u-du: Integrating Composite Functions

AP Calculus

Integrating Composite Functions(Chain Rule)

( 1)( ) = n( ) *n ndu u u

dx

Remember: Derivatives Rules

Remember: Layman’s Description of Antiderivatives

( 1)( ) n nn u du u c

*2nd meaning of “du” du is the derivative of an implicit “u”

u-du SubstitutionIntegrating Composite Functions

(Chain Rule)Revisit the Chain Rule

If let u = inside function

du = derivative of the inside

becomes

2 3( 1)d

xdx

2 3 2 2( 1) 3( 1) (2 )d

x x xdx

3 2( ) = 3( ) *d

u u dudx

Development

from the layman’s idea of antiderivative  

“The Family of functions that has the given derivative”

must have the derivative of the inside in order to find

---------- the antiderivative of the outside 

( ( )) '( ( ))* '( )d

f g x f g x g xdx

( ( )) '( ( ))* '( )d

f g x f g x g xdx

( ( )) '( ( ))* '( )f g x f g x g x dx

3( )d

udx

23( ) * u du

A Visual Aid

USING u-du Substitution a Visual AidREM: u = inside function du = derivative of the inside

let u =

becomes now only working with f , the outside function

2 23( 1) *2x xdx23u du

Working With Constants: Constant Property of Integration

With u-du Substitution

REM: u = inside function du = derivative of the inside

Missing Constant?

2 2 2 23( 1) *2 = 3 ( 1) *2x xdx x xdx 23 u du

Worksheet - Part 1

5cos 5 cosx dx x dx

4(1 2 )x dx u = du =

4 4 42 1 1(1 2 ) = (1 2 ) 2 = ( )

2 2 2x dx x dx u du

Example 1 : du given

Ex 1:2 3(5 1) *10x xdx

Example 2: du given

Ex 2:  

1 22 33 ( 1)x x dx

Example 3: du given

Ex 3:  

2

2*

1

xdx

x

Example 4: du given

Ex 4:  

2( ) sec ( )tan x x dx

Example 5: Regular Method

Ex 5:  

2

cos

sin

xdx

x

Working with Constants < multiplying by one>

Constant Property of Integration

 ILL. let u =

du = and

becomes =

 Or alternately = =

5cos 5 cosx dx x dx

4(1 2 )x dx (1 2 )x

4 1( )

2u du

41( )

2u du

2dx

42(1 2 )

2x dx 41

( )2

u du

1

2du dx

41(1 2 ) 2

2x dx

Example 6 : Introduce a Constant - my method

2* 9x x dx

Example 7 : Introduce a Constant

2sec (3 )x dx

Example 8 : Introduce a Constant << triple chain>>

4sin (2 )cos(2 )x x dx

Example 9 : Introduce a Constant - extra constant

<< extra constant>

5(3 4)x dx

Example 10: Polynomial

2 4

3 1

(3 2 1)

xdx

x x

Example 11: Separate the numerator

2

2 1

1

xdx

x

Formal Change of Variables << the Extra “x”>> 

Solve for x in terms of u

 ILL: Let

Solve for x in terms of u then

and  becomes

2 6 *2x x dx (2 6)u x

6

2

ux

2du dx

6* *

2

uu du

Formal Change of Variables << the Extra “x”>> 

Rewrite in terms of u - du

2 1

3

xdx

x

Complete Change of Variables << Changing du >>

At times it is required to even change the du as the u is changed above.

1cos

2x dx u x du dx

x

2

2

xdu dx

u du dx

cos 2u u du

We will solve this later in the course.

Development

  

must have the derivative of the inside in order to find

the antiderivative of the outside 

*2nd meaning of “dx” dx is the derivative of an implicit “x” more later if x = f then dx = f /

( ( )) '( ( ))* '( )d

f g x f g x g xdx

( ( )) '( ( ))* '( )d

f g x f g x g xdx

( ( )) [ '( ( ))* '( )]d f g x f g x g x dx

( ( )) '( ( ))* '( )f g x f g x g x dx