lesson 27: integration by substitution (slides)

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Sec on 5.5Integra on by Subs tu on

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

May 4, 2011

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ObjectivesI Given an integral and asubs tu on, transform theintegral into an equivalent oneusing a subs tu on

I Evaluate indefinite integralsusing the method ofsubs tu on.

I Evaluate definite integrals usingthe method of subs tu on.

Outline

Last Time: The Fundamental Theorem(s) of Calculus

Subs tu on for Indefinite IntegralsTheoryExamples

Subs tu on for Definite IntegralsTheoryExamples

Differentiation and Integration asreverse processes

Theorem (The Fundamental Theorem of Calculus)

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

ddx

∫ x

af(t) dt = f(x)

2. Let f be con nuous on [a, b] and f = F′ for some other func onF. Then ∫ b

af(x) dx = F(b)− F(a).

Techniques of antidifferentiation?So far we know only a few rules for an differen a on. Some aregeneral, like∫

[f(x) + g(x)] dx =∫

f(x) dx+∫

g(x) dx

Some are pre y par cular, like∫1

x√x2 − 1

dx = arcsec x+ C.

What are we supposed to do with that?

No straightforward system ofantidifferentiation

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 “an ” version of one of themost important rules of differen a on: the chain rule.

Outline




Substitution for IndefiniteIntegrals

Example

Find ∫x√

x2 + 1dx.

Solu onStare at this long enough and you no ce the the integrand is thederiva ve of the expression

√1+ x2.

Say what?Solu on (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?Solu on (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.

Leibnizian notation FTWSolu on (Same technique, new nota on)

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 theintegrand becomes completely transformed into∫

x dx√x2 + 1

=

∫ 12du√u=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.



√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.

Useful but unsavory variationSolu on (Same technique, new nota on, 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.

Mathema cians have serious issues with mixing the x and u like this.However, I can’t deny that it works.




dx =du2x


x2 + 1dx =

∫x√u· du2x

=

∫1

2√udu

=

∫12u

−1/2 du

=√u+ C =

√1+ x2 + C.





dx =du2x


x2 + 1dx =

∫x√u· du2x

=

∫1

2√udu

=

∫12u

−1/2 du =√u+ C =

√1+ x2 + C.


Theorem of the DayTheorem (The Subs tu on Rule)

If u = g(x) is a differen able func on whose range is an interval Iand f is con nuous on I, then∫

f(g(x))g′(x) dx =∫

f(u) du



f(g(x))g′(x) dx =∫

f(u) du

That is, if F is an an deriva ve for f, then∫f(g(x))g′(x) dx = F(g(x))



f(g(x))g′(x) dx =∫

f(u) du

In Leibniz nota on: ∫f(u)

dudx

dx =∫

f(u) du

A polynomial exampleExample

Use the subs tu on u = x2 + 3 to find∫

(x2 + 3)34x dx.

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

(x2 + 3)34x dx

=

∫u3 2du = 2

∫u3 du

=12u4 =

12(x2 + 3)4



(x2 + 3)34x dx.


(x2 + 3)34x dx =∫

u3 2du = 2∫

u3 du

=12u4 =

12(x2 + 3)4



(x2 + 3)34x dx.


(x2 + 3)34x dx =∫

u3 2du = 2∫

u3 du

=12u4

=12(x2 + 3)4



(x2 + 3)34x dx.


(x2 + 3)34x dx =∫

u3 2du = 2∫

u3 du

=12u4 =

12(x2 + 3)4

A polynomial example (brute force)Solu on

∫(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 subs tu on.

A polynomial example (brute force)Solu on∫

(x2 + 3)34x dx

=

∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

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



(x2 + 3)34x dx =∫ (

x6 + 9x4 + 27x2 + 27)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

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



(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 subs tu on.


(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 powers

I But for higher powers, it’s much easier to do subs tu on.


(x2 + 3)34x dx =∫ (

x6 + 9x4 + 27x2 + 27)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

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


CompareWe have the subs tu on method, which, when mul plied 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 there a difference? Is this a problem?

No, that’s what+Cmeans!

CompareWe have the subs tu on method, which, when mul plied 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 there a difference? Is this a problem? No, that’s what+Cmeans!

A slick exampleExample

Find∫

tan x dx.

