4029 u-du: Integrating Composite Functions

AP Calculus

(5 𝑥5+4 𝑥3+3𝑥+2 )5

Find the derivative

5 (5 𝑥5+4 𝑥3+3 𝑥+2 )4 (25 𝑥4+12 𝑥2+3 )

dx/du-part of the antiderivative

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 duu u dxdx

¿ 𝑥2+1

2 𝑥𝑑𝑥

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 xdx 23u du

𝑥2+12 𝑥𝑑𝑥

(𝑥2+1 )𝑑𝑢=¿ 2 𝑥𝑑𝑥

3 (𝑢3

3 )+𝑐𝑢3+𝑐

(𝑥2+1 )3+𝑐

Example 1 : du given

Ex 1:2 3(5 1) *10x xdx 𝑢=5 𝑥2+1

𝑑𝑢=10𝑥𝑑𝑥𝑢3𝑑𝑢𝑢4

4+𝑐

14

(5 𝑥2+1 )4+𝑐 proof

14

(5 𝑥2+1 )4+𝑐

4( 14 ) (5 𝑥2+1 )3 (10 𝑥 )

(5 𝑥2+1 )3 (10 𝑥 )

Example 2: du given

Ex 2:  

1 22 33 ( 1)x x dx 𝑢=𝑥3+1𝑑𝑢=3 𝑥2𝑑𝑥

(𝑥3+1 )12 3 𝑥2𝑑𝑥

𝑢12 𝑑𝑢

𝑢32

32

+𝑐

23𝑢

32+𝑐

𝑦=23

(𝑥3+1 )32 +𝑐

Example 3: du given

Ex 3:  

2

2*

1

xdx

x 𝑢=(𝑥2+1 )

𝑑𝑢=2𝑥𝑑𝑥

(𝑥2+1 )− 1

2 2 𝑥𝑑𝑥

𝑢− 1

2 𝑑𝑢𝑢

12

12

+𝑐

2𝑢12 +𝑐

2 (𝑥2+1 )12 +𝑐

Example 4: du given

Ex 4:  

2( ) sec ( )tan x x dx

Both ways !

𝑢=tan(𝑥¿)¿𝑑𝑢=𝑠𝑒𝑐2(𝑥)𝑑𝑥

𝑢𝑑𝑢𝑢2

2+𝑐

12𝑡𝑎𝑛2(𝑥)+𝑐

𝑢=sec (𝑥)𝑑𝑢=sec (𝑥 ) tan (𝑥 )

𝑢𝑑𝑢𝑢2

2+𝑐

12𝑠𝑒𝑐2 (𝑥 )+𝑐

1+𝑡𝑎𝑛2 (𝑥 )=𝑠𝑒𝑐2(𝑥)

Derivative only

𝑠𝑒𝑐2 (𝑥 )𝑑𝑥

Function and derivative

tan (𝑥 ) 𝑠𝑒𝑐2 (𝑥 )𝑑𝑥

Example 5: Regular Method

Ex 5:  

2

cos

sin

xdx

x cos (𝑥) (sin (𝑥 ))−2𝑑𝑥

(sin (𝑥 ))−2 cos (𝑥 )𝑑𝑥𝑢−2𝑑𝑢𝑢−1

−1+𝑐 −𝑢−1+𝑐

−sin (𝑥 )−1+𝑐

−1

sin (𝑥 )+𝑐=−csc (𝑥 )+𝑐

𝑢=sin(𝑥 )𝑑𝑢=cos (𝑥 )𝑑𝑥

cos (𝑥)sin (𝑥)

∗ 1sin (𝑥 )

= cot (𝑥 ) csc (𝑥 )𝑑𝑥

− csc (𝑥 )+𝑐

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

𝑢=9 −𝑥2

𝑑𝑢=−2 𝑥𝑑𝑥

−2−2𝑥 √9− 𝑥2𝑑𝑥

−12−2

−12 (9−𝑥2 )

