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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. Techniques 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.
x nϩ1
ϩC
nϩ1 yx n dx ye x ͑n Ϫ1͒ dx e x ϩ C y 1
dx ln x ϩ C
x ya Խ Խ x dx ax
ϩC
ln a y sin x dx Ϫcos x ϩ C y cos x dx sin x ϩ C y sec x dx tan x ϩ C y csc x dx Ϫcot x ϩ C y sec x tan x dx sec x ϩ C y csc x cot x dx Ϫcsc x ϩ C y sinh x dx cosh x ϩ C y cosh x dx sinh x ϩ C y tan x dx ln Խ sec x Խ ϩ C y cot x dx ln Խ sin x Խ ϩ C 2 yx 2 2 ͩͪ 1
1
x
dx tanϪ1
ϩ a2
a
a ϩC y sa 2 ͩͪ 1
x
dx sinϪ1
Ϫ x2
a ϩC 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 ͑x͒t͑x͔͒ f ͑x͒tЈ͑x͒ ϩ t͑x͒f Ј͑x͒
dx
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 ͓ f ͑x͒tЈ͑x͒ ϩ t͑x͒f Ј͑x͔͒ dx f ͑x͒t͑x͒
y f ͑x͒tЈ͑x͒ dx ϩ y t͑x͒f Ј͑x͒ dx f ͑x͒t͑x͒ or We can rearrange this equation as y f ͑x͒tЈ͑x͒ dx f ͑x͒t͑x͒ Ϫ y t͑x͒f Ј͑x͒ dx 1 Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f ͑x͒ and v t͑x͒. Then the differentials are
du f Ј͑x͒ dx and dv tЈ͑x͒ dx, so, by the Substitution Rule, the formula for integration
by parts becomes y u dv uv Ϫ y v du 2 EXAMPLE 1 Find y x sin x dx. SOLUTION USING FORMULA 1 Suppose we choose f ͑x͒ x and tЈ͑x͒ sin x. Then f Ј͑x͒ 1
and t͑x͒ Ϫcos x. (For t we can choose any antiderivative of tЈ.) Thus, using Formula
1, we have y x sin x dx f ͑x͒t͑x͒ Ϫ y t͑x͒f Ј͑x͒ dx
x͑Ϫcos x͒ Ϫ y ͑Ϫcos x͒ dx
Ϫx cos x ϩ y cos x dx
Ϫx cos x ϩ sin x ϩ 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:
uᮀ
dv ᮀ
du ᮀ
vᮀ ux
Then
and so dv sin x dx du dx v Ϫcos x u d√ u √ √ du y x sin x dx y x sin x dx x ͑Ϫcos x͒ Ϫ y ͑Ϫcos x͒ dx
Ϫx cos x ϩ y cos x dx
Ϫx cos x ϩ sin x ϩ C 5E08(pp 510519) 1/17/06 5:19 PM Page 513 SECTION 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 dx and expressed it in terms
of the simpler integral x cos x dx. If we had chosen u sin x and dv x dx, then
du cos x dx and v x 2͞2, so integration by parts gives
■ y x sin x dx ͑sin x͒ x2
1
Ϫ
2
2 yx 2 cos x dx Although this is true, x x 2 cos x dx 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 ͑x͒ to
be a function that becomes simpler when differentiated (or at least not more complicated)
as long as dv tЈ͑x͒ dx can be readily integrated to give v.
EXAMPLE 2 Evaluate y ln x dx. SOLUTION Here we don’t have much choice for u and dv. Let u ln x
du Then dv dx 1
dx
x vx Integrating by parts, we get y ln x dx x ln x Ϫ y x dx
x  It’s customary to write x 1 dx as x dx. x ln x Ϫ y dx  Check the answer by differentiating it. x ln x Ϫ x ϩ C
Integration by parts is effective in this example because the derivative of the function
f ͑x͒ ln x is simpler than f .
EXAMPLE 3 Find 2 t 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
u t2
Then dv e t dt du 2t dt v et Integration by parts gives
3 y t e dt t e
2 t 2 t Ϫ 2 y te t dt The integral that we obtained, x te 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 dv e t dt. Then du dt, v e t, and y te dt te
t t Ϫ y e t dt te t Ϫ e t ϩ C
Putting this in Equation 3, we get yt e dt t 2 e t Ϫ 2 y te t dt 2 t t 2 e t Ϫ 2͑te t Ϫ e t ϩ C͒
t 2 e t Ϫ 2te t ϩ 2e t ϩ C1
EXAMPLE 4 Evaluate
 An easier method, using complex numbers,
is given in Exercise 48 in Appendix G. ye x where C1 Ϫ2C sin x dx. SOLUTION Neither e x nor sin x becomes simpler when differentiated, but we try choosing u e x and dv sin x dx anyway. Then du e x dx and v Ϫcos x, so integration by
parts gives ye 4 x sin x dx Ϫe x cos x ϩ y e x cos x dx The integral that we have obtained, x e x cos x dx, 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 dx. Then du e x dx, v sin x, and ye 5 x cos x dx e x sin x Ϫ y e x sin x dx At ﬁrst glance, it appears as if we have accomplished nothing because we have arrived at
 Figure 1 illustrates Example 4 by
showing the graphs of f ͑x͒ e x sin x and
1
F͑x͒ 2 e x͑sin x Ϫ cos x͒. As a visual check
on our work, notice that f ͑x͒ 0 when F has
a maximum or minimum. we get ye x sin x dx Ϫe x cos x ϩ e x sin x Ϫ y e x sin x dx This can be regarded as an equation to be solved for the unknown integral. Adding 12 x e x sin x dx to both sides, we obtain F
f
_3 x e x sin x dx, which is where we started. However, if we put Equation 5 into Equation 4 2 y e x sin x dx Ϫe x cos x ϩ e x sin x
6 Dividing by 2 and adding the constant of integration, we get
_4 FIGURE 1 ye x sin x dx 1 e x ͑sin x Ϫ cos x͒ ϩ C
2 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 Ј and tЈ are continuous, and using the Fundamental 5E08(pp 510519) 1/17/06 5:20 PM Page 515 SECTION 8.1 INTEGRATION BY PARTS ❙❙❙❙ 515 Theorem, we obtain y 6 b a EXAMPLE 5 Calculate y 1 0 b b f ͑x͒tЈ͑x͒ dx f ͑x͒t͑x͒ a Ϫ y t͑x͒f Ј͑x͒ dx
a tanϪ1x dx. SOLUTION Let u tanϪ1x
du Then dv dx dx
1 ϩ x2 vx So Formula 6 gives y 1 0 1 tanϪ1x dx x tanϪ1x 0 Ϫ y 1 0 x
dx
1 ϩ x2 1 ؒ tanϪ1 1 Ϫ 0 ؒ tanϪ1 0 Ϫ y 1 0  Since tanϪ1x ജ 0 for x ജ 0, the integral in
Example 5 can be interpreted as the area of the
region shown in Figure 2.
1
x
Ϫy
2 dx
0 1 ϩ x
4 To evaluate this integral we use the substitution t 1 ϩ x 2 (since u has another meaning
in this example). Then dt 2x dx, so x dx dt͞2. When x 0, t 1; when x 1,
t 2; so y y=tan–!x y 0
1 x
dx
1 ϩ x2 1 0 x x
2 dt
dx 1 y
1 ln t
2
2
1 t
1 ϩ x2 Խ Խ] 2 1 1
1
2 ͑ln 2 Ϫ ln 1͒ 2 ln 2 FIGURE 2 Therefore y 1 0 tanϪ1x dx
1
x
ln 2
Ϫy
Ϫ
2 dx
0 1 ϩ x
4
4
2 EXAMPLE 6 Prove the reduction formula
 Equation 7 is called a reduction formula
because the exponent n has been reduced to
n Ϫ 1 and n Ϫ 2. 7 1 y sin x dx Ϫ n cos x sin
n nϪ1 xϩ nϪ1
n y sin nϪ2 x dx where n ജ 2 is an integer.
SOLUTION Let Then u sin nϪ1x dv sin x dx du ͑n Ϫ 1͒ sin nϪ2x cos x dx v Ϫcos x so integration by parts gives y sin x dx Ϫcos x sin
n nϪ1 x ϩ ͑n Ϫ 1͒ y sin nϪ2x cos 2x dx 5E08(pp 510519) ❙❙❙❙ 516 1/17/06 5:20 PM Page 516 CHAPTER 8 TECHNIQUES OF INTEGRATION Since cos 2x 1 Ϫ sin 2x, we have y sin x dx Ϫcos x sin
n nϪ1 x ϩ ͑n Ϫ 1͒ y sin nϪ2x dx Ϫ ͑n Ϫ 1͒ y sin n x dx 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 dx Ϫcos x sin nϪ1x ϩ ͑n Ϫ 1͒ y sin nϪ2x dx
1 y sin x dx Ϫ n cos x sin
n or nϪ1 xϩ nϪ1
n y sin nϪ2 x dx The reduction formula (7) is useful because by using it repeatedly we could eventually
express x sin n x dx in terms of x sin x dx (if n is odd) or x ͑sin x͒0 dx x dx (if n is even).  8.1 Exercises 1–2  Evaluate the integral using integration by parts with the
indicated choices of u and dv. 2. ■ 3–32 3.
5. u , dv sec 2 d 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ y ͞2 24. y cos Ϫ1x dx 26. y 27. u ln x, dv x dx y sec d ; ■ 25. y
dy
e 2y y cos x ln͑sin x͒ dx 28. y y cos͑ln x͒ dx 30. y 31. y x ln x dx; y 1 29. 1. 23. y x cos 5x dx
y re r͞2 dr 4.
6. Ϫx y xe y 32. y 1͞2 0 ■ Evaluate the integral.  0 2 sin x dx yx 9. y ln͑2x ϩ 1͒ dx 8. y t sin 2t dt 10. yx 2 cos mx dx 2 x 4͑ln x͒2 dx 1
■ ■ y arctan 4t dt 12. yp 5 33. ■ ■ ■ y sin sx dx
y ln p dp y ͑ln x͒ dx 14. y t e dt 15. y e sin 3 d 16. ye 17. y y sinh y dy 18. y y cosh ay dy 2 0 ■ e s sin͑t Ϫ s͒ ds ■ ■ s s ͞2 4 34. 36. 3 cos͑ 2 ͒ d y yx e 1 19.
21. y 0 y 2 1 t sin 3t dt
ln x
dx
x2 5 x2 22. ■ dx 3 t Ϫ ■ ■ ■ ■ ■ ■ ■ ■ ■ y 1 0 y 4 1 Ϫx ͑x ϩ 1͒e
2 ■ ■ ■ ; 37–40  Evaluate the indeﬁnite integral. Illustrate, and check that
your answer is reasonable, by graphing both the function and its
antiderivative (take C 0). cos 2 d y x cos x dx 38. yx 39. 20. ■ e sx dx 37.