(Hint: tan x =sin xcos x

)

Solu onLet 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


Find∫

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

)


tan x dx =∫

sin xcos x

dx

= −∫

1udu



Find∫


)


tan x dx =∫

sin xcos x

dx = −∫

1udu



Find∫


)


tan x dx =∫

sin xcos x

dx = −∫

1udu

= − ln |u|+ C

= − ln | cos x|+ C = ln | sec x|+ C


Find∫


)


tan x dx =∫

sin xcos x

dx = −∫

1udu


Can you do it another way?Example

Find∫


)

Solu on

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

.∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫u ducos2 x

=

∫u du

1− sin2 x=

∫u du1− u2


Find∫


)

Solu on


.

∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫u ducos2 x

=

∫u du

1− sin2 x=

∫u du1− u2


Find∫


)

Solu on


.∫tan x dx =

∫sin xcos x

dx =∫

ucos x

ducos x

=

∫u ducos2 x

=

∫u du

1− sin2 x=

∫u du1− u2

For those who really must know allSolu on (Con nued, with algebra help)

Let y = 1− u2, so dy = −2u du. Then∫tan x dx =

∫u du1− u2

=

∫uy

dy−2u

= −12

∫dyy

= −12ln |y|+ C = −1

2ln∣∣1− u2

∣∣+ C

= ln1√

1− u2+ C = ln

1√1− sin2 x

+ C

= ln1

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

There are other ways to do it, too.

Outline




Substitution for Definite IntegralsTheorem (The Subs tu on Rule for Definite Integrals)

If g′ is con nuous and f is con nuous 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 le happens in “x-land”I The integral on the right happens in “u-land”, so the limits needto be u-values

I To get from x to u, apply g

Example

Compute∫ π

0cos2 x sin x dx.

Solu on (Slow Way)

I 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.

I Therefore∫ π

0cos2 x sin x dx = −1

3cos3 x

∣∣∣∣π0

= −13((−1)3 − 13

)=

23.

Example

Compute∫ π

0cos2 x sin x dx.

Solu on (Slow Way)I First compute the indefinite integral

∫cos2 x sin x dx and then

evaluate.I 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.

I Therefore∫ π


3cos3 x

∣∣∣∣π0

= −13((−1)3 − 13

)=

23.

Example

Compute∫ π

0cos2 x sin x dx.




cos2 x sin x dx = −∫

u2 du

= −13u

3 + C = −13 cos

3 x+ C.

I Therefore∫ π


3cos3 x

∣∣∣∣π0

= −13((−1)3 − 13

)=

23.

Example

Compute∫ π

0cos2 x sin x dx.





u2 du = −13u

3 + C = −13 cos

3 x+ C.

I Therefore∫ π


3cos3 x

∣∣∣∣π0

= −13((−1)3 − 13

)=

23.

Example

Compute∫ π

0cos2 x sin x dx.

Solu on (Slow Way)I Let u = cos x . Then du = − sin x dx and∫


u2 du = −13u

3 + C = −13 cos

3 x+ C.

I Therefore∫ π


3cos3 x

∣∣∣∣π0

= −13((−1)3 − 13

)=

23.

Example

Compute∫ π

0cos2 x sin x dx.



u2 du = −13u

3 + C = −13 cos

3 x+ C.

I Therefore∫ π


3cos3 x

∣∣∣∣π0= −1

3((−1)3 − 13

)

=23.

Example

Compute∫ π

0cos2 x sin x dx.



u2 du = −13u

3 + C = −13 cos

3 x+ C.

I Therefore∫ π


3cos3 x

∣∣∣∣π0= −1

3((−1)3 − 13

)=

23.

Definite-ly QuickerSolu on (Fast Way)

Do both the subs tu on and the evalua on at the same me. 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 QuickerSolu on (Fast Way)

Do both the subs tu on and the evalua on at the same me. 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

Compare

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

I But the slow way is just as reliable.

