n sin nx + n 2 cos nx - = 1 p a3p2 cos np n 2 = 3 p ap2 n 2 (-1) n + 2(-1) n n 4 b. cos np = (-1) n L x sec x2 dx

p L 1

p 0

x³

cos nx dx

= 1 p c

x³

n sin nx +

3x

³

n²

cos nx - 6x

n³

sin nx - 6

n⁴

cos nx d

^p

0

= 1

p a 3 p

²

cos n p

n²

- 6 cos n p

n⁴

+ 6

n⁴

b

= 3

p a p

²n²

( -1)

ⁿ

+ 2(-1)

^{n+ 1}

+ 2

n⁴

b.

cos np = (-1)ⁿ

Integrals like those in Example 8 occur frequently in electrical engineering.

Integration by Parts

Evaluate the integrals in Exercises 1–24 using integration by parts.

1. Lx sinx

2dx 2.

Lu cos pu du

3. Lt²cos t dt 4.

Lx²sin x dx 5. L

2 1

x ln x dx 6.

L

e 1

x³ln x dx

7. Lxe^xdx 8.

Lxe^3xdx

9. Lx²e^-xdx 10.

L(x²- 2x + 1)e^2xdx

11. Ltan^-1y dy 12.

Lsin^-1y dy

13. Lx sec²x dx 14.

L4x sec²2x dx

15. Lx³e^xdx 16.

Lp⁴e^-pdp

17. L(x² - 5x)e^xdx 18.

L(r² + r + 1)e^rdr

19. Lx⁵e^xdx 20.

Lt²e^4tdt

21. Le^usinu du 22.

Le^-ycos y dy 23. Le^2xcos 3x dx 24.

Le^-2xsin 2x dx Using Substitution

Evaluate the integrals in Exercise 25–30 by using a substitution prior to integration by parts.

25. Le^23s+9ds 26.

L

1 0

x21- x dx

27. L

p>3 0

x tan²x dx 28.

Lln (x + x²) dx

29. Lsin (ln x) dx 30.

Lz(ln z)²dz Evaluating Integrals

Evaluate the integrals in Exercises 31–52. Some integrals do not require integration by parts.

31. Lx sec x²dx 32.

L cos2x

2x dx 33. Lx (ln x)²dx 34.

L 1 x (ln x)²dx

35. L ln x

x² dx 36.

L (ln x)³

x dx

37.

Lx³e^x4dx

38.

Lx⁵e^x3dx

39.

Lx³2x²+ 1 dx

40.

Lx²sin x³dx

41.

Lsin 3x cos 2x dx

42.

Lsin 2x cos 4x dx

43.

L²x ln x dx

44.

L e^2x 2xdx

45.

Lcos2x dx 46.

L²x e^2xdx

47. L

p>2 0

u²sin 2u du 48.

L

p>2 0

x³cos 2x dx

49. L

2 2>23

t sec^-1t dt 50.

L

1>22 0

2x sin^-1

(

x²

)

dx

51. Lx tan^-1x dx 52.

Lx²tan^-1x 2dx

Exercises 8.2

468

Chapter 8: Techniques of Integration

58. Finding volume Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y = x sin x, 0 … x … p, about

a. the y-axis.

b. the line x = p.

(See Exercise 53 for a graph.)

59. Consider the region bounded by the graphs of y = ln x, y = 0, andx = e.

a. Find the area of the region.

b. Find the volume of the solid formed by revolving this region about the x-axis.

c. Find the volume of the solid formed by revolving this region about the line x = -2.

d. Find the centroid of the region.

60. Consider the region bounded by the graphs of y = tan^-1x, y = 0, andx = 1.

a. Find the area of the region.

b. Find the volume of the solid formed by revolving this region about the y-axis.

61. Average value A retarding force, symbolized by the dashpot in the accompanying figure, slows the motion of the weighted spring so that the mass’s position at time t is

y = 2e^-tcos t, t Ú 0.

Find the average value of y over the interval 0 … t … 2p.

0 y Mass

Dashpot y

62. Average value In a mass-spring-dashpot system like the one in Exercise 61, the mass’s position at time t is

y = 4e^-t(sin t - cos t), t Ú 0.

Find the average value of y over the interval 0 … t … 2p.

Reduction Formulas

In Exercises 63–67, use integration by parts to establish the reduction formula.