■  First make a substitution and then use integration by parts
to evaluate the integral. x dx 13. 2 r3
dr
s4 ϩ r 2 1 t arctan͑1͞x͒ dx 33–36 35. 11. s3 0 x csc 2x dx x 5 x dx 1 Ϫ1 y sin 1 0 dx ■ 7. ͞4 y ͑2x ϩ 3͒e 40. yx e 3͞2 ln x dx dx
x dx 3 x2 dx st ln t dt
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E08(pp 510519) 1/17/06 5:21 PM Page 517 SECTION 8.1 INTEGRATION BY PARTS 41. (a) Use the reduction formula in Example 6 to show that y sin x dx
2 ❙❙❙❙ 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. x
sin 2x
Ϫ
ϩC
2
4 53. y x sin x, (b) Use part (a) and the reduction formula to evaluate
x sin 4x dx. y ͑x Ϫ 2͒2 54. y arctan 3x,
■ ■ ■ ■ y x͞2
■ ■ ■ ■ ■ ■ ■ ■ 42. (a) Prove the reduction formula 1
n1
y cos x dx n cos nϪ1x sin x ϩ n
n 55–58 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. nϪ2 x dx 55. y cos͑ x͞2͒, y 0, 0 ഛ x ഛ 1; (b) Use part (a) to evaluate x cos 2x dx.
(c) Use parts (a) and (b) to evaluate x cos 4x dx. 56. y e x, y eϪx, x 1; 43. (a) Use the reduction formula in Example 6 to show that y ͞2 0 nϪ1
n sin n x dx y ͞2 57. y e , y 0, x Ϫ1, x 0;
58. y e , x 0, y ; sin nϪ2x dx 0 ■ where n ജ 2 is an integer.
(b) Use part (a) to evaluate x0͞2 sin 3x dx and x0͞2 sin 5x dx.
(c) Use part (a) to show that, for odd powers of sine, y ͞2 sin 2nϩ1x dx 0 ͞2 0 45–48  46. yx e n n x n ■ ■ about the xaxis
■ ■ ■ ■ ■ ■ ■ 60. A rocket accelerates by burning its onboard fuel, so its mass 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 2 ؒ 4 ؒ 6 ؒ и и и ؒ 2n
3 ؒ 5 ؒ 7 ؒ и и и ؒ ͑2n ϩ 1͒ v͑t͒ Ϫtt Ϫ ve ln m Ϫ rt
m where t is the acceleration due to gravity and t is not too
large. If t 9.8 m͞s 2, m 30,000 kg, r 160 kg͞s, and
ve 3000 m͞s, ﬁnd the height of the rocket one minute
after liftoff. Use integration by parts to prove the reduction formula. y ͑ln x͒ dx x ͑ln x͒ ■ 1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒
2 ؒ 4 ؒ 6 ؒ и и и ؒ 2n
2 sin 2nx dx 45. ■ about x 1 59. Find the average value of f ͑x͒ x 2 ln x on the interval ͓1, 3͔. 44. Prove that, for even powers of sine, y about the yaxis Ϫx
x about the yaxis Ϫ n y ͑ln x͒nϪ1 dx 61. A particle that moves along a straight line has velocity
v͑t͒ t 2eϪt meters per second after t seconds. How far will dx x ne x Ϫ n y x nϪ1e x dx it travel during the ﬁrst t seconds?
47. y ͑x 2 62. If f ͑0͒ t͑0͒ 0 and f Љ and t Љ are continuous, show that ϩ a 2 ͒ n dx
2na 2
x ͑x 2 ϩ a 2 ͒ n
ϩ
2n ϩ 1
2n ϩ 1 y ͑x tan x sec nϪ2x
nϪ2
ϩ
48. y sec x dx
nϪ1
nϪ1
n ■ ■ ■ ■ ■ ■ ■ 2 ϩa ͒ 2 nϪ1 dx (n Ϫ 1
2 ) y a 0 a f ͑x͒t Љ͑x͒ dx f ͑a͒tЈ͑a͒ Ϫ f Ј͑a͒t͑a͒ ϩ y f Љ͑x͒t͑x͒ dx
0 63. Suppose that f ͑1͒ 2, f ͑4͒ 7, f Ј͑1͒ 5, f Ј͑4͒ 3, and y sec nϪ2 x dx ͑n 4
f Љ is continuous. Find the value of x1 x f Љ͑x͒ dx. 1͒ 64. (a) Use integration by parts to show that
■ ■ ■ ■ ■ y f ͑x͒ dx x f ͑x͒ Ϫ y x f Ј͑x͒ dx 49. Use Exercise 45 to ﬁnd x ͑ln x͒3 dx.
50. Use Exercise 46 to ﬁnd x x e dx.
4 x 51–52  Find the area of the region bounded by the given curves. 51. y xeϪ0.4x,
52. y 5 ln x,
■ ■ (b) If f and t are inverse functions and f Ј is continuous, prove
that ■ y 0, x5 y y x ln x
■ ■ b a ■ ■ ■ ■ ■ ■ ■ f ͑x͒ dx bf ͑b͒ Ϫ af ͑a͒ Ϫ y f ͑b͒ f ͑a͒ t͑ y͒ dy [Hint: Use part (a) and make the substitution y f ͑x͒.] 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 Ͼ a Ͼ 0, draw a diagram to give a geometric interpretation of part (b).
(d) Use part (b) to evaluate x1e ln x dx.
65. We arrived at Formula 6.3.2, V (c) Use parts (a) and (b) to show that
2n ϩ 1
I2nϩ1
ഛ1
ഛ
2n ϩ 2
I2n
and deduce that lim n l ϱ I2nϩ1͞I2n 1.
(d) Use part (c) and Exercises 43 and 44 to show that xab 2 x f ͑x͒ dx, 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 lim nlϱ This formula is usually written as an inﬁnite product: d V b 2d Ϫ a 2c Ϫ y ͓ t͑ y͔͒ 2 dy
2 2 4 4 6 6
ؒ ؒ ؒ ؒ ؒ ؒ иии
2
1 3 3 5 5 7 c Make the substitution y f ͑x͒ and then use integration by
parts on the resulting integral to prove that V xab 2 x f ͑x͒ dx.
y x=g(y) 2n
2n
2 2 4 4 6 6
ؒ ؒ ؒ ؒ ؒ ؒ иии ؒ
ؒ
1 3 3 5 5 7
2n Ϫ 1 2n ϩ 1
2 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 a b x x0͞2 sin n x dx. (a) Show that I2nϩ2 ഛ I2nϩ1 ഛ I2n.
(b) Use Exercise 44 to show that
I2nϩ2
2n ϩ 1
I2n
2n ϩ 2  8.2 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 dx. SOLUTION Simply substituting u cos x isn’t helpful, since then du Ϫsin 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 ϩ cos 2x 1: cos 3x cos 2x ؒ cos x ͑1 Ϫ sin 2x͒ cos x
We can then evaluate the integral by substituting u sin x, so du cos x dx and y cos x dx y cos x ؒ cos x dx y ͑1 Ϫ sin x͒ cos x dx
3 2 y ͑1 Ϫ u 2 ͒ du u Ϫ 1 u 3 ϩ C
3
sin x Ϫ 1 sin 3x ϩ C
3 2 5E08(pp 510519) 1/17/06 5:22 PM Page 519 SECTION 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 ϩ 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 dx SOLUTION We could convert cos 2x to 1 Ϫ 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 ͑sin2x͒2 cos 2x sin x ͑1 Ϫ cos 2x͒2 cos 2x sin x
 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 cos x, we have du Ϫsin x dx and so y sin x cos x dx y ͑sin x͒
5 2 2 2 cos 2x sin x dx 0.2 y ͑1 Ϫ cos 2x͒2 cos 2x sin x dx
_π π y ͑1 Ϫ u 2 ͒2 u 2 ͑Ϫdu͒ Ϫy ͑u 2 Ϫ 2u 4 ϩ u 6 ͒ du ͩ Ϫ _ 0.2 u3
u5
u7
Ϫ2
ϩ
3
5
7 ͪ ϩC Ϫ 1 cos 3x ϩ 2 cos 5x Ϫ 1 cos 7x ϩ C
3
5
7 FIGURE 1 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):
sin 2x 1 ͑1 Ϫ cos 2x͒
2  Example 3 shows that the area of the region
shown in Figure 2 is ͞2. EXAMPLE 3 Evaluate y 0 and cos 2x 1 ͑1 ϩ cos 2x͒
2 sin 2x dx. SOLUTION If we write sin 2x 1 Ϫ cos 2x, the integral is no simpler to evaluate. Using the halfangle formula for sin 2x, however, we have 1.5
y=sin@ x y 0 sin 2x dx 1 y ͑1 Ϫ cos 2x͒ dx
2
0 [ (x Ϫ
1
2 1
2
0 sin 2x) 1 ( Ϫ 1 sin 2) Ϫ 1 (0 Ϫ 1 sin 0) 1
2
2
2
2
2
0
_0.5 π Notice that we mentally made the substitution u 2x when integrating cos 2x. Another
method for evaluating this integral was given in Exercise 41 in Section 8.1. FIGURE 2 EXAMPLE 4 Find 4 y sin x dx. SOLUTION We could evaluate this integral using the reduction formula for x sin n x dx (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 2x͒2 and use a halfangle formula: y sin x dx y ͑sin x͒ dx
4 2 y ͩ 2 1 Ϫ cos 2x
2 ͪ 2 dx 1 y ͑1 Ϫ 2 cos 2x ϩ cos 2 2x͒ dx
4
Since cos 2 2x occurs, we must use another halfangle formula
cos 2 2x 1 ͑1 ϩ cos 4x͒
2
This gives y sin x dx y ͓1 Ϫ 2 cos 2x ϩ
1
4 4 1y
4 1
2 ͑1 ϩ cos 4x͔͒ dx ( 3 Ϫ 2 cos 2x ϩ 1 cos 4x) dx
2
2 1 ( 3 x Ϫ sin 2x ϩ 1 sin 4x) ϩ C
4 2
8
To summarize, we list guidelines to follow when evaluating integrals of the form x sin mx cos nx dx, where m ജ 0 and n ജ 0 are integers.
Strategy for Evaluating y sin m x cos nx dx (a) If the power of cosine is odd ͑n 2k ϩ 1͒, save one cosine factor and use
cos 2x 1 Ϫ sin 2x to express the remaining factors in terms of sine: y sin m x cos 2kϩ1x dx y sin m x ͑cos 2x͒k cos x dx
y sin m x ͑1 Ϫ sin 2x͒k cos x dx Then substitute u sin x.
(b) If the power of sine is odd ͑m 2k ϩ 1͒, save one sine factor and use
sin 2x 1 Ϫ cos 2x to express the remaining factors in terms of cosine: y sin 2kϩ1 x cos n x dx y ͑sin 2x͒k cos n x sin x dx
y ͑1 Ϫ cos 2x͒k cos n x sin x dx 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 ͑1 Ϫ cos 2x͒
2 cos 2x 1 ͑1 ϩ cos 2x͒
2 It is sometimes helpful to use the identity
sin x cos x 1 sin 2x
2 5E08(pp 520529) 1/17/06 5:24 PM Page 521 SECTION 8.2 TRIGONOMETRIC INTEGRALS ❙❙❙❙ 521 We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since
͑d͞dx͒ 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 ϩ tan 2x.
Or, since ͑d͞dx͒ 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 dx. 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 ϩ tan 2x. We can then evaluate the integral
by substituting u tan x with du sec 2x dx : y tan x sec x dx y tan x sec x sec x dx
6 4 6 2 2 y tan 6x ͑1 ϩ tan 2x͒ sec 2x dx
y u 6͑1 ϩ u 2 ͒ du y ͑u 6 ϩ u 8 ͒ du
u7
u9
ϩ
ϩC
7
9 1 tan 7x ϩ 1 tan 9x ϩ C
7
9
EXAMPLE 6 Find y tan sec d.
