Unformatted text preview: 5E08(pp 510519) 1/17/06 5:19 PM Page 510 CHAPTER 8
The techniques of this
chapter enable us to ﬁnd
the height of a rocket a
minute after liftoff and to
compute the escape
velocity of the rocket. T echniques of Integration 5E08(pp 510519) 1/17/06 5:19 PM Page 511 Because of the Fundamental Theorem of Calculus, we can
integrate a function if we know an antiderivative, that is,
an indeﬁnite integral. We summarize here the most important integrals that we have learned so far. yx n dx xn 1
n1 ye x dx ex 2 tan x y tan x d x dx 1
tan
a C sin x
cot x y cosh x d x C y cot x d x C
x
a C y csc x cot x d x C 1 C ax
ln a dx 2 ln sec x a2 x ln x y csc x dx sec x cosh x 1 1
dx
x y cos x d x C
C y sinh x d x 2 y ya cos x y sec x tan x d x yx 1 C y sin x d x
y sec x dx n C C y sa csc x sinh x x2 dx sin C C ln sin x 1
2 C C 1 x
a C In this chapter we develop techniques for using these basic integration formulas
to obtain indeﬁnite integrals of more complicated functions. We learned the most
important method of integration, the Substitution Rule, in Section 5.5. The other general technique, integration by parts, is presented in Section 8.1. Then we learn
methods that are special to particular classes of functions such as trigonometric functions and rational functions.
Integration is not as straightforward as differentiation; there are no rules that
absolutely guarantee obtaining an indeﬁnite integral of a function. Therefore, in
Section 8.5 we discuss a strategy for integration.  8.1 Integration by Parts
Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that
corresponds to the Product Rule for differentiation is called the rule for integration by
parts.
The Product Rule states that if f and t are differentiable functions, then
d
f xtx
dx f xt x txf x 511 5E08(pp 510519) 512 ❙❙❙❙ 1/17/06 5:19 PM Page 512 CHAPTER 8 TECHNIQUES OF INTEGRATION In the notation for indeﬁnite integrals this equation becomes y t x f x dx yf or f xt x x t x dx ytxf f xtx x dx f xtx We can rearrange this equation as yf 1 x t x dx f xtx ytxf x dx Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v t x . Then the differentials are
du f x d x and dv t x d x, so, by the Substitution Rule, the formula for integration
by parts becomes y u dv 2 EXAMPLE 1 Find y v du uv y x sin x d x. SOLUTION USING FORMULA 1 Suppose we choose f x and t x
1, we have x and t x
sin x. Then f x
1
cos x. (For t we can choose any antiderivative of t .) Thus, using Formula y x sin x d x ytxf f xtx
x cos x x dx y cos x d x x cos x y cos x d x x cos x sin x C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as
expected.
SOLUTION USING FORMULA 2 Let
 It is helpful to use the pattern:
u
dv
du
v u
Then
and so x dv sin x d x du dx v cos x u d√ y x sin x dx y x sin x dx u x √ cos x x cos x y cos x d x x cos x sin x C √ y du cos x dx 5E08(pp 510519) 1/17/06 5:19 PM Page 513 S ECTION 8.1 INTEGRATION BY PARTS NOTE ❙❙❙❙ 513 Our aim in using integration by parts is to obtain a simpler integral than the one
we started with. Thus, in Example 1 we started with x x sin x d x and expressed it in terms
of the simpler integral x cos x d x. If we had chosen u sin x and dv x d x, then
du cos x d x and v x 2 2, so integration by parts gives
■ y x sin x d x sin x x2
2 1
2 yx 2 cos x d x Although this is true, x x 2 cos x d x is a more difﬁcult integral than the one we started with.
In general, when deciding on a choice for u and dv, we usually try to choose u f x to
be a function that becomes simpler when differentiated (or at least not more complicated)
as long as dv t x d x can be readily integrated to give v.
EXAMPLE 2 Evaluate y ln x d x. SOLUTION Here we don’t have much choice for u and dv. Let u
Then du ln x dv 1
dx
x dx
x v Integrating by parts, we get y ln x d x dx
x x ln x yx  It’s customary to write x 1 dx as x dx. x ln x y dx  Check the answer by differentiating it. x ln x x C Integration by parts is effective in this example because the derivative of the function
fx
ln x is simpler than f .
EXAMPLE 3 Find 2t y t e dt. SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e t is unchanged when differentiated or integrated), so we choose
t2 u
Then du dv 2 t dt v e t dt
et Integration by parts gives
3 2t y t e dt t 2et 2 y t e t dt The integral that we obtained, x t e t dt, is simpler than the original integral but is still not
obvious. Therefore, we use integration by parts a second time, this time with u t and 5E08(pp 510519) 514 ❙❙❙❙ 1/17/06 5:20 PM Page 514 CHAPTER 8 TECHNIQUES OF INTEGRATION e t dt. Then du dv e t, and d t, v t tet y e dt tet y t e dt t et C Putting this in Equation 3, we get
t 2et 2t e dt 2 y t e t dt t 2et yt 2 tet 2t te
EXAMPLE 4 Evaluate
 An easier method, using complex numbers,
is given in Exercise 48 in Appendix G. ye x 2 te et t 2e C
t C1 where C1 2C sin x d x. SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing u e x and dv
parts gives sin x d x anyway. Then du ye 4 x e x dx and v e x cos x sin x d x ye x cos x, so integration by cos x d x The integral that we have obtained, x e x cos x d x, is no simpler than the original one, but
at least it’s no more difﬁcult. Having had success in the preceding example integrating
by parts twice, we persevere and integrate by parts again. This time we use u e x and
dv cos x d x. Then du e x dx, v sin x, and ye 5 x cos x d x e x sin x ye x sin x d x At ﬁrst glance, it appears as if we have accomplished nothing because we have arrived at
 Figure 1 illustrates Example 4 by
e x sin x and
showing the graphs of f x
1x
cos x . As a visual check
Fx
2 e sin x
on our work, notice that f x
0 when F has
a maximum or minimum. we get ye x sin x d x e x cos x e x sin x ye x sin x d x This can be regarded as an equation to be solved for the unknown integral. Adding 12 x e x sin x d x to both sides, we obtain F
f
_3 x e x sin x d x, which is where we started. However, if we put Equation 5 into Equation 4 2 y e x sin x d x e x cos x e x sin x 6 Dividing by 2 and adding the constant of integration, we get
_4 FIGURE 1 ye x sin x d x 1
2 e x sin x cos x C If we combine the formula for integration by parts with Part 2 of the Fundamental
Theorem of Calculus, we can evaluate deﬁnite integrals by parts. Evaluating both sides of
Formula 1 between a and b, assuming f and t are continuous, and using the Fundamental 5E08(pp 510519) 1/17/06 5:20 PM Page 515 S ECTION 8.1 INTEGRATION BY PARTS ❙❙❙❙ 515 Theorem, we obtain y 6 b a EXAMPLE 5 Calculate y 1 0 f x t x dx f xtx b y a b a t x f x dx tan 1x dx. SOLUTION Let tan 1x u
Then dv dx du x v x2 1 dx So Formula 6 gives y 1 0 1 tan 1x dx x tan 1x y 0 1 tan 1 1
 Since tan 1x 0 for x 0, the integral in
Example 5 can be interpreted as the area of the
region shown in Figure 2. 0 tan 1 0
x 1 0 dx x2 1 x 1 0 x2 1 dx dx x2 1 y To evaluate this integral we use the substitution t 1 x 2 (since u has another meaning
in this example). Then dt 2 x dx, so x dx dt 2. When x 0, t 1; when x 1,
t 2; so y y=tan–!x y 0
1 y 4 x 1 0 x x 1 0 x2 1 dx 1
2 y 1
2 FIGURE 2 y Therefore 1 0 tan 1x dx 4 2 1 dt
t 1
2 ln 2 y 1 x 2 2 1 1
2 ln 1
x 1 0 ln t dx ln 2
ln 2
2 4 EXAMPLE 6 Prove the reduction formula
 Equation 7 is called a reduction formula
because the exponent n has been reduced to
n 1 and n 2. where n n 1
n y sin n2 x dx 2 is an integer. SOLUTION Let Then 1
cos x sin n 1x
n n y sin x dx 7 u sin n 1x du n dv
v 1 sin n 2x cos x dx sin x d x
cos x so integration by parts gives
n y sin x d x cos x sin n 1x n 1 y sin n2 x cos 2x d x 5E08(pp 510519) ❙❙❙❙ 516 1/17/06 5:20 PM Page 516 CHAPTER 8 TECHNIQUES OF INTEGRATION Since cos 2x sin 2x, we have 1
n cos x sin n 1x y sin x dx n y sin 1 n2 x dx n n y sin x d x 1 As in Example 4, we solve this equation for the desired integral by taking the last term
on the right side to the left side. Thus, we have
n y sin n x d x cos x sin n 1x
1
cos x sin n 1x
n n y sin x d x or n 1
n y sin
1 n n2 y sin x dx n2 x dx The reduction formula (7) is useful because by using it repeatedly we could eventually
express x sin n x d x in terms of x sin x d x (if n is odd) or x sin x 0 d x x d x (if n is even).  8.1 Exercises 1–2  Evaluate the integral using integration by parts with the
indicated choices of u and dv. y
dy
e2y 1 ■ 3–32 5. ■ ■ ■ ■ ■ ■ ■ ■ y x cos 5 x d x
y re r2 4.
6. dr 2 yx 9. y ln 2 x sin x dx
1 dx y y cos x ln sin x 28. y y cos ln x dx 30. y y x 4 ln x 2 dx 32. y 12 0 x y xe cos 1x d x 8. y t sin 2 t d t 10. yx 2 2 1 cos m x d x ■ ■ ■ y arctan 4 t d t
ln x 2 d x 13. y 15. ye 17. 12. 5 21. s3 1 ■ ■ arctan 1 x d x
r3 1 t 0 ■ r2 s4 0 e s sin t ■ dr s ds ■ ■ ■ ■  First make a substitution and then use integration by parts
to evaluate the integral. x dx 33. y sin s x d x
y 4 e sx d x 34. y 36. yx e 1 ln p d p
s s 3
2 cos 2 d 5 x2 dx 3t 14. y t e dt sin 3 d 16. ye y y sinh y d y 18. y y cosh ay d y 2 ■ ■ ■ y 0 y 2 1 t sin 3 t d t
ln x
dx
x2 22. y 0 y 4 1 x 2 1e ■ ■ ■ ■ ■ ■ ■ ■  Evaluate the indeﬁnite integral. Illustrate, and check that
your answer is reasonable, by graphing both the function and its
antiderivative (take C 0). x y x cos 39. 20. 1 ■ ; 37–40 cos 2 d 37.
19. x 5x dx 1 y sin
yp 1 0 x csc 2x d x 33–36 35. 11. dx 4 dx ■ 7. 26. 2 y ■ Evaluate the integral.  24. 0 sec 2 d , dv
■ x dx y 31. ln x, dv sec 2 d ; u y ■ 3. u 25. 29. 2. y x ln x d x; y 27. 1. 23. y 32 x dx 38. yx 3 e x dx 40. yx e ln x dx dx
2x 3 x2 dx st ln t dt
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E08(pp 510519) 1/17/06 5:21 PM Page 517 S ECTION 8.1 INTEGRATION BY PARTS 41. (a) Use the reduction formula in Example 6 to show that x
2 2 y sin x d x sin 2 x
4 ❙❙❙❙ 517 ; 53–54  Use a graph to ﬁnd approximate xcoordinates of the
points of intersection of the given curves. Then ﬁnd (approximately) the area of the region bounded by the curves. C 53. y
54. y (b) Use part (a) and the reduction formula to evaluate
x sin 4x d x. x sin x, arctan 3 x, ■ ■ y ■ x
y ■ 2 2
x2 ■ ■ ■ ■ ■ ■ ■ ■ 42. (a) Prove the reduction formula
55–58 1
cos n 1x sin x
n n y cos x d x n 1
n y cos  Use the method of cylindrical shells to ﬁnd the volume
generated by rotating the region bounded by the given curves about
the speciﬁed axis. n2 x dx 55. y 2 y 0 n sin n x d x 1 y n 2 2 246
357 sin 2 n 1x d x 0 2 0 45–48 ln x n d x 45. y 46. yx e nx x ln x
x ne x dx n n y ln x 2n 1
2n n1 x e, x
■ 1; about the yaxis
1, x 0, y ■ 1; about the yaxis x 0; about x 1 ; about the xaxis ■ ■ ■ ■ ■ ■ ■ ■ ■ x 2 ln x on the interval 1, 3 . decreases with time. Suppose the initial mass of the rocket at
liftoff (including its fuel) is m, the fuel is consumed at rate r,
and the exhaust gases are ejected with constant velocity ve
(relative to the rocket). A model for the velocity of the rocket
at time t is given by the equation
vt tt m ve ln rt
m 2 where t is the acceleration due to gravity and t is not too
large. If t 9.8 m s 2, m 30,000 kg, r 160 kg s, and
ve
3000 m s, ﬁnd the height of the rocket one minute
after liftoff. Use integration by parts to prove the reduction formula.  e ,y 0, 0 60. A rocket accelerates by burning its onboard fuel, so its mass 2n
2n 1 135
246 sin 2 nx d x 0, x x 2, y 59. Find the average value of f x 44. Prove that, for even powers of sine, y x ■ where n 2 is an integer.
(b) Use part (a) to evaluate x0 2 sin 3x d x and x0 2 sin 5x d x.
(c) Use part (a) to show that, for odd powers of sine, y e x, x 58. y sin n 2x d x 0 e x, y 57. y 43. (a) Use the reduction formula in Example 6 to show that cos 56. y (b) Use part (a) to evaluate x cos 2x d x.
(c) Use parts (a) and (b) to evaluate x cos 4x d x. dx 61. A particle that moves along a straight line has velocity
vt
t 2e t meters per second after t seconds. How far will n y x n 1e x d x it travel during the ﬁrst t seconds?
47. x2 y x x2
2n
48. ■ a2
1 n 2na 2
2n 1 tan x sec n 2x
n1 n y sec x d x ■ 62. If f 0 a 2 n dx ■ ■ ■ y n
n ■ ■ x
2
1 2 a 2n1 dx (n 1
2 ) y a 0 0 and f and t are continuous, show that t0 f x t x dx f at a y f ata a 0 f x t x dx 63. Suppose that f 1 y sec n2 x dx n 2, f 4
7, f 1
5, f 4
4
f is continuous. Find the value of x1 x f x d x. 1 3, and 64. (a) Use integration by parts to show that
■ ■ ■ ■ ■ yf 49. Use Exercise 45 to ﬁnd x ln x 3 d x.
50. Use Exercise 46 to ﬁnd x x e d x. x dx xf x y xf x dx 4x 51–52  51. y ■ Find the area of the region bounded by the given curves.
xe 52. y 0.4 x , 5 ln x,
■ (b) If f and t are inverse functions and f is continuous, prove
that ■ y
y
■ 0, x 5 y x ln x
■ b a ■ ■ ■ ■ ■ ■ ■ f x dx bf b af a y fb fa t y dy [Hint: Use part (a) and make the substitution y f x .] 5E08(pp 510519) 518 ❙❙❙❙ 1/17/06 5:21 PM Page 518 CHAPTER 8 TECHNIQUES OF INTEGRATION (c) In the case where f and t are positive functions and
b a 0, draw a diagram to give a geometric interpretation of part (b).
(d) Use part (b) to evaluate x1e ln x d x. (c) Use parts (a) and (b) to show that
2n
2n x f x d x, by using
cylindrical shells, but now we can use integration by parts to
prove it using the slicing method of Section 6.2, at least for the
case where f is onetoone and therefore has an inverse function t. Use the ﬁgure to show that
b 2d V a 2c y d ty c 2 lim nl x=g(y) 1 2
1 2
3 4
3 4
5 6
5 6
7 2n
2n 1 2n
2n 1 2 This formula is usually written as an inﬁnite product:
dy Make the substitution y f x and then use integration by
parts on the resulting integral to prove that V xab 2 x f x d x.
y I2 n 1
I2 n and deduce that lim n l I2 n 1 I2 n 1.
(d) Use part (c) and Exercises 43 and 44 to show that xab 2 65. We arrived at Formula 6.3.2, V 1
2 2
1 2 2
3 4
3 4
5 6
5 6
7 and is called the Wallis product.
(e) We construct rectangles as follows. Start with a square of
area 1 and attach rectangles of area 1 alternately beside or
on top of the previous rectangle (see the ﬁgure). Find the
limit of the ratios of width to height of these rectangles. y=ƒ d x=b c x=a
0 66. Let In x0 2 a b x sin n x d x. (a) Show that I2 n 2 I2 n 1 I2 n.
(b) Use Exercise 44 to show that
I2 n 2
I2 n  8.2 2n
2n 1
2 Trigonometric Integrals
In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine.
EXAMPLE 1 Evaluate 3 y cos x d x. SOLUTION Simply substituting u
cos x isn’t helpful, since then du
sin x dx. In order
to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of
sine would require an extra cos x factor. Thus, here we can separate one cosine factor
and convert the remaining cos2x factor to an expression involving sine using the identity
sin 2x cos 2x 1: cos 3x cos 2x cos x 1 sin 2x cos x We can then evaluate the integral by substituting u sin x, so du 3 2 y cos x d x y cos x
y 1 sin x y cos x d x
u 2 du
1
3 sin 3x u
C 1 1
3 u3 cos x d x and sin 2x cos x d x
C 5E08(pp 510519) 1/17/06 5:22 PM Page 519 S ECTION 8.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 519 In general, we try to write an integrand involving powers of sine and cosine in a form
where we have only one sine factor (and the remainder of the expression in terms of
cosine) or only one cosine factor (and the remainder of the expression in terms of sine).
The identity sin 2x cos 2x 1 enables us to convert back and forth between even powers
of sine and cosine.
EXAMPLE 2 Find 5 2 y sin x cos x d x S OLUTION We could convert cos 2x to 1 sin 2x, but we would be left with an expression in
terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and
rewrite the remaining sin 4x factor in terms of cos x :
sin 5x cos 2x
 Figure 1 shows the graphs of the integrand
sin 5x cos 2x in Example 2 and its indeﬁnite integral (with C 0). Which is which? Substituting u sin2x 2 cos 2x sin x cos x, we have du
5 2 cos 2x 2 cos 2x sin x 1 sin x dx and so sin 2x 2 cos 2x sin x d x y sin x cos x d x y 0.2 y
_π cos 2x 2 cos 2x sin x d x y π 1
1 u2 2u2
u3
3 _ 0.2
1
3 FIGURE 1 2 cos 3x u5
5
2
5 u2 y du
u7
7 u 6 du C
1
7 cos 5x 2u 4 cos 7x C In the preceding examples, an odd power of sine or cosine enabled us to separate a
single factor and convert the remaining even power. If the integrand contains even powers
of both sine and cosine, this strategy fails. In this case, we can take advantage of the following halfangle identities (see Equations 17b and 17a in Appendix D):
1
2 sin 2x  Example 3 shows that the area of the region
shown in Figure 2 is 2. EXAMPLE 3 Evaluate y 0 1 cos 2 x 1
2 cos 2x and 1 cos 2 x sin 2x d x. SOLUTION If we write sin 2x 1 cos 2x, the integral is no simpler to evaluate. Using the
halfangle formula for sin x, however, we have
2 1.5
[email protected] x 0 0
_0.5 sin 2x d x 1
2 y 1
2 y ( 0 cos 2 x d x 1
1
2 [ (x ) 1
2 sin 2 1
2 (0 1
2 1
2 sin 2 x) sin 0) 0 1
2 π Notice that we mentally made the substitution u 2 x when integrating cos 2 x. Another
method for evaluating this integral was given in Exercise 41 in Section 8.1. FIGURE 2 EXAMPLE 4 Find 4 y sin x d x. SOLUTION We could evaluate this integral using the reduction formula for x sin n x d x (Equation 8.1.7) together with Example 1 (as in Exercise 41 in Section 8.1), but a better 5E08(pp 520529) 520 ❙❙❙❙ 1/17/06 5:23 PM Page 520 CHAPTER 8 TECHNIQUES OF INTEGRATION method is to write sin 4x sin 2x 2 and use a halfangle formula:
4 sin 2x 2 d x y sin x d x y y
1
4 y 2 1 cos 2 x
2 1 2 cos 2 x dx
cos 2 2 x d x Since cos 2 2 x occurs, we must use another halfangle formula
1
2 cos 2 2 x 1 cos 4 x This gives
1
4 y 1
4 4 y sin x d x y(
( 13
42 1
3
2 x 2 cos 2 x 1
2 2 cos 2 x 1
2 1
8 sin 2 x 1 cos 4 x d x cos 4 x) d x sin 4 x) C To summarize, we list guidelines to follow when evaluating integrals of the form
0 and n 0 are integers. x sin mx cos nx dx, where m
Strategy for Evaluating y sin m x cos nx dx (a) If the power of cosine is odd n 2 k 1 , save one cosine factor and use
cos 2x 1 sin 2x to express the remaining factors in terms of sine:
m x cos 2 k 1x d x y sin m x cos 2x k cos x d x y sin y sin m x1 sin 2x k cos x d x Then substitute u sin x.
