Lesson 27: Integration by Substitution, part II (Section 10 version)

. . . . . .

Section5.5IntegrationbySubstitution, PartDeux

V63.0121, CalculusI

April29, 2009

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Outline

Recall: Themethodofsubstitution

Multiplesubstitutions

OddandevenfunctionsExamples

Moreexamplesandadvice

CourseEvaluations

. . . . . .

LastTime: TheSubstitutionRule

TheoremIf u = g(x) isadifferentiablefunctionwhoserangeisaninterval Iand f iscontinuouson I, then∫

f(g(x))g′(x)dx =

∫f(u)du

or ∫f(u)

dudx

dx =

∫f(u)du

. . . . . .

LastTime: TheSubstitutionRuleforDefiniteIntegrals

TheoremIf g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

I Theintegralonthelefthappensin“x-land”, soitslimitsarevaluesof x

I Theintegralontherighthappensin“u-land”, soitslimitsneedtobevaluesof u

I Toconvert x to u, simplyapplythesubstitution u = g(x).

. . . . . .

LastTime: TheSubstitutionRuleforDefiniteIntegrals

TheoremIf g′ iscontinuousand f iscontinuousontherangeof u = g(x),then ∫ b

af(g(x))g′(x)dx =

∫ g(b)

g(a)f(u)du.

I Theintegralonthelefthappensin“x-land”, soitslimitsarevaluesof x

I Theintegralontherighthappensin“u-land”, soitslimitsneedtobevaluesof u

I Toconvert x to u, simplyapplythesubstitution u = g(x).

. . . . . .

Outline





CourseEvaluations

. . . . . .

Anexponentialexample

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

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

√8

ln√3

e2x√

e2x + 1dx =12

∫ 8

3

√u + 1du

Nowlet 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

. . . . . .


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

. . . . . .

Anotherwaytoskinthatcat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u = e2x + 1

, sothat du = 2e2x dx. Then∫ ln√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√udu

=13u3/2

∣∣∣∣94

=13

(27− 8) =193

. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u = e2x + 1, sothat du = 2e2x dx.

Then∫ ln√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√udu

=13u3/2

∣∣∣∣94

=13

(27− 8) =193

. . . . . .


Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

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

√8

ln√3

e2x√

e2x + 1dx =12

∫ 9

4

√udu

=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

√udu

=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

√udu

=13u3/2

∣∣∣∣94

=13

(27− 8) =193

. . . . . .

A thirdskinnedcat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, sothat

u2 = e2x + 1

=⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

e2x√

e2x + 1dx =

∫ 3

2u · udu =

13u3

∣∣∣∣32

=193

. . . . . .

A thirdskinnedcat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, sothat

u2 = e2x + 1 =⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

e2x√

e2x + 1dx =

∫ 3

2u · udu =

13u3

∣∣∣∣32

=193

. . . . . .

A thirdskinnedcat

Example

Find∫ ln

√8

ln√3

e2x√

e2x + 1dx

SolutionLet u =

√e2x + 1, sothat

u2 = e2x + 1 =⇒ 2udu = 2e2x dx

Thus ∫ ln√8

ln√3

e2x√

e2x + 1dx =

∫ 3

2u · udu =

13u3

∣∣∣∣32

=193

. . . . . .

Outline





CourseEvaluations

. . . . . .

Example

Find∫ π

−πsin(x)dx

Solution

∫ π

−πsin(x) = − cos(x)|π−π = cos(x)|−π

π = cos(−π) − cos(π) = 0

Thisisobviousfromthegraph:

. .x

.y

. . . . . .

Example

Find∫ π

−πsin(x)dx

Solution

∫ π


π = cos(−π) − cos(π) = 0


. .x

.y

. . . . . .

Example

Find∫ π

−πsin(x)dx

Solution

∫ π


π = cos(−π) − cos(π) = 0


. .x

.y

. . . . . .

EvenandOddFunctions

DefinitionA function f is even ifforall x,

f(−x) = f(x)

A function f is odd ifforall x,

f(−x) = −f(x).

Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis

. . . . . .

EvenandOddFunctions


f(−x) = f(x)


f(−x) = −f(x).


. . . . . .

EvenandOddFunctions


f(−x) = f(x)


f(−x) = −f(x).

Thesepropertiesarerevealedinthegraph.

I Anoddfunctionhasrotationalsymmetryabouttheorigin.I Anevenfunctionhasreflectivesymmetryinthe y-axis

. . . . . .

EvenandOddFunctions


f(−x) = f(x)


f(−x) = −f(x).

Thesepropertiesarerevealedinthegraph.I Anoddfunctionhasrotationalsymmetryabouttheorigin.

I Anevenfunctionhasreflectivesymmetryinthe y-axis

. . . . . .

EvenandOddFunctions


f(−x) = f(x)


f(−x) = −f(x).


. . . . . .

EvenandOddfunctionspictured

. .x

.y.odd

.x

.y.even

. . . . . .

Examplesofsymmetricfunctions

Evenandoddfunctionsabound.I x 7→ xn isoddwhen n isoddandevenwhen n iseven.Funny, that!

I sin isoddand cos iseven.

. . . . . .

Combiningsymmetricfunctions

Theorem

(a) Thesumofevenfunctionsiseven. Thesumofoddfunctionsisodd.

(b) Theproductofevenfunctionsiseven. Theproductofoddfunctionsiseven. Theproductofanoddfunctionandanevenfunctionisanoddfunction.

(c) If g iseven, then f ◦ g iseven. Thecompositionoftwooddfunctionsisodd. Thecompositionofanevenfunctionandanoddfunctioniseven.

. . . . . .

Integratingsymmetricfunctions

TheoremLet a beanynumber.

(a) If f isodd, then ∫ a

−af(x)dx = 0.

(b) If f iseven, then ∫ a

−af(x)dx = 2

∫ a

0f(x)dx.

. . . . . .

Integratingsymmetricfunctions

TheoremLet a beanynumber.

(a) If f isodd, then ∫ a

−af(x)dx = 0.

(b) If f iseven, then ∫ a

−af(x)dx = 2

∫ a

0f(x)dx.

. . . . . .

Proof(odd f).

Tocompute∫ a

−af(x)dx, let u = −x. Then du = −dx andwehave

∫ a

−af(x)dx = −

∫ −a

af(−u)du

=

∫ −a

af(u)du

= −∫ a

−af(u)du.

Theonlynumberwhichisequaltoitsownnegativeiszero.

. . . . . .

Proof(even f).Withthesamesubstitutionwehave∫ 0

−af(x)dx = −

∫ 0

af(−u)du

= −∫ 0

af(u)du

=

∫ a

0f(u)du.

So ∫ a

−af(x)dx =

∫ 0

−af(x)dx +

∫ a

0f(x)dx = 2

∫ a

0f(x)dx.

. . . . . .

ExampleCompute ∫ e

√π+1

−e√

π−1sin(x)

√1 + cos3(x)dx.

SolutionTheintegrandisodd! Sotheansweriszero.

. . . . . .

ExampleCompute ∫ e

√π+1

−e√

π−1sin(x)

√1 + cos3(x)dx.

SolutionTheintegrandisodd! Sotheansweriszero.

. . . . . .

ExampleCompute ∫ 2

−2

(x4 + x2 + 3

)dx.

. . . . . .

SolutionBecausetheintegrandisevenwecansimplifyourarithmetic. It’sespeciallynicetopluginzerosincetheresultisoftenzero.∫ 2

−2

(x4 + x2 + 3

)dx = 2

∫ 2

0

(x4 + x2 + 3

)dx

= 2[x5

5+

x3

3+ 3x

]20

= 2[25

5+

23

3+ 3(2)

]= 2

[325

+83

+ 6]

=215

(32 · 3 + 8 · 5 + 6 · 15)

=2 · 22615

. . . . . .

Outline





CourseEvaluations

. . . . . .

