


Notes

Notes

Notes

12


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))


Notes

Notes

Notes

13


lesson 26: integration by substitution (handout)

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