Respuestas Del Calculo De Larson

C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.1 Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8 Basic Integration Rules Integration by Parts . . . . . . . . . . . . . . . . . . . 50

. . . . . . . . . . . . . . . . . . . . . 55

Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 65 TrigonometricSubstitution . . . . . . . . . . . . . . . . . 74

Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . 84 Integration by Tables and Other Integration Techniques . . 90

Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . 96 Improper Integrals . . . . . . . . . . . . . . . . . . . . . 102

Review Exercises

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

C H A P T E R 7 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.1 Basic Integration Rules
Solutions to Odd-Numbered Exercises
1. (a) (b) (c) (d) d 2 x2 dx d dx d 1 dx 2 x2 x2 1 1 1 1 C C C C 2 1 2 x 2 1 1
1 2 1 2

2x x x2

2x x2 1 1 x 2 x2 1

1 2 x 2

2x
1 2

1 1 2 x 2 2 2x x2 11

2x

d ln x2 dx x x2 1

dx matches (b).

3. (a) (b) (c) (d)

d ln x2 dx d 2x dx x2 1 d arctan x dx d ln x2 dx 1 1 1

1

C C C C

1 2x 2 x2 1 x 1 1 x2 x2 2x 1
2

x x2 2
2

1 1 2x 21 x2 3x2 13

1

2

2

2x 2 x2 x 14

x2

dx matches (c).

5. u

3x 3x

2 4 dx 2, du 3 dx, n 4

7.

1 x1 u 1 2 x

dx 1 x

9. u Use

3 1 t, du t2

dt dt, a 1

2 x,du du . u

dx

Use

un du. Use

du a2 u2

11. u

t sin t 2 dt t 2, du 2t dt

13. u

cos xesin x dx sin x, du eu du. cos x dx

Use

sin u du.

Use

50

Section 7.1

Basic Integration Rules

51

15. Let u 2x

2x 5

5, du
3 2

2 dx. 1 2 1 5 2x 2x 5 5
3 2

17. Let u 5 2 dx C z 4

z
5

4, du dz 5 z

dz 4 5
5

dx

dx C

5

z

4 4

4

C5 2

4z

4

4

19. Let u t2 3 t3

t3

1, du 1 dt 1 3 1 3 t3

3t2 dt. t3 t3 1 1 4 3 1 4
1 3

21. 3t2 dt C C

v

1 3v 1
3

dv

v dv 1 2 v 2

1 3 1 6 3v

3v

1

3

3 dv

1

2

C

4 3

4 3

23. Let u

t3 t2 3 t 3 9t

9t 1

1, du dt 1 3

3t2

9 dt

3 t2

3 dt. 1 ln 3 t3 9t 1 C

3 t2 3 dt t 3 9t 1

25.

x2 x 1

dx

x 1 2 x 2

1 dxx ln x

1 x 1 1

dx C

27. Let u 1

1 ex ex

ex, du dx

ex dx. ex C

ln 1

29.

1

2x2 2 dx

4x 4

4x2

1 dx

4 5 x 5

4 3 x 3

x

C

x 12x 4 15

20x2

15

C

31. Let u

2 x2, du x cos 2 x2 dx

4 x dx. 1 4 cos 2 x2 4 x dx C

1 sin 2 x 2 4 33. Let u x, du csc x cot dx. x dx 1 csc

x cot

x

dx

1

csc x

C

35. Let u

5x, du e5x dx5 dx. 1 5x e 5 dx 5 1 5x e 5 C

37. Let u

1 2 e
x

e x, du dx 2 2

e x dx. 1 e 1
x

1

1 ex dx

ex dx ex 2 ln 1 ex C

ex

52

Chapter 7 ln x2 dx x 1

Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2

39.

2

ln x

1 dx x

2

ln x 2

C

ln x

2

C

41.

sin x dx cos x 1 cos 1

sec x

tan x dx

ln sec x

tan x

ln secx

C

ln sec x sec x

tan x

C

43.

1 cos csc 1

cos cos

1 1 csc2 csc2 C C

cos cos2

1 1

cos 1 sin2

cot

1 cos 1

d csc 1 sin 1

csc cot cot cos sin C

d

cos sin

45.

3z z2

2 dz 9

3 2z dz 2 z2 9 3 ln z2 2 9

2

dz z2 9 C

47. Let u

2t

1, du 1 2t
2

2 dt. dt 1 2 2 2t 1
2

2 z arctan 3 3

1

1

1

1

dt C

1 arcsin 2t 22 , du t 2 sin 2 t dt. t2 1 1 2 cos 2 t 1 2 ln cos 2 t 3 6x 1 x 2 sin 2 t t2 C dt

49. Let u

cos

tan 2 t dt t2

51.

x2 4 4x t 1

dx

3

9

3 1 1 2

2

dx

3 arcsin

x 3

3

C

53.

4x2 ds dt (a)

65

dx

x 1 2

2

16

dx

1 x arctan 4

1 2 4

C

1 2x 1 arctan 4 8

C

55.

t4
1

, 0,
s

(b) u

t 2, du t 1

2t dt t4 1 2 dt 1...