4029
u-du: Integrating Composite Functions
AP Calculus
Find the derivative
5𝑥 5 + 4𝑥 3 + 3𝑥 + 2
5
5 5𝑥 5 + 4𝑥 3 + 3𝑥 + 2
4
25𝑥 4 + 12𝑥 2 + 3
dx/du-part of the antiderivative
u-du Substitution
Integrating Composite Functions
(Chain Rule)
Revisit the Chain Rule
d 2
( x  1)3 
dx
If let u = inside function
= 𝑥2 + 1
du = derivative of the inside
d
(u )3
dx
=
3( u ) 2  du
becomes
d 2
( x  1)3  3( x 2  1) 2 (2 x)
dx
2𝑥𝑑𝑥
dx
A Visual Aid
USING u-du Substitution  a Visual Aid
2
𝑥
+1
REM: u = inside function
du = derivative of the inside 2𝑥𝑑𝑥
2
2
3(
x

1)
 2 xdx

let
u = 𝑥2 + 1
𝑑𝑢 =
becomes
 3u du
2
𝑢3
3
+𝑐
3
𝑢3 + 𝑐
𝑥2 + 1
3
+𝑐
2𝑥𝑑𝑥
now only working with
f , the outside
function
Example 1 :
Ex
1:
du given
2
3
(5
x

1)
*10 xdx

𝑢 = 5𝑥 2 + 1
𝑑𝑢 = 10𝑥𝑑𝑥
3
𝑢 𝑑𝑢
𝑢4
+𝑐
4
1
5𝑥 2 + 1
4
4
1
5𝑥 2 + 1
4
+𝑐
1
4
4
5𝑥 2 + 1
5𝑥 2 + 1
3
3
4
+𝑐
10𝑥
10𝑥
Example 2:
Ex 2:
du given
2
3
3
x
(
x
 1)

𝑥3
+1
1
𝑢2 𝑑𝑢
3
𝑢2
3
2
+𝑐
2 3
𝑢2 + 𝑐
3
1 2
dx
𝑢 = 𝑥3 + 1
𝑑𝑢 = 3𝑥 2 𝑑𝑥
1
2 3𝑥 2 𝑑𝑥
2 3
𝑦 = 𝑥 +1
3
3
2
+𝑐
Example 3:

Ex 3:
𝑥2
+1
1
−2
𝑢 𝑑𝑢
1
𝑢2
1
2
+𝑐
du given
2x
x 1
2
1
−2
2𝑥𝑑𝑥
𝑢 = 𝑥2 + 1
* dx
1
2𝑢2
𝑑𝑢 = 2𝑥𝑑𝑥
+𝑐
2 𝑥2 + 1
1
2
+𝑐
Example 4:
du given
  tan( x)  sec
Ex 4:
𝑢 = tan(𝑥)
𝑑𝑢 = 𝑠𝑒𝑐 2 (𝑥)𝑑𝑥
2
( x) dx
𝑢 = sec(𝑥)
𝑑𝑢 = sec 𝑥 tan(𝑥)
𝑢 𝑑𝑢
𝑢 𝑑𝑢
𝑢2
+𝑐
2
𝑢2
+𝑐
2
1
𝑡𝑎𝑛2 (𝑥) + 𝑐
2
1
𝑠𝑒𝑐 2 𝑥 + 𝑐
2
1 + 𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 (𝑥)
Both ways !
Derivative only
𝑠𝑒𝑐 2 𝑥 𝑑𝑥
Function and derivative
tan 𝑥 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
Example 5: Regular Method
Ex 5:
cos x
dx
 sin 2 x
cos(𝑥) sin(𝑥)
sin(𝑥)
−2
−2 cos
𝑑𝑥
cos(𝑥)
1
∗
=
sin(𝑥) sin 𝑥
𝑥 𝑑𝑥
− csc 𝑥 + 𝑐
𝑢 −2 𝑑𝑢
𝑢−1
+𝑐
−1
𝑢 = sin(𝑥)
𝑑𝑢 = cos 𝑥 𝑑𝑥
−𝑢−1 + 𝑐
− sin 𝑥 −1 + 𝑐
1
−
+ 𝑐 = − csc 𝑥 + 𝑐
sin 𝑥
cot 𝑥 csc 𝑥 𝑑𝑥
Working with Constants
< multiplying by one>
Constant Property of Integration
  5cos x dx
5  cos x  dx