(b) If the power of sine is odd m 2 k 1 , save one sine factor and use
sin 2x 1 cos 2x to express the remaining factors in terms of cosine:
2k 1 x cos n x d x y sin 2x k cos n x sin x d x y y sin 1 cos 2x k cos n x sin x d x Then substitute u cos x. [Note that if the powers of both sine and cosine are
odd, either (a) or (b) can be used.]
(c) If the powers of both sine and cosine are even, use the halfangle identities
sin 2x 1
2 1 cos 2 x cos 2x It is sometimes helpful to use the identity
sin x cos x 1
2 sin 2 x 1
2 1 cos 2 x 5E08(pp 520529) 1/17/06 5:24 PM Page 521 S ECTION 8.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 521 We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
d dx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x.
Or, since d dx sec x sec x tan x, we can separate a sec x tan x factor and convert the
remaining (even) power of tangent to secant.
EXAMPLE 5 Evaluate 6 4 y tan x sec x d x. SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in terms of tangent using the identity sec 2x 1
by substituting u tan x with du sec 2x d x :
6 4 tan 2x. We can then evaluate the integral 6 2 2 y tan x sec x d x y tan x sec x sec x d x
6 y tan x
yu 6 1
7 EXAMPLE 6 Find y tan 5 u 2 du 1 u7
7 tan 2x sec 2x d x 1 u9
9 u6 u 8 du C
1
9 tan 7x y tan 9x C sec 7 d . SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a
sec tan factor, we can convert the remaining power of tangent to an expression
sec 2
1. We can then evaluate the
involving only secant using the identity tan 2
integral by substituting u sec , so du sec tan d : y tan 5 sec 7 d y tan 4 y sec 2 y sec 6 sec tan d u2 u 11
11
1
11 sec 11 1 2 sec 6 sec 1 2 u 6 du 2 u9
9 u7
7
2
9 sec 9 tan d u 10 y 2u 8 u 6 du C
1
7 sec 7 C The preceding examples demonstrate strategies for evaluating integrals of the form x tan mx sec nx dx for two cases, which we summarize here. 5E08(pp 520529) 522 ❙❙❙❙ 1/17/06 5:24 PM Page 522 CHAPTER 8 TECHNIQUES OF INTEGRATION Strategy for Evaluating y tan m x sec nx dx (a) If the power of secant is even n 2 k, k 2 , save a factor of sec 2x and use
sec 2x 1 tan 2x to express the remaining factors in terms of tan x :
m y tan x sec 2 kx d x m x sec 2x y tan y tan m x1 k1 sec 2x d x tan 2x k1 sec 2x d x Then substitute u tan x.
(b) If the power of tangent is odd m 2 k 1 , save a factor of sec x tan x and
use tan 2x sec 2x 1 to express the remaining factors in terms of sec x :
2k 1 Then substitute u x sec n x d x y tan 2x k sec n 1x sec x tan x d x y y tan sec 2x 1 k sec n 1x sec x tan x d x sec x. For other cases, the guidelines are not as clearcut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to
integrate tan x by using the formula established in (5.5.5): y tan x d x ln sec x C We will also need the indeﬁnite integral of secant: 1 y sec x d x ln sec x tan x C We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x :
sec x y sec x d x y sec x sec x
y
If we substitute u
becomes x 1 u du sec x
ln u tan x
dx
tan x sec 2x sec x tan x
dx
sec x tan x
sec 2x d x, so the integral tan x, then du
sec x tan x
C. Thus, we have y sec x d x ln sec x tan x C 5E08(pp 520529) 1/17/06 5:25 PM Page 523 SECTION 8.2 TRIGONOMETRIC INTEGRALS E XAMPLE 7 Find ❙❙❙❙ 523 3 y tan x d x. SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in 2 terms of sec x :
3 2 y tan x d x y tan x tan x d x
y tan x sec 2x 1 dx 2 y tan x sec x d x y tan x d x
tan 2x
2 ln sec x In the ﬁrst integral we mentally substituted u C
sec 2x d x. tan x so that du If an even power of tangent appears with an odd power of secant, it is helpful to express
the integrand completely in terms of sec x. Powers of sec x may require integration by
parts, as shown in the following example.
EXAMPLE 8 Find 3 y sec x d x. SOLUTION Here we integrate by parts with u
du sec x
sec x tan x d x 3 tan x v 2 sec x tan x y sec x tan x d x sec x tan x y sec x sec x tan x Then y sec x d x sec 2x d x dv y sec x d x y sec x d x sec 2x 1 dx 3 Using Formula 1 and solving for the required integral, we get
3 y sec x d x 1
2 (sec x tan x ln sec x tan x ) C Integrals such as the one in the preceding example may seem very special but they
occur frequently in applications of integration, as we will see in Chapter 9. Integrals of
the form x cot m x csc n x d x can be found by similar methods because of the identity
1 cot 2x csc 2x.
Finally, we can make use of another set of trigonometric identities:
2 To evaluate the integrals (a) x sin m x cos n x d x, (b) x sin m x sin n x d x, or
(c) x cos m x cos n x d x, use the corresponding identity:
 These product identities are discussed in
Appendix D. (a) sin A cos B 1
2 sin A B sin A (b) sin A sin B 1
2 cos A B cos A B (c) cos A cos B 1
2 cos A B cos A B B 5E08(pp 520529) ❙❙❙❙ 524 1/17/06 5:25 PM Page 524 CHAPTER 8 TECHNIQUES OF INTEGRATION EXAMPLE 9 Evaluate y sin 4 x cos 5x d x. SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use
the identity in Equation 2(a) as follows: y sin 4 x cos 5x d x y 1
2 sin x sin 9x d x sin x sin 9x d x 1
2
1
2  8.2
1–47 (cos x 1
9 cos 9x C Exercises Evaluate the integral.  y 5 6 4 y 2 0 cos2 d 8. 4 9. y sin 3 t dt 11. y 1 13.
15. 0 y 4 0 d sin 4x cos 2x d x
3 y sin x scos x d x
2 3 17. y cos x tan x d x 19. y 1 sin x
dx
cos x
2 21. y sec x tan x d x 23.
25.
27.
29. y 2 0 y sin
y 2 0 3 sin 2 2
6 12. y csc x d x 40. y y sin 5x sin 2 x d x 42. y sin 3x cos x d x y cos 7 44. y y 46. y cos x y t sec 2
6 cot 2x d x 2
4 3 csc 3 d 3
6 y x cos2x d x 14.
16. 0 y cos 2 0 sin 2x cos 2x d x
cos5 sin y cos
5 y cot 20. y cos x sin 2 x d x 2 1 ■ sec 4 t 2 dt y tan x d x y sec 6 t dt 26. cos 5 d tan 2x
dx
sec 2x 4 2 x cot 6 x d x csc 3x d x cos x sin x
dx
sin 2 x
dx
1 t 2 tan 4 t 2 dt ■ ■ 48. If x0
4 x0
2 4 d sin 4 d 18. 0 cot 3x d x d 24. 0 y csc 2 d y y tan x d x y 38. d 4 mx dx y 3 y tan 3 cos 5x dx 22. 2 36. ay dy 3 y sin x cos x dx 10.
2 cos 6. y tan x sec x d x 47. 5 y cos x sin x dx 4. 34. 45. sin 5x cos 3x dx y tan 43. 4
2 32. 41. 3 6 y cot 39. 7. y 2. y 37. 5. 2 y cos 35. 3. 3 y sin x cos x dx y tan x d x 33. 1. 31. 4 ■ ■ tan 6 x sec x d x ■ ■ ■ ■ ■ I , express the value of tan x sec x d x in terms of I. ; 49–52  Evaluate the indeﬁnite integral. Illustrate, and check that
your answer is reasonable, by graphing both the integrand and its
antiderivative (taking C 0 .
5 4 4 tan 5 x sec 4 x d x
3 y tan x sec x dx 28.
30. 0 49. sec 4 y tan 3 y sin x d x 50. y sin x cos x d x y sin 3x sin 6 x d x 52. y sec tan 4 d 0 3 4 2 x sec 5 2 x d x
■ y ■ 8 51. y 4 ■ tan 5x sec6x d x ■ ■ ■ ■ ■ ■ ■ 53. Find the average value of the function f x the interval , . x
dx
2
■ ■ 2 ■ 3 sin x cos x on ■ 5E08(pp 520529) 1/17/06 5:26 PM Page 525 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION 54. Evaluate x sin x cos x d x by four methods: (a) the substitution  current that varies from 155 V to 155 V with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
Et
155 sin 120 t Find the area of the region bounded by the given curves. 55. y sin x, y sin 3x, 56. y sin x, y 2 sin 2x, ■ ■ ■ ■ ■ x x 0,
x 0, ■ where t is the time in seconds. Voltmeters read the RMS (rootmeansquare) voltage, which is the square root of the average
value of E t 2 over one cycle.
(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220 V.
Find the corresponding amplitude A needed for the voltage
Et
A sin 120 t . 2
x 2 ■ ■ ■ ■ ■ ■ ; 57–58  Use a graph of the integrand to guess the value of the
integral. Then use the methods of this section to prove that your
guess is correct. 57. y 2 cos 3x d x 0 ■ ■ 58. ■ ■ ■ ■ ■ y 2 0
■ 65–67 ■ ■  Prove the formula, where m and n are positive integers. 65. sin 2 x cos 5 x d x
■ ■ y sin m x cos n x d x 66. y sin m x sin n x d x 67. y cos m x cos n x d x 0
0 59–62  Find the volume obtained by rotating the region bounded
by the given curves about the speciﬁed axis. 59. y sin x, x 60. y 2 tan x, y 0, x 0, x 4; about the xaxis 61. y cos x, y 0, x 0, x 2; about y 2, x ,y ■ cos x, y
■ ■ 0, x
■ ■ 0, x
■ 2; about y
■ ■ ■ ■ ■ n
n ■ ■ ■ ■ ■ ■ ■ ■ N 1 ■ ■ a n sin n x fx
n1
■ a 1 sin x ■ a 2 sin 2 x a N sin Nx Show that the mth coefﬁcient a m is given by the formula
am 0.  8.3 if m
if m n
n 68. A ﬁnite Fourier series is given by the sum 1 63. A particle moves on a straight line with velocity function
vt
sin t cos 2 t. Find its position function s f t if f0 0 if m
if m 0; about the xaxis
■ 62. y 525 64. Household electricity is supplied in the form of alternating u cos x, (b) the substitution u sin x, (c) the identity
sin 2 x 2 sin x cos x, and (d) integration by parts. Explain the
different appearances of the answers.
55–56 ❙❙❙❙ 1 y f x sin m x d x Trigonometric Substitution
In ﬁnding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 d x arises,
where a 0. If it were x x sa 2 x 2 d x, the substitution u a 2 x 2 would be effective
but, as it stands, x sa 2 x 2 d x is more difﬁcult. If we change the variable from x to by
the substitution x a sin , then the identity 1 sin 2
cos 2 allows us to get rid of the
root sign because
sa 2 x2 sa 2 a 2 sin 2 sa 2 1 sin 2 sa 2 cos 2 a cos Notice the difference between the substitution u a 2 x 2 (in which the new variable is
a function of the old one) and the substitution x a sin (the old variable is a function of
the new one).
In general we can make a substitution of the form x t t by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is onetoone. In this case, if we replace u by x and x by t in the Substitution
Rule (Equation 5.5.4), we obtain yf x dx yf t t t t dt This kind of substitution is called inverse substitution. 5E08(pp 520529) 526 ❙❙❙❙ 1/17/06 5:27 PM Page 526 CHAPTER 8 TECHNIQUES OF INTEGRATION We can make the inverse substitution x a sin provided that it deﬁnes a onetoone
function. This can be accomplished by restricting to lie in the interval
2, 2 .
In the following table we list trigonometric substitutions that are effective for the given
radical expressions because of the speciﬁed trigonometric identities. In each case the restriction on is imposed to ensure that the function that deﬁnes the substitution is onetoone.
(These are the same intervals used in Appendix D in deﬁning the inverse functions.)
Table of Trigonometric Substitutions Expression Substitution sa 2 x2 x a sin , sa 2 x2 x a tan , sx 2 a2 x a sec , EXAMPLE 1 Evaluate
SOLUTION Let x x2 s9
x 2 x2 sec 2 2 3 cos s9 cos 2 3 cos 2 dx y cos 2
d
sin 2
csc 2 y cot œ„„„„„
9≈ d and 3 cos 2 d 1d cot ¨ tan 2 3 cos
3 cos d
9 sin 2 y x 1 2.) Thus, the Inverse Substitution Rule y 3 sec 2 d x. x2
x tan 2 3
2 or 2 2 s9 cos 2 2. Then d x 0 0 because y sin 2 1 2 9 sin 2 s9 1 2 2 3 sin , where s9
(Note that cos
gives y 2 Identity C Since this is an indeﬁnite integral, we must return to the original variable x. This can be
done either by using trigonometric identities to express cot in terms of sin
x 3 or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
Since sin
x 3, we label the opposite side and the hypotenuse as having lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2, so we
can simply read the value of cot from the ﬁgure: FIGURE 1 (Although
Since sin x2 s9 cot x
sin ¨ =
3 x 0 in the diagram, this expression for cot
x 3, we have
sin 1 x 3 and so y x2 s9
x 2 dx x2 s9
x sin is valid even when 1 x
3 C 0.) 5E08(pp 520529) 1/17/06 5:27 PM Page 527 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 527 EXAMPLE 2 Find the area enclosed by the ellipse x2
a2 (0, b) y2
b2 (a, 0) FIGURE 2 ¥
≈
+ =1
[email protected]
[email protected] 1 SOLUTION Solving the equation of the ellipse for y, we get y 0 y2
b2 x x2
a2 1 a2 x2
a or 2 b
sa 2
a y x2 Because the ellipse is symmetric with respect to both axes, the total area A is four times
the area in the ﬁrst quadrant (see Figure 2). The part of the ellipse in the ﬁrst quadrant is
given by the function
b
y
0xa
sa 2 x 2
a
1
4 and so y A a 0 b
sa 2
a x 2 dx To evaluate this integral we substitute x a sin . Then dx
limits of integration we note that when x 0, sin
0, so
sin
1, so
2. Also
sa 2 x2 since 0 sa 2 a 2 sin 2 sa 2 cos 2 a cos d . To change the
0; when x a, a cos a cos 2. Therefore
A 4 b
a y a 0 4 ab y sa 2
2 0 [ cos 2 d
1
2 2 ab x 2 dx sin 2 4 b
a 4 ab y 0 0 2 2 y 2 ab 0 21
2 a cos
1 2 a cos d cos 2
0 d 0 ab
We have shown that the area of an ellipse with semiaxes a and b is a b. In particular,
taking a b r, we have proved the famous formula that the area of a circle with
radius r is r 2.
NOTE Since the integral in Example 2 was a deﬁnite integral, we changed the limits of
integration and did not have to convert back to the original variable x.
■ EXAMPLE 3 Find 1 y x sx SOLUTION Let x sx 2 2 2 4 2 tan , d x.
2 2. Then d x s4 tan 2 4 1 s4 sec 2 2 sec 2 d and
2 sec 2 sec Thus, we have
dx y x sx
2 2 4 y 2 sec 2 d
4 tan 2 2 sec 1
4 sec y tan 2 d 5E08(pp 520529) 528 ❙❙❙❙ 1/17/06 5:28 PM Page 528 CHAPTER 8 TECHNIQUES OF INTEGRATION To evaluate this trigonometric integral we put everything in terms of sin
sec
tan 2
Therefore, making the substitution u y dx
x 2s x 2 cos
d
sin 2 y 1
4 1
u x y x sx
2 EXAMPLE 4 Find y sx x
2 4 2 du
u2
1
4 sin C 4 x and so
sx 2 4
4x dx x
2 y C
sx 2 We use Figure 3 to determine that csc FIGUR E 3 1
4
C csc
4 ¨
2 cos
sin 2 sin , we have
1
4 4 œ„„„„„
≈+4 tan ¨ = cos 2
sin 2 1
cos and cos : 4 C d x. SOLUTION It would be possible to use the trigonometric substitution x
2 tan here (as in
Example 3). But the direct substitution u x 2 4 is simpler, because then du 2 x d x
and y
NOTE x
sx 2 4 1
2 dx du y su su C sx 2 4 C Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method ﬁrst.
■ EXAMPLE 5 Evaluate y sx dx
2 a2 , where a 0. SOLUTION 1 We let x dx a sec a sec , where 0
tan d and sx 2 a2 2 or sa 2 sec 2 sa 2 tan 2 1 3 2. Then a tan a tan Therefore y sx dx
2 a y 2 a sec tan
a tan y sec x d d
ln sec tan C ≈[email protected]
œ„„„„„ The triangle in Figure 4 gives tan ¨
a
FIGU RE 4 x
sec ¨ =
a y sx dx
2 a2 sx 2
ln x
a ln x a 2 a, so we have
sx 2 a2
a sx 2 a2 C
ln a C 5E08(pp 520529) 1/17/06 5:28 PM Page 529 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION Writing C1 C 529 ln a, we have y sx 1
SOLUTION 2 For x
identity cosh 2 y dx
2 sx 2 ln x a2 a2 0 the hyperbolic substitution x
sinh 2 y 1, we have
sx 2 Since dx ❙❙❙❙ a2 sa 2 cosh 2 t C1 a cosh t can also be used. Using the sa 2 sinh 2 t 1 a sinh t a sinh t dt, we obtain y sx
Since cosh t dx
2 a x a, we have t cosh 1 sx 2 a y dt t C x a and dx y 2 a sinh t dt
a sinh t y 2 cosh 2 x
a 1 C Although Formulas 1 and 2 look quite different, they are actually equivalent by
Formula 7.6.4.
NOTE As Example 5 illustrates, hyperbolic substitutions can be used in place of trigonometric substitutions and sometimes they lead to simpler answers. But we usually use
trigonometric substitutions because trigonometric identities are more familiar than hyperbolic identities.
■ EXAMPLE 6 Find y x3 3 s3 2 0 4x 2 32 9 d x. 932
s4 x 2 9 )3 so trigonometric substitution
is appropriate. Although s4 x
9 is not quite one of the expressions in the table of
trigonometric substitutions, it becomes one of them if we make the preliminary substitution u 2 x. When we combine this with the tangent substitution, we have x 3 tan ,
2
which gives d x 3 sec 2 d and
2
SOLUTION First we note that 4 x 2
2 s4 x 2
When x 0, tan y 0, so 4x 2 s9 tan 2 0; when x x3 3 s3 2 0 9 9 32 dx y 3 0 so that du y 3
16 cos 3 sec 3 s3 2, tan 3
16 Now we substitute u
3, u 1.
2 9 y 27
8 tan3
27 sec3
3 0 0 3 tan 3
d
sec
1 3
2 s3, so 3. sec 2 d
3
16 y 3 0 sin3
d
cos2 cos 2
sin d
cos 2 sin d . When 0, u 1; when 5E08(pp 530539) 530 ❙❙❙❙ 1/17/06 6:03 PM Page 530 CHAPTER 8 TECHNIQUES OF INTEGRATION Therefore y x3 3 s3 2 4x 0 2 9 3
16 dx 32 3
16 x
2x y s3 EXAMPLE 7 Evaluate y u2 1 12 u 1 3
16 du y 12 1 1 u 2 du 12 1
u u 2 [( 1
2 3
16 2) 1 1 3
32 1 d x. x2 SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by ﬁrst completing the square under the root sign: 2x 3 x2 x2 4 3 x 2x 3 x
2x We now substitute u
 Figure 5 shows the graphs of the integrand
in Example 7 and its indeﬁnite integral (with
C 0 ). Which is which? x 2 1. Then du y dx 2 sin , giving du
x
2x x 2 cos u
s4 x 2 y dx 1 d x and x 2 sin
2 cos 1 2 sin u 1
du
u2
u2 d and s4 y y s3 2x 2 1 This suggests that we make the substitution u y s3 x2 1 2 cos , so 1d 2 cos d 3 2 _4 2 cos C s4 u2 sin s3 2x 1 x2 u
2 C _5 FIGURE 5  8.3 4–30  Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle. 2 s3 2. 3.
■ yx ■ x x dx ; x3
2 9
■ dx ;
■ x x y yx 9. dx ; y 7. 9
2 s9 y sx C x3
dx
s16 x 2 4. 1
x 2 sx 2
3 1
2 Evaluate the integral.  5. y x 1 Exercises 1–3 1. sin y sx 3 sec 3 sin 0 1 2 s2 3 t st
2 ■ ■ ■ ■ ■ ■ ■ 1 1
s25 3 tan
■ 2 x dx
2 16 2 dt 6. y dx 8. y 10. 2 0 x 3 sx 2 4 dx sx 2 a 2
dx
x4 y st t5
2 2 dt 1, so 5E08(pp 530539) 1/17/06 6:03 PM Page 531 S ECTION 8.3 TRIGONOMETRIC SUBSTITUTION 11. y s1 13. y 15. y 17. y sx 12. y sx 2 9
dx
x3 14. y u s5 16. yx 18. y a x2
x2 2 x 21. y 23. y s5 25. y s9x 27. y 29. dx 7 s1 y y x s1 0 ■ 2 9x 2 d x 22. y x 2 dx 24. y st 26.