5 7 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
involving only secant using the identity tan 2 sec 2 Ϫ 1. We can then evaluate the
integral by substituting u sec , so du sec tan d : y tan
5 sec 7 d y tan 4 sec 6 sec tan d
y ͑sec 2 Ϫ 1͒2 sec 6 sec tan d
y ͑u 2 Ϫ 1͒2 u 6 du y ͑u 10 Ϫ 2u 8 ϩ u 6 ͒ du
u 11
u9
u7
Ϫ2
ϩ
ϩC
11
9
7 1
11 sec 11 Ϫ 2 sec 9 ϩ 1 sec 7 ϩ C
9
7 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 2k, k ജ 2͒, save a factor of sec 2x and use
sec 2x 1 ϩ tan 2x to express the remaining factors in terms of tan x : y tan m x sec 2kx dx y tan m x ͑sec 2x͒kϪ1 sec 2x dx
y tan m x ͑1 ϩ tan 2x͒kϪ1 sec 2x dx Then substitute u tan x.
(b) If the power of tangent is odd ͑m 2k ϩ 1͒, save a factor of sec x tan x and
use tan 2x sec 2x Ϫ 1 to express the remaining factors in terms of sec x : y tan 2kϩ1 x sec n x dx y ͑tan 2x͒k sec nϪ1x sec x tan x dx
y ͑sec 2x Ϫ 1͒k sec nϪ1x sec x tan x dx Then substitute u 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 dx ln Խ sec x Խ ϩ C
We will also need the indeﬁnite integral of secant: 1 y sec x dx ln Խ sec x ϩ tan x Խ ϩ C We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x ϩ tan x :
sec x ϩ tan x y sec x dx y sec x sec x ϩ tan x dx
y sec 2x ϩ sec x tan x
dx
sec x ϩ tan x If we substitute u sec x ϩ tan x, then du ͑sec x tan x ϩ sec 2x͒ dx, so the integral
becomes x ͑1͞u͒ du ln u ϩ C. Thus, we have Խ Խ y sec x dx ln Խ sec x ϩ tan x Խ ϩ C 5E08(pp 520529) 1/17/06 5:25 PM Page 523 SECTION 8.2 TRIGONOMETRIC INTEGRALS EXAMPLE 7 Find ❙❙❙❙ 523 3 y tan x dx. SOLUTION Here only tan x occurs, so we use tan 2x sec 2x Ϫ 1 to rewrite a tan 2x factor in terms of sec 2x : y tan x dx y tan x tan x dx
3 2 y tan x ͑sec 2x Ϫ 1͒ dx
y tan x sec 2x dx Ϫ y tan x dx
tan 2x
Ϫ ln sec x ϩ C
2 Խ Խ In the ﬁrst integral we mentally substituted u tan x so that du sec 2x dx.
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 dx. SOLUTION Here we integrate by parts with u sec x
du sec x tan x dx
Then dv sec 2x dx
v tan x y sec x dx sec x tan x Ϫ y sec x tan x dx
3 2 sec x tan x Ϫ y sec x ͑sec 2x Ϫ 1͒ dx
sec x tan x Ϫ y sec 3x dx ϩ y sec x dx
Using Formula 1 and solving for the required integral, we get y sec x dx (sec x tan x ϩ ln Խ sec x ϩ tan x Խ) ϩ C
3 1
2 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 dx can be found by similar methods because of the identity
1 ϩ cot 2x csc 2x.
Finally, we can make use of another set of trigonometric identities:
2 To evaluate the integrals (a) x sin mx cos nx dx, (b) x sin mx sin nx dx, or
(c) x cos mx cos nx dx, use the corresponding identity:
 These product identities are discussed in
Appendix D. (a) sin A cos B 1 ͓sin͑A Ϫ B͒ ϩ sin͑A ϩ B͔͒
2
(b) sin A sin B 1 ͓cos͑A Ϫ B͒ Ϫ cos͑A ϩ B͔͒
2
(c) cos A cos B 1 ͓cos͑A Ϫ B͒ ϩ cos͑A ϩ B͔͒
2 5E08(pp 520529) ❙❙❙❙ 524 1/17/06 5:25 PM Page 524 CHAPTER 8 TECHNIQUES OF INTEGRATION EXAMPLE 9 Evaluate y sin 4x cos 5x dx. 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 4x cos 5x dx y 1
2 ͓sin͑Ϫx͒ ϩ sin 9x͔ dx 1 y ͑Ϫsin x ϩ sin 9x͒ dx
2
1 (cos x Ϫ 1 cos 9x͒ ϩ C
2
9  8.2
1–47 Exercises Evaluate the integral. 32. y tan ͑ay͒ dy 34. y tan x sec x dx cot 2x dx 36. y 37. y cot ␣ csc ␣ d␣ 38. y csc 39. y csc x dx 40. y 41. y sin 5x sin 2x dx 42. y sin 3x cos x dx 43. y cos 7 cos 5 d 44. y 45. y 1 Ϫ tan 2x
dx
sec 2x 46. y cos x Ϫ 1 47.  y t sec ͑t 5 3.
5.
7.
9.
11.
13.
15. 2 3͞4 y ͞2 sin 5x cos 3x dx
5 4 y cos x sin x dx
y ͞2 0 y 0 cos2 d sin ͑3t͒ dt
4 y ͑1 ϩ cos ͒2 d
y ͞4 0 sin 4x cos 2x dx
3 y sin x scos x dx 6.
8. y ͞2 0 y sin ͑mx͒ dx
y ͞2 0 y 12.
14.
16. y ͞2 0 y cos cos ͑sin ͒ d
5 y sec 6 t dt 26. y ͞3 0 3 sin 2x cos 2x dx y tan x dx 25. ͞2
͞4 3 ͞3
͞6 y x cos2x dx 0 24. 2 ͞6 cos d 10. y tan x dx 23. 2 ͞2 cot 3x dx 5 ͞2 x cot 6 x dx csc 3x dx 2 2 cos x ϩ sin x
dx
sin 2x
dx ͒ tan 4͑t 2 ͒ dt 4 2 0 4 6 y y sec x tan x dx 2 sin 2 ͑2 ͒ d 22. 21. 1 Ϫ sin x
dx
cos x 4 3 y cos x sin 2x dx y tan 3 cos 5x dx 20. 19. y 6 3 y cot sin d 3 y cos x tan x dx 29. 4. 6 y sin x cos x dx 18. 2 17. 27. 2. y cos d 35. 3 y sin x cos x dx y tan x dx 33. 1. 31. sec 4͑t͞2͒ dt
4 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ tan 5 x sec 4 x dx
3 y tan x sec x dx 28.
30. 0 x0͞4 tan 8 x sec x dx 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 0͒.
5 4 4 49. 3 ͞3 0 y sin x dx 50. y sin x cos x dx y sin 3x sin 6x dx 52. y sec sec 4 tan 4 d y tan ͑2x͒ sec ͑2x͒ dx
y ■
48. If x0 ͞4 tan 6 x sec x dx I , express the value of 51. y ͞4 ■ 5 ■ ■ ■ ■ ■ ■ ■ ■ 4 x
dx
2
■ ■ tan 5x sec6x dx ■ 53. Find the average value of the function f ͑x͒ sin x cos x on
2 the interval ͓Ϫ, ͔. 3 ■ 5E08(pp 520529) 1/17/06 5:26 PM Page 525 SECTION 8.3 TRIGONOMETRIC SUBSTITUTION 54. Evaluate x sin x cos x dx by four methods: (a) the substitution  current that varies from 155 V to Ϫ155 V with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
E͑t͒ 155 sin͑120 t͒ 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 ͞2 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 ͓E͑t͔͒ 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
E͑t͒ A sin͑120 t͒. x ͞2 x 0, ■ ■ ■ ■ ■ ■ ; 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 0 ■ ■ cos 3x dx
■ 58.
■ ■ ■ y ■ 2 0
■ 65–67 ■ ■ ■  Find the volume obtained by rotating the region bounded
by the given curves about the speciﬁed axis. 66. y sin mx sin nx dx 67. y cos mx cos nx dx about the xaxis 61. y cos x, y 0, x 0, x ͞2; about y Ϫ1 62. y cos x, y 0, x 0, x ͞2; about y 1 Ϫ Ϫ Ϫ about the xaxis 60. y tan x, y 0, x 0, x ͞4; Prove the formula, where m and n are positive integers. y sin mx cos nx dx 0 59–62 59. y sin x, x ͞2, x , y 0;  65. sin 2 x cos 5 x dx
■ ■ 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■  8.3 ■ ■ ■ ͭ
ͭ 0
0
■ ■ if m n
if m n
if m n
if m n
■ ■ ■ ■ ■ ■ 68. A ﬁnite Fourier series is given by the sum
N f ͑x͒ ͚a n sin nx n1
■ 63. A particle moves on a straight line with velocity function
v͑t͒ sin t cos 2 t. Find its position function s f ͑t͒ if f ͑0͒ 0. 525 64. Household electricity is supplied in the form of alternating u cos x, (b) the substitution u sin x, (c) the identity
sin 2x 2 sin x cos x, and (d) integration by parts. Explain the
different appearances of the answers.
55–56 ❙❙❙❙ ■ a 1 sin x ϩ a 2 sin 2x ϩ и и и ϩ a N sin Nx
Show that the mth coefﬁcient a m is given by the formula
am 1
y
Ϫ f ͑x͒ sin mx dx Trigonometric Substitution
In ﬁnding the area of a circle or an ellipse, an integral of the form x sa 2 Ϫ x 2 dx arises,
where a Ͼ 0. If it were x xsa 2 Ϫ x 2 dx, the substitution u a 2 Ϫ x 2 would be effective
but, as it stands, x sa 2 Ϫ x 2 dx is more difﬁcult. If we change the variable from x to by
the substitution x a sin , then the identity 1 Ϫ sin 2 cos 2 allows us to get rid of the
root sign because Խ sa 2 Ϫ x 2 sa 2 Ϫ a 2 sin 2 sa 2͑1 Ϫ sin 2 ͒ sa 2 cos 2 a cos Խ Notice the difference between the substitution u a 2 Ϫ 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 t͑t͒ 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 y f ͑x͒ dx y f ͑t͑t͒͒tЈ͑t͒ 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 ͓Ϫ͞2, ͞2͔.
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 Identity sa 2 Ϫ x 2 x a sin , Ϫ
ഛഛ
2
2 1 Ϫ sin 2 cos 2 sa 2 ϩ x 2 x a tan , Ϫ
ϽϽ
2
2 1 ϩ tan 2 sec 2 sx 2 Ϫ a 2 x a sec , 0ഛϽ EXAMPLE 1 Evaluate y
3
or ഛ Ͻ
2
2 sec 2 Ϫ 1 tan 2 s9 Ϫ x 2
dx.
x2 SOLUTION Let x 3 sin , where Ϫ͞2 ഛ ഛ ͞2. Then dx 3 cos d and Խ Խ s9 Ϫ x 2 s9 Ϫ 9 sin 2 s9 cos 2 3 cos 3 cos
(Note that cos ജ 0 because Ϫ͞2 ഛ ഛ ͞2.) Thus, the Inverse Substitution Rule
gives
3 cos
s9 Ϫ x 2
y x 2 dx y 9 sin 2 3 cos d
y cos 2
d y cot 2 d
sin 2 y ͑csc 2 Ϫ 1͒ d
Ϫcot Ϫ ϩ C 3
x
¨
œ„„„„„
9≈ 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 x͞3 or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
Since sin x͞3, 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 Ϫ x 2, so we
can simply read the value of cot from the ﬁgure: FIGURE 1 cot x
sin ¨=
3 s9 Ϫ x 2
x (Although Ͼ 0 in the diagram, this expression for cot is valid even when Ͻ 0.)
