 # Lesson 7-1 Integration by Parts.

## Presentation on theme: "Lesson 7-1 Integration by Parts."— Presentation transcript:

Lesson 7-1 Integration by Parts

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

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

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

∫ 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

∫ 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

∫ 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

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;

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

∫ 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

∫ 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

∫ 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

7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) ex dx Dif
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7-1 Example 7 Find the integral of ∫ (6x³ + 3x² - 5x – 7) ex dx Dif
36 ∫ (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

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