28. y 6x 8 dx
2x 2 dx 2 30. 4 x dx
■ ■ ■ ■ 9
b2 t y s4 x 1
2 1 t2 sx 2 0 5
2 y 0
■ ¨ 32 O dt dt
6t 2 y dx
4x x 2 dx
2 ln ( x a2 a2) sx 2 2 38. A charged rod of length L produces an electric ﬁeld at point 52 P a, b given by cos t
dt
s1 sin 2 t
■ dx
x 4 sx 2 ; 37. Use a graph to approximate the roots of the equation
x 2 s4 x 2 2 x. Then approximate the area bounded by
the curve y x 2 s4 x 2 and the line y 2 x. dx ■ x Graph the integrand and its indeﬁnite integral on the same
screen and check that your answer is reasonable. 13 x2 R ; 36. Evaluate the integral EP
■ y b La
a 4 0 x2 b2 32 dx ■ where is the charge density per unit length on the rod and 0
is the free space permittivity (see the ﬁgure). Evaluate the integral to determine an expression for the electric ﬁeld E P . 31. (a) Use trigonometric substitution to show that y sx Q 1 dx x2 ■ P dx
2 ax y s 25 4x y u2 dx
s16 x 2 531 equation x 2 y 2 r 2. Then A is the sum of the area of the
triangle POQ and the area of the region PQR in the ﬁgure.] 4 dx du 20. dx x 3s4 x2 x sx 2 0 2 x
23 dx 32 x
2 19. ■ 1 4x 2 dx ❙❙❙❙ y C P (a, b) (b) Use the hyperbolic substitution x a sinh t to show that
0 y sx dx
2 sinh a2 x
a 1 L x C
39. Find the area of the crescentshaped region (called a lune) These formulas are connected by Formula 7.6.3. bounded by arcs of circles with radii r and R. (See the ﬁgure.) 32. Evaluate y x 2 x2
a2 32 dx (a) by trigonometric substitution.
(b) by the hyperbolic substitution x
33. Find the average value of f x sx 2 r
R a sinh t.
1 x, 1 x 7. 34. Find the area of the region bounded by the hyperbola 9x 2 4y 2 36 and the line x
1
2 3. r 2 for the area of a sector of a circle
2
with radius r and central angle . [Hint: Assume 0
and place the center of the circle at the origin so it has the 35. Prove the formula A 40. A water storage tank has the shape of a cylinder with diameter 10 ft. It is mounted so that the circular crosssections are vertical. 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 x 2 y R2
about the xaxis. Find the volume enclosed by the torus. r2 5E08(pp 530539) 532 ❙❙❙❙ 1/17/06 6:03 PM Page 532 CHAPTER 8 TECHNIQUES OF INTEGRATION  8.4 Integration of Rational Functions by Partial Fractions
In this section we show how to integrate any rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called partial fractions, that we already know
how to integrate. To illustrate the method, observe that by taking the fractions 2 x 1
and 1 x 2 to a common denominator we obtain
2
x 1
1 2x x 2 2
x1
x 1x 2 x
x 2 5
x 2 If we now reverse the procedure, we see how to integrate the function on the right side of
this equation: yx x 5 2 x 2 dx 2 y x 1
1 2 ln x x 1 2
ln x dx
2 C To see how the method of partial fractions works in general, let’s consider a rational
function
Px
Qx fx where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions
provided that the degree of P is less than the degree of Q. Such a rational function is called
proper. Recall that if
Px an x n an 1x n 1 a1 x a0 where a n 0, then the degree of P is n and we write deg P
n.
If f is improper, that is, deg P
deg Q , then we must take the preliminary step
of dividing Q into P (by long division) until a remainder R x is obtained such that
deg R
deg Q . The division statement is
Px
Qx fx 1 Sx Rx
Qx where S and R are also polynomials.
As the following example illustrates, sometimes this preliminary step is all that is
required.
EXAMPLE 1 Find ≈+x +2
x1 ) ˛
+x
˛≈
≈+x
≈x
2x
2x2
2 y x3
x x
d x.
1 SOLUTION Since the degree of the numerator is greater than the degree of the denominator,
we ﬁrst perform the long division. This enables us to write y x3
x x
dx
1 y
x3
3 x2 x
x2
2 2
2x 2
x 1 2 ln x dx
1 C 5E08(pp 530539) 1/17/06 6:03 PM Page 533 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 533 The next step is to factor the denominator Q x as far as possible. It can be shown that
any polynomial Q can be factored as a product of linear factors (of the form ax b) and
irreducible quadratic factors (of the form ax 2 b x c, where b 2 4 ac 0). For instance, if Q x
x 4 16, we could factor it as
x2 Qx 4 x2 4 x 2 x2 2x 4 The third step is to express the proper rational function R x Q x (from Equation 1) as
a sum of partial fractions of the form
A
ax b Ax or i ax 2 B
bx c j A theorem in algebra guarantees that it is always possible to do this. We explain the details
for the four cases that occur.
CASE I ■ The denominator Q x is a product of distinct linear factors. This means that we can write
Qx a1 x b1 a 2 x b2 ak x bk where no factor is repeated (and no factor is a constant multiple of another). In this case
the partial fraction theorem states that there exist constants A1, A2 , . . . , Ak such that
Rx
Qx 2 A1
a 1 x b1 A2
a2 x b2 Ak
a k x bk These constants can be determined as in the following example.
EXAMPLE 2 Evaluate x2 y 2x 3 2x
3x 2 1
d x.
2x SOLUTION Since the degree of the numerator is less than the degree of the denominator,
we don’t need to divide. We factor the denominator as 2x 3 3x 2 2x x 2x 2 3x 2 x 2x 1x 2 Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
x2
x 2x 3  Another method for ﬁnding A, B, and C
is given in the note after this example. 2x 1
1x 2 A
x B
2x C
1 x 2 To determine the values of A, B, and C, we multiply both sides of this equation by the
product of the denominators, x 2 x 1 x 2 , obtaining
4 x2 2x 1 A 2x 1x 2 Bx x 2 Cx 2 x 1 Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get
5 x2 2x 1 2A B 2C x 2 3A 2B Cx 2A 5E08(pp 530539) 534 ❙❙❙❙ 1/17/06 6:04 PM Page 534 CHAPTER 8 TECHNIQUES OF INTEGRATION  Figure 1 shows the graphs of the integrand
in Example 2 and its indeﬁnite integral (with
K 0). Which is which?
2 The polynomials in Equation 5 are identical, so their coefﬁcients must be equal. The
coefﬁcient of x 2 on the right side, 2 A B 2C, must equal the coefﬁcient of x 2 on the
left side—namely, 1. Likewise, the coefﬁcients of x are equal and the constant terms are
equal. This gives the following system of equations for A, B, and C:
2A 2C 1 3A 2B C 2 2A _3 B 2B 2C 3 1 _2 FIGURE 1 1
2 Solving, we get A y x2
2x 3 1
5 ,B 2x
3x 2 1
dx
2x y
1
2  We could check our work by taking the terms
to a common denominator and adding them. 1
10 , and C , and so 11
2x 1
1
5 2x 1
1
10 ln x ln 2 x 1
1
10 x 2
1
10 1 ln x In integrating the middle term we have made the mental substitution u
gives du 2 d x and dx du 2.
NOTE dx
2
2x K
1, which We can use an alternative method to ﬁnd the coefﬁcients A, B, and C in
Example 2. Equation 4 is an identity; it is true for every value of x. Let’s choose values of
x that simplify the equation. If we put x 0 in Equation 4, then the second and third terms
on the right side vanish and the equation then becomes 2 A
1, or A 1 . Likewise,
2
1
1
1
1
x 2 gives 5B 4 4 and x
2 gives 10C
1, so B 5 and C
10 . (You may object
that Equation 3 is not valid for x 0, 1 , or 2, so why should Equation 4 be valid for those
2
values? In fact, Equation 4 is true for all values of x, even x 0, 1 , and 2. See Exercise 67
2
for the reason.)
■ EXAMPLE 3 Find yx dx
2 a2 , where a 0. SOLUTION The method of partial fractions gives 1
x2 a2 x 1
ax A
a x B
a x a and therefore
Ax a Bx a 1 Using the method of the preceding note, we put x a in this equation and get
A 2a
1, so A 1 2a . If we put x
a, we get B 2a
1, so B
1 2a .
Thus y dx
x2 a2 1
2a y 1 1 x a x a 1
(ln x
2a a ln x dx a ) C 5E08(pp 530539) 1/17/06 6:04 PM Page 535 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS Since ln x ❙❙❙❙ 535 ln x y , we can write the integral as ln y dx y 6 x 2 a 1
x
ln
2a
x 2 a
a C See Exercises 53–54 for ways of using Formula 6.
CASE II ■ Q x is a product of linear factors, some of which are repeated. Suppose the ﬁrst linear factor a 1 x b1 is repeated r times; that is, a 1 x b1 r occurs in
the factorization of Q x . Then instead of the single term A1 a 1 x b1 in Equation 2, we
would use
A1
a 1 x b1 7 A2 Ar a1 x b1 2 a1 x r b1 By way of illustration, we could write
x3 x 1
x2 x 1 3 A
x B
x2 C
x D
1 E x 1 2 x 1 3 but we prefer to work out in detail a simpler example. EXAMPLE 4 Find y x 4 2x 2
x3 x2 4x 1
d x.
x1 SOLUTION The ﬁrst step is to divide. The result of long division is x 4 2x 2
x3 x2 4x 1
x1 x x3 The second step is to factor the denominator Q x
we know that x 1 is a factor and we obtain
x3 x2 Since the linear factor x x 1 x 1 x2
2 4x 1. Since Q 1 x x x 1x 1x 1 1 occurs twice, the partial fraction decomposition is
4x
12x Ax
A  Another method for ﬁnding the coefﬁcients:
Put x 1 in (8): B 2.
Put x
1: C
1.
Put x 0: A B C 1. x2 1 1 A
1 x B
1 x Multiplying by the least common denominator, x
8 1 1 x x 4x
x2 x x3 1 1x 1 Bx C x2 B C
1 1
1 2C x 2 2 A B C 0 A B 2C 4 A B C 0 1 1 , we get x Cx
A Now we equate coefﬁcients: x 1
B 2 C 0, 5E08(pp 530539) 536 ❙❙❙❙ 1/17/06 6:04 PM Page 536 CHAPTER 8 TECHNIQUES OF INTEGRATION Solving, we obtain A y x 4 2x 2
x3 x2 1, B 2, and C 4x 1
dx
x1 y x
x x2
2
■ 1 1 x2
2 CASE III 1, so x x 2
1 ln x x x 2 1 x 2
x 1
2 1 ln 1 x
x ln x 1
1
1 dx 1
1 K K Q x contains irreducible quadratic factors, none of which is repeated. If Q x has the factor ax 2 b x c, where b 2 4 ac 0, then, in addition to the partial
fractions in Equations 2 and 7, the expression for R x Q x will have a term of the form
Ax 9 ax 2 B
bx c where A and B are constants to be determined. For instance, the function given by
fx
x x 2 x 2 1 x 2 4 has a partial fraction decomposition of the form
x
2 x2 x A
1 x2 4 x Bx
x2 2 C
1 Dx
x2 E
4 The term given in (9) can be integrated by completing the square and using the formula dx y 10 2x 2
x3 EXAMPLE 5 Evaluate y SOLUTION Since x 3 4x 2 x a 1
tan
a 2 x
a C x4
d x.
4x x x2 4 can’t be factored further, we write 2x 2 x 4
x x2 4
Multiplying by x x 2 1 A
x Bx
x2 C
4 4 , we have
2x 2 4 A x2 4 Bx Cx A x B x2 Cx 4A Equating coefﬁcients, we obtain
A
Thus A 1, B B 1, and C y 2x 2
x3 2 C 1 4A 4 1 and so
x4
dx
4x y 1
x x
x2 1
4 dx 5E08(pp 530539) 1/17/06 6:05 PM Page 537 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 537 In order to integrate the second term we split it into two parts: y x
x2 1
dx
4 x y 2 x yx dx 4 1
2 dx 4 We make the substitution u x 2 4 in the ﬁrst of these integrals so that du
We evaluate the second integral by means of Formula 10 with a 2: y 2x 2 x 4
dx
x x2 4 y 1
dx
x y
1
2 ln x
EXAMPLE 6 Evaluate 4x 2 y 4x 3x
4x 2 x
4 ln x 2 1 y dx x2 4 1
2 x2
1 tan 2 x d x. dx 4
x2 K 2
d x.
3 SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we ﬁrst divide and obtain 4x 2
4x 2 3x
4x 2
3 x 1 1
4x 4x 2 3 Notice that the quadratic 4 x 2 4 x 3 is irreducible because its discriminant is
b 2 4ac
32 0. This means it can’t be factored, so we don’t need to use the
partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x 2 4x 3 2x This suggests that we make the substitution u
x
u 1 2, so
4x 2
4x 2 3x
4x 2
dx
3 x y 1
4 yu 2 2 x 1
8 ln u 2 2 x ■ 1
2 x NOTE 1 1
8 ln 4 x 2 4x
1
2 2 u 2x 1
4x y
x y 1
u2 1 1
4 4x 2 d x and dx du
1
4 du 2 1. Then, du 3 2 u 2 1 y 1
4 x
1
u 2 1
s2
3 2
tan
1 4 s2 u yu 2 1
du
2 du
u 1 C s2
tan 1 2x
s2 1 C Example 6 illustrates the general procedure for integrating a partial fraction of the form
Ax B
ax 2 bx c where b 2 4 ac 0 5E08(pp 530539) 538 ❙❙❙❙ 1/17/06 6:05 PM Page 538 CHAPTER 8 TECHNIQUES OF INTEGRATION We complete the square in the denominator and then make a substitution that brings the
integral into the form y Cu
u2 D
du
a2 Cy u
u2 Dy du a2 1
u2 a2 du Then the ﬁrst integral is a logarithm and the second is expressed in terms of tan 1.
CASE IV ■ Q x contains a repeated irreducible quadratic factor. If Q x has the factor a x 2
partial fraction (9), the sum
A1 x B1
ax 2 bx c 11 c r, where b 2 bx A2 x
ax 2 4ac 0, then instead of the single B2
bx Ar x
c 2 ax Br 2 bx r c occurs in the partial fraction decomposition of R x Q x . Each of the terms in (11) can be
integrated by ﬁrst completing the square.
 It would be extremely tedious to work out by
hand the numerical values of the coefﬁcients in
Example 7. Most computer algebra systems,
however, can ﬁnd the numerical values very
quickly. For instance, the Maple command
convert f, parfrac, x EXAMPLE 7 Write out the form of the partial fraction decomposition of the function xx Apart[f]
gives the following values: E 1,
15
8 1
8 B , 1
8 F
I 1
2 , x3
1 x2 xx , C
, D G
J x2
x 1
1 x2 1 Ex
x2 F
1 Gx
x2 3 SOLUTION or the Mathematica command A x3
1 x2 H x2
x 1
1 x2 1 3 1,
3
4 A
x , 1
2 B
x Cx D
x2 x 1 1 EXAMPLE 8 Evaluate y 1 x 2x 2
x x2 1 x3 H
12 Ix
x2 J
13 d x. 2 SOLUTION The form of the partial fraction decomposition is 1 x 2x 2
x x2 1 Multiplying by x x 2
x3 2x 2 x3 Bx
x2 C
1 Bx C x x2 Dx
x2 E
12 1 2, we have
1 A x2 1 A x4 x A
x 2 2x 2 A 2 1 B x4 x2 Cx 3 B x4 1
2A B Dx
C x3 Ex
Dx 2 x D x2 C Ex Ex
A If we equate coefﬁcients, we get the system
A B 0 C which has the solution A 1 2A
1, B B 1, C D
1, D 2 C
1, and E E 1 0. Thus A 1 5E08(pp 530539) 1/17/06 6:05 PM Page 539 S ECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 1 x 2x 2
x x2 1 x3 y dx 2  In the second and fourth terms we made the
mental substitution u x 2 1. 1
x x
x2 y y dx
x yx
1
2 ln x 1
1 x
x2 x
2 dx dx yx 1 1 tan 1x ln x 2 539 dx 2 1 ❙❙❙❙ 2 y 1 x dx
x2 1 1
2x 2 1 2 K We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral
x2
x x2 y 1
dx
3 could be evaluated by the method of Case III, it’s much easier to observe that if
u x x2 3
x 3 3x, then du
3x 2 3 d x and so y x2
x x2 1
dx
3 1
3 ln x 3 3x C Rationalizing Substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form
n
n
st x , then the substitution u st x may be effective. Other instances appear in the
exercises. EXAMPLE 9 Evaluate
SOLUTION Let u y sx 4
x d x. 4. Then u 2 sx x u2 4, so x 4 and dx 2u du. Therefore y sx 4
x dx yu
2 u
2 y 4
1 4
u 2 sx 4
x dx 2 y du
2u 2 sx 8y 22
4 4 du 4 as u 2u 2 and using du
u2 1 8 u2 du 4 We can evaluate this integral either by factoring u 2
partial fractions or by using Formula 6 with a 2: y u2 2y 2u du ln 2 ln 4
u
u
sx
sx 2
2
4
4 C
2
2 C 5E08(pp 540549) ❙❙❙❙ 540 1/17/06 5:30 PM Page 540 CHAPTER 8 TECHNIQUES OF INTEGRATION  8.4 Exercises 1–6  Write out the form of the partial fraction decomposition
of the function (as in Example 7). Do not determine the numerical
values of the coefﬁcients. 1. (a) 2x
3 3x x x
2. (a) 3
x
3. (a) x (b) 1 x
(b) 3
x 1
x2
2
3x 2 x3 (b) 4 4. (a) x 6. (a) 4 x x4
x x2 3
■ 7. y 11. y ■ x yx 9. x
■ ■ 37. 2 x x
1 x2 x 2x 1
1 3 x2
t4 t x t2
1 t2 2 6 x9
5x x 1 3 x 2 1 1
4 dx dx 1 x 13.
15. yx
y r2 yr 10. y
y 0 17. y 19. y 21. y 23.
25.
27.
29. 1 y
y
y 4y
yy 14. dx bx 2x
x 1 2 2 3
dx
12
2 16. 7y 12
dy
2y 3
1 x 5 2 x 1 dx 5x 2 3x 2
dx
x 3 2x 2
x 3 1 y y sx 45. y sx dt 46. y sx dx 47. ye ■ 0 x 1 2 3x 2 x 10
1 x2 9 dx x 3 x 2 2x 1
dx
x2 1 x2 2 yx x
2 1 y
y x
1 0 ax 10
dx
6 x2 1 y 20. y 22. ys 4
2x 5 dx 26.
28.