Since sin x͞3, we have sinϪ1͑x͞3͒ and so y ͩͪ x
s9 Ϫ x 2
s9 Ϫ x 2
dx Ϫ
Ϫ sinϪ1
2
x
x
3 ϩC 5E08(pp 520529) 1/17/06 5:27 PM Page 527 SECTION 8.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 527 EXAMPLE 2 Find the area enclosed by the ellipse x2
y2
ϩ 2 1
a2
b
SOLUTION Solving the equation of the ellipse for y, we get y y2
x2
a2 Ϫ x2
1Ϫ 2
b2
a
a2 (0, b)
(a, 0)
0 FIGURE 2 ¥
≈
+ =1
b@
a@ x yϮ or b
sa 2 Ϫ x 2
a 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 sa 2 Ϫ x 2
0ഛxഛa
a
1
4 and so Ay a 0 b
sa 2 Ϫ x 2 dx
a To evaluate this integral we substitute x a sin . Then dx a cos d. To change the
limits of integration we note that when x 0, sin 0, so 0; when x a,
sin 1, so ͞2. Also Խ Խ sa 2 Ϫ x 2 sa 2 Ϫ a 2 sin 2 sa 2 cos 2 a cos a cos
since 0 ഛ ഛ ͞2. Therefore
A4 b
a y a 0 4ab y sa 2 Ϫ x 2 dx 4 ͞2 0 b
a cos 2 d 4ab y y 0 ͞2 1
2 0 [ 2ab ϩ 1 sin 2
2 ͞2
0 ͞2 ͩ 2ab a cos ؒ a cos d ͑1 ϩ cos 2 ͒ d ͪ
ϩ0Ϫ0
2 ab
We have shown that the area of an ellipse with semiaxes a and b is ab. 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
2 2 ϩ4 dx. SOLUTION Let x 2 tan , Ϫ͞2 Ͻ Ͻ ͞2. Then dx 2 sec 2 d and Խ Խ sx 2 ϩ 4 s4͑tan 2 ϩ 1͒ s4 sec 2 2 sec 2 sec
Thus, we have
dx y x sx
2 2 ϩ4 y 2 sec 2 d
1
4 tan 2 ؒ 2 sec
4 sec y tan d
2 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 and cos :
sec
1
cos 2
cos
ؒ
2
tan
cos sin 2
sin 2
Therefore, making the substitution u sin , we have y dx
1
x 2sx 2 ϩ 4
4
œ„„„„„
≈+4
x Ϫ ¨
2 ͩ ͪ
Ϫ 1
u y ϩCϪ du
u2
1
ϩC
4 sin csc
ϩC
4 We use Figure 3 to determine that csc sx 2 ϩ 4͞x and so FIGUR E 3 tan ¨= 1
4 cos
1
d
sin 2
4 y dx x
2 y x sx
2 EXAMPLE 4 Find y sx 2 ϩ4 Ϫ sx 2 ϩ 4
ϩC
4x x
dx.
ϩ4 2 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 ϩ 4 is simpler, because then du 2x dx
and y
NOTE x
1
dx
2
sx 2 ϩ 4 du y su su ϩ C sx 2 ϩ 4 ϩ 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
, where a Ͼ 0.
Ϫ a2 2 SOLUTION 1 We let x a sec , where 0 Ͻ dx a sec tan d and Ͻ ͞2 or Ͻ Ͻ 3͞2. Then Խ Խ sx 2 Ϫ a 2 sa 2͑sec 2 Ϫ 1͒ sa 2 tan 2 a tan a tan
Therefore y sx dx
a sec tan
y
d
2
Ϫa
a tan 2 Խ ≈a@
œ„„„„„ The triangle in Figure 4 gives tan sx 2 Ϫ a 2͞a, so we have ¨
a
FIGU RE 4 x
sec ¨=
a Խ y sec d ln sec ϩ tan ϩ C x y sx Ϳ Ϳ dx
x
sx 2 Ϫ a 2
ln
ϩ
ϩC
2 Ϫ a2
a
a Խ Խ ln x ϩ sx 2 Ϫ a 2 Ϫ ln a ϩ C 5E08(pp 520529) 1/17/06 5:28 PM Page 529 SECTION 8.3 TRIGONOMETRIC SUBSTITUTION ❙❙❙❙ 529 Writing C1 C Ϫ ln a, we have y sx 1 dx
ln x ϩ sx 2 Ϫ a 2 ϩ C1
Ϫ a2 Խ 2 Խ SOLUTION 2 For x Ͼ 0 the hyperbolic substitution x a cosh t can also be used. Using the
identity cosh 2 y Ϫ sinh 2 y 1, we have sx 2 Ϫ a 2 sa 2 ͑cosh 2 t Ϫ 1͒ sa 2 sinh 2 t a sinh t
Since dx a sinh t dt, we obtain y sx dx
a sinh t dt
y
y dt t ϩ C
2
Ϫa
a sinh t 2 Since cosh t x͞a, we have t coshϪ1͑x͞a͒ and ͩͪ dx
x
coshϪ1
2
a
sx Ϫ a y 2 2 ϩC 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 3 s3͞2 0 x3
dx.
͑4x ϩ 9͒3͞2
2 SOLUTION First we note that ͑4x 2 ϩ 9͒3͞2 ͑s4x 2 ϩ 9 )3 so trigonometric substitution is appropriate. Although s4x ϩ 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 2x. When we combine this with the tangent substitution, we have x 3 tan ,
2
which gives dx 3 sec 2 d and
2
2 s4x 2 ϩ 9 s9 tan 2 ϩ 9 3 sec
When x 0, tan 0, so 0; when x 3s3͞2, tan s3, so ͞3. y 3 s3͞2 0 27
3
x3
͞3 8 tan
3͞2 dx y
0
͑4x ϩ 9͒
27 sec3
2 3
16 y ͞3 0 3
16 y ͞3 0 3
2 sec 2 d 3
tan 3
͞3 sin
3
d 16 y
d
0
sec
cos2 1 Ϫ cos 2
sin d
cos 2 Now we substitute u cos so that du Ϫsin d. When 0, u 1; when
͞3, u 1.
2 5E08(pp 530539) 530 ❙❙❙❙ 1/17/06 6:03 PM Page 530 CHAPTER 8 TECHNIQUES OF INTEGRATION Therefore y 3 s3͞2 0 2
x3
1͞2 1 Ϫ u
1͞2
3
3
dx Ϫ 16 y
du 16 y ͑1 Ϫ u Ϫ2 ͒ du
2
3͞2
2
1
1
͑4x ϩ 9͒
u ͫ ͬ 3
16 u ϩ EXAMPLE 7 Evaluate x y s3 Ϫ 2x Ϫ x 2 1
u 1͞2
3
3
16 [( 1 ϩ 2) Ϫ ͑1 ϩ 1͒] 32
2 1 dx. SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by ﬁrst completing the square under the root sign: 3 Ϫ 2x Ϫ x 2 3 Ϫ ͑x 2 ϩ 2x͒ 3 ϩ 1 Ϫ ͑x 2 ϩ 2x ϩ 1͒
4 Ϫ ͑x ϩ 1͒2
This suggests that we make the substitution u x ϩ 1. Then du dx and x u Ϫ 1, so
x y s3 Ϫ 2x Ϫ x 2 dx y uϪ1
du
s4 Ϫ u 2 We now substitute u 2 sin , giving du 2 cos d and s4 Ϫ u 2 2 cos , so  Figure 5 shows the graphs of the integrand
in Example 7 and its indeﬁnite integral (with
C 0 ). Which is which? x y s3 Ϫ 2x Ϫ x 2 dx y 2 sin Ϫ 1
2 cos d
2 cos 3 y ͑2 sin Ϫ 1͒ d
2 _4 Ϫ2 cos Ϫ ϩ C
Ϫs4 Ϫ u 2 Ϫ sinϪ1 _5 4–30  Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle. 2. 3.
■ 1
dx ;
x 2 sx 2 Ϫ 9 yx 3 s9 Ϫ x dx ; y sx
■ ͩ ͪ
xϩ1
2 ϩC 2 x3
dx ;
ϩ9 2 ■ ■ 4. x 3 sec Evaluate the integral.  y 5. y ϩC Exercises 1–3 1. u
2 Ϫs3 Ϫ 2x Ϫ x 2 Ϫ sinϪ1 FIGURE 5  8.3 ͩͪ y 2 s3 0 x3
dx
s16 Ϫ x 2 x 3 sin 2 1 s2 t st Ϫ 1
3 yx 9. y sx x 3 tan
■ ■ ■ ■ ■ ■ ■ ■ dt 1
dx
s25 Ϫ x 2 7. 2 2 dx
ϩ 16 2 6. y 8. y 10. 2 0 x 3 sx 2 ϩ 4 dx
sx 2 Ϫ a 2
dx
x4 y st t5
dt
ϩ2 2 5E08(pp 530539) 1/17/06 6:03 PM Page 531 SECTION 8.3 TRIGONOMETRIC SUBSTITUTION 11. y s1 Ϫ 4x 13. y 15. y ͑a 17. y sx 1 12. y sx 2 Ϫ 9
dx
x3 14. y u s5 Ϫ u x2
dx
Ϫ x 2 ͒3͞2 16. yx x
dx
Ϫ7 18. y ͓͑ax͒ s1 ϩ x
dx
x 20. y s25 Ϫ t 22. y 24. y st 26. y s4x Ϫ x 28. y ͑5 Ϫ 4x Ϫ x 2 2 0 du 2 y 21. y 23. y s5 ϩ 4x Ϫ x 25. y s9x 27. y ͑x 29. y x s1 Ϫ x 2͞3 0 x 3s4 Ϫ 9x 2 dx 2 2 2 2 dx 1
dx
ϩ 6x Ϫ 8 dx
ϩ 2x ϩ 2͒2
4 0 2 ¨
O dt Q 2 y dt
Ϫ 6t ϩ 13
2 ͞2 0 ■ ■ ■ ■ ■ ■ ■ 38. A charged rod of length L produces an electric ﬁeld at point ͒ 2 5͞2 P͑a, b͒ given by cos t
dt
s1 ϩ sin 2 t E͑P͒ y ■ ■ ■ 31. (a) Use trigonometric substitution to show that y sx dx
x 4 sx 2 Ϫ 2 ; 37. Use a graph to approximate the roots of the equation
x 2 s4 Ϫ x 2 2 Ϫ x. Then approximate the area bounded by
the curve y x 2 s4 Ϫ x 2 and the line y 2 Ϫ x. dx ■ x Graph the integrand and its indeﬁnite integral on the same
screen and check that your answer is reasonable. LϪa Ϫa ■ R ; 36. Evaluate the integral dx y P sx 2 ϩ 1 dx x2 30. dx dx
Ϫ b 2 ͔ 3͞2 t 1 y 2 dx
s16x 2 Ϫ 9 2 19. 531 equation x 2 ϩ 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.] x sx 2 ϩ 4 dx dx 2 ❙❙❙❙ dx
ln ( x ϩ sx 2 ϩ a 2 ) ϩ C
ϩ a2 2 b
dx
4 0 ͑x 2 ϩ b 2 ͒3͞2 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 E͑P͒.
y P (a, b) (b) Use the hyperbolic substitution x a sinh t to show that y ͩͪ dx
x
sinhϪ1
a
sx 2 ϩ a 2 0 L x ϩC These formulas are connected by Formula 7.6.3. 39. Find the area of the crescentshaped region (called a lune) bounded by arcs of circles with radii r and R. (See the ﬁgure.) 32. Evaluate y ͑x 2 x2
dx
ϩ a 2 ͒3͞2 (a) by trigonometric substitution.
