1. Lesson 7-1
Integration by Parts

2. Used to make integrals simpler
Integration by Parts
Derived from the Product Rule in Differentiation
D(uv) = v × du + u × dv
∫ d(uv) = ∫v du + ∫u dv
uv = ∫v du + ∫u dv
∫u dv = uv – ∫v du
Used to make integrals simpler

3. Integration by Parts Strategies
Select u so that taking its derivative makes a simpler function
Let dv be something that can be integrated
Use derivative to drive a polynomial function to zero
Reduce polynomials to get a u-substitution
Use derivative to get the original integral and the simplify using addition/subtraction

4. Walk through Example Solve:  x cos x dx
If we let u = x and dv = cos x dx,
then du = dx and v = sin x
∫u dv = uv – ∫v du
 x cos x dx = x sin x -  sin x dx
 x cos x dx = x sin x + cos x + C

5. ∫ x ex dx 7-1 Example 1 = x ex – ∫ ex dx = = x ex – ex + C
let u = x and dv = ex dx
du = dx and v = ex
= x ex – ex + C
= ex (x + 1) + C 5

6. ∫ x ln x dx 7-1 Example 2 = ½x² ln x - ∫ ½x² dx/x = ½x² ln x - ∫ ½x dx
let u = ln x and dv = x dx
du = dx/x and v = ½ x²
= ½x² ln x - ∫ ½x dx
= ½x² ln x - ¼x² + C
= ¼x² (2ln x – 1) + C 6

7. ∫ x sin 3x dx 7-1 Example 3 = -⅓x cos 3x - ∫ -⅓ cos 3x dx
let u = x and dv = sin 3x dx
du = dx and v = -⅓cos 3x
= -⅓x cos 3x +⅓ ∫ cos 3x dx
= -⅓x cos 3x +⅓(⅓ sin 3x) + C
= (1/9) (sin 3x – 3x cos 3x) + C 7

8. Summary & Homework Summary: Homework:
Integration by parts allows us to solve some previously unsolvable integrals
Methods:
Use derivative to drive a polynomial function to zero
Reduce polynomials to get a u-substitution
Use derivative to get the original integral and the simplify using addition/subtraction
Homework:
pg 480 – 482: Day 1: 3, 4, 7, 9, 36;

9. Integration by Parts – Repeated Use
Sometimes we have to use the method of integration by parts several times to get an integral that we can solve or to get it to repeat
Using a table to record our differentiations and integrations can help keep things straight

10. ∫ sin x ex dx 7-1 Example 4 = sin x ex – ∫ cos x ex dx =
Remember to keep the () in the problem!
let u = sin x and dv = ex dx
du = cos x dx and v = ex
= sin x ex – (cos x ex - ∫ - sin x ex dx ) =
let u = cos x and dv = ex dx
du = - sin x dx and v = ex
2 ∫ sin x ex dx = sin x ex – cos x ex
= ½ (sin x ex – cos x ex) + c 10

11. ∫ cos 2x ex dx = cos 2x ex + 2sin 2x ex - 4∫ cos 2x ex dx
7-1 Example 5
∫ cos 2x ex dx
= cos 2x ex – ∫ -2sin 2x ex dx =
Remember to keep the () in the problem!
let u = cos 2x and dv = ex dx
du = -2sin 2x dx and v = ex
= cos 2x ex + 2 (sin 2x ex - ∫ 2cos 2x ex dx )
let u = sin 2x and dv = ex dx
du = 2cos 2x dx and v = ex
∫ cos 2x ex dx = cos 2x ex + 2sin 2x ex - 4∫ cos 2x ex dx
5 ∫ cos 2x ex dx = cos 2x ex + 2sin 2x ex + C
= (ex/5) (cos 2x + 2sin 2x ) + C 11

12. ∫ x² sin x dx = -x² cos x + 2(x sin x – (-cos x)) + C
7-1 Example 6
∫ x² sin x dx
= -x² cos x – ∫ -cos x 2x dx =
Remember to keep the () in the problem!
let u = x² and dv = sin x dx
du = 2x dx and v = -cos x
= -x² cos x + 2(x sin x - ∫ sin x dx )
let u = x and dv = cos x dx
du = dx and v = sin x
∫ x² sin x dx = -x² cos x + 2(x sin x – (-cos x)) + C
= -x² cos x + 2x sin x + 2cos x + C
= (2 - x²) cos x + 2x sin x + C 12

13. 7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) ex dx Dif13

14. 7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) ex dx Dif36
∫ (6x³ + 3x² - 5x – 7) ex dx =
= ex (6x³ + 3x² - 5x - 7) – ex (18x² + 6x – 5) + ex (36x + 6) – ex (36) + C
= ex (6x³ - 15x² + 25x - 28) + C 14

15. Summary & Homework Summary: Homework:
Integration by parts allows us to solve some previously unsolvable integrals
Methods:
Use derivative to drive a polynomial function to zero
Reduce polynomials to get a u-substitution
Use derivative to get the original integral and the simplify using addition/subtraction
Homework:
pg 480 – 482: Day 1: 3, 4, 7, 9, 36; Day 2: 1, 14, 19, 51