63. Lxⁿcos x dx = xⁿsin x - n

Lx^n-1sin x dx 64. Lxⁿsin x dx = -xⁿcos x + n

Lx^n-1cos x dx Theory and Examples

53. Finding area Find the area of the region enclosed by the curve y = x sin x and the x-axis (see the accompanying figure) for a. 0 … x … p.

b. p … x … 2p.

c. 2p … x … 3p.

d. What pattern do you see here? What is the area between the curve and the x-axis for np … x … (n + 1)p, n an arbitrary nonnegative integer? Give reasons for your answer.

x y

0 p 2p

5

y= x sin x 10

−5

3p

54. Finding area Find the area of the region enclosed by the curve y = x cos x and the x-axis (see the accompanying figure) for a. p>2 … x … 3p>2.

b. 3p>2 … x … 5p>2.

c. 5p>2 … x … 7p>2.

d. What pattern do you see? What is the area between the curve and the x-axis for

a2n - 12 bp … x … a2n + 12 bp,

n an arbitrary positive integer? Give reasons for your answer.

0 10

−10

y= x cos x

x y

p

2 7p

2 5p

2 3p

2

55. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordi- nate axes, the curve y = e^x, and the line x = ln 2 about the line x = ln 2.

56. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordi- nate axes, the curve y = e^-x, and the line x = 1

a. about the y-axis.

b. about the line x = 1.

57. Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordi- nate axes and the curve y = cos x, 0 … x … p>2, about a. the y-axis.

b. the line x = p>2.

For the integral of cos^-1x we get

Lcos ^-1x dx = x cos^-1x -

Lcos y dy y = cos^-1x

= x cos^-1x - sin y + C

= x cos^-1x - sin (cos^-1x) + C.

Use the formula

Lƒ^-1(x) dx = xƒ^-1(x) -

Lƒ(y) dy y = ƒ^-1(x) (4) to evaluate the integrals in Exercises 71–74. Express your answers in terms of x.

71. Lsin^-1x dx 72.

Ltan^-1x dx

73. Lsec^-1x dx 74.

Llog₂x dx

Another way to integrate ƒ^-1(x) (when ƒ^-1 is integrable, of course) is to use integration by parts with u = ƒ^-1(x) and dy = dx to rewrite the integral of ƒ^-1 as

Lƒ^-1(x) dx = xƒ^-1(x) -

Lxa ddx ƒ^-1(x)b dx. (5) Exercises 75 and 76 compare the results of using Equations (4) and (5).

75. Equations (4) and (5) give different formulas for the integral of cos^-1x:

a. Lcos^-1x dx = x cos^-1x - sin (cos^-1x) + C ^{Eq. (4)}

b. Lcos^-1x dx = x cos^-1x - 21 - x²+ C ^{Eq. (5)} Can both integrations be correct? Explain.

76. Equations (4) and (5) lead to different formulas for the integral of tan^-1x:

a. Ltan^-1x dx = x tan^-1x - ln sec (tan^-1x) + C ^{Eq. (4)}

b. Ltan^-1x dx = x tan^-1x - ln 21 + x² + C ^{Eq. (5)} Can both integrations be correct? Explain.

Evaluate the integrals in Exercises 77 and 78 with (a) Eq. (4) and (b) Eq. (5). In each case, check your work by differentiating your answer with respect to x.

77. L sinh^-1x dx 78.

Ltanh^-1x dx 65. Lxⁿe^axdx = xⁿe^ax

a - n

aLx^n-1e^axdx, a ≠ 0 66. L(ln x)ⁿdx = x(ln x)ⁿ- n

L(ln x)^n-1dx 67. Lx^m(ln x)ⁿdx= x^m+1

m+ 1 (ln x)ⁿ- n m+ 1

#

Lx^m(ln x)^n-1dx, m≠ -1 68. Use Example 5 to show that

L

p>2 0

sinⁿx dx = L

p>2 0

cosⁿx dx

= μap2 b

1

#

3

#

5g(n - 1) 2

#

4

#

6gn , n even 2

#

4

#

6g(n - 1)

1

#

3

#

5gn , n odd 69. Show that

L

b a

aL

b x

ƒ(t) dtb dx = L

b a

(x - a)ƒ(x) dx.

70. Use integration by parts to obtain the formula

L²1 - x²dx = 1

2x21 - x² + 1 2L

1 21 - x²dx.