(b) by the hyperbolic substitution x a sinh t. r
R 33. Find the average value of f ͑x͒ sx 2 Ϫ 1͞x, 1 ഛ x ഛ 7.
34. Find the area of the region bounded by the hyperbola 9x 2 Ϫ 4y 2 36 and the line x 3.
35. Prove the formula A 2 r 2 for the area of a sector of a circle
1 with radius r and central angle . [Hint: Assume 0 Ͻ Ͻ ͞2
and place the center of the circle at the origin so it has the 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 ϩ ͑ y Ϫ R͒2 r 2 about the xaxis. Find the volume enclosed by the torus. 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 2͑͞x Ϫ 1͒
and 1͑͞x ϩ 2͒ to a common denominator we obtain
2
1
2͑x ϩ 2͒ Ϫ ͑x Ϫ 1͒
xϩ5
Ϫ
2
xϪ1
xϩ2
͑x Ϫ 1͒͑x ϩ 2͒
x ϩxϪ2
If we now reverse the procedure, we see how to integrate the function on the right side of
this equation: yx 2 xϩ5
dx
ϩxϪ2 y ͩ 2
1
Ϫ
xϪ1
xϩ2 Խ Խ ͪ dx Խ Խ 2 ln x Ϫ 1 Ϫ ln x ϩ 2 ϩ C
To see how the method of partial fractions works in general, let’s consider a rational
function
f ͑x͒ P͑x͒
Q͑x͒ 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
P͑x͒ a n x n ϩ a nϪ1 x nϪ1 ϩ и и и ϩ a 1 x ϩ a 0
where a n 0, then the degree of P is n and we write deg͑P͒ n.
If f is improper, that is, deg͑P͒ ജ deg͑Q͒, then we must take the preliminary step
of dividing Q into P (by long division) until a remainder R͑x͒ is obtained such that
deg͑R͒ Ͻ deg͑Q͒. The division statement is
f ͑x͒ 1 P͑x͒
R͑x͒
S͑x͒ ϩ
Q͑x͒
Q͑x͒ 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 ϩ x
dx.
xϪ1 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 ϩ x
dx
xϪ1
y ͩ x2 ϩ x ϩ 2 ϩ 2
xϪ1 ͪ dx x3
x2
ϩ
ϩ 2x ϩ 2 ln x Ϫ 1 ϩ C
3
2 Խ Խ 5E08(pp 530539) 1/17/06 6:03 PM Page 533 SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 533 The next step is to factor the denominator Q͑x͒ 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 ϩ b) and
irreducible quadratic factors (of the form ax 2 ϩ bx ϩ c, where b 2 Ϫ 4ac Ͻ 0). For instance, if Q͑x͒ x 4 Ϫ 16, we could factor it as
Q͑x͒ ͑x 2 Ϫ 4͒͑x 2 ϩ 4͒ ͑x Ϫ 2͒͑x ϩ 2͒͑x 2 ϩ 4͒
The third step is to express the proper rational function R͑x͒͞Q͑x͒ (from Equation 1) as
a sum of partial fractions of the form
A
͑ax ϩ b͒i or Ax ϩ B
͑ax 2 ϩ bx ϩ c͒ 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 Q͑x͒ is a product of distinct linear factors. This means that we can write
Q͑x͒ ͑a 1 x ϩ b1 ͒͑a 2 x ϩ b 2 ͒ и и и ͑a k x ϩ 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
2 R͑x͒
A1
A2
Ak
ϩ
ϩ иии ϩ
Q͑x͒
a 1 x ϩ b1
a2 x ϩ b2
a k x ϩ bk These constants can be determined as in the following example.
EXAMPLE 2 Evaluate x 2 ϩ 2x Ϫ 1
dx.
3
ϩ 3x 2 Ϫ 2x y 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 ϩ 3x 2 Ϫ 2x x͑2x 2 ϩ 3x Ϫ 2͒ x͑2x Ϫ 1͒͑x ϩ 2͒
Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
3  Another method for ﬁnding A, B, and C
is given in the note after this example. x 2 ϩ 2x Ϫ 1
A
B
C
ϩ
ϩ
x͑2x Ϫ 1͒͑x ϩ 2͒
x
2x Ϫ 1
xϩ2 To determine the values of A, B, and C, we multiply both sides of this equation by the
product of the denominators, x͑2x Ϫ 1͒͑x ϩ 2͒, obtaining
4 x 2 ϩ 2x Ϫ 1 A͑2x Ϫ 1͒͑x ϩ 2͒ ϩ Bx͑x ϩ 2͒ ϩ Cx͑2x Ϫ 1͒ Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get
5 x 2 ϩ 2x Ϫ 1 ͑2A ϩ B ϩ 2C͒x 2 ϩ ͑3A ϩ 2B Ϫ C͒x Ϫ 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, 2A ϩ B ϩ 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 ϩ B ϩ 2C 1 _3 3 3A ϩ 2B Ϫ C 2
Ϫ2A ϩ 2B Ϫ 2C Ϫ1 _2 FIGURE 1 1
Solving, we get A 1 , B 1 , and C Ϫ 10 , and so
2
5 y x 2 ϩ 2x Ϫ 1
dx
2x 3 ϩ 3x 2 Ϫ 2x y ͩ 1 1
1
1
1
1
ϩ
Ϫ
2 x
5 2x Ϫ 1
10 x ϩ 2 Խ Խ Խ Խ Խ ͪ dx Խ 1
1
1 ln x ϩ 10 ln 2x Ϫ 1 Ϫ 10 ln x ϩ 2 ϩ K
2  We could check our work by taking the terms
to a common denominator and adding them. In integrating the middle term we have made the mental substitution u 2x Ϫ 1, which
gives du 2 dx and dx du͞2.
NOTE 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 Ϫ2A Ϫ1, or A 1 . Likewise,
2
1
x 1 gives 5B͞4 1 and x Ϫ2 gives 10C Ϫ1, so B 1 and C Ϫ 10 . (You may object
2
4
5
that Equation 3 is not valid for x 0, 1 , or Ϫ2, 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 Ϫ2. See Exercise 67
2
for the reason.)
■ EXAMPLE 3 Find yx 2 dx
, where a
Ϫ a2 0. SOLUTION The method of partial fractions gives 1
1
A
B
ϩ
x2 Ϫ a2
͑x Ϫ a͒͑x ϩ a͒
xϪa
xϩa
and therefore
A͑x ϩ a͒ ϩ B͑x Ϫ a͒ 1
Using the method of the preceding note, we put x a in this equation and get
A͑2a͒ 1, so A 1͑͞2a͒. If we put x Ϫa, we get B͑Ϫ2a͒ 1, so B Ϫ1͑͞2a͒.
Thus y dx
1
x2 Ϫ a2
2a
y ͩ 1
1
Ϫ
xϪa
xϩa ͪ dx 1
(ln x Ϫ a Ϫ ln x ϩ a
2a Խ Խ Խ Խ) ϩ C 5E08(pp 530539) 1/17/06 6:04 PM Page 535 SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 535 Since ln x Ϫ ln y ln͑x͞y͒, we can write the integral as y 6 Ϳ Ϳ dx
1
xϪa
ln
ϩC
2
x Ϫa
2a
xϩa
2 See Exercises 53–54 for ways of using Formula 6.
CASE II ■ Q͑x͒ is a product of linear factors, some of which are repeated. Suppose the ﬁrst linear factor ͑a 1 x ϩ b1 ͒ is repeated r times; that is, ͑a 1 x ϩ b1 ͒r occurs in
the factorization of Q͑x͒. Then instead of the single term A1͑͞a 1 x ϩ b1 ͒ in Equation 2, we
would use
A1
A2
Ar
ϩ
ϩ иии ϩ
a 1 x ϩ b1
͑a 1 x ϩ b1 ͒2
͑a 1 x ϩ b1 ͒r 7 By way of illustration, we could write
x3 Ϫ x ϩ 1
A
B
C
D
E
ϩ 2 ϩ
ϩ
ϩ
x 2͑x Ϫ 1͒3
x
x
xϪ1
͑x Ϫ 1͒2
͑x Ϫ 1͒3
but we prefer to work out in detail a simpler example. EXAMPLE 4 Find y x 4 Ϫ 2x 2 ϩ 4x ϩ 1
dx.
x3 Ϫ x2 Ϫ x ϩ 1 SOLUTION The ﬁrst step is to divide. The result of long division is x 4 Ϫ 2x 2 ϩ 4x ϩ 1
4x
xϩ1ϩ 3
x3 Ϫ x2 Ϫ x ϩ 1
x Ϫ x2 Ϫ x ϩ 1
The second step is to factor the denominator Q͑x͒ x 3 Ϫ x 2 Ϫ x ϩ 1. Since Q͑1͒ 0,
we know that x Ϫ 1 is a factor and we obtain
x 3 Ϫ x 2 Ϫ x ϩ 1 ͑x Ϫ 1͒͑x 2 Ϫ 1͒ ͑x Ϫ 1͒͑x Ϫ 1͒͑x ϩ 1͒
͑x Ϫ 1͒2͑x ϩ 1͒
Since the linear factor x Ϫ 1 occurs twice, the partial fraction decomposition is
4x
A
B
C
ϩ
ϩ
͑x Ϫ 1͒2͑x ϩ 1͒
xϪ1
͑x Ϫ 1͒2
xϩ1
Multiplying by the least common denominator, ͑x Ϫ 1͒2͑x ϩ 1͒, we get
8 4x A͑x Ϫ 1͒͑x ϩ 1͒ ϩ B͑x ϩ 1͒ ϩ C͑x Ϫ 1͒2
͑A ϩ C͒x 2 ϩ ͑B Ϫ 2C͒x ϩ ͑ϪA ϩ B ϩ C ͒  Another method for ﬁnding the coefﬁcients:
Put x 1 in (8): B 2.
Put x Ϫ1: C Ϫ1.