30. y
y
y
y dx b x 3 4x
x2 x
2x
x3 x
x2
3x x dx 2 2 dx s
x x 1 dx 3 x2 x 6
dx
x 3 3x
x2
x
x 3 2x 1
dx
1 2 x2 1
2 x4 ■ ■ yx dx 1 2 x 3 5x
dx
x
5x 2 4
4 0 ■ dx x4
x2 1
1 ■ dx 2 ■ ■ ■ ■ x 40. y 44. dx 4 y 1
sx yx 42. dx 1 1 3
sx 1 0 2 dx dx ■ 3 x3
2 1
1
3
sx 1 3 4
sx dx
dx [Hint: Substitute u dx [Hint: Substitute u e 2x
3e x 2x ■ 3 13 ■ 2
■ 48. dx
■ ■ ■ sx
dx
x2 x
6
sx .] sx .] 12 cos x
dx
sin x sin x y 2 ■ ■ ■ ■ ■ 49–50  Use integration by parts, together with the techniques of
this section, to evaluate the integral. 49. y ln x 2 ■ x
■ 50. 2 dx
■ ■ ■ ■ y x tan
■ 1 x dx ■ ■ ■ ■ 1 x 2 2 x 3 to decide whether
; 51. Use a graph of f x
2
x0 f x d x is positive or negative. Use the graph to give a rough
estimate of the value of the integral and then use partial
fractions to ﬁnd the exact value. 2 3 1 1 sx 16 9 ■ ds
2 ■ y x sx 1 x 1 18. 24. dx 1 ■ 39. 2 ■ 1
4t t 4 dx 1 x3
3 13  Make a substitution to express the integrand as a rational
function and then evaluate the integral. dr 4 2 x 2x 2 y 38. 2 3 2 ■ ■ ax x 41. ■ 4 yx x2 0 39–48 2 4 34. dx dx
4 x ■ x3 ■ 12. 2 y 3x 2 y 36. 2x 3 x 2 1
x6 ■ 8. dx yx x2 5 x
4x 1 32. dx 1 43. (b) 3 35. Evaluate the integral.  2 (b) 1 ■ 7–38 (b) 3 x4 5. (a) ■ x
4x x2 y 1
3 x 1
x 3 yx 33. 1
2x 2 31. 2x
x1
dx
5x 2 4 1 x3 ; 52. Graph both y
same screen. 2 x 2 and an antiderivative on the 53–54  Evaluate the integral by completing the square and using
Formula 6. 53.
■ yx
■ dx
2 54. 2x
■ ■ ■ ■ ■ y 4x
■ 2x
2
■ 1
12 x
■ 7 dx
■ 55. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution t tan x 2 will convert any ■ 5E08(pp 540549) 1/17/06 5:31 PM Page 541 S ECTION 8.5 STRATEGY FOR INTEGRATION x
2 1 and t2 s1 sin x
2 t t2 (b) Show that
t2
t2 1
1 cos x and 2t sin x 2 56–59 t2 1 dt CAS 58. y dx
5 sin x ■ 1
sin x 2
3 1 ■ 60–61 ■  57. cos x ■ ■ 59. dx
■ y 3 sin x
y 2 sin x ■ ■ ■ 1 sin 2 x
■ fx dx 4 cos x
1 dx
■ ■ Find the area of the region under the given curve from x
x
x 61. y
■ 1
6x 2 ■ 1
,
1
■ 8
a a 5, b 2, b CAS 4 x 3 27x 2 5 x 32
13x 4 50 x 3 286 x 2 299x 100 x 6 12 x 5 7x 3 13x 2 8
80 x
116 x 4 80 x 3 41x 2
5 20 x 4 (b) Use part (a) to ﬁnd x f x d x and graph f and its indeﬁnite
integral on the same screen.
(c) Use the graph of f to discover the main features of the
graph of x f x d x. 10 67. Suppose that F, G, and Q are polynomials and
■ ■ ■ ■ ■ ■ ■ ■ ■ Fx
Qx 62. Find the volume of the resulting solid if the region under the curve y 1 x 2 3x 2 from x 0 to x
about (a) the xaxis and (b) the yaxis. 1 is rotated 63. One method of slowing the growth of an insect population without using pesticides is to introduce into the population a
number of sterile males that mate with fertile females but produce no offspring. If P represents the number of female insects
in a population, S the number of sterile males introduced each
generation, and r the population’s natural growth rate, then the  8.5 70 66. (a) Find the partial fraction decomposition of the function fx 3 , 30 x 5 (b) Use part (a) to ﬁnd x f x d x (by hand) and compare with
the result of using the CAS to integrate f directly.
Comment on any discrepancy. a to b.
60. y dP decomposition of the function Use the substitution in Exercise 55 to transform the integrand into a rational function of t and then evaluate the integral. y3 S 65. (a) Use a computer algebra system to ﬁnd the partial fraction  56. S
1P 1 as a difference of squares by ﬁrst adding and
subtracting the same quantity. Use this factorization to evaluate
x 1 x 4 1 d x. (c) Show that
dx r 64. Factor x 4 t2 1 P yP Suppose an insect population with 10,000 females grows at a
rate of r 0.10 and 900 sterile males are added. Evaluate the
integral to give an equation relating the female population to
time. (Note that the resulting equation can’t be solved explicitly for P.) t
s1 541 female population is related to time t by rational function of sin x and cos x into an ordinary rational
function of t.
x
(a) If t tan x 2 ,
, sketch a right triangle or use
trigonometric identities to show that
cos ❙❙❙❙ Gx
Qx 0. Prove that F x
for all x except when Q x
all x. [Hint: Use continuity.]
68. If f is a quadratic function such that f 0 y fx
xx 1
2 3 G x for 1 and dx is a rational function, ﬁnd the value of f 0 . Strategy for Integration
As we have seen, integration is more challenging than differentiation. In ﬁnding the derivative of a function it is obvious which differentiation formula we should apply. But it may
not be obvious which technique we should use to integrate a given function.
Until now individual techniques have been applied in each section. For instance, we
usually used substitution in Exercises 5.5, integration by parts in Exercises 8.1, and partial
fractions in Exercises 8.4. But in this section we present a collection of miscellaneous integrals in random order and the main challenge is to recognize which technique or formula
to use. No hard and fast rules can be given as to which method applies in a given situation,
but we give some advice on strategy that you may ﬁnd useful. 5E08(pp 540549) 542 ❙❙❙❙ 1/17/06 5:32 PM Page 542 CHAPTER 8 TECHNIQUES OF INTEGRATION A prerequisite for strategy selection is a knowledge of the basic integration formulas.
In the following table we have collected the integrals from our previous list together with
several additional formulas that we have learned in this chapter. Most of them should be
memorized. It is useful to know them all, but the ones marked with an asterisk need not be
memorized since they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20.
Table of Integration Formulas Constants of integration have been omitted.
1. yx n dx xn 1
n1 3. ye x dx ex 5. y sin x d x 7. y sec x d x 9. y sec x tan x d x 2 1
dx
x 2. tan x
sec x y cos x d x 8. cos x ya 6. 1 y 4. n y csc x d x x ln x
ax
ln a dx sin x 2 cot x 10. y csc x cot x d x 12. y csc x d x ln csc x
ln sin x 11. y sec x d x ln sec x 13. y tan x d x ln sec x 14. y cot x d x 15. y sinh x d x cosh x 16. y cosh x d x 17. y 18. y *19. y *20. y dx
x 2 a 2 dx
x2 a2 1
tan
a tan x x
a 1 1
x
ln
2a
x a
a sa x cot x sinh x dx
2 csc x 2 dx
sx 2 a 2 sin 1 ln x x
a
sx 2 a2 Once you are armed with these basic integration formulas, if you don’t immediately see
how to attack a given integral, you might try the following fourstep strategy.
1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples: y s x (1
tan y sec 2 d sx ) dx
sin y cos
y sin y (s x x) d x cos2 d
cos d 1
2 y sin 2 d 5E08(pp 540549) 1/17/06 5:32 PM Page 543 SECTION 8.5 STRATEGY FOR INTEGRATION sin x cos x 2 d x y sin 2x y y 1 ❙❙❙❙ 543 cos 2x d x 2 sin x cos x 2 sin x cos x d x 2. Look for an Obvious Substitution Try to ﬁnd some function u t x in the intet x d x also occurs, apart from a constant factor. grand whose differential du
For instance, in the integral y x
x 2 1 dx we notice that if u x 2 1, then du 2 x d x. Therefore, we use the substitution u x 2 1 instead of the method of partial fractions.
3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f x .
(a) Trigonometric functions. If f x is a product of powers of sin x and cos x,
of tan x and sec x, or of cot x and csc x, then we use the substitutions recommended in Section 8.2.
(b) Rational functions. If f is a rational function, we use the procedure of Section 8.4 involving partial fractions.
(c) Integration by parts. If f x is a product of a power of x (or a polynomial)
and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 8.1. If you look at the functions in Exercises 8.1, you will see that most of them are the type just described.
(d) Radicals. Particular kinds of substitutions are recommended when certain
radicals appear.
(i) If s x 2 a 2 occurs, we use a trigonometric substitution according to the
table in Section 8.3.
n
n
(ii) If sax b occurs, we use the rationalizing substitution u sax b.
n
More generally, this sometimes works for st x .
4. Try Again If the ﬁrst three steps have not produced the answer, remember that
there are basically only two methods of integration: substitution and parts.
(a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration
or ingenuity (or even desperation) may suggest an appropriate substitution.
(b) Try parts. Although integration by parts is used most of the time on products
of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 8.1, we see that it works on tan 1x, sin 1x, and ln x,
and these are all inverse functions.
(c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the
denominator or using trigonometric identities) may be useful in transforming
the integral into an easier form. These manipulations may be more substantial
than in Step 1 and may involve some ingenuity. Here is an example: y1 dx
cos x y
y 1
1 1
cos x 1
1 cos x
dx
sin 2x cos x
dx
cos x y csc 2x y 1
1 cos x
sin 2x cos x
dx
cos 2x
dx 5E08(pp 540549) 544 ❙❙❙❙ 1/17/06 5:32 PM Page 544 CHAPTER 8 TECHNIQUES OF INTEGRATION (d) Relate the problem to previous problems. When you have built up some experience in integration, you may be able to use a method on a given integral that
is similar to a method you have already used on a previous integral. Or you
may even be able to express the given integral in terms of a previous one. For
instance, x tan 2x sec x d x is a challenging integral, but if we make use of the
identity tan 2x sec 2x 1, we can write
2 3 y tan x sec x d x y sec x d x y sec x d x
and if x sec 3x d x has previously been evaluated (see Example 8 in Section 8.2),
then that calculation can be used in the present problem.
(e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions
of different types, or it might combine integration by parts with one or more
substitutions.
In the following examples we indicate a method of attack but do not fully work out the
integral.
EXAMPLE 1 tan 3x y cos x d x
3 In Step 1 we rewrite the integral: y tan 3x
dx
cos 3x 3 3 y tan x sec x d x The integral is now of the form x tan m x sec n x d x with m odd, so we can use the advice in
Section 7.2.
Alternatively, if in Step 1 we had written y tan 3x
dx
cos 3x sin 3x y cos x
3 sin 3x 1
dx
cos 3x y cos x d x
6 then we could have continued as follows with the substitution u y sin 3x
dx
cos 6x y
y EXAMPLE 2 ye sx 1 cos 2x
sin x d x
cos 6x u2 1
u 6 du y u y
4 u2 1
u
u cos x : 6 6 du du dx According to Step 3(d)(ii) we substitute u ye sx dx s x. Then x u 2, so dx 2u du and 2 y ue u du The integrand is now a product of u and the transcendental function e u so it can be integrated by parts. 5E08(pp 540549) 1/17/06 5:33 PM Page 545 SECTION 8.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 545 x5 1
dx
3
3x 2 10x
No algebraic simpliﬁcation or substitution is obvious, so Steps 1 and 2 don’t apply here.
The integrand is a rational function so we apply the procedure of Section 8.4, remembering that the ﬁrst step is to divide.
E XAMPLE 3 yx EXAMPLE 4 y x sln x dx Here Step 2 is all that is needed. We substitute u
du d x x, which occurs in the integral. ln x because its differential is 1x
dx
1x
Although the rationalizing substitution
EXAMPLE 5 y 1
1 u x
x works here [Step 3(d)(ii)], it leads to a very complicated rational function. An easier
method is to do some algebraic manipulation [either as Step 1 or as Step 4(c)]. Multiplying numerator and denominator by s1 x, we have y 1
1 x
dx
x 1 x
dx
x2 y s1
y s1 1
x2 sin 1x dx s1 y s1
x2 x
x2 dx C Can We Integrate All Continuous Functions?
The question arises: Will our strategy for integration enable us to ﬁnd the integral of every
2
continuous function? For example, can we use it to evaluate x e x d x ? The answer is no, at
least not in terms of the functions that we are familiar with.
The functions that we have been dealing with in this book are called elementary functions. These are the polynomials, rational functions, power functions x a , exponential
functions a x , logarithmic functions, trigonometric and inverse trigonometric functions,
hyperbolic and inverse hyperbolic functions, and all functions that can be obtained from
these by the ﬁve operations of addition, subtraction, multiplication, division, and composition. For instance, the function
fx x2
x 3 1
2x ln cosh x 1 xe sin 2 x is an elementary function.
If f is an elementary function, then f is an elementary function but x f x d x need not
2
be an elementary function. Consider f x
e x . Since f is continuous, its integral exists,
and if we deﬁne the function F by
Fx y x 0 2 e t dt 5E08(pp 540549) ❙❙❙❙ 546 1/17/06 5:34 PM Page 546 CHAPTER 8 TECHNIQUES OF INTEGRATION then we know from Part 1 of the Fundamental Theorem of Calculus that
ex Fx 2 2 Thus, f x
e x has an antiderivative F , but it has been proved that F is not an elementary function. This means that no matter how hard we try, we will never succeed in evalu2
ating x e x d x in terms of the functions we know. (In Chapter 11, however, we will see how
2
to express x e x d x as an inﬁnite series.) The same can be said of the following integrals: y ex
dx
x y sx 3 y sin x y cos e dx 1
dx
ln x y 1 dx 2 y x dx sin x
dx
x In fact, the majority of elementary functions don’t have elementary antiderivatives. You
may be assured, though, that the integrals in the following exercises are all elementary
functions.  8.5
1–80 1.
3.
5. Evaluate the integral.  sin x sec x
dx
tan x y
y 2t 2 t 0 y 1
3 2 9. yx 8. r ln r dr
x
2 1 3 x
x2 16. y 18. y1 20. ye dt 22. y x sin s x )8 d x 24. y ln x 17. y x sin x d x 19. ye 21. yt e 23. y (1 32 x
s1 x 2 dx 2 1 5 y y 0 4x x ex dx 4 s1 3 dx 1 x2 s2 2 0 x 2 dx e2t
dt
e4t sx 31. y 33. dx s1 ln x
dx
x ln x 15. 2t x2 14. y 3 1 y sin x cos cos x dx
1 x2 yx 28. y sin sat dt 1
dw
2 30. y x
dx
x 32. y y s3 34. y 35. y 36. y sin 4 x cos 3x dx 37. y 38. y 39. y 40. y s4y 41. y 42. yx y e s1 44. y s1 45. yx e dx 46. y1 dx 2 47. yx a
dx
a2 48. y 3w 5 w 0 cot x d x x 4 0 y 43. x4 y x csc x
y 29. 3x 2 2
dx
2x 8 26. 3 dx dx 12. 5 13. 0 x y s3 y cot x ln sin x 2 d yx dx cos 5 d 12 3 10. 4x y sin y tan yx 27. 2. 6. 4 y 1 3 3x 2 2
dx
2x 8 25. 4. dt e arctan y
dy
1 y2 1 7. 11. Exercises 1 x dx
1 dx dx 1
1 1 x 2 dx 2x x 8 sin x dx 1 4 0 cos2 tan2 d
x
x2 1 s1 tan2 d
x 5 x
2 e x dx x3 x2 dx 2 x2 2 4x dx s2 x
2x
1
4 2
4 4 0 2 1
dx
3
4 cot x
dx
cot x tan 5 sec 3 d
1 2 4y tan 1x d x
e x dx
ex
dx
ex 1
x
x4 a4 dx 3 dy 5E08(pp 540549) 1/17/06 5:35 PM Page 547 ❙❙❙❙ S ECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS 49. 1 y x s4 x 1 1 51. y x s4 x 53. yx 55. yx 57. y x sx 59. ye 3x 61. yx 10 63. y sx e 65. 2 50. dx 1
s4 x 54. y 56. y sx c dx 58. yx dx 60. yx dx 62. y 64. y dx 1
4 sx 3 1
ex
x4
16 1 dx
1
1  8.6 sx dx dx 66. sin x 2 d x x x ln x y 2 2 1 ln 1 4
3 2 dx
x dx 1
3
sx x3
x1
3 dx dx 10 dx ln tan x
dx
sin x cos x
3 u
u3 arctan st
dt
st 3 68. y1 e 2x
dx
ex 70. y 72. y1 74. ye 76. y 78. y sin x 80. y sin 1
du
u2 67. y 69. 1 dx
x4 1 yx 1 4 y sx 2 52. 2 sinh mx dx sx yx y1 71. yx 73. y 75. y sin x sin 2 x sin 3x d x 77. y1 79. y x sin ■ 1 x
4x 2 4 3 1
2 x2 x dx 4 dx sx
dx
x3
2 x cos x d x ■ ■ ■ ■ ■ 2 e x and y
antiderivatives, but y
2x 2
2
x2
x 2 x 1 e d x. 81. The functions y 1
2e x 547 e x dx ln x 1
dx
x2
st
dt
3
st
x dx
e x2 x bx sin 2 x dx sec x cos 2 x
dx
sec x
sin x cos x
dx
4
x cos 4 x ■ ■ ■ ■ ■ ■ 2 x 2e x don’t have elementary
2
1 e x does. Evaluate Integration Using Tables and Computer Algebra Systems
In this section we describe how to use tables and computer algebra systems to integrate
functions that have elementary antiderivatives. You should bear in mind, though, that even
the most powerful computer algebra systems can’t ﬁnd explicit formulas for the antideriv2
atives of functions like e x or the other functions described at the end of Section 8.5. Tables of Integrals
Tables of indeﬁnite integrals are very useful when we are confronted by an integral that is
difﬁcult to evaluate by hand and we don’t have access to a computer algebra system. A relatively brief table of 120 integrals, categorized by form, is provided on the Reference Pages
at the back of the book. More extensive tables are available in CRC Standard Mathematical Tables and Formulae, 30th ed. by Daniel Zwillinger (Boca Raton, FL: CRC
Press, 1995) (581 entries) or in Gradshteyn and Ryzhik’s Table of Integrals, Series,
and Products, 6e (New York: Academic Press, 2000), which contains hundreds of pages
of integrals. It should be remembered, however, that integrals do not often occur in exactly
the form listed in a table. Usually we need to use substitution or algebraic manipulation to
transform a given integral into one of the forms in the table.
EXAMPLE 1 The region bounded by the curves y
arctan x, y
about the yaxis. Find the volume of the resulting solid. 0, and x SOLUTION Using the method of cylindrical shells, we see that the volume is V y 1 0 2 x arctan x d x 1 is rotated 5E08(pp 540549) 548 ❙❙❙❙ 1/17/06 5:35 PM Page 548 CHAPTER 8 TECHNIQUES OF INTEGRATION  The Table of Integrals appears on the
Reference Pages at the back of the book. In the section of the Table of Integrals entitled Inverse Trigonometric Forms we locate
Formula 92:
1 y u tan u2 u du 1 u
2 tan 1u 2 C Thus, the volume is
V y 2 1 0 x x tan 1x d x 2 2 1 1 tan x
4 x2 2
x
1
2 1 EXAMPLE 2 Use the Table of Integrals to ﬁnd 1
2 1
0 x
2 tan 1x 2 tan 1 1 1 0 1 2 y x2 d x. 4x 2 s5 SOLUTION If we look at the section of the table entitled Forms involving sa 2 u 2, we see that the closest entry is number 34:
u2 y sa 2 u2 u
sa 2
2 du a2
sin
2 u2 u
a 1 C This is not exactly what we have, but we will be able to use it if we ﬁrst make the substitution u 2 x : y x2
4x 2 s5 y dx 5 (so a Then we use Formula 34 with a 2 y s5 x2
4x 2 1
8 dx y s5 u2 x
s5
8 u 2 2 du
s5 u 2 2 u2 4x 2 u2 du s5 u2 u2 5
sin
2 u
s5
2 5
sin
16 EXAMPLE 3 Use the Table of Integrals to ﬁnd y s5 ):
1
8 du 1
8 yx 3 1 2x
s5 1 u
s5 C C sin x d x. SOLUTION If we look in the section called Trigonometric Forms, we see that none of
the entries explicitly includes a u 3 factor. However, we can use the reduction formula
in entry 84 with n 3: yx
85. yu n cos u du
u n sin u n y un 1 sin u du 3 sin x d x x 3 cos x 3 y x 2 cos x d x We now need to evaluate x x 2 cos x dx. We can use the reduction formula in entry 85
with n 2, followed by entry 82: yx 2 cos x d x x 2 sin x 2 y x sin x d x x 2 sin x 2 sin x x cos x K 5E08(pp 540549) 1/17/06 5:36 PM Page 549 SECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 549 Combining these calculations, we get yx
where C 3 x 3 cos x sin x d x 3x 2 sin x 6 x cos x y x sx 2 2x SOLUTION Since the table gives forms involving sa 2 not sax 2 bx If we make the substitution u
pattern sa 2 u 2 : y x sx 2 2x
x 2x 4 x 2, sa 2 y u su
u
sa 2
2 u2 a2
ln (u
2 sa 2 u2) a 2, but u 1 (so x y 4 dx 2 1 1), the integrand will involve the 2 1
2 3 du 2 1
2 y st dt y su 2 u
su 2
2 3 du 3 du 3 du t 2 3 du 3:
1
3 u2 3 32 s3 :
3
2 3 y su u2 2 32
3 For the second integral we use Formula 21 with a
C 3 1 su 2 u The ﬁrst integral is evaluated using the substitution t u 2 du x 2, and s x 2 x y u su 2 4 d x. c , we ﬁrst complete the square:
x2 y sa C 3K . EXAMPLE 4 Use the Table of Integrals to ﬁnd 21. 6 sin x ln(u su 2 3) Thus y x sx 2 2x
1
3 x2 4 dx
2x 4 32 x 1
2 sx 2 2x 4 3
2 ln( x 1 sx 2 2x 4) C Computer Algebra Systems
We have seen that the use of tables involves matching the form of the given integrand with
the forms of the integrands in the tables. Computers are particularly good at matching patterns. And just as we used substitutions in conjunction with tables, a CAS can perform
substitutions that transform a given integral into one that occurs in its stored formulas. So
it isn’t surprising that computer algebra systems excel at integration. That doesn’t mean
that integration by hand is an obsolete skill. We will see that a hand computation sometimes produces an indeﬁnite integral in a form that is more convenient than a machine
a
n
s
w
e
r
.