4
(1

2
x
)
dx
ILL. 
let
u = (1  2 x )
du =
1 
becomes  (u )  du 
2 
4
2dx
1
du  dx
and
2
1
4
(
u
)
 du 
=

2
2
1
4
(1

2
x
)
dx


Or alternately 
=  (1  2 x)4  2dx 
2
2
=
1
4
(
u
)
 du 

2
Example 6 : Introduce a Constant
−2
𝑥 9 − 𝑥 2 𝑑𝑥
−2
x
*

9−
1
−
2
1
−
2
3
𝑢2
3
2
1
2
𝑥 2
𝑢 = 9 − 𝑥2
𝑑𝑢 = −2𝑥𝑑𝑥
9  x dx
2
1 −2
−
2
1
−
2
- my method
− 2𝑥𝑑𝑥
1
𝑢2 𝑑𝑢
+𝑐
1 3
− 𝑢2 + 𝑐
3
3
1
2
− 9−𝑥 2+𝑐
3
Example 7 : Introduce a Constant
sec
(3
x
)
dx

2
3
3
𝑠𝑒𝑐 2 3𝑥 𝑑𝑥
1
𝑠𝑒𝑐 2 3𝑥 3𝑑𝑥
3
1
𝑠𝑒𝑐 2 𝑢 𝑑𝑢
3
1
tan 𝑢 + 𝑐
3
1
tan 3𝑥 + 𝑐
3
𝑢 = 3𝑥
𝑑𝑢 = 3𝑑𝑥
sec 𝑥 tan 𝑥 𝑑𝑥
sec 𝑥
sec 𝑥
1
sec 𝑥
1
sec 𝑥
𝑢 = sec 𝑥
𝑑𝑢 = sec 𝑥 tan 𝑥
sec 𝑥 tan 𝑥 𝑑𝑥
sec 𝑥 tan 𝑥 sec 𝑥 𝑑𝑥
𝑢 𝑑𝑢
1 𝑢2
+𝑐
sec 𝑥 2
1
𝑠𝑒𝑐 2 𝑥
+𝑐
sec 𝑥
2
1
sec 𝑥 + 𝑐
2
Example 8 : Introduce a Constant
<< triple chain>>
sin
(2
x
)
cos(2
x
)
dx