An exponential exampleExample

Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu on


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

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

√8

ln√3

e2x√e2x + 1 dx =

12

∫ 8

3

√u+ 1 du

About those limits

Since

e2(ln√3) = eln

√32

= eln 3 = 3

we have ∫ ln√8

ln√3

e2x√e2x + 1 dx =

12

∫ 8

3

√u+ 1 du


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onNow let y = u+ 1, dy = du. So

12

∫ 8

3

√u+ 1 du =

12

∫ 9

4

√y dy =

12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193

About those fractional powers

We have

93/2 = (91/2)3 = 33 = 2743/2 = (41/2)3 = 23 = 8

so12

∫ 9

4y1/2 dy =

12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onNow let y = u+ 1, dy = du. So

12

∫ 8

3

√u+ 1 du =

12

∫ 9

4

√y dy =

12· 23y3/2

∣∣∣∣94=

13(27− 8) =

193

Another way to skin that catExample

Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onLet u = e2x + 1,

so that du = 2e2x dx. Then∫ ln√8

ln√3

e2x√e2x + 1 dx =

12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94=

13(27− 8) =

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

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

Then∫ ln√8

ln√3

e2x√e2x + 1 dx =

12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94=

13(27− 8) =

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

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

√8

ln√3

e2x√e2x + 1 dx =

12

∫ 9

4

√u du

=13u3/2

∣∣∣∣94=

13(27− 8) =

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx


√8

ln√3

e2x√e2x + 1 dx =

12

∫ 9

4

√u du =

13u3/2

∣∣∣∣94

=13(27− 8) =

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx


√8

ln√3

e2x√e2x + 1 dx =

12

∫ 9

4

√u du =

13u3/2

∣∣∣∣94=

13(27− 8) =

193

A third skinned catExample

Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onLet u =

√e2x + 1, so that u2 = e2x + 1

=⇒ 2u du = 2e2x dx Thus∫ ln√8

ln√3

e2x√e2x + 1 dx =

∫ 3

2u · u du =

13u3∣∣∣∣32=

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onLet u =

√e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx

Thus∫ ln√8

ln√3

e2x√e2x + 1 dx =

∫ 3

2u · u du =

13u3∣∣∣∣32=

193


Find∫ ln

√8

ln√3

e2x√e2x + 1 dx

Solu onLet u =

√e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus∫ ln

√8

ln√3

e2x√e2x + 1 dx =

∫ 3

2u · u du =

13u3∣∣∣∣32=

193

A Trigonometric Example

Example

Find ∫ 3π/2

π

cot5(θ

6

)sec2

(θ

6

)dθ.

Before we dive in, think about:I What “easy” subs tu ons might help?I Which of the trig func ons suggests a subs tu on?

Solu on

Let φ =θ

6. Then dφ =

16dθ.

∫ 3π/2

π

cot5(θ

6

)sec2

(θ

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φ dφ

= 6∫ π/4

π/6

sec2 φ dφtan5 φ

Solu on

Let φ =θ

6. Then dφ =

16dθ.

∫ 3π/2

π

cot5(θ

6

)sec2

(θ

6

)dθ = 6

∫ π/4

π/6cot5 φ sec2 φ dφ

= 6∫ π/4

π/6

sec2 φ dφtan5 φ

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

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

6∫ π/4

π/6

sec2 φ dφtan5 φ

= 6∫ 1

1/√3u−5 du

= 6(−14u−4

)∣∣∣∣11/

√3=

32[9− 1] = 12.

The limits explained

tanπ

4=

sin(π/4)cos(π/4)

=

√2/2√2/2

= 1

tanπ

6=

sin(π/6)cos(π/6)

=1/2√3/2

=1√3

The limits explained

6(−14u−4

)∣∣∣∣11/

√3=

32[−u−4]1

1/√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

π

cot5(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/√36u−5 du

..1√3

..1

The areas of these two regions are the same.

u/du pairs

When deciding on a subs tu on, look for sub-expressions whereone is (a constant mul ple of) the deriva ve of the other. Such as:

u xn ln x sin x cos x tan x√x ex

constant× du xn−1 1x

cos x sin x sec2 x1√x

ex

SummaryI If F is an an deriva ve for f, then:∫

f(g(x))g′(x) dx = F(g(x))

I If F is an an deriva ve for f, which is con nuous on the rangeof g, then:∫ b

af(g(x))g′(x) dx =

∫ g(b)

g(a)f(u) du = F(g(b))− F(g(a))

I An differen a on in general and subs tu on in par cular is a“nonlinear” problem that needs prac ce, intui on, andperserverance.

I The whole an differen a on story is in Chapter 6.

lesson 27: integration by substitution (slides)

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