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

2. ∫ ∫ ∫ Quiz Homework Problem: (3ex + 7sec2x) dx = 3ex + 7tan x + C
Reading questions: Fill in the squares below ∫
= 3ex + 7tan x + C ∫ ∫
e7x dx =
U substitution: u = ______
sec² (3x) dx =
U substitution: 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. ∫ 2cos (2x) dx Example 0 = ∫ cos (2x) 2 dx = ∫ cos u du = sin u + C
If we let u = 2x then du = 2dx
or ½du = dx
To get du we need to move
2 back to the dx
= ∫ 2cos (u) ½du
= ∫ cos u du = sin u + C
2 and ½ cancel out
= sin (2x) + C

6. ∫(2x + 1)² dx Example 1 = ∫(2x + 1)² 2/2 dx = ½ ∫(2x + 1)² 2 dx
If we let u = 2x +1 then it becomes u² and du = 2dx
or ½du = dx
we are missing a 2 from dx so we multiple by 1 (2/2)
= ½ ∫(2x + 1)² 2 dx
= ∫ (u)² ½du
½ goes outside ∫ and 2 stays with dx
= ½ ∫u² du = 1/6 u³ + C
= 1/6 (2x + 1)³ + C

7. ∫ ∫ Your Turn (5x² + 1)² (10x) dx Let u = 5x² + 1 then du = 10x dx
So it becomes
u² du
= u² du = ⅓ u³ + C ∫
= ⅓ (5x² + 1)³ + C

8. ∫xcos(4x²) dx Example 2 = ∫cos(4x²) 8/8 xdx = 1/8∫cos(4x²) 8xdx
If we let u = 4x², then we get cos u and du = 8xdx
or du/8x = dx
we are missing an 8 from dx so we multiple by 1 (8/8)
= 1/8∫cos(4x²) 8xdx
= ∫ xcos(u) du/8x
= 1/8∫cos(u) du = 1/8 sin(u) + C
= 1/8 sin(4x²) + C

9. ∫ ∫ ∫ ∫ Example 3 e1/x x-2 dx = e1/x (-1/-1)x-2dx = - e1/x (-x-2)dx
If we let u = 1/x = x-1 then we have eu and du = -x-2dx
we are missing an -1 from dx so we multiple by 1 (-1/-1) ∫ 2 1
= - e1/x (-x-2)dx ∫
u=1/2
u=1
to save steps change integrand from x = to u=
= eu du
u = 1/x so at x = 1, u = 1
at x = 2, u = ½
= - [ eu ]
u=1/2
u=1
= - (e½ - e1) = (e1 - e½) =

10. U Substitution Technique
Recognize 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

11. ∫ ∫ Example Problems Find the derivative of each of the following:
0) (2x +3) cos(x² + 3x) dx ∫
Let u = x² + 3x
then du = 2x + 3 dx
So it becomes
cos u du
= cos u du = sin (u) + C ∫
= sin (x² + 3x) + C

12. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
2) (1 + 2x)4 (2)dx ∫
(x² - 1)3 (2x)dx ∫
Let u = 1 + 2x
then du = 2 dx
So it becomes
u4 du
Let u = x² - 1
then du = 2x dx
So it becomes
u3 du
= u4 du = 1/5 u5 + C ∫
= u3 du = ¼ u4 + C ∫
= 1/5 (1 + 2x)5 + C
= ¼ (x² - 1)4 + C

13. ∫ ∫ ∫ ∫ Example Problems cont
Find the derivative of each of the following:
4) √9 – x2 (-2x)dx ∫
0) x (x2 + 1)² dx ∫
Let u = 9 - x²
then du = -2x dx
So it becomes
u½ du
Let u = x² + 1
then du = 2x dx
So it becomes
½ u² du
= ½ u² du = 1/6 u³ + C ∫
= u½ du = ⅔ u3/2 + C ∫
= ⅔ (9 - x²)3/2 + C
= 1/6 (x² + 1)³ + C

14. ∫ ∫ ∫ ∫ 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

15. 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
We can change a definite integral into a u= problem instead of an x= problem
Homework:
Day One: pg : 1, 2, 6, 8, 13, 21,
Day Two: pg : 35, 42, 51, 58, 59, 76