To begin, let’s see what happens when we ask a machine to integrate the relatively
simple function y 1 3x 2 . Using the substitution u 3x 2, an easy calculation
by hand gives
1
y 3x 2 d x 13 ln 3x 2 C 5E08(pp 550559) 550 ❙❙❙❙ 1/17/06 6:00 PM Page 550 CHAPTER 8 TECHNIQUES OF INTEGRATION whereas Derive, Mathematica, and Maple all return the answer
1
3 ln 3x 2 The ﬁrst thing to notice is that computer algebra systems omit the constant of integration. In other words, they produce a particular antiderivative, not the most general one.
Therefore, when making use of a machine integration, we might have to add a constant.
Second, the absolute value signs are omitted in the machine answer. That is ﬁne if our
problem is concerned only with values of x greater than 2 . But if we are interested in other
3
values of x, then we need to insert the absolute value symbol.
In the next example we reconsider the integral of Example 4, but this time we ask a
machine for the answer. y x sx EXAMPLE 5 Use a computer algebra system to ﬁnd 2 2x 4 d x. SOLUTION Maple responds with the answer
1
3 x2 2x 4 1
4 32 2 sx 2 2x 2x 3
s3
arcsinh
1
2
3 4 x This looks different from the answer we found in Example 4, but it is equivalent because
the third term can be rewritten using the identity
ln( x sx 2 1) s3
1
3 arcsinh x  This is Equation 7.6.3. x s1 1
3 Thus
arcsinh s3
1
3 x ln
ln 1
1
s3 ln 1
s3 [ x  x s1 ln( x 1 2 x 2 1
3 sx 2 4) 2x The resulting extra term 3 ln(1 s3 ) can be absorbed into the constant of integration.
2
Mathematica gives the answer
5
6 x2
3 x
6 sx 2 2x 4 3
1x
arcsinh
2
s3 Mathematica combined the ﬁrst two terms of Example 4 (and the Maple result) into a
single term by factoring.
Derive gives the answer
1
6 sx 2 2x 4 2x 2 x 5 3
2 ln(sx 2 2x 4 x 1) The ﬁrst term is like the ﬁrst term in the Mathematica answer, and the second term is
identical to the last term in Example 4.
EXAMPLE 6 Use a CAS to evaluate yx x2 5 8 d x. SOLUTION Maple and Mathematica give the same answer:
1
18 x 18 5
2 x 16 50 x 14 1750
3 x 12 4375x 10 21875 x 8 218750
3 x6 156250 x 4 390625
2 x2 5E08(pp 550559) 1/17/06 6:00 PM Page 551 ❙❙❙❙ S ECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS 551 It’s clear that both systems must have expanded x 2 5 8 by the Binomial Theorem and
then integrated each term.
If we integrate by hand instead, using the substitution u x 2 5, we get
x2 yx  Derive and the TI89/92 also give this answer. 1
18 5 8 dx x2 5 9 C For most purposes, this is a more convenient form of the answer.
EXAMPLE 7 Use a CAS to ﬁnd 5 2 y sin x cos x d x. SOLUTION In Example 2 in Section 8.2 we found that
5 1
3 2 y sin x cos x dx 1 2
5 cos 3x 1
7 cos7x 8
105 cos 5x cos 3x C Derive and Maple report the answer
1
7 sin 2x cos 3x whereas Mathematica produces 10 5
64 5 0 cos x 1
192 3
320 cos 3x cos 5x 1
448 cos 7x We suspect that there are trigonometric identities which show these three answers are
equivalent. Indeed, if we ask Derive, Maple, and Mathematica to simplify their expressions using trigonometric identities, they ultimately produce the same form of the answer
as in Equation 1. F EXAMPLE 8 If f x FIGURE 1 Graph F for 0 x x 60 sin 4x cos 5x, ﬁnd the antiderivative F of f such that F 0
5. Where does F have extreme values and inﬂection points? 0. SOLUTION The antiderivative of f produced by Maple is 10
fª F x2 20
3 _7 FIGURE 2  Use the indicated entry in the Table of Integrals on the
Reference Pages to evaluate the integral. 2x 2 s7
x sin x cos 6x 4
7 cos 4x sin x 16
21 cos 2x sin x 32
21 sin x Exercises 1–4 y 20
7 sin 3x cos 6x and we note that F 0
0. This expression could probably be simpliﬁed, but there’s no
need to do so because a computer algebra system can graph this version of F as easily as
any other version. A graph of F is shown in Figure 1. To locate the extreme values of F,
we graph its derivative F
f in Figure 2 and observe that F has a local maximum
when x 2.3 and a local minimum when x 2.5. The graph of F
f in Figure 2
shows that F has inﬂection points when x 0.7, 1.3, 1.8, 2.4, 3.3, and 3.9. 5  8.6 1
2 Fx f 0 1. 4
35 sin 4x cos 3x 2 dx ; entry 33 2. y s3 2x y sec 4. 3x 3. ye 2 3 x dx ; entry 71 sin 3 d ; entry 98 dx ; entry 55
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E08(pp 550559) 552 ❙❙❙❙ 1/17/06 6:01 PM Page 552 CHAPTER 8 TECHNIQUES OF INTEGRATION 5–30  Use the Table of Integrals on the Reference Pages to
evaluate the integral. 5. y 1 2 x cos 1x dx 0 y 6. 1
x 2 s4 x 2 3 2 7 CAS 35–42  Use a computer algebra system to evaluate the integral.
Compare the answer with the result of using tables. If the answers
are not the same, show that they are equivalent. dx 9.
11. 3x ye dx
x 2s4x 2 y
y 0
1 t 2e t dt y x2 1 x 3 4 dx y sin x cos 2x d x 38. y tan x 2 sec 4x d x 39. 3 y x s1 2x dx 40. y sin 4x d x y tan5x d x 42. y x 5s x 3 dy x 2 cos 3x d x y 12. 36. x 2 dx s2y 2
y2 y 10. 9 3 x 2 dx 41. y csc 8. cos 4 x d x y x 2s5 37. 7. 35. 0 2 1 dx ■ ■ 3 tan 1 z
dz
z2 13. y 15. ye 17. y y s6 19. y sin x cos x ln sin x 21.
23. x ex
dx
3 e2x
5 25. y 27. y se 29. y sx 0 ■ ■ ■ CAS cos 3 x 2 d x x s4 x
6 x 4e y ye t ■ x s4 x 2, 0 x 32. The region under the curve y 2 x 1 dx 2x 1 dx instead. Why do you think it was successful with this form of
the integrand?
CAS 44. Try to evaluate y ■ 1 x ln x 2 dx ln x s1 with a computer algebra system. If it doesn’t return an answer,
make a substitution that changes the integral into one that the
CAS can evaluate. 3 dt CAS
■ ■ 45–48  Use a CAS to ﬁnd an antiderivative F of f such
that F 0
0. Graph f and F and locate approximately the
xcoordinates of the extreme points and inﬂection points of F . 45. f x x2
x4 2 bu. 46. f x
47. f x xe 48. f x
■ ■ 1
x2 x 1 sin x, 5 x sin x cos x, 0 x 4 3 34. Verify Formula 31 (a) by differentiation and (b) by substituting a sin . ■ x dx 33. Verify Formula 53 in the Table of Integrals (a) by differentiation u ■ 4 2, is rotated about the a ■ 43. Computer algebra systems sometimes need a helping hand dx sin t ■ ■ y 2 x s2 2x dx
x ■ If it doesn’t return an answer, ask it to try tan x from 0 to
4 is rotated
about the xaxis. Find the volume of the resulting solid.
and (b) by using the substitution t ■ y 2 x s4 31. Find the volume of the solid obtained when the region under the curve y
yaxis. ■ dx sec 2 tan 2
d
tan 2 ■ ■ from human beings. Ask your CAS to evaluate y s9 0 ■ 2e x 3 30. 2
■ ■ dx
1 y sin
1 2 s2 28. 1 dx 10 y 2 sx d x x5 26. dx x ■ x 2 x 4 dx ■ ye 24. ln x 2x 2 22. y sec x d x
s4 yx 20. dx y x sin x 18. 4y2 dy 4y y sin 16. sech e x d x 2 y 1 14. x
x6
■ 6 5 x
1
■ ■ ■ ■ ■ ■ ■ ■ ■ 5E08(pp 550559) 1/17/06 6:01 PM Page 553 D ISCOVERY PROJECT PATTERNS IN INTEGRALS ❙❙❙❙ 553 DISCOVERY PROJECT
CAS Patterns in Integrals
In this project a computer algebra system is used to investigate indeﬁnite integrals of families of
functions. By observing the patterns that occur in the integrals of several members of the family,
you will ﬁrst guess, and then prove, a general formula for the integral of any member of the
family.
1. (a) Use a computer algebra system to evaluate the following integrals. (i) y (iii) y x 1
2x 3 x 1
2x 5 dx (ii) y dx (iv) y 1
1x x
1
x 2 2 5 dx dx (b) Based on the pattern of your responses in part (a), guess the value of the integral y x 1
ax b dx if a b. What if a b?
(c) Check your guess by asking your CAS to evaluate the integral in part (b). Then prove it
using partial fractions.
2. (a) Use a computer algebra system to evaluate the following integrals. (i) y sin x cos 2 x dx (ii) y sin 3 x cos 7x dx (iii) y sin 8 x cos 3 x dx (b) Based on the pattern of your responses in part (a), guess the value of the integral y sin ax cos bx dx
(c) Check your guess with a CAS. Then prove it using the techniques of Section 7.2. For
what values of a and b is it valid?
3. (a) Use a computer algebra system to evaluate the following integrals. (i)
(iv) y ln x dx (ii) y x ln x dx y x 3 ln x dx (v) y x 2 ln x dx y x 7 ln x dx (iii) (b) Based on the pattern of your responses in part (a), guess the value of y x n ln x dx
(c) Use integration by parts to prove the conjecture that you made in part (b). For what
values of n is it valid?
4. (a) Use a computer algebra system to evaluate the following integrals. (i) y xe x dx (ii) y x 2e x dx (iv) y x 4e x dx (v) y x 5e x dx (iii) y x 3e x dx (b) Based on the pattern of your responses in part (a), guess the value of x x 6e x d x. Then use
your CAS to check your guess.
(c) Based on the patterns in parts (a) and (b), make a conjecture as to the value of the integral
nx yx e dx when n is a positive integer.
(d) Use mathematical induction to prove the conjecture you made in part (c). 5E08(pp 550559) 554 ❙❙❙❙ 1/17/06 6:01 PM Page 554 CHAPTER 8 TECHNIQUES OF INTEGRATION  8.7 Approximate Integration
There are two situations in which it is impossible to ﬁnd the exact value of a deﬁnite
integral.
The ﬁrst situation arises from the fact that in order to evaluate xab f x d x using the
Fundamental Theorem of Calculus we need to know an antiderivative of f . Sometimes,
however, it is difﬁcult, or even impossible, to ﬁnd an antiderivative (see Section 8.5). For
example, it is impossible to evaluate the following integrals exactly: y 1 2 e x dx 0 y 1 x 3 dx s1 1 The second situation arises when the function is determined from a scientiﬁc experiment through instrument readings or collected data. There may be no formula for the function (see Example 5).
In both cases we need to ﬁnd approximate values of deﬁnite integrals. We already know
one such method. Recall that the deﬁnite integral is deﬁned as a limit of Riemann sums,
so any Riemann sum could be used as an approximation to the integral: If we divide a, b
into n subintervals of equal length x
b a n, then we have y y b a n f x*
i f x dx x i1 where x * is any point in the ith subinterval x i 1, x i . If x * is chosen to be the left endpoint
i
i
of the interval, then x * x i 1 and we have
i
0 x¸ ⁄ x™ x£ x¢ x y 1 f x dx a (a) Left endpoint approximation Ln f xi 1 x i1 If f x
0, then the integral represents an area and (1) represents an approximation of this
area by the rectangles shown in Figure 1(a). If we choose x * to be the right endpoint, then
i
x * x i and we have
i y y 2 0 n b x¸ ⁄ x™ x£ x¢ x (b) Right endpoint approximation
y b a n f x dx Rn f xi x i1 [See Figure 1(b).] The approximations L n and Rn deﬁned by Equations 1 and 2 are called
the left endpoint approximation and right endpoint approximation, respectively.
In Section 5.2 we also considered the case where x * is chosen to be the midpoint xi of
i
the subinterval x i 1, x i . Figure 1(c) shows the midpoint approximation Mn , which appears
to be better than either L n or Rn.
Midpoint Rule y b a f x dx where
0 x
–¡ x™
– –
x£ –
x¢ (c) Midpoint approximation
FIGURE 1 x and xi Mn
b x f x1 f x2 a
n x
1
2 xi 1 xi midpoint of x i 1, x i f xn 5E08(pp 550559) 1/17/06 6:01 PM Page 555 S ECTION 8.7 APPROXIMATE INTEGRATION ❙ ❙❙❙ 555 Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2: y b a n 1
2 f x dx n f xi x 1 f xi i1 x
2 i1 f x0 f x1 2 f x1 x
f x0
2 f x1 n x
2 x 2 f x2 f xi f xi 1 i1 f x2 f xn
2 f xn 1 1 f xn f xn y Trapezoidal Rule y b a f x dx where x
0 x¸ ⁄ x™ x£ x¢ x x
f x0
2 Tn
b a n and xi a 2 f x2 2 f xn 1 f xn i x. The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates
the case f x
0. The area of the trapezoid that lies above the ith subinterval is FIGURE 2 Trapezoidal approximation y= 2 f x1 x f xi f xi 1 x
f xi
2 2 f xi 1 and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal
Rule. 1
x EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n
approximate the integral x12 1 x d x. 5 to SOLUTION (a) With n 5, a
zoidal Rule gives y
1 2 F IGURE 3 2 1 1
dx
x T5
0.1 1, and b 2, we have x 0.2
f1
2
1
1 2 f 1.2 2
1.2 2
1.4 2 2 f 1.4 2
1.6 15 2 f 1.6 2
1.8 0.2, and so the Trape 2 f 1.8 f2 1
2 0.695635
y= This approximation is illustrated in Figure 3.
(b) The midpoints of the ﬁve subintervals are 1.1, 1.3, 1.5, 1.7, and 1.9, so the Midpoint
Rule gives
21
y1 x d x x f 1.1 f 1.3 f 1.5 f 1.7 f 1.9 1
x 1
5
1 FIGURE 4 2 1
1.1 1
1.3 0.691908
This approximation is illustrated in Figure 4. 1
1.5 1
1.7 1
1.9 5E08(pp 550559) ❙❙❙❙ 556 1/17/06 6:01 PM Page 556 CHAPTER 8 TECHNIQUES OF INTEGRATION In Example 1 we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are. By the
Fundamental Theorem of Calculus, y 2 1 y b a f x dx approximation error 1
dx
x ln x]1
2 ln 2 0.693147 . . . The error in using an approximation is deﬁned to be the amount that needs to be added to
the approximation to make it exact. From the values in Example 1 we see that the errors
in the Trapezoidal and Midpoint Rule approximations for n 5 are
ET 0.002488 and EM 0.001239 In general, we have
ET
Module 5.1/5.2/8.7 allows you to compare approximation methods. Approximations to y 2 y b a f x dx Tn and y EM b a f x dx Mn The following tables show the results of calculations similar to those in Example 1, but
for n 5, 10, and 20 and for the left and right endpoint approximations as well as the
Trapezoidal and Midpoint Rules. 1
dx
x n Rn Tn Mn 0.745635
0.718771
0.705803 0.645635
0.668771
0.680803 0.695635
0.693771
0.693303 0.691908
0.692835
0.693069 n Corresponding errors Ln 5
10
20 1 EL ER ET EM 5
10
20 0.052488
0.025624
0.012656 0.047512
0.024376
0.012344 0.002488
0.000624
0.000156 0.001239
0.000312
0.000078 We can make several observations from these tables:
1. In all of the methods we get more accurate approximations when we increase the 2.
 It turns out that these observations are true
in most cases. 3.
4.
5. value of n. (But very large values of n result in so many arithmetic operations that
we have to beware of accumulated roundoff error.)
The errors in the left and right endpoint approximations are opposite in sign and
appear to decrease by a factor of about 2 when we double the value of n.
The Trapezoidal and Midpoint Rules are much more accurate than the endpoint
approximations.
The errors in the Trapezoidal and Midpoint Rules are opposite in sign and appear
to decrease by a factor of about 4 when we double the value of n.
The size of the error in the Midpoint Rule is about half the size of the error in the
Trapezoidal Rule. Figure 5 shows why we can usually expect the Midpoint Rule to be more accurate than
the Trapezoidal Rule. The area of a typical rectangle in the Midpoint Rule is the same as
the trapezoid ABCD whose upper side is tangent to the graph at P. The area of this trapezoid is closer to the area under the graph than is the area of the trapezoid AQRD used in 5E08(pp 550559) 1/17/06 6:01 PM Page 557 SECTION 8.7 APPROXIMATE INTEGRATION C 557 the Trapezoidal Rule. [The midpoint error (shaded red) is smaller than the trapezoidal error
(shaded blue).]
These observations are corroborated in the following error estimates, which are proved
in books on numerical analysis. Notice that Observation 4 corresponds to the n 2 in each
denominator because 2 n 2 4n 2. The fact that the estimates depend on the size of the
second derivative is not surprising if you look at Figure 5, because f x measures how
much the graph is curved. [Recall that f x measures how fast the slope of y f x
changes.] P
B A ❙❙❙❙ D
x i1 xi
– xi K for a
3 Error Bounds Suppose f x
in the Trapezoidal and Midpoint Rules, then C
R Kb a
12n 2 ET P x b. If ET and EM are the errors 3 and Kb a
24n 2 EM 3 B Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If
fx
1 x, then f x
1 x 2 and f x
2 x 3. Since 1 x 2, we have 1 x 1, so Q
A D 2
x3 fx 2
13 2 FIGURE 5 Therefore, taking K
 K can be any number larger than all the
values of f x , but smaller values of K
give better error bounds. 2, a
ET 1, b 2, and n 22 1
12 5 2 5 in the error estimate (3), we see that 3 1
150 0.006667 Comparing this error estimate of 0.006667 with the actual error of about 0.002488, we see
that it can happen that the actual error is substantially less than the upper bound for the
error given by (3).
EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and
Midpoint Rule approximations for x12 1 x d x are accurate to within 0.0001?
SOLUTION We saw in the preceding calculation that f x
2 for 1 x 2, so we can
take K 2, a 1, and b 2 in (3). Accuracy to within 0.0001 means that the size of
the error should be less than 0.0001. Therefore, we choose n so that 213
12n 2 0.0001 Solving the inequality for n, we get
n2  It’s quite possible that a lower value for n
would sufﬁce, but 41 is the smallest value for
which the error bound formula can guarantee us
accuracy to within 0.0001. or
Thus, n 2
12 0.0001 n 1
s0.0006 41 will ensure the desired accuracy. 40.8 5E08(pp 550559) 558 ❙❙❙❙ 1/17/06 6:01 PM Page 558 CHAPTER 8 TECHNIQUES OF INTEGRATION For the same accuracy with the Midpoint Rule we choose n so that
213
24n 2
which gives
y 0.0001
1
s0.0012 n 29 EXAMPLE 3 2 (a) Use the Midpoint Rule with n 10 to approximate the integral x01 e x dx.
