Deber1 Integrales Por Sustitucion Trigonometrica Stewart

7/31/2019 Deber1 Integrales Por Sustitucion Trigonometrica Stewart

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430 |||| Evaluate the integral.

4.

5. 6.

8.

9. 10. y t5

st2 2dty dxsx 2 16

ysx2 a 2

x 4dxy 1

x 2s25 x 2dx7.

y20x 3sx 2 4 dxy2

s2

1

t3st2 1dt

y2s30

x 3

s16 x 2dx

13 |||| Evaluate the integral using the indicated trigonometric

substitution. Sketch and label the associated right triangle.

1. ;

2. ;

;

s s s s s s s s s s s s

x 3 tan y x3

sx 2 9dx3.

x 3 sin yx 3s9 x 2dx

x

3 sec y1

x 2sx 2 9 dx

|||| 7.3 Exercises

Therefore

EXAMPLE 7 Evaluate .

SOLUTION We can transform the integrand into a function for which trigonometric substi-

tution is appropriate by first completing the square under the root sign:

This suggests that we make the substitution . Then and , so

We now substitute , giving and , so

s3 2x x 2 sin1x 12 C

s4 u 2 sin1

u

2

C

2 cos C

y2 sin 1d

y xs3 2x x 2 dx y2 sin 1

2 cos 2 cos d

s4 u 2 2 cos du 2 cos du 2 sin

y xs3 2x x 2 dx yu 1s4 u 2

du

x u 1du dxu x 1

4 x 12

3 2x x 2 3 x 2 2x 3 1 x 2 2x 1

y xs3 2x x 2 dx

316

u

1

u112

316 [(

12 2) 1 1]

332

y3s320

x 3

4x 2 932dx

3

16y121

1 u 2

u 2du

3

16y121

1 u2du

|||| Figure 5 shows the graphs of the integrand

in Example 7 and its indefinite integral (with

). Which is which?C 0

_4

_5

3

2

F IGURE 5

Resolver los ejercicios impares


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equation . Then is the sum of the area of the

triangle and the area of the region in the figure.]

; 36. Evaluate the integral

Graph the integrand and its indefinite integral on the same

screen and check that your answer is reasonable.

; 37. Use a graph to approximate the roots of the equation. Then approximate the area bounded by

the curve and the line .

38. A charged rod of length produces an electric field at point

given by

where is the charge density per unit length on the rod and

is the free space permittivity (see the figure). Evaluate the inte-

gral to determine an expression for the electric field .

39. Find the area of the crescent-shaped region (called a lune)

bounded by arcs of circles with radii and . (See the figure.)

40. A water storage tank has the shape of a cylinder with diameter

10 ft. It is mounted so that the circular cross-sections are verti-

cal. If the depth of the water is 7 ft, what percentage of the

total capacity is being used?

41. A torus is generated by rotating the circle

about the -axis. Find the volume enclosed by the torus.x

x 2 y R2 r2

R

r

Rr

0 x

y

L

P(a, b)

EP

0

EP yLaa

b

40x2 b 2 32

dx

Pa, bL

y 2 xy x 2s4 x 2x 2s4 x 2 2 x

y dxx 4sx 2 2

O x

y

RQ

P

PQRPOQ

Ax 2 y 2 r211. 12.

14.

15. 16.

18.

19. 20.

21.

23. 24.

25. 26.

27. 28.

29. 30.

(a) Use trigonometric substitution to show that

(b) Use the hyperbolic substitution to show that

These formulas are connected by Formula 3.9.3.

32. Evaluate

(a) by trigonometric substitution.

(b) by the hyperbolic substitution .

33. Find the average value of , .

34. Find the area of the region bounded by the hyperbola

and the line .

35. Prove the formula for the area of a sector of a circle

with radius and central angle . [Hint: Assume

and place the center of the circle at the origin so it has the

0 2rA

12 r

2

x 39x 2 4y 2 36

1 x 7fx sx 2 1x

x a sinh t

y x2

x 2 a 232dx

y dxsx 2 a 2 sinh1x

a C

x a sinh t

y dxsx 2 a 2 ln(x sx2 a 2) C

31.

s s s s s s s s s s s s

y20

cos t

s1 sin2tdtyxs1 x 4dx

y dx5 4x x 252

y dxx 2 2x 22

yx 2

s4x x 2 dxy1

s9x 2 6x 8 dx

y dtst2 6t 13ys5 4x x2dx

y10sx 2 1 dx22.y23

0x 3s4 9x 2dx

y ts25 t2 dtys1 x 2

xdx

y dxax2 b 232

y xsx 2 7 dx17.

y dxx 2s16x 2 9y x2

a 2 x 232dx

y duus5 u 2y

sx 2 9x 3

dx13.

y10xsx 2 4 dxys1 4x 2dx


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Chapter 7

Exercises 7.1 s page 480

1. 3.