Put x 0: A B ϩ C 1. Now we equate coefﬁcients:
AϩBϩ C0
A Ϫ B Ϫ 2C 4
ϪA ϩ B ϩ C 0 5E08(pp 530539) 536 ❙❙❙❙ 1/17/06 6:04 PM Page 536 CHAPTER 8 TECHNIQUES OF INTEGRATION Solving, we obtain A 1, B 2, and C Ϫ1, so y x 4 Ϫ 2x 2 ϩ 4x ϩ 1
dx
x3 Ϫ x2 Ϫ x ϩ 1 y ͫ xϩ1ϩ 1
2
1
ϩ
Ϫ
xϪ1
͑x Ϫ 1͒2
xϩ1 ͬ dx
CASE III ■ x2
2
ϩ x ϩ ln x Ϫ 1 Ϫ
Ϫ ln x ϩ 1 ϩ K
2
xϪ1
x2
2
xϪ1
ϩxϪ
ϩ ln
ϩK
2
xϪ1
xϩ1 Խ Խ Ϳ Ϳ Խ Խ Q͑x͒ contains irreducible quadratic factors, none of which is repeated. If Q͑x͒ has the factor ax 2 ϩ bx ϩ c, where b 2 Ϫ 4ac Ͻ 0, then, in addition to the partial
fractions in Equations 2 and 7, the expression for R͑x͒͞Q͑x͒ will have a term of the form
Ax ϩ B
ax ϩ bx ϩ c 9 2 where A and B are constants to be determined. For instance, the function given by
f ͑x͒ x͓͑͞x Ϫ 2͒͑x 2 ϩ 1͒͑x 2 ϩ 4͔͒ has a partial fraction decomposition of the form
x
A
Bx ϩ C
Dx ϩ E
ϩ 2
ϩ 2
͑x Ϫ 2͒͑x 2 ϩ 1͒͑x 2 ϩ 4͒
xϪ2
x ϩ1
x ϩ4
The term given in (9) can be integrated by completing the square and using the formula y 10 EXAMPLE 5 Evaluate y ͩͪ dx
1
x
tanϪ1
2
x ϩa
a
a
2 ϩC 2x 2 Ϫ x ϩ 4
dx.
x 3 ϩ 4x SOLUTION Since x 3 ϩ 4x x͑x 2 ϩ 4͒ can’t be factored further, we write 2x 2 Ϫ x ϩ 4
A
Bx ϩ C
ϩ 2
2
x͑x ϩ 4͒
x
x ϩ4
Multiplying by x͑x 2 ϩ 4͒, we have
2x 2 Ϫ x ϩ 4 A͑x 2 ϩ 4͒ ϩ ͑Bx ϩ C͒x
͑A ϩ B͒x 2 ϩ Cx ϩ 4A
Equating coefﬁcients, we obtain
AϩB2 C Ϫ1 Thus A 1, B 1, and C Ϫ1 and so y 2x 2 Ϫ x ϩ 4
dx
x 3 ϩ 4x y ͩ 4A 4 1
xϪ1
ϩ 2
x
x ϩ4 ͪ dx 5E08(pp 530539) 1/17/06 6:05 PM Page 537 SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ❙❙❙❙ 537 In order to integrate the second term we split it into two parts: y xϪ1
x
1
dx y 2
dx Ϫ y 2
dx
2
x ϩ4
x ϩ4
x ϩ4 We make the substitution u x 2 ϩ 4 in the ﬁrst of these integrals so that du 2x dx.
We evaluate the second integral by means of Formula 10 with a 2: y 2x 2 Ϫ x ϩ 4
1
x
1
dx y dx ϩ y 2
dx Ϫ y 2
dx
x͑x 2 ϩ 4͒
x
x ϩ4
x ϩ4 Խ Խ ln x ϩ 1 ln͑x 2 ϩ 4͒ Ϫ 1 tanϪ1͑x͞2͒ ϩ K
2
2
EXAMPLE 6 Evaluate 4x 2 Ϫ 3x ϩ 2
dx.
2
Ϫ 4x ϩ 3 y 4x SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we ﬁrst divide and obtain 4x 2 Ϫ 3x ϩ 2
xϪ1
1ϩ
4x 2 Ϫ 4x ϩ 3
4x 2 Ϫ 4x ϩ 3
Notice that the quadratic 4x 2 Ϫ 4x ϩ 3 is irreducible because its discriminant is
b 2 Ϫ 4ac Ϫ32 Ͻ 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 Ϫ 4x ϩ 3 ͑2x Ϫ 1͒2 ϩ 2
This suggests that we make the substitution u 2x Ϫ 1. Then, du 2 dx and
x ͑u ϩ 1͒͞2, so y 4x 2 Ϫ 3x ϩ 2
dx
4x 2 Ϫ 4x ϩ 3 y ͩ 1ϩ xϩ1y
2
xϩ1y
4 xϪ1
4x Ϫ 4x ϩ 3
2 1
2 ͪ dx ͑u ϩ 1͒ Ϫ 1
uϪ1
du x ϩ 1 y 2
du
4
u2 ϩ 2
u ϩ2 u
1
du Ϫ 1 y 2
du
4
u ϩ2
u ϩ2
2 x ϩ 1 ln͑u 2 ϩ 2͒ Ϫ
8 ϩC 1 2x Ϫ 1 x ϩ 1 ln͑4x 2 Ϫ 4x ϩ 3͒ Ϫ
8
NOTE ■ ͩ ͪ
ͩ ͪ 1
1
u
ؒ
tanϪ1
4 s2
s2
4 s2 tanϪ1 s2 ϩC Example 6 illustrates the general procedure for integrating a partial fraction of the form
Ax ϩ B
ax 2 ϩ bx ϩ c where b 2 Ϫ 4ac Ͻ 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 ϩ D
u
1
du C y 2
du ϩ D y 2
du
u2 ϩ a2
u ϩ a2
u ϩ a2 Then the ﬁrst integral is a logarithm and the second is expressed in terms of tanϪ1.
CASE IV ■ Q͑x͒ contains a repeated irreducible quadratic factor. If Q͑x͒ has the factor ͑ax 2 ϩ bx ϩ c͒r, where b 2 Ϫ 4ac Ͻ 0, then instead of the single
partial fraction (9), the sum
A1 x ϩ B1
A2 x ϩ B2
Ar x ϩ Br
ϩ
2
2
2 ϩ иии ϩ
ax ϩ bx ϩ c
͑ax ϩ bx ϩ c͒
͑ax 2 ϩ bx ϩ c͒r 11 occurs in the partial fraction decomposition of R͑x͒͞Q͑x͒. 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
convert͑f, parfrac, x͒
or the Mathematica command
Apart[f]
gives the following values:
A Ϫ1,
E 15
8 , B 1,
8 C D Ϫ1, FϪ , G H 3,
4 1
8 I Ϫ 1,
2 EXAMPLE 7 Write out the form of the partial fraction decomposition of the function x3 ϩ x2 ϩ 1
x͑x Ϫ 1͒͑x 2 ϩ x ϩ 1͒͑x 2 ϩ 1͒3
SOLUTION x3 ϩ x2 ϩ 1
x͑x Ϫ 1͒͑x 2 ϩ x ϩ 1͒͑x 2 ϩ 1͒3
J1
2 A
B
Cx ϩ D
Ex ϩ F
Gx ϩ H
Ix ϩ J
ϩ
ϩ 2
ϩ 2
ϩ 2
ϩ 2
x
xϪ1
x ϩxϩ1
x ϩ1
͑x ϩ 1͒2
͑x ϩ 1͒3 EXAMPLE 8 Evaluate y 1 Ϫ x ϩ 2x 2 Ϫ x 3
dx.
x͑x 2 ϩ 1͒2 SOLUTION The form of the partial fraction decomposition is 1 Ϫ x ϩ 2x 2 Ϫ x 3
A
Bx ϩ C
Dx ϩ E
ϩ 2
ϩ 2
x͑x 2 ϩ 1͒2
x
x ϩ1
͑x ϩ 1͒2
Multiplying by x͑x 2 ϩ 1͒2, we have
Ϫx 3 ϩ 2x 2 Ϫ x ϩ 1 A͑x 2 ϩ 1͒2 ϩ ͑Bx ϩ C͒x͑x 2 ϩ 1͒ ϩ ͑Dx ϩ E͒x
A͑x 4 ϩ 2x 2 ϩ 1͒ ϩ B͑x 4 ϩ x 2 ͒ ϩ C͑x 3 ϩ x͒ ϩ Dx 2 ϩ Ex
͑A ϩ B͒x 4 ϩ Cx 3 ϩ ͑2A ϩ B ϩ D͒x 2 ϩ ͑C ϩ E͒x ϩ A
If we equate coefﬁcients, we get the system
AϩB0 C Ϫ1 2A ϩ B ϩ D 2 C ϩ E Ϫ1 which has the solution A 1, B Ϫ1, C Ϫ1, D 1, and E 0. Thus A1 5E08(pp 530539) 1/17/06 6:05 PM Page 539 SECTION 8.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS y 1 Ϫ x ϩ 2x 2 Ϫ x 3
dx
x͑x 2 ϩ 1͒2 y y ͩ 1
xϩ1
x
Ϫ 2
ϩ 2
x
x ϩ1
͑x ϩ 1͒2 ͪ 539 dx dx
x
dx
x dx
Ϫy 2
dx Ϫ y 2
ϩy 2
x
x ϩ1
x ϩ1
͑x ϩ 1͒2 Խ Խ ln x Ϫ 1 ln͑x 2 ϩ 1͒ Ϫ tanϪ1x Ϫ
2  In the second and fourth terms we made the
mental substitution u x 2 ϩ 1. ❙❙❙❙ 1
ϩK
2͑x ϩ 1͒
2 We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral y x2 ϩ 1
dx
x͑x 2 ϩ 3͒ could be evaluated by the method of Case III, it’s much easier to observe that if
u x͑x 2 ϩ 3͒ x 3 ϩ 3x, then du ͑3x 2 ϩ 3͒ dx and so y x2 ϩ 1
dx 1 ln x 3 ϩ 3x ϩ C
3
x͑x 2 ϩ 3͒ Խ Խ 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
st͑x͒, then the substitution u st͑x͒ may be effective. Other instances appear in the
exercises. EXAMPLE 9 Evaluate y sx ϩ 4
dx.
x SOLUTION Let u sx ϩ 4. Then u 2 x ϩ 4, so x u 2 Ϫ 4 and dx 2u du. Therefore y u
u2
sx ϩ 4
dx y 2
2u du 2 y 2
du
x
u Ϫ4
u Ϫ4
2 y ͩ 1ϩ 4
u Ϫ4
2 ͪ du We can evaluate this integral either by factoring u 2 Ϫ 4 as ͑u Ϫ 2͒͑u ϩ 2͒ and using
partial fractions or by using Formula 6 with a 2: y du
sx ϩ 4
dx 2 y du ϩ 8 y 2
x
u Ϫ4
2u ϩ 8 ؒ Ϳ Ϳ 1
uϪ2
ln
ϩC
2ؒ2
uϩ2 2sx ϩ 4 ϩ 2 ln Ϳ Ϳ sx ϩ 4 Ϫ 2
ϩC
sx ϩ 4 ϩ 2 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. xϪ1
2. (a) 3
x ϩ x2 33. xϪ1
(b) 3
x ϩx
2 2
x 2 ϩ 3x Ϫ 4 3. (a) (b) x
͑x Ϫ 1͒͑x 2 ϩ x ϩ 1͒ y yx 37. 1
(b) 3
x ϩ 2x 2 ϩ x yx 35. 2x
1. (a)
͑x ϩ 3͒͑3x ϩ 1͒ 31. y ͑x 1
dx
Ϫ1 4 ■ ■ ■ ■ ■ y 38. xϪ3
dx
ϩ 2x ϩ 4͒2 yx 36. dx
Ϫ x2
2 x
dx
x 2 ϩ 4x ϩ 13 y 34. 3 2 1 32. x 2 ϩ 2x
dx
x ϩ 3x 2 ϩ 4 5 ■ 3 y x ͑x 0 1 3 x3
dx
ϩ1 2x 3 ϩ 5x
dx
x ϩ 5x 2 ϩ 4
4 0 x4 ϩ 1
dx
2
ϩ 1͒2 ■ ■ 4. (a) x
x 2 ϩ 4x ϩ 3 (b) 2x ϩ 1
͑x ϩ 1͒ 3͑x 2 ϩ 4͒ 2 5. (a) x4
x Ϫ1 (b) t4 ϩ t2 ϩ 1
͑t ϩ 1͒͑t 2 ϩ 4͒2 39. y x sx ϩ 1 dx 6. (a) x4
͑x ϩ x͒͑x 2 Ϫ x ϩ 3͒ (b) 1
x6 Ϫ x3 41. y 43. y sx 45. 1
y sx Ϫ sx dx [Hint: Substitute u ■ ■ ■ ■ 39–48 3 ■ 4 3 ■ 7–38 ■ ■ ■ x y x Ϫ 6 dx 9. y ͑x ϩ 5͒͑x Ϫ 2͒ dx
3 2 ■ ■ 8. xϪ9 y 2 ■ ■ ■ Evaluate the integral.  7. 11. ■  Make a substitution to express the integrand as a rational
function and then evaluate the integral. 1
dx
x Ϫ1
2 r2 y r ϩ 4 dr
1 1 16 9 13.
