1. U-Substitution or The Chain Rule of Integration
Lesson 5-5b
U-Substitution or The Chain Rule of Integration

2. ∫ ∫ ∫ Practice Quiz Homework Problem: (2x - ex) dx = x² - ex + C |
Reading question: Fill in the squares below
= x² - ex + C |
= (0-1) – (1- 1/e)
= /e =
x= -1
x=0 ∫ -1
x=1
u= █ 1 2 ∫ ∫
f(x) dx = g(u) du
With u = x² + 1
x=0
u= █

3. Objectives Recognize when to try ‘u’ substitution techniques
Solve integrals of algebraic, exponential, logarithmic, and trigonometric functions using ‘u’ substitution technique
Use symmetry to solve integrals about x = 0 (y-axis)

4. Vocabulary
Change of Variable – substitution of one variable for another in an integral (sort of reverse of the chain rule)
Even Functions – when f(-x) = f(x); even functions are symmetric to the y-axis
Odd Functions – when f(-x) = -f(x); odd functions are symmetric to the origin

5. U Substitution Technique
Recognize in a problem that the integral in its present form is one that we cannot evaluate!
See if changing the variable by letting u = g(x) (applying the “anti-chain rule”) will yield an integral that we can evaluate.
Usually we have to multiple by a form of 1 (k/k) to get the du portion of the integral and the other part of the constant fraction is moved out in front of the integral. ∫ b a ∫
x=b
x=a ∫
u=d
u=c
f(x) dx = k g(u) du or k g(u) du

6. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
1) x2√x3 + 1 dx ∫ ∫
2) sec 2x tan 2x dx
Let u = x3 + 1
then du = 3x² dx
So it becomes
⅓ u½ du
Let u = 2x
then du = 2 dx
So it becomes
½ sec u tan u du
= ½ sec u tan u du = ½ sec u + C ∫
= ⅓ u½ du = 2/9 u3/2 + C ∫
= 2/9 (x3 + 1)3/2 + C
= ½ sec (2x) + C

7. ∫ ∫ ∫ ∫ Example Problems Find the derivative of each of the following:
3) x3√ x4 + 2 dx ∫
(1 + 1/t) t-2 dt ∫
Let u = x4 + 2
then du = 4x3 dx
So it becomes
¼ u½ du
Let u = 1 + 1/t
then du = -1/t² dx
So it becomes
-u du
= ¼ u½ du = 1/6 (u)3/2 + C ∫
= - u du = -½ u² + C ∫
= (1/6) (x4 + 2)3/2 + C
= -½ (1 + 1/t)² + C

8. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following: 1 dx
1 + (2x)² ∫ ∫ 1
6) dx
 1 – 9x²
Let u = 2x
then du = 2 dx
So it becomes
du / (1 + u²)
Let u = 3x
then du = 3 dx
So it becomes
du / (1 – u²)½ du
= ½
1 + u² ∫ du
= ⅓
1 + u² ∫
= ½ tan-1 (u) + C
= ⅓ sin-1 (u) + C
= ½ tan-1 (2x) + C
= ⅓ sin-1 (3x) + C

9. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
7) x sin (x²) dx ∫
8) sin (√x) dx / √x ∫
Let u = x²
then du = 2x dx
So it becomes
sin u du
Let u = x½
then du = ½x-½ dx
So it becomes
sin u du
= ½ sin u du ∫
= 2 sin u du ∫
= -½ cos (u) + C
= -2 cos u + C
= -½ cos (x²) + C
= -2 cos (x) + C

10. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
9) xsin³(x²) cos(x²) dx ∫ ∫
10) e2x+1 dx
Let u = sin (x²)
then du = cos(x²) 2xdx
So it becomes
u³ du
Let u = 2x + 1
then du = 2 dx
So it becomes
eu du
= ½ u³ du = 1/8u4 + C ∫
= ½ eu du = eu + C ∫
= 1/8 sin4 (x²) + C
= ⅓ (5x² + 1)³ + C

11. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
11) tan x dx ∫ ∫
12) cot x dx
Let u = cos x
then du = - sin x dx
So it becomes
du / u
Let u = sin x
then du = cos x dx
So it becomes
du / u
= - du / u = ln u + C ∫
= du / u = ln u + C ∫
= - ln (cos x) + C
= ln (sin x) + C

12. ∫ ∫ Example Problems cont
Find the derivative of each of the following:
13) sec x dx ∫ ∫
14) csc x dx
Form of 1 unique: sec x + tan x
sec x + tan x
Let u = sec x + tan x
then du = sec x tan x + sec² x
Yields a du / u form
Form of 1 unique: csc x + cot x
csc x + cot x
Let u = csc x + cot x
then du = -csc x tan x - csc² x
Yields a du / u form
= ln |sec x + tan x| + C
= - ln |csc x + cot x| + C

13. Summary & Homework Summary: Homework:
U substitution is the reverse of the chain rule
We can only change things by multiplying by another form of 1
Homework:
Day One: pg : 1, 2, 6, 8, 13, 21
Day Two: pg : 35, 42, 51, 58, 59, 76