5.

7.

9.

11.

13.

15.

17.

19. 21. 23.

25. 27.

29. 31.

33. 35.

37.

39.

41. (b)

43. (b) 49.

51. 53.

55. 57.

59. 61. 63. 2


1. 3.

5. 7. 9.

11. 13.

15.17. 19.

21. 23.

25. 27.

29.

31.

33. 35. s3 316 tan6

1

4 tan4 C

1

4 sec4x tan2x ln secx C

1

3 sec3x sec x C

117

8

1

5 tan5t

2

3 tan3t tan t C

tanx x C1

2tan2x C

ln1 sinx C1

2 cos2x ln cos x C

(2

7 cos3

x 2

3 cos x)scos x C

3 41923

2 2 sin 1

4 sin 2 C

3841

5 sin5x

2

7 sin7x

1

9 sin9x C

11

384

1

5 cos5x

1

3 cos3x C

2 ett2 2t 2 m9

2 ln 3 13

9

2e4 8

1.0475, 2.8731; 2.182825

4 75

4 e2

x lnx3 3lnx2 6 lnx 6 C2

3,8

15

1

4 cos x sin3x

3

8x 3

16 sin 2x C

7

1

3.5 1.5

F

2x 1ex C

1.2

1.2

2 2

F

x sin x cos x2 C

1

2 42(sin sx sx cos sx ) C

32

5 ln 22

64

25 ln 2 62

125

1

2x cos lnx sin ln x C

sinxln sin x 1 C( 6 3s3 )6

1

4 3

4e21

2 1

2 ln 23

y coshy sinhy C

1

13e22 sin 3 3 cos 3 C

x lnx2 2x ln x 2x C

t arctan 4t1

8 ln1 16t2 C

1

22x 1 ln2x 1 x C

1

x 2 cos x

2

2x sin x

2

3

cos x C

2r 2er2 C

1

5

x sin 5x 1

25

cos 5x C1

2

x 2 lnx 1

4

x 2 C

37. 39.

41. 43.

45. 47.

49.

51.

53. 0 55. 57. 0 59. 61.

63.


1. 3.

5. 7.

9. 11.

13.

15. 17.

19. 21.

23.25.

27.

29.

33. 37. 0.81, 2; 2.10

39. 41.


1. (a) (b)

3. (a) (b)

5. (a)

(b)

7.

9. 11.

13. 15.

17.27

5 ln 2 9

5 ln 3 (or9

5 ln8

3)

2 ln 2 1

2a ln x b C

1

2 ln3

22 ln x 5 ln x 2 Cx 6 lnx 6 C

At B

t2 1

Ct D

t2 4

Et F

t2 42

1 A

x 1 B

x 1 Cx D

x 2 1

A

x 1

Bx C

x 2 x 1

A

x 4

B

x 1

A

x

B

x 1

C

x 12A

x 3

B

3x 1

22Rr2rsR 2 r2 r22 R 2 arcsinrR

1

6 (s48 sec1 7)

1

4 sin1x 2

1

4x2s1 x 4 C

1

2 tan1x 1 x 1x 2 2x 2 C

1

3 ln 3x 1 s9x2 6x 8 C

9

2 sin

1

x

23

1

2x

2s5

4xx

2 C

64

1215ln (s1 x 2 1)x s1 x 2 Csx 2 7 C(xsa 2 x 2 ) sin1xa C

1

6 sec1x3 sx 2 92x 2 C

1

4 sin12x

1

2xs1 4x 2 Cln(sx 2 16 x) C

s25 x 225x C24 s38 14

1

3 x2 18sx 2 9 Csx 2 99x C

s 1 cos3t3

2 24241

3

1

1

_2 2

F

1

6 sin 3x 1

18 sin 9x C

1.1

1.1

_2 2

F

1

5 cos5x

2

3 cos3x cos x C

1

10 tan5t2 C

1

2 sin 2x C

1

4 sin 21

24 sin 12 C1

6 sin 3x 1

14 sin 7x C

ln cscx cotx C1

3 csc3

1

5csc5 C

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Deber1 Integrales Por Sustitucion Trigonometrica Stewart