Lesson 27: Integration by Substitution (Section 041 slides)

Section 5.5Integration by Substitution

V63.0121.041, Calculus I

New York University

December 13, 2010

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V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37

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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).


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?




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

∫f (x) dx +

∫g(x) dx


x√






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

∫f (x) dx +

∫g(x) dx


x√




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.


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





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.


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 .


Say what?


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 .


Say what?


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 .




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 .





1 + x2 =√


x dx√x2 + 1

=

∫ 12du√

u=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .





1 + x2 =√


x dx√x2 + 1

=

∫ 12du√

u=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .





1 + x2 =√


x dx√x2 + 1

=

∫ 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)


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.





1 + x2 =√


dx =du

2x


x2 + 1dx =

∫x√u· du

2x

=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .






1 + x2 =√


dx =du

2x


x2 + 1dx =

∫x√u· du

2x=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .






1 + x2 =√


dx =du

2x


x2 + 1dx =

∫x√u· du

2x=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .






1 + x2 =√


dx =du

2x


x2 + 1dx =

∫x√u· du

2x=

∫1

2√

udu

=

∫12u−1/2 du

=√

u + C =√

1 + x2 + C .






1 + x2 =√


dx =du

2x


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.V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37

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



Example


∫(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



Example


∫(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



Example


∫(x2 + 3)34x dx .

Solution


∫u3 2du = 2

∫u3 du

=1

2u4

=1

2(x2 + 3)4



Example


∫(x2 + 3)34x dx .

Solution


∫u3 2du = 2

∫u3 du

=1

2u4 =

1

2(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

=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 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







∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=1

2x8 + 6x6 + 27x4 + 54x2







∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=1

2x8 + 6x6 + 27x4 + 54x2







∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=1

2x8 + 6x6 + 27x4 + 54x2







∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=1

2x8 + 6x6 + 27x4 + 54x2







∫ (x6 + 9x4 + 27x2 + 27

)4x dx

=

∫ (4x7 + 36x5 + 108x3 + 108x

)dx

=1

2x8 + 6x6 + 27x4 + 54x2





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 there a difference? 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 =

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 there a difference? 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


A slick example

Example

Find

∫tan x dx . (Hint: tan x =

sin x

cos x)

Solution


∫sin x

cos xdx

= −∫

1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx

= −∫

1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx

= −∫

1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx

= −∫

1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx = −

∫1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx = −

∫1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


A slick example

Example

Find


sin x

cos x)

Solution


∫sin x

cos xdx = −

∫1

udu

= − ln |u|+ C

= − ln | cos x |+ C = ln | sec x |+ C


Can you do it another way?

Example

Find


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).



Example

Find


sin x

cos x)

Solution


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




Example

Find


sin x

cos x)

Solution


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



For those who really must know all

Solution (Continued, with algebra help)

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

∫u du

1− u2=

∫u

y

dy

−2u

= −1

2

∫dy

y= −1

2ln |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 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


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


∫ 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.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




= −∫

u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




= −∫

u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




= −∫

u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)



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


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




∫u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




∫u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)=

2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




∫u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


3cos3 x

∣∣∣∣π0

= −1

3

((−1)3 − 13

)

=2

3.


Example

Compute

∫ π

0cos2 x sin x dx .

Solution (Slow Way)




∫u2 du

= −13u3 + C = −1

3 cos3 x + C .

Therefore∫ π


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.


Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx

=

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx

=

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx =

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx =

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx =

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




Definite-ly Quicker

Solution (Fast Way)


0cos2 x sin x dx =

∫ −11−u2 du =

∫ 1

−1u2 du

=1

3u3

∣∣∣∣1−1

=1

3

(1− (−1)

)=

2

3




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



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


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


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



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



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



Example

Find

∫ ln√8

ln√3

e2x√

e2x + 1 dx

Solution


ln√3

e2x√

e2x + 1 dx =1

2

∫ 9

4

√u du

=1

3u3/2

∣∣∣∣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

∫ 9

4

√u du

=1

3u3/2

∣∣∣∣94

=1

3(27− 8) =

19

3


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 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 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?


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.


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.


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.


Summary

I If F is an antiderivative for f , then:∫f (g(x))g ′(x) dx = F (g(x))

I If F is an antiderivative for f , which is continuous on the range of g ,then:∫ b


∫ g(b)

g(a)f (u) du = F (g(b))− F (g(a))

I Antidifferentiation in general and substitution in particular is a“nonlinear” problem that needs practice, intuition, and perserverance

I The whole antidifferentiation story is in Chapter 6


Lesson 27: Integration by Substitution (Section 041 slides)

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