Lesson 5-5a

U-Substitution or The Chain Rule of Integration

Quiz• Homework Problem:

(3ex + 7sec2x) dx

• Reading questions: Fill in the squares below

e7x dx =

U substitution: u = ______

∫

∫ = 3ex + 7tan x + C

sec² (3x) dx =

U substitution: u = ______

∫

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)

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

Example 0

∫ 2cos (2x) dx

= ∫ cos u du = sin u + C

= sin (2x) + C

= ∫ cos (2x) 2 dx

If we let u = 2xthen du = 2dx

or ½du = dx

= ∫ 2cos (u) ½du

To get du we need to move2 back to the dx

2 and ½ cancel out

Example 1

∫(2x + 1)² dx = ∫(2x + 1)² 2/2 dx

= ½ ∫u² du = 1/6 u³ + C

= 1/6 (2x + 1)³ + C

= ½ ∫(2x + 1)² 2 dx

If we let u = 2x +1then it becomes u² and du = 2dx

or ½du = dx

we are missing a 2 from dxso we multiple by 1 (2/2)

½ goes outside ∫ and 2 stays with dx = ∫ (u)² ½du

Your Turn

1) (5x² + 1)² (10x) dx∫Let u = 5x² + 1then du = 10x dx

So it becomesu² du

= ⅓ (5x² + 1)³ + C

= u² du = ⅓ u³ + C∫

Example 2

∫xcos(4x²) dx = ∫cos(4x²) 8/8 xdx

= 1/8∫cos(u) du = 1/8 sin(u) + C

= 1/8 sin(4x²) + C

= 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 dxso we multiple by 1 (8/8)

= ∫ xcos(u) du/8x

Example 3

e1/x x-2 dx = e1/x (-1/-1)x-2dx

= - eu du

= - [ eu ]

= - 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 dxso we multiple by 1 (-1/-1)

to save steps change integrand from x = to u=

= - (e½ - e1) = (e1 - e½) = 1.0696

∫2

1

∫2

1

∫2

1

∫u=1/2

u=1

u = 1/x so at x = 1, u = 1 at x = 2, u = ½

u=1/2

u=1

U Substitution Technique

1. Recognize that the integral in its present form is one that we cannot evaluate!

2. See if changing the variable by letting u = g(x) (applying the “anti-chain rule”) will yield an integral that we can evaluate.

3. 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.

f(x) dx = k g(u) du or k g(u) du ∫b

a∫

x=b

x=a∫

u=d

u=c

Example Problems

Find the derivative of each of the following:

0) (2x +3) cos(x² + 3x) dx∫Let u = x² + 3xthen du = 2x + 3 dx

So it becomescos u du

= sin (x² + 3x) + C

= cos u du = sin (u) + C∫

Example Problems cont

Find the derivative of each of the following:

2) (1 + 2x)4 (2)dx∫ 3) (x² - 1)3 (2x)dx∫Let u = 1 + 2xthen du = 2 dx

So it becomes u4 du

= 1/5 (1 + 2x)5 + C

= u4 du = 1/5 u5 + C∫

Let u = x² - 1then du = 2x dx

So it becomesu3 du

= ¼ (x² - 1)4 + C

= u3 du = ¼ u4 + C∫

Example Problems cont

Find the derivative of each of the following:

0) x (x2 + 1)² dx∫4) √9 – x2 (-2x)dx∫Let u = 9 - x²then du = -2x dx

So it becomesu½ du

= ⅔ (9 - x²)3/2 + C

= u½ du = ⅔ u3/2 + C∫

Let u = x² + 1then du = 2x dx

So it becomes½ u² du

= 1/6 (x² + 1)³ + C

= ½ u² du = 1/6 u³ + C∫

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 + 1then du = 3x² dx

So it becomes⅓ u½ du

= 2/9 (x3 + 1)3/2 + C

= ⅓ u½ du = 2/9 u3/2 + C∫

Let u = 2xthen du = 2 dx

So it becomes½ sec u tan u du

= ½ sec (2x) + C

= ½ sec u tan u du

= ½ sec u + C

∫

Summary & Homework• Summary:

– 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 420 - 422: 1, 2, 6, 8, 13, 21, – Day Two: pg 420 - 422: 35, 42, 51, 58, 59, 76