(b) Give an upper bound for the error involved in this approximation.
y=e x SOLUTION 2 (a) Since a 0, b y 1 0 1, and n 2 e x dx 10, the Midpoint Rule gives x f 0.05
0.1 e 0.0025 f 0.15
e 0.0225 e 0.4225
0 1 x f 0.85
e 0.0625 e 0.5625 e 0.1225 e 0.7225 f 0.95
e 0.2025 e 0.3025 e 0.9025 1.460393
Figure 6 illustrates this approximation.
2
2
(b) Since f x
e x , we have f x
2 xe x and f x
0 x 1, we have x 2 1 and so FIGURE 6 0
Taking K 6e, a 0, b
bound for the error is  Error estimates are upper bounds for
the error. They give theoretical, worstcase
scenarios. The actual error in this case turns
out to be about 0.0023. fx 1, and n
6e 1 3
24 10 2 2 4x 2 e x 2 2 2 4 x 2 e x . Also, since 6e 10 in the error estimate (3), we see that an upper
e
400 0.007 Simpson ’s Rule
Another rule for approximate integration results from using parabolas instead of straight
line segments to approximate a curve. As before, we divide a, b into n subintervals of
equal length h
x
b a n, but this time we assume that n is an even number. Then
on each consecutive pair of intervals we approximate the curve y f x
0 by a parabola
as shown in Figure 7. If yi f x i , then Pi x i , yi is the point on the curve lying above x i.
A typical parabola passes through three consecutive points Pi , Pi 1 , and Pi 2 .
y y P¸ P¡ P∞ P™
P£ 0 a=x¸ FIGURE 7 ⁄ x™ x£ P¸ (_h, y¸) Pß P¡ (0, › ) P™ (h, ﬁ) P¢ x¢ x∞ xß=b x _h FIGURE 8 0 h x 5E08(pp 550559) 1/17/06 6:01 PM Page 559 SECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 559 To simplify our calculations, we ﬁrst consider the case where x 0
h, x 1 0, and
x 2 h. (See Figure 8.) We know that the equation of the parabola through P0 , P1 , and P2
is of the form y Ax 2 Bx C and so the area under the parabola from x
h to
x h is
 Here we have used Theorem 5.5.6. Notice
that Ax 2 C is even and Bx is odd. y h
h Ax 2 Bx h 2 y Ax 2 C dx C dx 0 h x3
2A
3 Cx h3
3 Ch 2A
But, since the parabola passes through P0
y0 A h y1 Ah 2
y0 and therefore B h
2 Ah 2
3 6C h, y0 , P1 0, y1 , and P2 h, y2 , we have Bh h Ah 2 C y2 2 0 C Bh C C 4y1 2 Ah 2 y2 6C Thus, we can rewrite the area under the parabola as
h
y0
3 4y1 y2 Now, by shifting this parabola horizontally we do not change the area under it. This means
that the area under the parabola through P0 , P1 , and P2 from x x 0 to x x 2 in Figure 7
is still
h
y0 4y1 y2
3
Similarly, the area under the parabola through P2 , P3 , and P4 from x
h
y2
3 4y3 x 2 to x x 4 is y4 If we compute the areas under all the parabolas in this manner and add the results, we get
b a f x dx h
y0
3 4y1 y2 h
y2
3 h
y0
3 y 4y1 2y2 4y3 4y3
2y4 h
yn
3 y4
2yn 2 4yn 1 4yn 2 1 yn yn Although we have derived this approximation for the case in which f x
0, it is a reasonable approximation for any continuous function f and is called Simpson’s Rule after
the English mathematician Thomas Simpson (1710–1761). Note the pattern of coefﬁcients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1. 5E08(pp 560569) 560 ❙❙❙❙ 1/17/06 5:57 PM Page 560 CHAPTER 8 TECHNIQUES OF INTEGRATION Simpson’s Rule  Thomas Simpson was a weaver who taught
himself mathematics and went on to become one
of the best English mathematicians of the 18th
century. What we call Simpson’s Rule was
actually known to Cavalieri and Gregory in the
17th century, but Simpson popularized it in his
bestselling calculus textbook, entitled A New
Treatise of Fluxions. y b a f x dx x
f x0
3 Sn 4 f x1
2 f xn where n is even and x b y 2 1 1
dx
x 1 x, n 4 f x3 4 f xn 2 f xn 1 a n.
10 to approximate x12 1 x d x. EXAMPLE 4 Use Simpson’s Rule with n
SOLUTION Putting f x 2 f x2 10, and x 0.1 in Simpson’s Rule, we obtain S10
x
f1
3
0.1
3 4 f 1.1 1
1 4
1.1 2 f 1.2 2
1.2 4
1.3 4 f 1.3
2
1.4 2 f 1.8 4
1.5 2
1.6 4
1.7 4 f 1.9
2
1.8 f2
4
1.9 1
2 0.693150
Notice that, in Example 4, Simpson’s Rule gives us a much better approximation
S10 0.693150 to the true value of the integral ln 2 0.693147. . . than does the
Trapezoidal Rule T10 0.693771 or the Midpoint Rule M10 0.692835 . It turns out
(see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of
those in the Trapezoidal and Midpoint Rules:
1
3 S2 n Tn 2
3 Mn (Recall that ET and EM usually have opposite signs and EM is about half the size of ET .)
In many applications of calculus we need to evaluate an integral even if no explicit formula is known for y as a function of x. A function may be given graphically or as a table
of values of collected data. If there is evidence that the values are not changing rapidly,
then the Trapezoidal Rule or Simpson’s Rule can still be used to ﬁnd an approximate value
for xab y d x, the integral of y with respect to x.
E XAMPLE 5 Figure 9 shows data trafﬁc on the link from the United States to SWITCH,
the Swiss academic and research network, on February 10, 1998. D t is the data
throughput, measured in megabits per second Mb s . Use Simpson’s Rule to estimate
the total amount of data transmitted on the link up to noon on that day.
D
8
6
4
2 FIGURE 9 0 3 6 9 12 15 18 21 24 t (hours) 5E08(pp 560569) 1/17/06 5:57 PM Page 561 S ECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 561 SOLUTION Because we want the units to be consistent and D t is measured in megabits
per second, we convert the units for t from hours to seconds. If we let A t be the
amount of data (in megabits) transmitted by time t, where t is measured in seconds, then
At
D t . So, by the Net Change Theorem (see Section 5.4), the total amount of data
transmitted by noon (when t 12 60 2 43,200) is A 43,200 y 43,200 0 D t dt We estimate the values of D t at hourly intervals from the graph and compile them in
the table.
t hours t seconds Dt t hours t seconds Dt 0
1
2
3
4
5
6 0
3,600
7,200
10,800
14,400
18,000
21,600 3.2
2.7
1.9
1.7
1.3
1.0
1.1 7
8
9
10
11
12 25,200
28,800
32,400
36,000
39,600
43,200 1.3
2.8
5.7
7.1
7.7
7.9 Then we use Simpson’s Rule with n y 43,200 0 A t dt t
3 D0 12 and t 4D 3600 3600 to estimate the integral: 2D 7200 4D 39,600 D 43,200 3600
3.2 4 2.7
2 1.9
4 1.7
2 1.3
4 1.0
3
2 1.1
4 1.3
2 2.8
4 5.7
2 7.1
4 7.7 7.9 143,880
Thus, the total amount of data transmitted up to noon is about 144,000 megabits, or
144 gigabits.
In Exercises 27 and 28 you are asked to demonstrate, in particular cases, that the error
in Simpson’s Rule decreases by a factor of about 16 when n is doubled. That is consistent
with the appearance of n 4 in the denominator of the following error estimate for Simpson’s
Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but
it uses the fourth derivative of f . f4 x
K for a
If ES is the error involved in using Simpson’s Rule, then
4 Error Bound for Simpson’s Rule Suppose that ES Kb a
180 n 4 5 x b. 5E08(pp 560569) 562 ❙❙❙❙ 1/17/06 5:57 PM Page 562 CHAPTER 8 TECHNIQUES OF INTEGRATION EXAMPLE 6 How large should we take n in order to guarantee that the Simpson’s Rule
approximation for x12 1 x d x is accurate to within 0.0001?
SOLUTION If f x 4 1 x, then f 24 x 5. Since x x
4 f
 Many calculators and computer algebra systems have a builtin algorithm that computes an
approximation of a deﬁnite integral. Some of
these machines use Simpson’s Rule; others use
more sophisticated techniques such as adaptive
numerical integration. This means that if a function ﬂuctuates much more on a certain part of
the interval than it does elsewhere, then that
part gets divided into more subintervals. This
strategy reduces the number of calculations
required to achieve a prescribed accuracy. Therefore, we can take K
choose n so that 1, we have 1 x 24
x5 x 1 and so 24 24 in (4). Thus, for an error less than 0.0001 we should
24 1 5
180n 4 0.0001 n4 or 24
180 0.0001 n This gives 1
4
s0.00075 6.04 Therefore, n 8 (n must be even) gives the desired accuracy. (Compare this with
Example 2, where we obtained n 41 for the Trapezoidal Rule and n 29 for the
Midpoint Rule.)
E XAMPLE 7 2 (a) Use Simpson’s Rule with n 10 to approximate the integral x01 e x d x.
(b) Estimate the error involved in this approximation.
SOLUTION (a) If n
 Figure 10 illustrates the calculation in
Example 7. Notice that the parabolic arcs are
2
so close to the graph of y e x that they are
practically indistinguishable from it. y 1 0 10, then x
2 e x dx 0.1 and Simpson’s Rule gives x
f0
3 4 f 0.1 0.1 0
e
3 y 4e 0.01
4e 0.49 2 f 0.2 2e 0.04
2e 0.64 2 f 0.8 4e 0.09
4e 0.81 2e 0.16 4e 0.25 4 f 0.9 f1 2e 0.36 e1 1.462681
y=e ≈ f
and so, since 0 x FIGURE 10 1 x Therefore, putting K
most 4 x 12 48 x 2 16 x 4 e x 2 1, we have
0 0 2 e x is (b) The fourth derivative of f x f 76e, a 4 x 12 0, b 48 1, and n 76e 1 5
180 10 4 16 e 1 76e 10 in (4), we see that the error is at 0.000115 (Compare this with Example 3.) Thus, correct to three decimal places, we have y 1 0 2 e x dx 1.463 5E08(pp 560569) 1/17/06 5:57 PM Page 563 ❙❙❙❙ S ECTION 8.7 APPROXIMATE INTEGRATION  8.7
1. Let I Exercises
x04 f (Round your answers to six decimal places.) Compare your results
to the actual value to determine the error in each approximation. x d x, where f is the function whose graph is
shown.
(a) Use the graph to ﬁnd L 2 , R2, and M2.
(b) Are these underestimates or overestimates of I ?
(c) Use the graph to ﬁnd T2. How does it compare with I ?
(d) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in
increasing order.
y
3 5. 0 ■ ■ ■ ■ ■ y 1 0 ■ sx e ■ dx, n
■ 6 ■ ■ ■ f y 1 9. y 11. y 13. y 15. y 17. y 1 2 3 4x 2. The left, right, Trapezoidal, and Midpoint Rule approximations were used to estimate x02 f x d x, where f is the function whose
graph is shown. The estimates were 0.7811, 0.8675, 0.8632,
and 0.9540, and the same number of subintervals were used in
each case.
(a) Which rule produced which estimate?
(b) Between which two approximations does the true value of
x02 f x d x lie? y 4
s1 x 2 d x, ln x
d x,
1x 2 1 12 0
2 e 1 x d x, 1 n cos x
d x,
x 5 1 1 3 1 0 n ■ y 5 ■ 8.
10.
12. y
y 18. 8
n ■ y 16. 8 4 dy, y y 8
10 n y 14. n sin e t 2 dt, n x02 e 6
■ ■ ■ 12 0 sin x 2 dx, n
dt
t2 3 4 0
4 0 6 4
4 2
■ s1 , n 6 sx d x, 1 0 4 n 8 t4 s x sin x d x, n
ln x 3 2 d x, ex
d x,
x n ■ 8
n 10 10
■ ■ ■ x2 d x.
(b) Estimate the errors in the approximations of part (a).
(c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to
within 0.00001? y=ƒ 0 2 0 19. (a) Find the approximations T10 and M10 for the integral 1 2 1
20. (a) Find the approximations T8 and M8 for x0 cos x 2 d x. x (b) Estimate the errors involved in the approximations of
part (a).
(c) How large do we have to choose n so that the approximations Tn and Mn to the integral in part (a) are accurate to
within 0.00001? 1
2
; 3. Estimate x0 cos x d x using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with n 4. From a graph of the
integrand, decide whether your answers are underestimates or
overestimates. What can you conclude about the true value of
the integral? 1
21. (a) Find the approximations T10 and S10 for x0 e x d x and the corresponding errors ET and ES.
(b) Compare the actual errors in part (a) with the error estimates given by (3) and (4).
(c) How large do we have to choose n so that the approximations Tn , Mn , and Sn to the integral in part (a) are accurate to
within 0.00001? sin x 2 2 in the viewing rectangle
0, 1 by 0, 0.5 and let I x01 f x d x.
(a) Use the graph to decide whether L 2 , R2 , M2, and T2 underestimate or overestimate I .
(b) For any value of n, list the numbers L n , Rn , Mn , Tn , and I in
increasing order.
(c) Compute L 5 , R5 , M5, and T5. From the graph, which do you
think gives the best estimate of I ? ; 4. Draw the graph of f x 22. How large should n be to guarantee that the Simpson’s Rule
2 approximation to x01 e x dx is accurate to within 0.00001?
CAS Use (a) the Midpoint Rule and (b) Simpson’s Rule to
approximate the given integral with the speciﬁed value of n.
 ■ 6. 8 7–18  Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and
(c) Simpson’s Rule to approximate the given integral with the
speciﬁed value of n. (Round your answers to six decimal places.) 7. 0 x 2 sin x dx, n y 2 5–6 563 23. The trouble with the error estimates is that it is often very difﬁ cult to compute four derivatives and obtain a good upper bound
K for f 4 x by hand. But computer algebra systems have no 5E08(pp 560569) 564 ❙❙❙❙ 1/17/06 5:57 PM Page 564 CHAPTER 8 TECHNIQUES OF INTEGRATION problem computing f 4 and graphing it, so we can easily ﬁnd
a value for K from a machine graph. This exercise deals
with approximations to the integral I x02 f x d x, where
fx
e cos x.
(a) Use a graph to get a good upper bound for f x .
(b) Use M10 to approximate I .
(c) Use part (a) to estimate the error in part (b).
(d) Use the builtin numerical integration capability of your
CAS to approximate I .
(e) How does the actual error compare with the error estimate
in part (c)?
(f) Use a graph to get a good upper bound for f 4 x .
(g) Use S10 to approximate I .
(h) Use part (f) to estimate the error in part (g).
(i) How does the actual error compare with the error estimate
in part (h)?
(j) How large should n be to guarantee that the size of the
error in using Sn is less than 0.0001?
CAS 1 24. Repeat Exercise 23 for the integral y s4 Use Simpson’s Rule to estimate the area of the pool. 6.2 x 1 x 3 dx 0 ■ ■ y 26.
■ ■ ■ ■ 2 ■ ■ ■ ■ ■ 27–28  Find the approximations Tn , Mn , and Sn for n
6 and 12.
Then compute the corresponding errors ET, EM, and ES. (Round
your answers to six decimal places. You may wish to use the sum
command on a computer algebra system.) What observations can
you make? In particular, what happens to the errors when n is
doubled? 27.
■ y 4
1
■ ■ y 28. sx dx
■ ■ ■ ■ 2
1
■ fx 6.8
6.5
6.3
6.4
6.9 2.0
2.4
2.8
3.2 7.6
8.4
8.8
9.0 32. A radar gun was used to record the speed of a runner during the ﬁrst 5 seconds of a race (see the table). Use Simpson’s
Rule to estimate the distance the runner covered during those
5 seconds.
t (s) v (m s) t (s) v (m s) 0
0.5
1.0
1.5
2.0
2.5 0
4.67
7.34
8.86
9.73
10.22 3.0
3.5
4.0
4.5
5.0 10.51
10.67
10.76
10.81
10.81 33. The graph of the acceleration a t of a car measured in ft s2 is shown. Use Simpson’s Rule to estimate the increase in the
velocity of the car during the 6second time interval.
a
12 xe x d x
■ x 1 for all x, estimate the
(b) If it is known that 4 f x
error involved in the approximation in part (a). e x dx 0 ■ fx 0.0
0.4
0.8
1.2
1.6  Find the approximations L n , Rn , Tn , and Mn for n
4, 8,
and 16. Then compute the corresponding errors EL , ER, ET , and EM.
(Round your answers to six decimal places. You may wish to use
the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when
n is doubled? y 4.8 value of the integral x03.2 f x d x. 25–26 25. 5.6 5.0 4.8 31. (a) Use the Midpoint Rule and the given data to estimate the x 3 d x. 1 7
.2 6.8 ■ ■ ■ 29. Estimate the area under the graph in the ﬁgure by using (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson’s
Rule, each with n 4. 8
4
0 2 4 6t y 34. Water leaked from a tank at a rate of r t liters per hour, where the graph of r is as shown. Use Simpson’s Rule to estimate the
total amount of water that leaked out during the ﬁrst six hours.
r
4 1
0 1 2 3 4x 30. The widths (in meters) of a kidneyshaped swimming pool were measured at 2meter intervals as indicated in the ﬁgure. 2 0 2 4 6t 5E08(pp 560569) 1/17/06 5:57 PM Page 565 S ECTION 8.7 APPROXIMATE INTEGRATION 35. The table (supplied by San Diego Gas and Electric) gives the power consumption in megawatts in San Diego County from
midnight to 6:00 A.M. on December 8, 1999. Use Simpson’s
Rule to estimate the energy used during that time period. (Use
the fact that power is the derivative of energy.)
t P t 1814
1735
1686
1646
1637
1609
1604 3:30
4:00
4:30
5:00
5:30
6:00 565 39. The region bounded by the curves y 3
s1 x 3, y 0, x 0,
and x 2 is rotated about the xaxis. Use Simpson’s Rule with
n 10 to estimate the volume of the resulting solid. CAS 40. The ﬁgure shows a pendulum with length L that makes a maxi mum angle 0 with the vertical. Using Newton’s Second Law
it can be shown that the period T (the time for one complete
swing) is given by P 0:00
0:30
1:00
1:30
2:00
2:30
3:00 ❙❙❙❙ 1611
1621
1666
1745
1886
2052 T
where k sin( 1
2
If L 1 m and
ﬁnd the period. 4 0 L
t y 2 0 s1 dx
k 2 sin 2x ) and t is the acceleration due to gravity.
42 , use Simpson’s Rule with n 0 10 to 36. Shown is the graph of trafﬁc on an Internet service provider’s T1 data line from midnight to 8:00 A.M. D is the data throughput, measured in megabits per second. Use Simpson’s Rule to
estimate the total amount of data transmitted during that time
period. ¨¸ D
0.8 41. The intensity of light with wavelength traveling through
a diffraction grating with N slits at an angle is given by
I
N 2 sin 2k k 2, where k
Nd sin
and d is the
distance between adjacent slits. A heliumneon laser with
wavelength
632.8 10 9 m is emitting a narrow band of
light, given by 10 6
10 6, through a grating with
4
10,000 slits spaced 10 m apart. Use the Midpoint Rule with
n 10 to estimate the total light intensity x10 0 I
d emerg1
ing from the grating. 0.4 6 0 2 4 6 8 t (hours) 6 37. If the region shown in the ﬁgure is rotated about the yaxis to form a solid, use Simpson’s Rule with n
volume of the solid. 8 to estimate the 42. Use the Trapezoidal Rule with n 10 to approximate
x dx. Compare your result to the actual value. Can
you explain the discrepancy? x020 cos 43. Sketch the graph of a continuous function on 0, 2 for which
y
4 the Trapezoidal Rule with n
Midpoint Rule. 2 is more accurate than the 44. Sketch the graph of a continuous function on 0, 2 for which 2 the right endpoint approximation with n
than Simpson’s Rule.
0 2 4 6 10 x 8 45. If f is a positive function and f x 2 is more accurate 0 for a x b, show that
38. The table shows values of a force function f x where x is measured in meters and f x in newtons. Use Simpson’s Rule
to estimate the work done by the force in moving an object a
distance of 18 m. Tn y b a f x dx Mn 46. Show that if f is a polynomial of degree 3 or lower, then
b
Simpson’s Rule gives the exact value of xa f x d x.