15. y
y 0 17. y 19.
21. y 23.
25.
27.
29. 1 2 2x ϩ 3
dx
͑x ϩ 1͒2
4y Ϫ 7y Ϫ 12
dy
y͑ y ϩ 2͒͑ y Ϫ 3͒ 3 3 xϪ1
dx
x ϩ 3x ϩ 2 47. ye 0 2 14.
16. 1
dx
͑x ϩ a͒͑x ϩ b͒ y
y 1 0 x 3 Ϫ 4x Ϫ 10
dx
x2 Ϫ x Ϫ 6
x 2 ϩ 2x Ϫ 1
dx
x3 Ϫ x y ͑x Ϫ 3͒͑x ϩ 2͒ 5x 2 ϩ 3x Ϫ 2
dx
x 3 ϩ 2x 2 22. y s ͑s Ϫ 1͒ dx y 10
dx
͑x Ϫ 1͒͑x 2 ϩ 9͒ y x 3 ϩ x 2 ϩ 2x ϩ 1
dx
͑x 2 ϩ 1͒͑x 2 ϩ 2͒ yx 2 x2 2 3 xϩ4
dx
ϩ 2x ϩ 5 24.
26.
28. 2 x 3 y ͑x ϩ 1͒ dx 3 2x e 2x
dx
ϩ 3e x ϩ 2
■ ■ ■ 48.
■ ■ sx
dx
x2 ϩ x
6
sx .] sx .] 12 cos x
dx
sin x ϩ sin x y 2 ■ ■ ■ ■  Use integration by parts, together with the techniques of
this section, to evaluate the integral. 49. y ln͑x
■ 2 Ϫ x ϩ 2͒ dx
■ ■ ■ 50.
■ ■ Ϫ1 y x tan
■ x dx ■ ■ ■ ■ 2
; 51. Use a graph of f ͑x͒ 1͑͞x Ϫ 2x Ϫ 3͒ to decide whether
2
x0 f ͑x͒ dx 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.
3
2
; 52. Graph both y 1͑͞x Ϫ 2x ͒ and an antiderivative on the
same screen. 53–54  Evaluate the integral by completing the square and using
Formula 6. y 53. y x 2 Ϫ 2x Ϫ 1
dx
͑x Ϫ 1͒2͑x 2 ϩ 1͒ y x Ϫ 2x ϩ x ϩ 1
dx
x 4 ϩ 5x 2 ϩ 4 55. The German mathematician Karl Weierstrass (1815–1897) 2 ■ 49–50 x2 Ϫ x ϩ 6
dx
x 3 ϩ 3x 3 30. dx 4 ■ ■ ds 2 2
3 3 y 1 1
dx
3
1 ϩ sx 1͞3 12. 20. x y 1
y sx ϩ sx dx [Hint: Substitute u y ͑x ϩ 5͒ ͑x Ϫ 1͒ dx y ͑x ϩ 1͒ x3
dx
2 ϩ 1 46. y 2 44. 1 0 y ͑t ϩ 4͒͑t Ϫ 1͒ dt 2 1 y 10. 18. 1 42. sx
dx
xϪ4 y x Ϫ sx ϩ 2 dx ■ ■ ax
dx
x 2 Ϫ bx 1 40. ■ yx
■ 2 dx
Ϫ 2x
■ 54.
■ ■ ■ ■ y 4x
■ 2 2x ϩ 1
dx
ϩ 12x Ϫ 7
■ ■ ■ noticed that the substitution t tan͑x͞2͒ will convert any ■ 5E08(pp 540549) 1/17/06 5:31 PM Page 541 SECTION 8.5 STRATEGY FOR INTEGRATION ͩͪ
x
2 1
s1 ϩ t 2 ͩͪ and sin x
2 ty (b) Show that
cos x 1 Ϫ t2
1 ϩ t2 2t
1 ϩ t2 sin x and 64. Factor x 4 ϩ 1 as a difference of squares by ﬁrst adding and subtracting the same quantity. Use this factorization to evaluate
x 1͑͞x 4 ϩ 1͒ dx. (c) Show that
dx
56–59 2
dt
1 ϩ t2 CAS 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.
dx y 3 Ϫ 5 sin x 58. y ͞2
͞3 ■ ■  ■ ■ ■ f ͑x͒ 1 57. 1
dx
1 ϩ sin x Ϫ cos x ■ 60–61 y 3 sin x Ϫ 4 cos x dx 59. y 2 sin x ϩ sin 2x ■ 1 ■ ■ ■ dx
■ ■ 1
,
x Ϫ 6x ϩ 8 61. y xϩ1
,
xϪ1 ■ ■ 66. (a) Find the partial fraction decomposition of the function 12x 5 Ϫ 7x 3 Ϫ 13x 2 ϩ 8
100x Ϫ 80x ϩ 116x 4 Ϫ 80x 3 ϩ 41x 2 Ϫ 20x ϩ 4
6 5 (b) Use part (a) to ﬁnd x f ͑x͒ dx 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 ͑x͒ dx. a 5, b 10 2 CAS f ͑x͒ Find the area of the region under the given curve from 60. y 4x 3 Ϫ 27x 2 ϩ 5x Ϫ 32
30x Ϫ 13x 4 ϩ 50x 3 Ϫ 286x 2 Ϫ 299x Ϫ 70
5 (b) Use part (a) to ﬁnd x f ͑x͒ dx (by hand) and compare with
the result of using the CAS to integrate f directly.
Comment on any discrepancy. a to b. ■ 65. (a) Use a computer algebra system to ﬁnd the partial fraction  56. PϩS
dP
P͓͑r Ϫ 1͒P Ϫ S͔ 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 ϩ t 2 a 2, b 3 67. Suppose that F, G, and Q are polynomials and
■ ■ ■ ■ ■ ■ ■ ■ ■ F͑x͒
G͑x͒
Q͑x͒
Q͑x͒ 62. Find the volume of the resulting solid if the region under the curve y 1͑͞x 2 ϩ 3x ϩ 2͒ from x 0 to x 1 is rotated
about (a) the xaxis and (b) the yaxis.
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  8.5 541 female population is related to time t by rational function of sin x and cos x into an ordinary rational
function of t.
(a) If t tan͑x͞2͒, Ϫ Ͻ x Ͻ , sketch a right triangle or use
trigonometric identities to show that
cos ❙❙❙❙ for all x except when Q͑x͒ 0. Prove that F͑x͒ G͑x͒ for
all x. [Hint: Use continuity.]
68. If f is a quadratic function such that f ͑0͒ 1 and y f ͑x͒
dx
x ͑x ϩ 1͒3
2 is a rational function, ﬁnd the value of f Ј͑0͒. 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. x nϩ1
nϩ1 1. yx n dx 3. ye x dx e x 5. ͑n Ϫ1͒ 1
dx ln x
x Խ Խ 2. y 4. ya y sin x dx Ϫcos x 6. y cos x dx sin x 7. y sec x dx tan x 8. y csc x dx Ϫcot x 9. y sec x tan x dx sec x 10. y csc x cot x dx Ϫcsc x 11. y sec x dx ln Խ sec x ϩ tan x Խ 12. y csc x dx ln Խ csc x Ϫ cot x Խ 13. y tan x dx ln Խ sec x Խ 14. y cot x dx ln Խ sin x Խ 15. y sinh x dx cosh x 16. y cosh x dx sinh x 17. y dx
1
x
tanϪ1
2
x ϩa
a
a 18. y dx
x
sinϪ1
2
a
sa Ϫ x *19. y dx
1
xϪa
ln
x2 Ϫ a2
2a
xϩa *20. y dx
ln x ϩ sx 2 Ϯ a 2
sx 2 Ϯ a 2 2 ͩͪ 2 Ϳ Ϳ x dx ax
ln a 2 ͩͪ 2 Խ Խ 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 sx (1 ϩ sx ) dx y (sx ϩ x) dx
tan sin y sec d y cos cos d
2 2 y sin cos d 1 y sin 2 d
2 5E08(pp 540549) 1/17/06 5:32 PM Page 543 SECTION 8.5 STRATEGY FOR INTEGRATION ❙❙❙❙ 543 y ͑sin x ϩ cos x͒ dx y ͑sin x ϩ 2 sin x cos x ϩ cos x͒ dx
2 2 2 y ͑1 ϩ 2 sin x cos x͒ dx
2. Look for an Obvious Substitution Try to ﬁnd some function u t͑x͒ in the inte grand whose differential du tЈ͑x͒ dx also occurs, apart from a constant factor.
For instance, in the integral y x
dx
x Ϫ1
2 we notice that if u x 2 Ϫ 1, then du 2x dx. Therefore, we use the substitution u x 2 Ϫ 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 ͑x͒.
(a) Trigonometric functions. If f ͑x͒ 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 ͑x͒ 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 sϮx 2 Ϯ a 2 occurs, we use a trigonometric substitution according to the
table in Section 8.3.
n
n
(ii) If sax ϩ b occurs, we use the rationalizing substitution u sax ϩ b.
n
More generally, this sometimes works for st͑x͒.
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 tanϪ1x, sinϪ1x, 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: dx y 1 Ϫ cos x y
y 1
1 ϩ cos x
1 ϩ cos x
ؒ
dx y
dx
1 Ϫ cos x 1 ϩ cos x
1 Ϫ cos 2x
1 ϩ cos x
dx
sin 2x y ͩ csc 2x ϩ cos x
sin 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 dx is a challenging integral, but if we make use of the
identity tan 2x sec 2x Ϫ 1, we can write y tan x sec x dx y sec x dx Ϫ y sec x dx
2 3 and if x sec 3x dx 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 dx
3 In Step 1 we rewrite the integral: y tan 3x
dx y tan 3x sec 3x dx
cos 3x The integral is now of the form x tan m x sec n x dx with m odd, so we can use the advice in
Section 7.2.
Alternatively, if in Step 1 we had written y tan 3x
sin 3x 1
sin 3x
dx
3 dx y
3
3 dx y
cos x
cos x cos x
cos 6x then we could have continued as follows with the substitution u cos x : y sin 3x
1 Ϫ cos 2x
1 Ϫ u2
sin x dx y
͑Ϫdu͒
6 dx y
6
cos x
cos x
u6
y EXAMPLE 2 ye sx u2 Ϫ 1
du y ͑u Ϫ4 Ϫ u Ϫ6 ͒ du
u6 dx According to Step 3(d)(ii) we substitute u sx. Then x u 2, so dx 2u du and ye sx dx 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 ϩ 1
dx
3
Ϫ 3x 2 Ϫ 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.
