MAT 271 Recitation. MAT 271 Recitation. Sections 7.1 and 7.2. Lindsey K. Gamard, ASU SoMSS. 30 August 2013

Transcription

1 MAT 271 Recitation Sections 7.1 and 7.2 Lindsey K. Gamard, ASU SoMSS 30 August 2013

2 Agenda Today s agenda: 1. Review 2. Review Section 7.2 (Trigonometric Integrals) 3. (If time) Start homework in pairs

3 Theorems Integration by Parts for Indefinite Integrals Suppose that u and v are differentiable functions. Then, u dv = uv v du.

4 Theorems Integration by Parts for Definite Integrals Let u and v be differentiable. Then, b a u(x)v (x) dx = u(x)v(x) b a b v(x)u (x) dx. a

5 Discussion Questions 1. Q: On which derivative rule is integration by parts based?

6 Discussion Questions 1. Q: On which derivative rule is integration by parts based? A: The Product Rule.

7 Discussion Questions 1. Q: On which derivative rule is integration by parts based? A: The Product Rule. 2. Q: For what type of integrand is integration by parts useful?

8 Discussion Questions 1. Q: On which derivative rule is integration by parts based? A: The Product Rule. 2. Q: For what type of integrand is integration by parts useful? A: When the integrand is a product of functions, and no easier method is available.

9 Discussion Questions 1. Q: On which derivative rule is integration by parts based? A: The Product Rule. 2. Q: For what type of integrand is integration by parts useful? A: When the integrand is a product of functions, and no easier method is available. 3. Q: How would you choose u and dv to simplify x 4 e 2x dx?

10 Discussion Questions 1. Q: On which derivative rule is integration by parts based? A: The Product Rule. 2. Q: For what type of integrand is integration by parts useful? A: When the integrand is a product of functions, and no easier method is available. 3. Q: How would you choose u and dv to simplify x 4 e 2x dx? A: Either way would really be fine, but in general, we want to pick u to be something easy to differentiate, and dv to be something easy to integrate. In this particular case, it might be slightly easier to pick u = x 4 and dv = e 2x dx.

11 xe x dx

12 xe x dx Choose u = x and dv = e x dx.

13 xe x dx Choose u = x and dv = e x dx. Then du = 1 dx = dx and v = e x.

14 xe x dx Choose u = x and dv = e x dx. Then du = 1 dx = dx and v = e x. Therefore, xe x dx = xe x e x dx = xe x e x + C = e x (x 1) + C.

15 π 0 x sin x dx

16 π 0 x sin x dx Choose u = x and dv = sin x dx.

17 π 0 x sin x dx Choose u = x and dv = sin x dx. Then du = 1 dx = dx and v = cos x.

18 π 0 x sin x dx Choose u = x and dv = sin x dx. Then du = 1 dx = dx and v = cos x. Therefore, π 0 x sin x dx = x cos x π 0 π 0 cos x dx = ( π cos π + 0 cos 0) + sin x = π + (sin π sin 0) = π. π 0

19 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx?

20 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx? A: One would compute this integral by writing cos 2 x sin 3 x as cos 2 x sin 2 x sin x = cos 2 x(1 cos 2 x) sin x and then performing the substitution u = cos x.

21 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx? A: One would compute this integral by writing cos 2 x sin 3 x as cos 2 x sin 2 x sin x = cos 2 x(1 cos 2 x) sin x and then performing the substitution u = cos x. 2. Q: How would you evaluate tan 10 x sec 2 x dx?

22 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx? A: One would compute this integral by writing cos 2 x sin 3 x as cos 2 x sin 2 x sin x = cos 2 x(1 cos 2 x) sin x and then performing the substitution u = cos x. 2. Q: How would you evaluate tan 10 x sec 2 x dx? A: One would compute this integral by letting u = tan x, so that du = sec 2 x dx. This substitution leads to the integral u 10 du, which can easily be evaluated.

23 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx? A: One would compute this integral by writing cos 2 x sin 3 x as cos 2 x sin 2 x sin x = cos 2 x(1 cos 2 x) sin x and then performing the substitution u = cos x. 2. Q: How would you evaluate tan 10 x sec 2 x dx? A: One would compute this integral by letting u = tan x, so that du = sec 2 x dx. This substitution leads to the integral u 10 du, which can easily be evaluated. 3. Q: How would you evaluate sec 12 x tan x dx?

24 Section 7.2 (Trigonometric Integrals) 1. Q: How would you evaluate cos 2 x sin 3 x dx? A: One would compute this integral by writing cos 2 x sin 3 x as cos 2 x sin 2 x sin x = cos 2 x(1 cos 2 x) sin x and then performing the substitution u = cos x. 2. Q: How would you evaluate tan 10 x sec 2 x dx? A: One would compute this integral by letting u = tan x, so that du = sec 2 x dx. This substitution leads to the integral u 10 du, which can easily be evaluated. 3. Q: How would you evaluate sec 12 x tan x dx? A: One would compute this integral by letting u = sec x, so that du = sec x tan x dx. This substitution leads to the integral u 11 du, which can easily be evaluated.

25 Section 7.2 (Trigonometric Integrals) Evaluate the integral sin 2 x cos 2 x dx. ( ) ( ) 1 cos 2x 1 + cos 2x sin 2 x cos 2 x dx = 2 2 = 1 1 cos 2 2x dx 4 = cos 4x 1 dx 4 2 = cos 4x dx 2 = 1 ( ) x sin 4x + C dx

26 Section 7.2 (Trigonometric Integrals) Homework Evaluate the following integrals: e x cos x dx e2 1 x 2 ln x dx.