1
2 x 0 3 6 9 12 15 18 47. Show that fx 9.8 9.1 8.5 8.0 7.7 7.5 7.4 48. Show that 3 Tn 1 Tn Mn
2
3 Mn T2 n.
S2 n. 5E08(pp 560569) 566 ❙❙❙❙ 1/17/06 5:57 PM Page 566 CHAPTER 8 TECHNIQUES OF INTEGRATION  8.8 Improper Integrals
In deﬁning a deﬁnite integral xab f x d x we dealt with a function f deﬁned on a ﬁnite interval a, b and we assumed that f does not have an inﬁnite discontinuity (see Section 5.2).
In this section we extend the concept of a deﬁnite integral to the case where the interval is
inﬁnite and also to the case where f has an inﬁnite discontinuity in a, b . In either case
the integral is called an improper integral. One of the most important applications of this
idea, probability distributions, will be studied in Section 9.5. Type I: Infinite Intervals
Consider the inﬁnite region S that lies under the curve y 1 x 2, above the xaxis, and to
the right of the line x 1. You might think that, since S is inﬁnite in extent, its area must
be inﬁnite, but let’s take a closer look. The area of the part of S that lies to the left of the
line x t (shaded in Figure 1) is Try painting a fence that never ends.
Resources / Module 6
/ How To Calculate
/ Start of Improper Integrals y At
Notice that A t t 1 1
dx
x2 1
x t 1
t 1
1 1 no matter how large t is chosen.
y y= 1
≈
area=1  x=1
0 FIGURE 1 1
t t 1 x We also observe that
lim A t 1
t lim 1 tl tl The area of the shaded region approaches 1 as t l
of the inﬁnite region S is equal to 1 and we write y 1 y 1 2 x t 1 1 3 x (see Figure 2), so we say that the area 1
dx
x2 1
y area= 4
5 area= 2
3
0 lim y tl y y
1
area= 2 0 1
dx
x2 1 0 1 area=1
5x 0 1 x FIGURE 2 Using this example as a guide, we deﬁne the integral of f (not necessarily a positive
function) over an inﬁnite interval as the limit of integrals over ﬁnite intervals. 5E08(pp 560569) 1/17/06 5:57 PM Page 567 SECTION 8.8 IMPROPER INTEGRALS 1 ❙❙❙❙ 567 Definition of an Improper Integral of Type 1 (a) If xat f x d x exists for every number t y a a, then
t lim y f x d x f x dx tl a provided this limit exists (as a ﬁnite number).
(b) If xtb f x d x exists for every number t b, then y b f x dx lim tl y b t f x dx provided this limit exists (as a ﬁnite number).
The improper integrals xa f x d x and xb f x d x are called convergent if the
corresponding limit exists and divergent if the limit does not exist.
(c) If both xa f x d x and xa f x d x are convergent, then we deﬁne y f x dx y a f x dx y a f x dx In part (c) any real number a can be used (see Exercise 74).
Any of the improper integrals in Deﬁnition 1 can be interpreted as an area provided that
f is a positive function. For instance, in case (a) if f x
0 and the integral xa f x d x
is convergent, then we deﬁne the area of the region S
x, y x a, 0 y f x in
Figure 3 to be y AS a f x dx y y=ƒ S FIGURE 3 0 a x This is appropriate because xa f x d x is the limit as t l
f from a to t.
EXAMPLE 1 Determine whether the integral x1 of the area under the graph of 1 x d x is convergent or divergent. SOLUTION According to part (a) of Deﬁnition 1, we have y 1 1
dx
x lim y tl t 1 1
dx
x lim ln t tl lim ln x tl ln 1 t 1 lim ln t tl The limit does not exist as a ﬁnite number and so the improper integral x1 1 x d x is
divergent. 5E08(pp 560569) 568 ❙❙❙❙ 1/17/06 5:57 PM Page 568 CHAPTER 8 TECHNIQUES OF INTEGRATION y y= Let’s compare the result of Example 1 with the example given at the beginning of this
section: 1
≈ 1
d x converges
x2 y 1 ﬁnite area
0 x 1 FIGURE 4 y= 1
x 1 Geometrically, this says that although the curves y 1 x 2 and y 1 x look very similar
for x 0, the region under y 1 x 2 to the right of x 1 (the shaded region in Figure 4)
has ﬁnite area whereas the corresponding region under y 1 x (in Figure 5) has inﬁnite
area. Note that both 1 x 2 and 1 x approach 0 as x l but 1 x 2 approaches 0 faster than
1 x. The values of 1 x don’t decrease fast enough for its integral to have a ﬁnite value.
EXAMPLE 2 Evaluate y y 0 xe x d x. SOLUTION Using part (b) of Deﬁnition 1, we have y 0 xe x d x We integrate by parts with u
1 y lim tl inﬁnite area 0 1
d x diverges
x y 0 xe x d x t e x d x so that du x, dv e x: d x, v x y 0 xe x d x xe x 0 FIGURE 5 We know that e t l 0 as t l t y te t t 0 1 t e x dx
et , and by l’Hospital’s Rule we have lim te t lim tl tl t
et
et lim tl 1
e lim tl t 0 Therefore y 0 xe x d x te t lim tl 0
EXAMPLE 3 Evaluate 1 y x2 1 1
1 x 2 0 1 d x. SOLUTION It’s convenient to choose a y 1 et 1 dx 0 in Deﬁnition 1(c): y 1 0 1 x 2 dx y 0 1
x2 1 dx We must now evaluate the integrals on the right side separately: y 0 1
1 x 2 dx lim y tl dx t 0 1 lim tan 1t tl x 2 lim tan 1x tl tan 1 0 t
0 lim tan 1t tl 2 5E08(pp 560569) 1/17/06 5:58 PM Page 569 SECTION 8.8 IMPROPER INTEGRALS y 1 0 1 x2 dx dx 0 y lim tl tl lim tan 1 0 569 0
t tan 1t tl 0 lim tan 1x x2 1 t ❙❙❙❙ 2 2 Since both of these integrals are convergent, the given integral is convergent and
1
y=
1+≈ y 0 FIGURE 6 1 y area=π
x dx x2 1 2 2 0, the given improper integral can be interpreted as the area of
Since 1 1 x 2
the inﬁnite region that lies under the curve y 1 1 x 2 and above the xaxis (see
Figure 6).
E XAMPLE 4 For what values of p is the integral y 1 convergent? 1
dx
xp 1, then the integral is divergent, so let’s
SOLUTION We know from Example 1 that if p
assume that p 1. Then
1
t
im
y1 x p d x tll y1 x p d x
tl lim tl If p 1, then p 1 0, so as t l , t p y 1 1
dx
xp 1
tp 1 t1 1
1 1
pt l 1 if p 1 1 1, then p
l 1 p1 and 1 t p 1 p x1 1 p and so the integral converges. But if p xt xp1
p1 lim 1 l 0. Therefore 0 and so as t l and the integral diverges.
We summarize the result of Example 4 for future reference: 2 y 1 1
d x is convergent if p
xp 1 and divergent if p 1. Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function deﬁned on a ﬁnite interval a, b but has
a vertical asymptote at b. Let S be the unbounded region under the graph of f and above
the xaxis between a and b. (For Type 1 integrals, the regions extended indeﬁnitely in a 5E08(pp 570579) 570 ❙❙❙❙ 1/17/06 5:40 PM Page 570 CHAPTER 8 TECHNIQUES OF INTEGRATION y horizontal direction. Here the region is inﬁnite in a vertical direction.) The area of the part
of S between a and t (the shaded region in Figure 7) is
y=ƒ 0 x=b a x tb y At t a f x dx If it happens that A t approaches a deﬁnite number A as t l b , then we say that the
area of the region S is A and we write FIGURE 7 y b a f x dx lim y t a tlb f x dx We use this equation to deﬁne an improper integral of Type 2 even when f is not a positive function, no matter what type of discontinuity f has at b.
3 Definition of an Improper Integral of Type 2 (a) If f is continuous on a, b and is discontinuous at b, then
 Parts (b) and (c) of Deﬁnition 3 are illustrated
in Figures 8 and 9 for the case where f x
0
and f has vertical asymptotes at a and c,
respectively. y b a f x dx lim y t a tlb f x dx if this limit exists (as a ﬁnite number).
(b) If f is continuous on a, b and is discontinuous at a, then y y b a f x dx lim tla y b t f x dx if this limit exists (as a ﬁnite number).
0 at b The improper integral xab f x d x is called convergent if the corresponding limit
exists and divergent if the limit does not exist. x y y b a EXAMPLE 5 Find
0 b, and both xac f x d x and (c) If f has a discontinuity at c, where a c
xcb f x d x are convergent, then we deﬁne FIGURE 8 a bx c y 1 5 2 sx y c a f x dx y b c f x dx d x. 2 SOLUTION We note ﬁrst that the given integral is improper because f x
1 sx 2
has the vertical asymptote x 2. Since the inﬁnite discontinuity occurs at the left endpoint of 2, 5 , we use part (b) of Deﬁnition 3: FIGURE 9 y 5 2 y y= f x dx 1
œ„„„„
x2 dx
sx 2 lim t l2 y t 5 dx
sx 2 5 lim 2 s x 2 lim 2(s3 st t l2 t l2 3
area=2œ„
0 1 FIGURE 10 2 3 4 t 2) 2 s3
5 x Thus, the given improper integral is convergent and, since the integrand is positive, we
can interpret the value of the integral as the area of the shaded region in Figure 10. 5E08(pp 570579) 1/17/06 5:41 PM Page 571 S ECTION 8.8 IMPROPER INTEGRALS y EXAMPLE 6 Determine whether 0 /2 /2 0 sec x d x y lim tl /2 t 0 tan x lim and tan t l E XAMPLE 7 Evaluate y dx 3 x 0 tan t /2 tl ln sec t /2 as t l . Using sec x d x lim ln sec x tl because sec t l
divergent. 571 sec x d x converges or diverges. SOLUTION Note that the given integral is improper because lim x l / 2 sec x
part (a) of Deﬁnition 3 and Formula 14 from the Table of Integrals, we have y ❙❙❙❙ t 0 ln 1 2 . Thus, the given improper integral is if possible. 1 1 is a vertical asymptote of the integrand. Since it
SOLUTION Observe that the line x
occurs in the middle of the interval 0, 3 , we must use part (c) of Deﬁnition 3 with
c 1:
3
dx
1
dx
3
dx
y0 x 1 y0 x 1 y1 x 1 y where dx 1 0 x lim 1 t l1 y dx t 0 x lim ln x 1 lim (ln t 1 t l1 lim ln 1 1 t l1 ln 1 t 0 ) t t l1 because 1 t l 0 as t l 1 . Thus, x01 d x x 1 is divergent. This implies that
x03 d x x 1 is divergent. [We do not need to evaluate x13 d x x 1 .]  WARNING If we had not noticed the asymptote x 1 in Example 7 and had instead
confused the integral with an ordinary integral, then we might have made the following
erroneous calculation:
■ y dx 3 0 x ln x 1 1 3 0 ln 2 ln 1 ln 2 This is wrong because the integral is improper and must be calculated in terms of limits.
From now on, whenever you meet the symbol xab f x d x you must decide, by looking
at the function f on a, b , whether it is an ordinary deﬁnite integral or an improper
integral.
EXAMPLE 8 Evaluate y 1 0 ln x d x. SOLUTION We know that the function f x lim x l 0 ln x ln x has a vertical asymptote at 0 since
. Thus, the given integral is improper and we have y 1 0 ln x d x lim tl0 y t 1 ln x d x 5E08(pp 570579) 572 ❙❙❙❙ 1/17/06 5:42 PM Page 572 CHAPTER 8 TECHNIQUES OF INTEGRATION Now we integrate by parts with u y 1 t ln x, dv ln x d x d x, du x ln x 1 y t 1 t ln t t ln t x: dx t 1 ln 1 d x x, and v 1 1 t t To ﬁnd the limit of the ﬁrst term we use l’Hospital’s Rule:
lim t ln t ln t
1t lim tl0 tl0 y 1t
1 t2 lim tl0 0 1 lim x t tl0 area=1 y Therefore 1 0 ln x d x lim t ln t tl0 0 y=ln x 0 1 1 0 t
1 Figure 11 shows the geometric interpretation of this result. The area of the shaded region
above y ln x and below the xaxis is 1. FIGURE 11 A C omparison Test for Improper Integrals
Sometimes it is impossible to ﬁnd the exact value of an improper integral and yet it
is important to know whether it is convergent or divergent. In such cases the following theorem is useful. Although we state it for Type 1 integrals, a similar theorem is true for
Type 2 integrals.
Comparison Theorem Suppose that f and t are continuous functions with fx tx 0 for x a. (a) If xa f x d x is convergent, then xa t x d x is convergent.
(b) If xa t x d x is divergent, then xa f x d x is divergent. y f
g 0 a FIGURE 12 x We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible.
If the area under the top curve y f x is ﬁnite, then so is the area under the bottom curve
y t x . And if the area under y t x is inﬁnite, then so is the area under y f x .
[Note that the reverse is not necessarily true: If xa t x dx is convergent, xa f x dx may
or may not be convergent, and if xa f x dx is divergent, xa t x dx may or may not be
divergent.]
EXAMPLE 9 Show that y 0 e x2 d x is convergent. SOLUTION We can’t evaluate the integral directly because the antiderivative of e an elementary function (as explained in Section 8.5). We write y 0 e x2 dx y 1 0 e x2 dx y 1 e x2 dx x2 is not 5E08(pp 570579) 1/17/06 5:42 PM Page 573 S ECTION 8.8 IMPROPER INTEGRALS y y=e 573 and observe that the ﬁrst integral on the righthand side is just an ordinary deﬁnite integral. In the second integral we use the fact that for x 1 we have x 2 x, so x 2
x
2
and therefore e x
e x. (See Figure 13.) The integral of e x is easy to evaluate: _x 2 y=e_x t e x dx y 1 0 ❙❙❙❙ lim y e x d x tl x 1 1 lim e 1 e tl FIGURE 13 t e 1 2 Thus, taking f x
e x and t x
e x in the Comparison Theorem, we see that
2
x2
x1 e d x is convergent. It follows that x0 e x d x is convergent.
2 In Example 9 we showed that x0 e x d x is convergent without computing its value. In
Exercise 70 we indicate how to show that its value is approximately 0.8862. In probability theory it is important to know the exact value of this improper integral, as we will see
in Section 9.5; using the methods of multivariable calculus it can be shown that the exact
value is s 2. Table 1 illustrates the deﬁnition of an improper integral by showing how
2
the (computergenerated) values of x0t e x d x approach s 2 as t becomes large. In fact,
2
these values converge quite quickly because e x l 0 very rapidly as x l . TABLE 1 x0t e t
1
2
3
4
5
6 x2 dx 0.7468241328
0.8820813908
0.8862073483
0.8862269118
0.8862269255
0.8862269255 EXAMPLE 10 The integral because y 1 1 e x d x is divergent by the Comparison Theorem x 1 e x x 1
x and x1 1 x d x is divergent by Example 1 [or by (2) with p 1]. Table 2 illustrates the divergence of the integral in Example 10. It appears that the
values are not approaching any ﬁxed number.
TABLE 2 x1t t
2
5
10
100
1000
10000  8.8
y (c) y 1 x 4e 2 0 e x x dx 0.8636306042
1.8276735512
2.5219648704
4.8245541204
7.1271392134
9.4297243064 Exercises 1. Explain why each of the following integrals is improper. (a) 1 x2 x 4 (b) dx x
5x 6 dx y (d) y 2 0 sec x dx
1 0 x2 5 dx 2. Which of the following integrals are improper? Why? (a) y (c) y 2 1 1
dx
2x 1
sin x
dx
1 x2 (b) y (d) y 1 1 0
2 1 2x
ln x 1 dx 1 dx 5E08(pp 570579) ❙❙❙❙ 574 1/17/06 5:44 PM Page 574 CHAPTER 8 TECHNIQUES OF INTEGRATION 1 x 3 from x 1 to x t
and evaluate it for t 10, 100, and 1000. Then ﬁnd the total
area under this curve for x 1. 3. Find the area under the curve y 1 x 1.1 and t x
1 x 0.9 in the
viewing rectangles 0, 10 by 0, 1 and 0, 100 by 0, 1 .
(b) Find the areas under the graphs of f and t from x 1 to
x t and evaluate for t 10, 100, 10 4, 10 6, 10 10, and 10 20.
(c) Find the total area under each curve for x 1, if it exists. ; 4. (a) Graph the functions f x 39. y 2 z 2 ln z dz 0 ■ ■ 41–46 ■  40.
■ ■ ■ y 1 0 ■ ln x
dx
sx ■ ■ 5. 1 y 3x 1 1 2 1 1 7. y 9. y 11. y 13. y xe 15. y 17. dx 6. y 1 0 2x 5 y 10. y 12. y 2 14. y x 2e sin d 16. y y x
x2 1
dx
2x 18. y 19. y se ds 20. y 21. y ln x
dx
x 22. y 23. s2 x 1 0 1 29.
31.
33. y 35. y 0
1
3
2
33 y 37. dx x6 24. dx 0 0 1
dx
x4
x 32. 1 15 ex
e x 1 dx x, y x 0, 0 y x, y 0 x 2, 0 y sec 2x 0, 0 y 1 sx ■ ■ dx dz
3z 2 dx y 1 53. y 0 y 9 1 y
y
y 38. y 2 0 9 ■ 2}
■ ■ ■ ■ cos 2x
dx
1 x2 1 50. y dx
e 2x 52. y dx
x sin x 54. y x 1 /2 0
■ ■ ■ ■ ■ ■ 2 e ■ x dx x 1 x
x6 s1 1
1 0
■ dx ex
dx
sx
■ ■ ■ 55. The integral y 0 1 dy 4y 1 x2 1
x x
2x 3
dx
3 4 0 dx x2 s1 1 0 9
dx 1 0 34. 3
sx x x2  Use the Comparison Theorem to determine whether the
integral is convergent or divergent. 51. 3 ■ 9 49–54 1
dx
x sx 0 2 x2 x x /2 e 1 (s x 1), use your calculator or computer to
make a table of approximate values of x2t t x d x for t 5,
10, 100, 1000, and 10,000. Does it appear that x2 t x d x is
convergent or divergent?
(b) Use the Comparison Theorem with f x
1 s x to show
that x2 t x d x is divergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 2 x 20. Use your graph to explain intuitively why
x2 t x d x is divergent. re r 3 dr
x 2
■ y ; 48. (a) If t x cos 2 d e 2, 0 sin 2 x x 2, use your calculator or computer to
make a table of approximate values of x1t t x d x for t 2,
5, 10, 100, 1000, and 10,000. Does it appear that x1 t x d x
is convergent?
(b) Use the Comparison Theorem with f x
1 x 2 to show
that x1 t x d x is convergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 1 x 10. Use your graph to explain intuitively why
x1 t x d x is convergent. v 4 dv z2 ■ y ; 47. (a) If t x dt x3 1, 0 { x, y
■ y 36. dx ; 44. S 49. y 30. y x arctan x
dx
1 x2 2 1 28. 1
dx
x2 x, y 0 y
y 1
1 6 x, y x ; 46. S dx 2 2 2t e 42. S ln x
dx
x3 26. sec x d x 0 1 0 1
dx
sx 3 0 y 5s dx x2 0 ln x
dx
x2 1 y x2 9 y
y x2 x2 y 27. dy 1 2 25. y2 e 4 w x, y x ■ x 8. dw dx ■ ex 41. S ; 45. S Determine whether each integral is convergent or
divergent. Evaluate those that are convergent.
 ■ Sketch the region and ﬁnd its area (if the area is ﬁnite). ; 43. S
5–40 ■ 6 dx 1
sx 1 x dx is improper for two reasons: The interval 0,
is inﬁnite and
the integrand has an inﬁnite discontinuity at 0. Evaluate it by
expressing it as a sum of improper integrals of Type 2 and
Type 1 as follows:
1
1
1
1
y0 s x 1 x d x y0 s x 1 x d x y1 s x 1 x d x ■ 5E08(pp 570579) 1/17/06 5:45 PM Page 575 S ECTION 8.8 IMPROPER INTEGRALS 56. Evaluate 1
x sx 2 y 2 by the same method as in Exercise 55.
