4.5 Integration by Substitution Outside Function Inside Function Derivative of Inside Function.

4.5 Integration by Substitution

( ( )) ' )) ( )( (f g x g x d F gx x C Outside Function

Inside FunctionDerivative of

Inside Function

Antidifferentiation of a Composite Function

Official Description

Antidifferentiation of a Composite Function

Let be a continuous function whose range is an interval ,

and let be a function that is continuous on . If is

differentiable on its domain and is an antiderivative of

on , then

g I

f I g

F f

I

( ( )) '( ) ( ( ))f g x g x dx F g x C If ( ), then '( ) and u g x du g x dx

( ) ( ) .f u du F u C Problem

4.5 Integration by Substitution2 2Evaluate ( 1) (2 )x x dx

Composite Function?

Derivative of Inside Function?

3

2 2 2

2

32( 1) (2 )3

1

11

3

2

ux x dx u d x C

u x

du xdx

u C

Substitution


:Check

22 1 2x x 2 31( 1)

3

dx C

dx

2 2 2 3( ( ))1

( 1) (2 ) ( 1)3

F g xx Cx dx x C


Evaluate 5cos5xdx5

5

u x

du dx

5cos5 cos sin 5sinxdx udu x Cu C

2

32 2

3232

1

1( 1

2

1)

2 6

1

632

u x

du xdx

u ux x dx u du

xCC C


2 2Evaluate ( 1)x x dx


Evaluate 2 1x dx

3/ 2 3/ 3/ 221/ 2

2 1

12 1

2 3/

11

2 32 3

2

2

u x

du dx

u ux dx u du CC

xC

1/ 2

5/ 2 3/ 21/ 2 3/ 2 1/ 2

5/ 2 3/ 2 5/ 2 3 5/ 2 3// 2

Change of Variables or Write in terms o

2 1

?????

1

2 2

1 1 1

2 2 4 4 5 / 2 3/ 2

1 2 2

4 5 3 1

2

2 1 2 1

f

10

06

u x

du dx

u du

u dux dx

u du u uu u u du

x x

x

C

u u u uC C

u

2

6C


Evaluate 2 1x x dx

4.5 Integration by Substitution2sin 3Evaluate cos3x xdx

??????

sin 3 co 3 3s

u

u x du x dx

2 2

3 3 3

1sin 3 cos3

3

1

3 3

si

99

n 3x

x xdx u du

u uC C C

We do not need change of variables!

4.5 Integration by SubstitutionTHM 4.13 The General Power Rule for Integration

is a differentiable function of g x

1( )

( ) '( ) , 11

nn g x

g x g x dx C nn

Equivalently, ( )u g x

1

, 11

nn uu du C n

n


Substitution and the General Power Rule4(a) 3(3 1) x dx

5(3 1)

5

xC

2(b) (2 1)( )x x x dx 2 2( )

2

x xC

2 3(c) 3 2x x dx 3 / 232 2

3

xC

2 2

4(d)

(1 2 )

xdx

x

2 1(1 2 )x C

2(e) cos sinx xdx 3cos

3

xC

2cos ( sin )x x dx

22( 2)(f) ?x dx General Power Rule won't work because

the integrand lacks a factor of .xGOTO (f)


Substitution and the General Power Rule

4 513(a) 1

5 3(3 1)

3 1

3

x d x Cx

u x

du dx



2

2

222

(b) (2 1)( )

2 1

1 1

22

x x x dx

u x x

du

x x

x dx

udu C Cu



2 3

3

2

1/ 2 3 / 2 3 / 23

(c) 3 2

2

22

3

3

2

3

x x dx

u x

du x dx

u du u C x C



2

2 2

2

2 12

1

2

4(d)

(1 2 )

1 2

4

1

1 2 1

xdx

x

u x

du xdx

Cu du u C Cx x



3

2

2 3

(e) cos sin

cos

sin

1cos

3

1

3

x xdx

u x

du xdx

u du Cu C x


2 2( 2)x dx 4 2( 4 4)x x dx 5 34

45 3

x xx C

5 344

5 3

x xx C

So what do we do when substitution

doesn't w

Write the integrand in another for

ork?

m!!!

4.5 Integration by Substitution1

0

2 3Evaluate ( 1)x x dx2 1 2u x du xdx

Now determine the new upper and lower bounds.

Lower Bound Upper Bound0 1x u 1 2x u

1 2 3

0( 1)x x dx

1 2 3

0

1( 1) 2

2x x dx

2 3

1

1

2u du

24

4 4

1

1 12 1

2 4 8

u

15

8

Change of Variables

You can change the bounds and

keep everything in terms of u!

4.5 Integration by Substitution5

1Evaluate

2 1

xA x

xd

2 1 2u x du dx Won't work!

2 1u x 2 2 1u x

dx udu

2 1

2

ux

Differential of x?

Lower Bound Upper Bound1 1x u 5 3x u

5

1 2 -1

xdx

x

23

1

1 1

2

uudu

u

3 2

1

1( 1)

2u du

33

1

1

2 3

uu

1 19 3 1

2 3

16

3

Change of Variables


THM 4.15 Integration of Even and Odd Functions

- 0

-

Let be integrable on , .

1. is an even function ( ) 2 ( ) .

2. is an odd function ( ) 0

a a

a

a

a

f a a

f f x dx f x dx

f f x dx

/2 3

- /2(sin cos sin cos 0)x x x x dx

4.5 Integration by Substitution/2 3

- /2Evaluate (sin cos sin cos )x x x x dx

3( ) sin ( ) cos( ) sin( ) cos( )f x x x x x

Note: cos( ) cos ,x x 3 3sin ( ) sin ,x x sin( ) sinx x

3( ) sin cos sin cosf x x x x x ( )f x

Even or Odd??


HW 4.5/2-6,7-29odd,33-43odd,49,51,59,61,65,67,71,79,80,90,91-94

Welcome to the first abstract part of integration. In today's assignment there

will be times when is appropriate and times when y

r

ou simply need

to or use a ewrite the integrand trigonome

substitution

tric identity

"differential equations

You will also be asked to solve (another branch of mathematics

that engineers study.) That is: giv

(they never

en , solve for

go

.

.

"

away )

dyy

dx

recipes critical

thinking ski

If you are dependent upon , you will struggle. If you have developed the

that are needed to achieve success at the college level, you will struggle on

a few problems

lls

, but still be well on your way to receiving a on the "5" AP Exam.

Who are you?

4.5 Integration by Substitution Outside Function Inside Function Derivative of Inside Function.

Documents

Transcript of 4.5 Integration by Substitution Outside Function Inside Function Derivative of Inside Function.