Example∫x3

(5x4 + 2)2dx

SolutionLet u = 5x4 + 2, so du = 20x3 dx. Then∫

x3

(5x4 + 2)2dx =

120

∫1u2

du

= − 120

· 1u

+ C

= − 120(5x4 + 2)

+ C

. . . . . .

Example∫x3

(5x4 + 2)2dx

SolutionLet u = 5x4 + 2, so du = 20x3 dx. Then∫

x3

(5x4 + 2)2dx =

120

∫1u2

du

= − 120

· 1u

+ C

= − 120(5x4 + 2)

+ C

. . . . . .

Example∫sin(sin(θ)) cos(θ)dθ

SolutionLet u = sin(θ), so du = cos(θ)dθ. Then∫

sin(sin(θ)) cos(θ)dθ =

∫sin(u)du

= − cos(u) + C

= − cos(sin(θ)) + C

. . . . . .

Example∫sin(sin(θ)) cos(θ)dθ

SolutionLet u = sin(θ), so du = cos(θ)dθ. Then∫

sin(sin(θ)) cos(θ)dθ =

∫sin(u)du

= − cos(u) + C

= − cos(sin(θ)) + C

. . . . . .

Example∫ex + e−x

ex − e−x dx

SolutionThenumeratoristhederivativeofthedenominator! Letu = ex − e−x, so du =

(ex + e−x) dx. Then∫

ex + e−x

ex − e−x dx =

∫1udu

= ln |u| + C = ln∣∣ex − e−x

∣∣ + C

. . . . . .

Example∫ex + e−x

ex − e−x dx

SolutionThenumeratoristhederivativeofthedenominator! Letu = ex − e−x, so du =

(ex + e−x) dx. Then∫

ex + e−x

ex − e−x dx =

∫1udu

= ln |u| + C = ln∣∣ex − e−x

∣∣ + C

. . . . . .

Example∫3x

1 + 9xdx

SolutionNotice 9x = (32)x = 32x = (3x)2. Solet u = 3x,du = (ln 3) · 3x dx. Then∫

3x

(1 + 9x)dx =

1ln 3

∫1

1 + u2du

=1ln 3

arctan(u) + C

=1ln 3

arctan(3x) + C

. . . . . .

Example∫3x

1 + 9xdx

SolutionNotice 9x = (32)x = 32x = (3x)2. Solet u = 3x,du = (ln 3) · 3x dx. Then∫

3x

(1 + 9x)dx =

1ln 3

∫1

1 + u2du

=1ln 3

arctan(u) + C

=1ln 3

arctan(3x) + C

. . . . . .

Example∫sec2

√x√

xdx

SolutionLet u =

√x, so du =

12√xdu. Then

∫sec2

√x√

xdx = 2

∫sec2(u)du

= 2 tan(u) + C

= 2 tan(√

x)

+ C

. . . . . .

Example∫sec2

√x√

xdx

SolutionLet u =

√x, so du =

12√xdu. Then

∫sec2

√x√

xdx = 2

∫sec2(u)du

= 2 tan(u) + C

= 2 tan(√

x)

+ C

. . . . . .

Example∫dx

x ln x

SolutionLet u = ln x, so du =

1xdx. Then∫dx

x ln x=

∫1udu

= ln |u| + C

= ln |ln x| + C

. . . . . .

Example∫dx

x ln x

SolutionLet u = ln x, so du =

1xdx. Then∫dx

x ln x=

∫1udu

= ln |u| + C

= ln |ln x| + C

. . . . . .

Whatdowesubstitute?

I Linearfactors (ax + b) areeasysubstitutions: u = ax + b,du = a dx

I Lookfor function/derivativepairs intheintegrand, onetomake u andonetomake du:

I xn and xn−1 (fudgethecoefficient)I sineandcosine(fudgetheminussign)I ex and exI ax and ax (fudgethecoefficient)

I√x and

1√x(fudgethefactorof 2)

I ln x and1x

. . . . . .

Outline





CourseEvaluations

. . . . . .

CourseEvaluations

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Lesson 27: Integration by Substitution, part II (Section 10 version)

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