4029 u-du : Integrating Composite Functions
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Transcript of 4029 u-du : Integrating Composite Functions
4029 u-du: Integrating Composite Functions
AP Calculus
(5 π₯5+4 π₯3+3π₯+2 )5
Find the derivative
5 (5 π₯5+4 π₯3+3 π₯+2 )4 (25 π₯4+12 π₯2+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
ΒΏ π₯2+1
2 π₯ππ₯
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
π₯2+12 π₯ππ₯
(π₯2+1 )ππ’=ΒΏ 2 π₯ππ₯
3 (π’3
3 )+ππ’3+π
(π₯2+1 )3+π
Example 1 : du given
Ex 1: 2 3(5 1) *10x xdx π’=5 π₯2+1ππ’=10π₯ππ₯π’3ππ’
π’4
4+π
14
(5 π₯2+1 )4+π proof
14
(5 π₯2+1 )4+π
4( 14 ) (5 π₯2+1 )3 (10 π₯ )
(5 π₯2+1 )3 (10 π₯ )
Example 2: du given
Ex 2:
1 22 33 ( 1)x x dxπ’=π₯3+1ππ’=3 π₯2ππ₯
(π₯3+1 )12 3 π₯2ππ₯
π’12 ππ’
π’32
32
+π
23π’
32+π
π¦=23
(π₯3+1 )32 +π
Example 3: du given
Ex 3:
2
2 *1
x dxx
π’=(π₯2+1 )
ππ’=2π₯ππ₯
(π₯2+1 )β 1
2 2 π₯ππ₯
π’β 1
2 ππ’π’
12
12
+π
2π’12 +π
2 (π₯2+1 )12 +π
Example 4: du given
Ex 4:
2( ) sec ( )tan x x dx
Both ways !
π’=tan (π₯ΒΏ)ΒΏππ’=π ππ2(π₯)ππ₯
π’ππ’π’2
2+π
12 π‘ππ
2(π₯)+π
π’=sec (π₯)ππ’=sec (π₯ ) tan (π₯ )
π’ππ’π’2
2+π
12 π ππ
2 (π₯ )+π
1+π‘ππ2 (π₯ )=π ππ2(π₯)
Derivative only
π ππ2 (π₯ )ππ₯
Function and derivative
tan (π₯ ) π ππ2 (π₯ )ππ₯
Example 5: Regular Method
Ex 5:
2
cossin
x dxx
cos (π₯) (sin (π₯ ))β2ππ₯ (sin (π₯ ))β2 cos (π₯ )ππ₯π’β2ππ’π’β1
β1+π βπ’β1+π
βsin (π₯ )β1+πβ 1
sin (π₯ )+π=βcsc (π₯ )+π
π’=sin(π₯ )ππ’=cos (π₯ )ππ₯
cos (π₯)sin (π₯)
β 1sin (π₯ )
= cot (π₯ ) csc (π₯ )ππ₯
β csc (π₯ )+π
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( )2
u du 41 ( )
2u du
2dx
42 (1 2 )2
x dx 41 ( )2
u du
12
du dx
41 (1 2 ) 22
x dx
Example 6 : Introduce a Constant - my method
2* 9x x dx
π’=9 βπ₯2
ππ’=β2 π₯ππ₯β2β2π₯ β9β π₯2ππ₯
β 12β2
β 12 (9 βπ₯2 )
12 β 2π₯ππ₯
β 12π’
12 ππ’
β 12 (π’
32
32 )+π
β 13π’
32 +π
β 13
(9 βπ₯2 )32 +π
Example 7 : Introduce a Constant
2sec (3 )x dxπ’=3 π₯ππ’=3ππ₯
33 π ππ2 (3 π₯ ) ππ₯13 π ππ2 (3 π₯ ) 3ππ₯13 π ππ2 (π’)ππ’13 tan (π’)+π
13 tan (3 π₯ )+π
sec (π₯ ) tan (π₯ )ππ₯ π’=sec π₯ππ’=sec π₯ tanπ₯sec π₯
sec π₯ sec (π₯ ) tan (π₯ )ππ₯1
sec π₯ sec π₯ tanπ₯ sec π₯ ππ₯1
sec π₯π’ππ’1
sec π₯π’2
2+π
( 1sec π₯ ) π ππ
2π₯2
+π
12 sec π₯+π
Example 8 : Introduce a Constant << triple chain>>
4sin (2 )cos(2 )x x dx π’=sin(2π₯)ππ’=cos (2 π₯ )2ππ₯
12 π ππ4 (2π₯ ) cos (2π₯ )β2ππ₯12π’4ππ’12 (π’
5
5 )+ππ’5
10+π
110 π ππ
5 (2π₯ )+π
Example 9 : Introduce a Constant - extra constant
<< extra constant>
You is what You is inside5 (3 π₯+4 )5ππ₯
π’=3 π₯+4ππ’=3ππ₯
13 5 (3 π₯+4 )5 3ππ₯53π’5ππ’
53 (π’
6
6 )+π5
18(3 π₯+4 )6+π
Example 10: Polynomial
2 4
3 1(3 2 1)
x dxx x
π’=(3 π₯2β 2π₯+1 )ππ’=(6 π₯β2)ππ₯
12
2(3 π₯β1)
(3 π₯2β 2π₯+1 )4ππ₯
12π’β 4ππ’
12 (π’
β3
β 3 )+πβ 1
6(3 π₯2β 2π₯+1 )β 3
+π
Example 11: Separate the numerator
2
2 11
x dxx
π’=π₯2+1ππ’=2π₯ππ₯
2 π₯π₯2+1
ππ₯+ 1π₯2+1
ππ’π’ + 1
π₯+12
ln|π’|+arctan(π₯)+π
π’β1ππ’+ 1π₯2+1
ΒΏπ’0
0
Formal Change of Variables << the Extra βxβ>>
ILL: Let
Solve for x in terms of u then
and becomes
2 6 *2x x dx (2 6)u x
62
u x
2du dx6 * *
2u u du
ΒΏ 1
2 (π’32 β 6π’
12 )ππ’
12π’3 /2ππ’β 1
26π’1/2ππ’=12 (π’
5/2
52 )
β
β( 12 )6(π’
3 / 2
32 )=1
5π’5 /2 β2π’3 /2
15
(2 π₯+6 )52 β2 (2π₯β6 )3/2+π
Formal Change of Variables << the Extra βxβ>>
Rewrite in terms of u - du
2 13
x dxx
(2π’β7 )π’β 1
2 ππ’
π’=π₯+3ππ’=ππ₯π₯=π’β3
2 π₯=2π’β 62 π₯β1=2π’β 7
(2π’β 32 β7π’
β 12)ππ’
2β 25π’
52 β7β2π’
12 +π
45π’
52 β14π’
12+π
45
(π₯+3 )52 β14 (π₯+3 )
12 +π
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 nd u u udx
Remember: Derivatives Rules
Remember: Laymanβs Description of Antiderivatives
( 1)( ) n nn u du u c
*2nd meaning of βduβ du is the derivative of an implicit βuβ
Development
must have the derivative of the inside in order to find
the antiderivative of the outside
*2nd meaning of βdxβ dx is the derivative of an implicit βxβ 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 laymanβs idea of antiderivative
βThe Family of functions that has the given derivativeβ
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 2
x dx x dx u du