Integrating Inverses of Functions

Integration by parts leads to a rule for integrating inverses that usually gives good results:

Lƒ^-1(x) dx =

Lyƒ′(y)dy _dx^{y = ƒ}= ƒ′( y) dy^{-1(x), x = ƒ( y)}

= yƒ(y) -

Lƒ(y) dy Integration by parts with u = y, dy = ƒ′( y) dy

= xƒ^-1(x) -

Lƒ(y) dy

The idea is to take the most complicated part of the integral, in this case ƒ^-1(x), and simplify it first. For the integral of ln x, we get

Lln x dx =

Lye^ydy ^y = ln x, x = e^y dx = e^ydy

= ye^y - e^y + C

= x ln x - x + C.

8.3 Trigonometric Integrals

Trigonometric integrals involve algebraic combinations of the six basic trigonometric functions. In principle, we can always express such integrals in terms of sines and cosines, but it is often simpler to work with other functions, as in the integral

L sec

²x dx

= tan x + C.

474

Chapter 8: Techniques of Integration

These identities come from the angle sum formulas for the sine and cosine functions (Section 1.3). They give functions whose antiderivatives are easily found.

EXAMPLE 8 Evaluate

L sin 3x cos 5x dx.

Solution

From Equation (4) with m = 3 and n = 5, we get

L sin 3x cos 5x dx = 1

2L 3sin(-2x) + sin 8x4 dx

= 1

2L (sin 8x - sin 2x)dx

= - cos 8x

16 + cos 2x 4 + C.

Powers of Sines and Cosines

Evaluate the integrals in Exercises 1–22.

1. Lcos 2x dx 2.

L

p 0

3 sin x 3dx 3. Lcos³x sin x dx 4.

Lsin⁴2x cos 2x dx

5. Lsin³x dx 6.

Lcos³4x dx

7. Lsin⁵x dx 8.

L

p 0

sin⁵x 2dx

9. Lcos³x dx 10.

L

p>6 0

3 cos⁵3x dx

11. Lsin³x cos³x dx 12.

Lcos³2x sin⁵2x dx

13. Lcos²x dx 14.

L

p>2 0

sin²x dx

15. L

p>2 0

sin⁷y dy 16.

L7 cos⁷t dt 17. L

p 0

8 sin⁴x dx 18.

L8 cos⁴2px dx 19. L16 sin²x cos²x dx 20.

L

p 0

8 sin⁴y cos²y dy

21. L8 cos³2u sin 2u du 22.

L

p>2 0

sin²2u cos³2u du

Integrating Square Roots

Evaluate the integrals in Exercises 23–32.

23. L

2p

0 A

1 - cos x

2 dx 24.

L

p 0

21 - cos 2x dx

25. L

p 0

21 - sin²t dt 26.

L

p 0

21 - cos²u du

27. L

p>2 p>3

sin²x

21 - cos xdx 28.

L

p/6 0

21 + sin x dx

aHint: Multiply by B

1 - sin x 1 - sin x.b 29. L

p 5p>6

cos⁴x

21 - sin xdx 30.

L

3p>4

p>2 21 - sin 2x dx 31. L

p>2 0

u21 - cos 2u du 32.

L

p -p

(1 - cos²t)^3>2dt

Powers of Tangents and Secants Evaluate the integrals in Exercises 33–50.

33. Lsec²x tan x dx 34.

Lsec x tan²x dx 35. Lsec³x tan x dx 36.

Lsec³x tan³x dx 37. Lsec²x tan²x dx 38.

Lsec⁴x tan²x dx 39. L

0 -p>3

2 sec³x dx 40.

Le^xsec³e^xdx

41. Lsec⁴u du 42.

L3 sec⁴3x dx 43. L

p>2

p>4 csc⁴u du 44.

Lsec⁶x dx

45. L4 tan³x dx 46.

L

p>4

-p>46 tan⁴x dx

47. Ltan⁵x dx 48.

Lcot⁶2x dx 49. L

p>3

p>6 cot³x dx 50.

L8 cot⁴t dt

Exercises 8.3

8.4 Trigonometric Substitutions

Trigonometric substitutions occur when we replace the variable of integration by a trigo- nometric function. The most common substitutions are x = a tan u, x = a sin u, and

x

= a sec u. These substitutions are effective in transforming integrals involving

2a²

+ x

²

,

₂a²

- x

²

, and

₂x²

- a

²

into integrals we can evaluate directly since they come from the reference right triangles in Figure 8.2.

Products of Sines and Cosines

Evaluate the integrals in Exercises 51–56.

51. Lsin 3x cos 2x dx 52.

Lsin 2x cos 3x dx 53. L

p

-psin 3x sin 3x dx 54.