57–59  Find the values of p for which the integral converges and
evaluate the integral for those values of p. 57. y 59. y ■ 1
dx
xp 1 0
1 1
x ln x y 58. e p ■ ■ y ys R s 2r
x r dr
sr 2 s 2 If the actual density of stars in a cluster is x r
ﬁnd the perceived density y s . dx 1
2 R r 2, 67. A manufacturer of lightbulbs wants to produce bulbs that last x p ln x d x 0 575 a photograph. Suppose that in a spherical cluster of radius R
the density of stars depends only on the distance r from the
center of the cluster. If the perceived star density is given by
y s , where s is the observed planar distance from the center of
the cluster, and x r is the actual density, it can be shown that dx 4 ❙❙❙❙ ■ ■ ■ ■ ■ ■ ■ ■ 60. (a) Evaluate the integral x0 x ne x d x for n 0, 1, 2, and 3.
(b) Guess the value of x0 x ne x d x when n is an arbitrary positive integer.
(c) Prove your guess using mathematical induction. 61. (a) Show that x x d x is divergent. about 700 hours but, of course, some bulbs burn out faster than
others. Let F t be the fraction of the company’s bulbs that burn
out before t hours, so F t always lies between 0 and 1.
(a) Make a rough sketch of what you think the graph of F
might look like.
(b) What is the meaning of the derivative r t
F t?
(c) What is the value of x0 r t dt ? Why?
68. As we will see in Section 10.4, a radioactive substance decays (b) Show that
t lim y x d x tl ■ exponentially: The mass at time t is m t
m 0 e kt, where
m 0 is the initial mass and k is a negative constant. The mean
life M of an atom in the substance is 0 t This shows that we can’t deﬁne k y te kt dt M y f x dx lim y f x d x tl t 62. The average speed of molecules in an ideal gas is
v 4
s M
2RT 0 t 32 y 0 v 3e Mv 2 2RT dv where M is the molecular weight of the gas, R is the gas
constant, T is the gas temperature, and v is the molecular
speed. Show that For the radioactive carbon isotope, 14 C, used in radiocarbon
dating, the value of k is 0.000121. Find the mean life of a
14
C atom.
69. Determine how large the number a has to be so that y 1 a x2 1 v 63. We know from Example 1 that the region x, y x
that by rotating
ﬁnite volume. 1, 0 y 1 x has inﬁnite area. Show
about the xaxis we obtain a solid with 65. Find the escape velocity v0 that is needed to propel a rocket of mass m out of the gravitational ﬁeld of a planet with mass
M and radius R. Use Newton’s Law of Gravitation (see Exercise 29 in Section 6.4) and the fact that the initial kinetic
energy of 1 mv 2 supplies the needed work.
0
2
66. Astronomers use a technique called stellar stereography to determine the density of stars in a star cluster from the
observed (twodimensional) density that can be analyzed from x2 d x by writing it as
the sum of x04 e x d x and x4 e x d x. Approximate the ﬁrst integral by using Simpson’s Rule with n 8 and show that the
second integral is smaller than x4 e 4 x d x, which is less than
0.0000001.
2 71. If f t is continuous for t 0, the Laplace transform of f is the function F deﬁned by 64. Use the information and data in Exercises 29 and 30 of Sec tion 6.4 to ﬁnd the work required to propel a 1000kg satellite
out of Earth’s gravitational ﬁeld. 0.001 70. Estimate the numerical value of x0 e
2 8RT
M dx Fs y 0 f te st dt and the domain of F is the set consisting of all numbers s for
which the integral converges. Find the Laplace transforms of
the following functions.
(a) f t
(b) f t
(c) f t
1
et
t
ft
Me at for t 0, where M and a are
constants, then the Laplace transform F s exists for s a. 72. Show that if 0 ft
Me at and 0 f t
Ke at for t 0,
where f is continuous. If the Laplace transform of f t is F s 73. Suppose that 0 5E08(pp 570579) ❙❙❙❙ 576 1/17/06 5:46 PM Page 576 CHAPTER 8 TECHNIQUES OF INTEGRATION 77. Find the value of the constant C for which the integral and the Laplace transform of f t is G s , show that
Gs sF s f0 s a y 74. If x f x d x is convergent and a and b are real numbers,
show that y a y f x dx a x2 75. Show that x0 x 2e
76. Show that x0 e 1
2 dx x2 dx
integrals as areas.  y f x dx x0 x01 s e b x2 y f x dx b f x dx y
0 ln y d y by interpreting the ■ CONCEPT CHECK 2. How do you evaluate x sin mx cos nx dx if m is odd? What if n is odd? What if m and n are both even?
x 2 occurs in an integral, what substitution might you try? What if sa 2 x 2 occurs? What if
sx 2 a 2 occurs? 4. What is the form of the partial fraction expansion of a rational function P x Q x if the degree of P is less than the degree of
Q and Q x has only distinct linear factors? What if a linear
factor is repeated? What if Q x has an irreducible quadratic
factor (not repeated)? What if the quadratic factor is repeated? ■ 4
A
can be put in the form
4
x2 B
x x
x x2 4
A
can be put in the form
4
x x x2
xx 4
A
can be put in the form 2
4
x x 4. x2
x x2 A
4
can be put in the form
4
x x2 5. y 3. 6. 2 0 y x 4 1 x 2 1 dx 1
2 dx x
x2 C
1 3x dx 1 ■ b
5. State the rules for approximating the deﬁnite integral xa f x dx with the Midpoint Rule, the Trapezoidal Rule, and Simpson’s
Rule. Which would you expect to give the best estimate? How
do you approximate the error for each rule? 2 B
4 x 2 . . B
4 a f x dx (b) y b f x dx (c) y f x dx b
7. Deﬁne the improper integral xa f x d x for each of the follow ing cases.
(a) f has an inﬁnite discontinuity at a.
(b) f has an inﬁnite discontinuity at b.
(c) f has an inﬁnite discontinuity at c, where a c b. 8. State the Comparison Theorem for improper integrals. ■ 8. The Midpoint Rule is always more accurate than the Trapezoidal Rule.
(b) Every elementary function has an elementary antiderivative. C
2 y 9. (a) Every elementary function has an elementary derivative. . B (a) TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. 2. 2 6. Deﬁne the following improper integrals. 3. If the expression sa 2 2 x converges. Evaluate the integral for this value of C. use it? xx
x2 C
4 78. Find the value of the constant C for which the integral 1. State the rule for integration by parts. In practice, how do you 1. 1
sx 2 converges. Evaluate the integral for this value of C. d x. 8 Review 2 0 10. If f is continuous on 0, x0 and x1 f x d x is convergent, then f x d x is convergent. 11. If f is a continuous, decreasing function on 1, lim x l f x
. and 0 , then x1 f x d x is convergent. 12. If xa f x d x and xa t x d x are both convergent, then xa fx t x d x is convergent. ln 15
13. If xa f x d x and xa t x dx are both divergent, then xa 1
d x is convergent.
x s2 7. If f is continuous, then x f x dx lim t l xt t f x d x. fx 14. If f x diverges. t x d x is divergent.
t x and x0 t x dx diverges, then x0 f x dx also 5E08(pp 570579) 1/17/06 5:48 PM Page 577 CHAPTER 8 REVIEW ■ EXERCISES Note: Additional practice in techniques of integration is provided
in Exercises 7.5.
1–40 1. 3. y x 5 x 0 10 cos
1 sin 2 y 0 7 4. d 3 5. y tan x sec x d x 7. sin ln t
dt
t 9. y
y 4 1 6.
8. x 3 2 ln x d x
sx 2
x 2 11. y 13. yx 15. y sin 1 1 10. cos 5 d 17. y x sec x tan x d x 19. y 9x x
2 21. y sx 23. y csc 25. y 2 2 4 yy y sin x
dx
1 x2 1 y
y x2
x y sec 6
d
tan 2 1 20. y sin t
y
ye 26. y1 y 31. y dx
ex 33. y 35. y sx 37. y 0 y e s1 32. y dx
dx cos x 32 sin x 2 cos 2 x d x y 48. y 50. 4x ■ ■ ■ 0 ■ ■ 1
dt
1
y 6 sy 2 1 1 2 0
1
1 dy 2
3x dx x1
dx
3
sx 4
tan 1x
dx
x2 1
■ ■ ■ 52. 2 dx
■ ■ ■ ■ y sx x3
2 1 ■ dx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■  Use the Table of Integrals on the Reference Pages to
evaluate the integral.
x x sin x
dx
cos 3 x 57. y sx 34. y arcsin x 2 d x 36. y1 38. y x tan 1 2x ■ 0 tan
tan 2 ■ ye 1 5
■ ■ 55. 4 dx
4x 2 ■ 2x e 2 ■ y ln x dx x t2
t2 1 y dx
x x2 dx 3 55–58 ■ 32 1
x 3 46. ■ ■ 1
dx
1 30. e xse x 1
dx
ex 8 x2
4 x2 sx ■ 2x
d x by hand?
(Don’t actually carry out the integration.)
(b) How would you evaluate x x 5e 2 x d x using tables?
(Don’t actually do it.)
(c) Use a CAS to evaluate x x 5e 2 x d x.
(d) Graph the integrand and the indeﬁnite integral on the same
screen. cos x dx y sx ■ 54. (a) How would you evaluate x x 5e dx x 28. x 5 sec x d x ln 10 10 y 1 3 0 ■ cos2 x sin3 x and use the graph to
guess the value of the integral x02 f x d x. Then evaluate the
integral to conﬁrm your guess. CAS 1 44. ln x
dx
sx 4 ■ ; 53. Graph the function f x cos 2t
3 x y dx
x ln x 0 ■ dt x 24. cos 3x sin 2 x d x 51. 2 ■ Evaluate the integral or show that it is divergent. 2x 2 ■ ■ stan
d
sin 2  Evaluate the indeﬁnite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C 0). x 2 8x 3
dx
x 3 3x 2 y y ■ 4 ; 51–52 2
dx
2 18. 22. x 2 6x 4
dx
1 x2 2 49. sarctan x
dx
1 x2 1 0 y ■ 3 y 42. 1 dy dx
x 2s1 x 2 y y 47. 12 40. dx ■ 1 y 45.  y 43. 1
4y 2 ■ 3 1 2 ■ 41. 2t 1 xe 2x
1 2x 12 0 41–50 dy dt 4 3 29. 1 dx 4x dx y 1 5 4x 27. 0 1
6x dx 3x 3
x2 y 0.6 y ye 16. x
2 5 0 14. dx
3 y 12. dx y ■ 2. dx 577 ■ 39. Evaluate the integral.  ❙❙❙❙ ■ 2 x
■ ■ ■ y csc t dt 58. 1 dx
■ 5 56. e 2x dx s1 y s1 ■ cot x
dx
2 sin x ■ ■ ■ 59. Verify Formula 33 in the Table of Integrals (a) by differentia tion and (b) by using a trigonometric substitution.
d x 2 dx ■ 60. Verify Formula 62 in the Table of Integrals.
61. Is it possible to ﬁnd a number n such that x0 x n d x is convergent? ■ 5E08(pp 570579) 578 ❙❙❙❙ 1/17/06 5:49 PM Page 578 CHAPTER 8 TECHNIQUES OF INTEGRATION 62. For what values of a is x0 e ax cos x dx convergent? Evaluate the integral for those values of a. point to be 53 cm. The circumference 7 cm from each end is
45 cm. Use Simpson’s Rule to make your estimate. 63–64  Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and
(c) Simpson’s Rule with n 10 to approximate the given integral.
Round your answers to six decimal places. 63. y 1 64. x 4 dx s1 0 2 y 0 ssin x d x
28 cm ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 65. Estimate the errors involved in Exercise 63, parts (a) and (b). How large should n be in each case to guarantee an error of
less than 0.00001? 71. Use the Comparison Theorem to determine whether the integral
x3 y x
is convergent or divergent.
1 66. Use Simpson’s Rule with n e x x from x curve y 6 to estimate the area under the
1 to x 4. 67. The speedometer reading (v) on a car was observed at 1minute intervals and recorded in the chart. Use Simpson’s
Rule to estimate the distance traveled by the car.
t (min) v (mi h) t (min) v (mi h) 0
1
2
3
4
5 40
42
45
49
52
54 6
7
8
9
10 56
57
57
55
56 week, where the graph of r is as shown. Use Simpson’s Rule
with six subintervals to estimate the increase in the bee population during the ﬁrst 24 weeks.
r
12000 4000 C AS dx y2 x2 1 and the line y 3. 73. Find the area bounded by the curves y between x 0 and x cos x and y cos 2x . 74. Find the area of the region bounded by the curves y 1 (2 s x ), y 1 (2 s x ), and x 75. The region under the curve y 1. 2 cos x, 0 x
2, is rotated
about the xaxis. Find the volume of the resulting solid. 76. The region in Exercise 75 is rotated about the yaxis. Find the volume of the resulting solid. y and lim x l f x f x dx 0 0, show that f0 78. We can extend our deﬁnition of average value of a continuous function to an inﬁnite interval by deﬁning the average value of
to be
f on the interval a,
1
t
lim
y f x dx
tl
t aa
(a) Find the average value of y tan 1x on the interval 0, .
(b) If f x
0 and xa f x dx is divergent, show that the average value of f on the interval a,
is lim x l f x , if this
limit exists.
(c) If xa f x dx is convergent, what is the average value of f
on the interval a, ?
(d) Find the average value of y sin x on the interval 0, . 8000 0 2 72. Find the area of the region bounded by the hyperbola 77. If f is continuous on 0, 68. A population of honeybees increased at a rate of r t bees per 5 4 8 12 16 20 t
24
(weeks) 69. (a) If f x sin sin x , use a graph to ﬁnd an upper bound
for f 4 x .
(b) Use Simpson’s Rule with n 10 to approximate
x0 f x d x and use part (a) to estimate the error.
(c) How large should n be to guarantee that the size of the
error in using Sn is less than 0.00001? 70. Suppose you are asked to estimate the volume of a football. You measure and ﬁnd that a football is 28 cm long. You use a
piece of string and measure the circumference at its widest 79. Use the substitution u y 0 1 x to show that
1 ln x
dx
x2 0 80. The magnitude of the repulsive force between two point charges with the same sign, one of size 1 and the other of size q, is
q
F
4 0r 2
where r is the distance between the charges and 0 is a
constant. The potential V at a point P due to the charge q is
deﬁned to be the work expended in bringing a unit charge to P
from inﬁnity along the straight line that joins q and P. Find a
formula for V . 5E08(pp 570579) 1/17/06 5:49 PM Page 579 PROBLEMS
PLUS Cover up the solution to the example and try it yourself ﬁrst.
EXAMPLE
(a) Prove that if f is a continuous function, then
a y 0 y f x dx a 0 fa x dx (b) Use part (a) to show that
2 y 0 sin n x
dx
sin x cos n x
n 4 for all positive numbers n.
SOLUTION
 The principles of problem solving are
discussed on page 58. (a) At ﬁrst sight, the given equation may appear somewhat bafﬂing. How is it possible
to connect the left side to the right side? Connections can often be made through one of
the principles of problem solving: introduce something extra. Here the extra ingredient is
a new variable. We often think of introducing a new variable when we use the Substitution Rule to integrate a speciﬁc function. But that technique is still useful in the present
circumstance in which we have a general function f .
Once we think of making a substitution, the form of the right side suggests that it
should be u a x. Then du
d x. When x 0, u a; when x a, u 0. So y a 0 fa y x dx 0 a y f u du a 0 f u du But this integral on the right side is just another way of writing x0a f x d x. So the given
equation is proved.
(b) If we let the given integral be I and apply part (a) with a
2, we get
I
 The computer graphs in Figure 1 make it
seem plausible that all of the integrals in the
example have the same value. The graph of each
integrand is labeled with the corresponding
value of n.
1 3
4 2 y 0 2 sin n x
dx
sin n x cos n x FIGURE 1 0 sin n sin n
2x A wellknown trigonometric identity tells us that sin
cos
2x
sin x, so we get
I y 0 2 2 2x
cos n
x 2 x dx cos x and cos n x
dx
cos n x sin n x Notice that the two expressions for I are very similar. In fact, the integrands have the
same denominator. This suggests that we should add the two expressions. If we do so,
we get 1 2I
0 y 2 π
2 Therefore, I y 0 2 sin n x
sin n x cos n x
dx
cos n x y 0 2 1 dx 2 4. 579 5E08(pp 580581) 1/17/06 5:38 PM Page 580 ; 1. Three mathematics students have ordered a 14inch pizza. Instead of slicing it in the tradi P RO B L E M S tional way, they decide to slice it by parallel cuts, as shown in the ﬁgure. Being mathematics
majors, they are able to determine where to slice so that each gets the same amount of pizza.
Where are the cuts made?
1
d x.
x7 x
The straightforward approach would be to start with partial fractions, but that would be brutal.
Try a substitution. 2. Evaluate y 1 3
3. Evaluate y (s1 0 x 3 ) dx. 7
s1 x7 4. A man initially standing at the point O walks along a pier pulling a rowboat by a rope of 14 in length L. The man keeps the rope straight and taut. The path followed by the boat is a curve
called a tractrix and it has the property that the rope is always tangent to the curve (see the
ﬁgure).
(a) Show that if the path followed by the boat is the graph of the function y f x , then FIGURE FOR PROBLEM 1 y pier (b) Determine the function y
L sL 2
x dy
dx fx x2 f x. (x, y) 5. A function f is deﬁned by
(L, 0)
O FIGURE FOR PROBLEM 4 y fx x cos t cos x 0 0 t dt x 2 Find the minimum value of f .
6. If n is a positive integer, prove that y 1 0 ln x n d x 1 n n! 7. Show that y 1 0 x 2 n dx 1 2 2 n n! 2
2n 1 ! Hint: Start by showing that if In denotes the integral, then
Ik 2k
2k 1 2
Ik
3 ; 8. Suppose that f is a positive function such that f is continuous.
(a) How is the graph of y f x sin n x related to the graph of y
as n l ?
(b) Make a guess as to the value of the limit
lim nl y 1 0 f x ? What happens f x sin nx dx based on graphs of the integrand.
(c) Using integration by parts, conﬁrm the guess that you made in part (b). [Use the fact that,
since f is continuous, there is a constant M such that f x
M for 0 x 1.]
9. If 0 580 a b, ﬁnd lim
tl0 y 1 0 1t bx a1 x t dx . 5E08(pp 580581) 1/17/06 5:39 PM Page 581 sin e x and use the graph to estimate the value of t such that xtt
maximum. Then ﬁnd the exact value of t that maximizes this integral. ; 10. Graph f x
y 11. The circle with radius 1 shown in the ﬁgure touches the curve y 1 f x d x is a 2 x twice. Find the area of the region that lies between the two curves.
12. A rocket is ﬁred straight up, burning fuel at the constant rate of b kilograms per second. Let
v
v t be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M M t be the mass of the rocket at time t and note that M decreases
as the fuel burns. If we neglect air resistance, it follows from Newton’s Second Law that y= 2x 
0 F IGURE FOR PROBLEM 11 x F
where the force F M dv
dt ub M t. Thus 1 M dv
dt ub Mt Let M1 be the mass of the rocket without fuel, M2 the initial mass of the fuel, and
M0 M1 M2 . Then, until the fuel runs out at time t M2 b, the mass is M M0 bt.
(a) Substitute M M0 b t into Equation 1 and solve the resulting equation for v. Use the
initial condition v 0
0 to evaluate the constant.
(b) Determine the velocity of the rocket at time t M2 b. This is called the burnout velocity.
(c) Determine the height of the rocket y y t at the burnout time.
(d) Find the height of the rocket at any time t.
13. Use integration by parts to show that, for all x 0 y 0 ln 1 sin t
x 0,
t dt 2
ln 1 x ; 14. The Chebyshev polynomials Tn are deﬁned by
Tn x cos n arccos x 0, 1, 2, 3, . . . n (a) What are the domain and range of these functions?
(b) We know that T0 x
1 and T1 x
x. Express T2 explicitly as a quadratic polynomial
and T3 as a cubic polynomial.
(c) Show that, for n 1,
Tn 1 x 2 x Tn x Tn 1 x (d) Use part (c) to show that Tn is a polynomial of degree n.
(e) Use parts (b) and (c) to express T4 , T5 , T6 , and T7 explicitly as polynomials.
(f) What are the zeros of Tn ? At what numbers does Tn have local maximum and minimum
values?
(g) Graph T2 , T3 , T4 , and T5 on a common screen.
(h) Graph T5 , T6 , and T7 on a common screen.
(i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the
zeros of Tn 1 ? What about the xcoordinates of the maximum and minimum values?
(j) Based on your graphs in parts (g) and (h), what can you say about x1 1 Tn x d x when n is
odd and when n is even?
(k) Use the substitution u arccos x to evaluate the integral in part (j).
(l) The family of functions f x
cos c arccos x are deﬁned even when c is not an integer
(but then f is not a polynomial). Describe how the graph of f changes as c increases. 581 ...
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 hamrick
 Sin, Cos, dx

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