Lesson 5-5b

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

2. Practice Quiz 0 x=0 • Homework Problem: (2x - ex) dx • Reading question: Fill in the squares below ∫ = x² - ex + C | = (0-1) – (1- 1/e) = -2 + 1/e = -1.632 x= -1 -1 u= █ 2 x=1 ∫ ∫ f(x) dx = g(u) du With u = x² + 1 x=0 u= █ 1

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 x=b u=d ∫ ∫ ∫ a x=a u=c f(x) dx = k g(u) du or k g(u) du

6. Example Problems cont ∫ 1) x2√x3 + 1 dx Find the derivative of each of the following: ∫ 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. ∫ 3) x3√ x4 + 2 dx Example Problems Find the derivative of each of the following: ∫ (1 + 1/t) t-2dt 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 6) ------------dx  1 – 9x² ∫ • 1 • ------------- dx • 1 + (2x)² ∫ Let u = 3x then du = 3 dx So it becomes du / (1 – u²)½ Let u = 2x then du = 2 dx So it becomes du / (1 + u²) 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. ∫ 8) sin (√x) dx / √x Example Problems cont ∫ 7) x sin (x²) dx Find the derivative of each of the following: 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 = -2 cos (x) + C = -½ cos (x²) + C

10. Example Problems cont ∫ 9) xsin³(x²) cos(x²) dx Find the derivative of each of the following: ∫ 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 ∫ 11) tan x dx Find the derivative of each of the following: ∫ 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 ∫ 13) sec x dx Find the derivative of each of the following: ∫ 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: • 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 420 - 422: 1, 2, 6, 8, 13, 21 • Day Two: pg 420 - 422: 35, 42, 51, 58, 59, 76