Lesson 7 1 Integration by Parts Integration by

Lesson 7 -1 Integration by Parts

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 = 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 1) Use derivative to drive a polynomial function to zero 2) Reduce polynomials to get a u-substitution 3) 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

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

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

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

Summary & Homework • Summary: – 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

7 -1 Example 4 ∫ sin x e x dx ∫ = 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 e x dx = sin x ex – cos x ex = ½ (sin x ex – cos x ex) + c

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

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

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

7 -1 Example 7 Find the integral of ∫ (6 x³ + 3 x² - 5 x – 7) ex dx Dif 6 x³ + 3 x² - 5 x - 7 18 x² + 6 x - 5 36 x + 6 + + Int ex ex ex 36 0 + ex ex ∫ (6 x³ + 3 x² - 5 x – 7) ex dx = = ex (6 x³ + 3 x² - 5 x - 7) – ex (18 x² + 6 x – 5) + ex (36 x + 6) – ex (36) + C = ex (6 x³ - 15 x² + 25 x - 28) + C

Summary & Homework • Summary: – 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