L

p>2 0

sin x cos x dx

55. Lcos 3x cos 4x dx 56.

L

p>2 -p>2

cos x cos 7x dx

Exercises 57–62 require the use of various trigonometric identities before you evaluate the integrals.

57. Lsin²u cos 3u du 58.

Lcos²2u sin u du 59. Lcos³u sin 2u du 60.

Lsin³u cos 2u du 61. Lsinu cos u cos 3u du 62.

Lsinu sin 2u sin 3u du Assorted Integrations

Use any method to evaluate the integrals in Exercises 63–68.

63. L sec³x

tan x dx 64.

L sin³x cos⁴xdx

65. L tan²x

csc x dx 66.

L cot x cos²xdx

67. Lx sin²x dx 68.

Lx cos³x dx

Applications

69. Arc length Find the length of the curve y = ln (sec x), 0 … x … p>4.

70. Center of gravity Find the center of gravity of the region bounded by the x-axis, the curve y = sec x, and the lines x = -p>4, x = p>4.

71. Volume Find the volume generated by revolving one arch of the curve y = sin x about the x-axis.

72. Area Find the area between the x-axis and the curve y = 21 + cos 4x, 0 … x … p.

73. Centroid Find the centroid of the region bounded by the graphs ofy = x + cos x and y = 0 for 0 … x … 2p.

74. Volume Find the volume of the solid formed by revolving the region bounded by the graphs of y = sin x + sec x, y = 0, x = 0, and x = p>3 about the x@axis.

With x = a tan u,

a²

+ x

²

= a

²

+ a

²

tan

²

u = a

²

(1 + tan

²

u) = a

²

sec

²

u.

With x = a sin u,

a²

- x

²

= a

²

- a

²

sin

²

u = a

²

(1 - sin

²

u) = a

²

cos

²

u.

FIGURE 8.2 Reference triangles for the three basic substitutions identifying the sides labeled x and a for each substitution.

u u u

a

a

a x x

x

"a²− x²

x= a tan u x= a sin u x= a sec u

"x²− a²

"a²+ x²

"a²+ x²= a 0 sec u 0 "a²− x²= a 0 cos u 0 "x²− a²= a 0 tan u 0

8.4  Trigonometric Substitutions

479

Using Trigonometric Substitutions Evaluate the integrals in Exercises 1–14.

1. L dx

29 + x² 2.

L 3 dx 21 + 9x² 3. L

2 -2

dx

4 + x² 4.

L

2 0

dx 8 + 2x² 5. L

3>2 0

dx

29 - x² 6.

L

1>222 0

2 dx 21- 4x²

7. L²25 - t²dt 8.

L²1 - 9t²dt 9. L

dx

24x² - 49, x 7 7

2 10.

L 5 dx

225x²- 9, x 7 3 5

11. L

2y²- 49

y dy, y 7 7 12.

L

2y² - 25

y³ dy, y 7 5 13. L

dx

x²2x²- 1, x 7 1 14.

L 2 dx

x³2x² - 1, x 7 1 Assorted Integrations

Use any method to evaluate the integrals in Exercises 15–34. Most will require trigonometric substitutions, but some can be evaluated by other methods.

15. L x

29 - x²dx 16.

L x² 4 + x²dx 17. L

x³dx

2x²+ 4 18.

L dx x²2x² + 1 19. L

8 dw

w²24 - w² 20.

L

29 - w² w² dw

21. LA x + 1

1- xdx 22.

Lx2x² - 4 dx 23. L

23>2 0

4x²dx

(

1 - x²

)

^{3>2 24.} L

1 0

dx

(

4 - x²

)

^3>2

25. L dx

(

x²- 1

)

^{3>2, x 7 1 26.} L x²dx

(

x² - 1

)

^{5>2, x 7 1}

27. L

(

1- x²

)

^3>2

x⁶ dx 28.

L

(

1 - x²

)

^1>2

x⁴ dx

29. L 8 dx

(

4x²+ 1

)

^{2 30.} L

6 dt

(

9t² + 1

)

²

31. L x³dx

x²- 1 32.

L x dx 25 + 4x² 33. L

y²dy

(

1- y²

)

^{5>2 34.} L

(

1 - r²

)

^5>2

r⁸ dr

In Exercises 35–48, use an appropriate substitution and then a trigono- metric substitution to evaluate the integrals.

35. L

ln 4 0

e^tdt

2e^2t + 9 36.

