Lesson 5-5a

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

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

6. Example 1 = ∫(2x + 1)² 2/2 dx ∫(2x + 1)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 ∫(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. Example 2 ∫xcos(4x²) dx = ∫cos(4x²) 8/8 xdx If we let u = 4x²,then we get cos uand du = 8xdx or du/8x = dx we are missing an 8 from dxso 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. u=1/2 2 2 2 ∫ ∫ ∫ ∫ u=1 1 1 1 Example 3 e1/x x-2 dx = e1/x (-1/-1)x-2dx If we let u = 1/x = x-1then we have euand du = -x-2dx we are missing an -1 from dxso we multiple by 1 (-1/-1) = - e1/x (-x-2)dx to save steps change integrand from x = to u= u = 1/x so at x = 1, u = 1 at x = 2, u = ½ = - eu du u=1/2 = - [ eu ] u=1 = - (e½ - e1) = (e1 - e½) = 1.0696

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

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

14. 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

15. 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