12 − 2𝑥𝑑𝑥

−12𝑢

12 𝑑𝑢

−12 (𝑢

32

32

)+𝑐−

13𝑢

32 +𝑐

−13

(9−𝑥2 )32 +𝑐

Example 7 : Introduce a Constant

2sec (3 )x dx𝑢=3 𝑥𝑑𝑢=3𝑑𝑥

33 𝑠𝑒𝑐2 (3 𝑥 ) 𝑑𝑥

13 𝑠𝑒𝑐2 (3 𝑥 ) 3𝑑𝑥13 𝑠𝑒𝑐2 (𝑢)𝑑𝑢13

tan (𝑢)+𝑐

13

tan (3 𝑥 )+𝑐

sec (𝑥 ) tan (𝑥 )𝑑𝑥 𝑢=sec 𝑥𝑑𝑢=sec 𝑥 tan𝑥

sec 𝑥sec 𝑥 sec (𝑥 ) tan (𝑥 )𝑑𝑥

1sec 𝑥 sec 𝑥 tan𝑥 sec 𝑥 𝑑𝑥

1sec 𝑥𝑢𝑑𝑢

1sec 𝑥

𝑢2

2+𝑐

( 1sec𝑥 ) 𝑠𝑒𝑐

2𝑥2

+𝑐

12

sec𝑥+𝑐

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

4sin (2 )cos(2 )x x dx 𝑢=sin(2𝑥)𝑑𝑢=cos (2 𝑥 )2𝑑𝑥

12 𝑠𝑖𝑛4 (2𝑥 ) cos (2𝑥 )∗2𝑑𝑥12𝑢4𝑑𝑢12 (𝑢

5

5 )+𝑐𝑢5

10+𝑐

110

𝑠𝑖𝑛5 (2𝑥 )+𝑐

Example 9 : Introduce a Constant - extra constant

<< extra constant>

You is what You is inside5 (3 𝑥+4 )5𝑑𝑥

𝑢=3 𝑥+4𝑑𝑢=3𝑑𝑥

13

5 (3 𝑥+4 )5 3𝑑𝑥53𝑢5𝑑𝑢

53 (𝑢

6

6 )+𝑐5

18(3 𝑥+4 )6+𝑐

Example 10: Polynomial

2 4

3 1

(3 2 1)

xdx

x x

𝑢=(3 𝑥2− 2𝑥+1 )𝑑𝑢=(6 𝑥−2)𝑑𝑥

12 2(3 𝑥−1)

(3 𝑥2− 2𝑥+1 )4𝑑𝑥

12𝑢− 4𝑑𝑢  

12 (𝑢

−3

− 3 )+𝑐−

16

(3 𝑥2− 2𝑥+1 )− 3+𝑐

Example 11: Separate the numerator

2

2 1

1

xdx

x

𝑢=𝑥2+1𝑑𝑢=2𝑥𝑑𝑥

2 𝑥𝑥2+1

𝑑𝑥+ 1

𝑥2+1

𝑑𝑢𝑢

+ 1

𝑥+12

ln|𝑢|+arctan(𝑥)+𝑐

𝑢−1𝑑𝑢+ 1

𝑥2+1

¿ 𝑢0

0

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

 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 dx6

* *2

uu du

¿ 1

2 (𝑢

32 − 6𝑢

12 )𝑑𝑢

12𝑢3 /2𝑑𝑢−

126𝑢1/2𝑑𝑢= 1

2 (𝑢5/2

52 )

❑

−( 12 )6(𝑢

3 /2

32 )= 1

5𝑢5 /2 −2𝑢3 /2

15

(2 𝑥+6 )52 −2 (2𝑥−6 )3/2+𝑐

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

Rewrite in terms of u - du

2 1

3

xdx

x

(2𝑢−7 )𝑢− 1

2 𝑑𝑢

𝑢=𝑥+3𝑑𝑢=𝑑𝑥

𝑥=𝑢−3

2 𝑥=2𝑢− 6

2 𝑥−1=2𝑢− 7

(2𝑢−32 −7𝑢

−12)𝑑𝑢

2∗ 25𝑢

52 −7∗2𝑢

12 +𝑐

45𝑢

52 −14𝑢

12+𝑐

45

(𝑥+3 )52 −14 (𝑥+3 )

12 +𝑐

Assignment

Day 1 Worksheet Larson HW 4029

Day 2 Basic Integration Rules Wksht

extra x Larson 4029 58f

anti for tan /cot Text p. 338 # 18 - 52 (3x)

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”

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

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

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