L

ln (4>3) ln (3>4)

e^tdt

(

1 + e^2t

)

^3>2

37. L

1>4 1>12

2 dt

2t + 4t2t 38.

L

e 1

dy y21 + (ln y)² 39. L

dx

x2x²- 1 40.

L dx 1 + x² 41. L

x dx

2x²- 1 42.

L dx 21 - x² 43. L

x dx

21 + x⁴ 44.

L

21 - (ln x)² x ln x dx

45. LB 4 - x

x dx 46.

LA x 1 - x³dx

(

Hint: Let x = u².

) (

Hint: Let u = x^3>2.

)

47. L²x21 - x dx 48.

L 2x - 2 2x - 1dx

Initial Value Problems

Solve the initial value problems in Exercises 49–52 for y as a function of x.

49. x dy

dx= 2x²- 4, x Ú 2, y(2) = 0 50. 2x² - 9dy

dx = 1, x 7 3, y(5) = ln 3 51. (x²+ 4) dy

dx = 3, y(2) = 0 52. (x²+ 1)²dy

dx= 2x² + 1, y(0) = 1 Applications and Examples

53. Area Find the area of the region in the first quadrant that is enclosed by the coordinate axes and the curve y = 29- x²>3.

54. Area Find the area enclosed by the ellipse x²

a² + y² b² = 1.

55. Consider the region bounded by the graphs of y = sin^-1x, y = 0, andx = 1>2.

a. Find the area of the region.

b. Find the centroid of the region.

56. Consider the region bounded by the graphs of y = 2x tan^-1x and y = 0 for 0 … x … 1. Find the volume of the solid formed by revolving this region about the x-axis (see accompanying figure).

x y

0 1

y= "x tan⁻¹x

Exercises 8.4

57. Evaluate 1^x321 - x²dx using a. integration by parts.

b. a u-substitution.

c. a trigonometric substitution.

58. Path of a water skier Suppose that a boat is positioned at the origin with a water skier tethered to the boat at the point (30, 0) on a rope 30 ft long. As the boat travels along the positive y-axis, the skier is pulled behind the boat along an unknown path y = ƒ(x), as shown in the accompanying figure.

a. Show that ƒ′(x) = -²900 - x²

x .

(Hint: Assume that the skier is always pointed directly at the boat and the rope is on a line tangent to the path y = ƒ(x).)

b. Solve the equation in part (a) for ƒ(x), using ƒ(30) = 0.

NOT TO SCALE

x y

0 x (30, 0)

f (x) (x, f (x)) skier 30 ft rope y= f(x) path of skier boat

8.5 Integration of Rational Functions by Partial Fractions

This section shows how to express a rational function (a quotient of polynomials) as a sum of simpler fractions, called partial fractions, which are easily integrated. For instance, the rational function (5x - 3)>(x

²

- 2x - 3) can be rewritten as

5x - 3

x²

- 2x - 3 = 2

x

+ 1 + 3

x

- 3 .

You can verify this equation algebraically by placing the fractions on the right side over a common denominator (x + 1)(x - 3). The skill acquired in writing rational functions as such a sum is useful in other settings as well (for instance, when using certain transform methods to solve differential equations). To integrate the rational function (5x - 3)>(x

²

- 2x - 3) on the left side of our previous expression, we simply sum the integrals of the fractions on the right side:

L

5x - 3

(x + 1)(x - 3)

dx

= L

2

x

+ 1

dx

+

L 3

x

- 3

dx

= 2 ln 0

^x

+ 1 0 + 3 ln 0

^x

- 3 0 + C.

The method for rewriting rational functions as a sum of simpler fractions is called the method of partial fractions. In the case of the preceding example, it consists of finding constants A and B such that

5x - 3

x²

- 2x - 3 =

A

x

+ 1 +

B

x

- 3 . (1)

(Pretend for a moment that we do not know that A = 2 and B = 3 will work.) We call the fractions A >(x + 1) and B>(x - 3) partial fractions because their denominators are only part of the original denominator x

²

- 2x - 3. We call A and B undetermined coef- ficients until suitable values for them have been found.

To find A and B, we first clear Equation (1) of fractions and regroup in powers of x, obtaining

5x - 3 = A(x - 3) + B(x + 1) = (A + B)x - 3A + B.

This will be an identity in x if and only if the coefficients of like powers of x on the two sides are equal:

A

+ B = 5, -3A + B = -3.

Solving these equations simultaneously gives A = 2 and B = 3.