4
1
2
𝑠𝑖𝑛4 2𝑥 cos 2𝑥 ∗ 2𝑑𝑥
1
2
𝑢4 𝑑𝑢
1 𝑢5
+𝑐
2 5
𝑢5
+𝑐
10
1
𝑠𝑖𝑛5 2𝑥 + 𝑐
10
𝑢 = sin(2𝑥)
𝑑𝑢 = cos 2𝑥 2𝑑𝑥
Example 9 : Introduce a Constant
5 3𝑥 + 4 5 𝑑𝑥
1
5 3𝑥 + 4 5 3𝑑𝑥
3
5
𝑢5 𝑑𝑢
3
5 𝑢6
+𝑐
3 6
5
3𝑥 + 4
18
6
+𝑐
- extra constant
You is what
You is inside
<< extra constant>
𝑢 = 3𝑥 + 4
𝑑𝑢 = 3𝑑𝑥
𝑢 = 3𝑥 2 − 2𝑥 + 1
Example 10
: Polynomial
3x  1
dx
 (3x2  2 x  1)4
1
2
2(3𝑥 − 1)
3𝑥 2 − 2𝑥 + 1
1
2
4
𝑑𝑥
𝑢−4 𝑑𝑢
1 𝑢−3
+𝑐
2 −3
1
− 3𝑥 2 − 2𝑥 + 1
6
−3
+𝑐
𝑑𝑢 = (6𝑥 − 2)𝑑𝑥
𝑢 = 𝑥2 + 1
Example 11: Separate the numerator
𝑑𝑢 = 2𝑥𝑑𝑥
2x 1
dx
 x2  1
2𝑥
𝑑𝑥 +
2
𝑥 +1
𝑑𝑢
+
𝑢
1
𝑥2 + 1
1
𝑥 + 12
𝑢−1 𝑑𝑢
+
𝑢0
=
0
ln 𝑢 + arctan(𝑥) + 𝑐
1
𝑥2 + 1
Formal Change of Variables
ILL:
x
<< the Extra “x”>>
2 x  6 * 2dx
Solve for x in terms of u
becomes
1
2
1
−
2
1
2𝑥 + 6
5
6𝑢 1 2 𝑑𝑢
5
2
1 𝑢5 2
=
5
2
2
− 2 2𝑥 − 6
u 6
x
2
then
and
u6
  2  * u * du
𝑢3 2 𝑑𝑢
Let
u  (2 x  6)
1
=
2
du  2dx
3
𝑢2
−
1
6𝑢2
1
𝑢3 2
1 5
−
6
= 𝑢
3
2
5
2
3 2
+𝑐
2
𝑑𝑢
− 2𝑢3
2
Formal Change of Variables
<< the Extra “x”>>
Rewrite in terms of u - du

2x 1
dx
x3
2𝑢 − 7
3
−2
2𝑢
1
−2
𝑢 𝑑𝑢
1
−2
− 7𝑢
5
2
𝑑𝑢 = 𝑑𝑥
𝑥 =𝑢−3
2𝑥 = 2𝑢 − 6
2𝑥 − 1 = 2𝑢 − 7
𝑑𝑢
1
2 5
2 ∗ 𝑢2 − 7 ∗ 2𝑢2 + 𝑐
5
1
4 5
𝑢2 − 14𝑢2 + 𝑐
5
4
𝑥+3
5
𝑢 =𝑥+3
− 14 𝑥 + 3
1
2
+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)
Remember:
Derivatives Rules
d
(u ) n
dx
n( u ) ( n 1) * u
=
Remember: Layman’s Description of Antiderivatives
 n (u )
( n 1)
du  u  c
*2nd meaning of “du”
n
du is the derivative of an implicit “u”
Development
d
 f ( g ( x))  f '( g ( x)) * g '( x)
dx
d
 f ( g ( x))  f '( g ( x)) * g '( x)
dx
d  f ( g ( x))  [ f '( g ( x))* g '( x)]dx
f ( g ( x))   f '( g ( x)) * g '( x)dx
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”
if x = f then dx = f /
more later
Development
d
 f ( g ( x))  f '( g ( x)) * g '( x)
dx
from the layman’s idea of antiderivative
d
 f ( g ( x))  f '( g ( x)) * g '( x)
dx
“The Family of functions that has the given derivative”
f ( g ( x))   f '( g ( x)) * g '( x)dx
---------d
(u )3
dx

3(u ) 2 * du
must have the derivative of
the inside in order to find
the antiderivative of the outside
Working With Constants:
  5cos x dx
With
u-du
Constant Property of Integration
5  cos x  dx

Substitution
2
2
2
2
3(
x

1)
*
2
xdx
=
3
(
x

1)
* 2 xdx


REM:
u = inside function
du = derivative of the inside
Missing Constant?
4
(1

2
x
)
dx

u
du
3 u 2 du
=
=
2
1
1
4
4
4
(1

2
x
)
dx
=
(1

2
x
)
2
dx
=
(
u
)




 du 



2
2
2
Worksheet - Part 1