EXAMPLE 3 yx EXAMPLE 4 y xsln x dx Here Step 2 is all that is needed. We substitute u ln x because its differential is
du dx͞x, which occurs in the integral. ͱ 1Ϫx
dx
1ϩx
Although the rationalizing substitution
EXAMPLE 5 y u ͱ 1Ϫx
1ϩx 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 Ϫ x, we have y ͱ 1Ϫx
1Ϫx
dx y
dx
1ϩx
s1 Ϫ x 2
y 1
x
dx Ϫ y
dx
s1 Ϫ x 2
s1 Ϫ x 2 sinϪ1x ϩ s1 Ϫ x 2 ϩ 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 dx ? 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
f ͑x͒ ͱ x2 Ϫ 1
ϩ ln͑cosh x͒ Ϫ xe sin 2x
x ϩ 2x Ϫ 1
3 is an elementary function.
If f is an elementary function, then f Ј is an elementary function but x f ͑x͒ dx need not
2
be an elementary function. Consider f ͑x͒ e x . Since f is continuous, its integral exists,
and if we deﬁne the function F by
x 2 F͑x͒ y e t dt
0 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
FЈ͑x͒ e x 2 2 Thus, f ͑x͒ 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 dx in terms of the functions we know. (In Chapter 11, however, we will see how
2
to express x e x dx as an inﬁnite series.) The same can be said of the following integrals: y ex
dx
x y sx 3 y sin͑x ϩ 1 dx y 2 ͒ dx 1
dx
ln x y cos͑e
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. sin x ϩ sec x
dx
tan x y 2t
dt
͑t Ϫ 3͒2 2 0 y e arctan y
dy
1 ϩ y2 1 Ϫ1
3 4 7. y 9. yx 11. 1 xϪ1
dx
2
Ϫ 4x ϩ 5 y sin
3 cos 5 d dx
͑1 Ϫ x 2 ͒ 3͞2 13. y 15. y 17. y x sin x dx 19. ye 21. yt e 23. y (1 ϩ sx ) dx 1͞2 26. yx 4. y cot x ln͑sin x͒ dx 28. y sin sat dt 29. y 30. y Խx 6. xϪ1
dx
x 2 Ϫ 4x Ϫ 5 31. y 32. y x
dx
ϩ x2 ϩ 1 33. y s3 Ϫ 2x Ϫ x 34. y 35. y 36. y sin 4x cos 3x dx 37. y 38. y 39. y 40. y s4y 41. y tan d 42. yx 43. y e s1 ϩ e 44. y s1 ϩ e 45. yx e 46. y 1Ϫe 47. yx 48. y y tan d
x
dx
s3 Ϫ x 4 y y x csc x
y 4 0 yx 12. y sin x cos͑cos x͒ dx 4 s1 ϩ ln x
dx
x ln x dt
8 14. y 16. y y 1ϩe
ye y x sin 24. y ln͑x x2
dx
s1 Ϫ x 2 s2͞2 0 e 2t 3 sx 4t dt dx
Ϫ1 2 x dx Ϫ 1͒ dx 3w Ϫ 1
dw
wϩ2 5 0 cot x dx 10. 22. dx 2 3 20. xϩe x 3 Ϫ2t 1 3x 2 Ϫ 2
dx
Ϫ 2x Ϫ 8 18. x
dx
s1 Ϫ x 2
2 0 yx 27. 2. 8. r ln r dr 0 25. Evaluate the integral.  y Exercises ͱ
1 1ϩx
dx
1Ϫx
2 dx x 8 sin x dx Ϫ1 ͞4 0 cos2 tan2 d x
dx
1 Ϫ x 2 ϩ s1 Ϫ x 2
2 x x dx 5 Ϫx 3 dx xϩa
dx
2
ϩ a2 3 2 3x 2 Ϫ 2
dx
Ϫ 2x Ϫ 8 2 Ϫ2 Խ Ϫ 4x dx s2x Ϫ 1
dx
2x ϩ 3
͞2
͞4 ͞4 0 2 1 ϩ 4 cot x
dx
4 Ϫ cot x tan 5 sec 3 d 2 1
dy
Ϫ 4y Ϫ 3 tanϪ1x dx 1 ϩ ex
x x dx dx x
dx
x4 Ϫ a4 5E08(pp 540549) 1/17/06 5:35 PM Page 547 ❙❙❙❙ SECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS 49. 1 y x s4x ϩ 1 dx
1 51. y x s4x 53. yx 55. 50. yx 1
dx
s4x ϩ 1 y 69. y 1ϩe 71. yx x ln x
dx
2
Ϫ1 73. y ͑x Ϫ 2͒͑x ln͑1 ϩ x͒ dx 75. y sin x sin 2x sin 3x dx 77. y 1ϩx 79. y x sin dx
ϩ 1͒ 52. y x ͑x 54. y ͑x ϩ sin x͒ dx y x ϩ 4 ϩ 4 sx ϩ 1 dx 56. y sx 57. y x sx ϩ c dx 58. yx 59. ye 3x 1
dx
Ϫ ex 60. y x ϩ sx dx 61. yx 10 x4
dx
ϩ 16 62. y ͑x ϩ 1͒ 63. y sxe 64. y 65. 2 2 ϩ1 dx sinh mx dx
1 3 sx dx
1 y sx ϩ 1 ϩ sx dx  8.6 66. arctanst
dt
st 3 67. 2 4 2 2 1 3 x3 10 ͞3
͞4 y 3 2 dx ln͑tan x͒
dx
sin x cos x u ϩ1
du
u3 Ϫ u2
3 ■ 1 e 2x 4 x sx ■ 2 3 ϩ 4͒ 74. x cos x dx
■ ■ ■ ■ y ͑x y sin x ϩ sec x dx 80. dx ye 76. dx y 1 ϩ st dt 78. 1 y 72. x
dx
ϩ 4x 2 ϩ 3 y 1 ϩ 2e 70. dx 2 1 68. 547 y sin Ϫ eϪx dx ln͑x ϩ 1͒
dx
x2
st 3 x dx
Ϫ e Ϫx 2 Ϫ bx͒ sin 2x dx sec x cos 2x sin x cos x
dx
4
x ϩ cos 4 x ■ 2 x ■ ■ ■ ■ ■ 2 81. The functions y e x and y x 2e x don’t have elementary antiderivatives, but y ͑2x ϩ 1͒e
2
x ͑2x 2 ϩ 1͒e x dx.
2 x2 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 0, and x 1 is rotated
about the yaxis. Find the volume of the resulting solid.
SOLUTION Using the method of cylindrical shells, we see that the volume is
1 V y 2 x arctan x dx
0 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: y u tan Ϫ1 u du u2 ϩ 1
u
tanϪ1u Ϫ ϩ C
2
2 Thus, the volume is
1 V 2 y x tanϪ1x dx 2
0 ͫ x2 ϩ 1
x
tanϪ1x Ϫ
2
2 ͬ 1 0 ͓͑x 2 ϩ 1͒ tanϪ1x Ϫ x͔1 ͑2 tanϪ1 1 Ϫ 1͒
0
͓2͑͞4͒ Ϫ 1͔ 1 2 Ϫ
2
EXAMPLE 2 Use the Table of Integrals to ﬁnd y x2
dx.
s5 Ϫ 4x 2 SOLUTION If we look at the section of the table entitled Forms involving sa 2 Ϫ u 2, we see that the closest entry is number 34: ͩͪ u2
u
a2
u
du Ϫ sa 2 Ϫ u 2 ϩ
sinϪ1
2
2
a
sa 2 Ϫ u 2 y ϩC This is not exactly what we have, but we will be able to use it if we ﬁrst make the substitution u 2x : y x2
͑u͞2͒2 du
1
dx y
8
s5 Ϫ 4x 2
s5 Ϫ u 2 2 y u2
du
s5 Ϫ u 2 Then we use Formula 34 with a 2 5 (so a s5 ):
x2 y s5 Ϫ 4x 2 dx 1
8 Ϫ u2 y s5 Ϫ u 2 du 1
8 ͩ Ϫ u
5
u
s5 Ϫ u 2 ϩ sinϪ1
2
2
s5 ͩ ͪ x
5
2x
sinϪ1
s5 Ϫ 4x 2 ϩ
8
16
s5 EXAMPLE 3 Use the Table of Integrals to ﬁnd yx 3 ͪ ϩC ϩC sin x dx. 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 Ϫ n y u nϪ1 sin u du 3 sin x dx Ϫx 3 cos x ϩ 3 y x 2 cos x dx 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 dx x 2 sin x Ϫ 2 y x sin x dx
x 2 sin x Ϫ 2͑sin x Ϫ x cos x͒ ϩ 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 3 sin x dx Ϫx 3 cos x ϩ 3x 2 sin x ϩ 6x cos x Ϫ 6 sin x ϩ C where C 3K .
EXAMPLE 4 Use the Table of Integrals to ﬁnd y xsx 2 ϩ 2x ϩ 4 dx. SOLUTION Since the table gives forms involving sa 2 ϩ x 2, sa 2 Ϫ x 2, and sx 2 Ϫ a 2, but
not sax 2 ϩ bx ϩ c , we ﬁrst complete the square: x 2 ϩ 2x ϩ 4 ͑x ϩ 1͒2 ϩ 3
If we make the substitution u x ϩ 1 (so x u Ϫ 1), the integrand will involve the
pattern sa 2 ϩ u 2 : y xsx 2 ϩ 2x ϩ 4 dx y ͑u Ϫ 1͒ su 2 ϩ 3 du
y usu 2 ϩ 3 du Ϫ y su 2 ϩ 3 du The ﬁrst integral is evaluated using the substitution t u 2 ϩ 3: y usu
21. y sa 2 ϩ u 2 du ϩ u
sa 2 ϩ u 2
2 2 ϩ 3 du 1 y st dt 1 ؒ 2 t 3͞2 1 ͑u 2 ϩ 3͒3͞2
2
2
3
3 For the second integral we use Formula 21 with a s3 : a2
ln (u ϩ sa 2 ϩ u 2 ) ϩ C
2 y su 2 ϩ 3 du u
3
su 2 ϩ 3 ϩ 2 ln(u ϩ su 2 ϩ 3 )
2 Thus y xsx 2 ϩ 2x ϩ 4 dx 1͑x 2 ϩ 2x ϩ 4͒3͞2 Ϫ
3 xϩ1
3
sx 2 ϩ 2x ϩ 4 Ϫ 2 ln( x ϩ 1 ϩ sx 2 ϩ 2x ϩ 4 ) ϩ C
2 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 1͑͞3x Ϫ 2͒. Using the substitution u 3x Ϫ 2, an easy calculation
by hand gives
1
y 3x Ϫ 2 dx 13 ln 3x Ϫ 2 ϩ C Խ Խ 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 ln͑3x Ϫ 2͒ 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.
EXAMPLE 5 Use a computer algebra system to ﬁnd y xsx 2 ϩ 2x ϩ 4 dx. SOLUTION Maple responds with the answer
1
3 3
s3
arcsinh
͑1 ϩ x͒
2
3 ͑x 2 ϩ 2x ϩ 4͒3͞2 Ϫ 1 ͑2x ϩ 2͒sx 2 ϩ 2x ϩ 4 Ϫ
4 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
arcsinh x ln( x ϩ sx 2 ϩ 1 )  This is Equation 7.6.3. Thus
arcsinh ͫ ͬ s3
s3
͑1 ϩ x͒ ln
͑1 ϩ x͒ ϩ s 1 ͑1 ϩ x͒2 ϩ 1
3
3
3
 ln 1
1 ϩ x ϩ s͑1 ϩ x͒2 ϩ 3
s3 ln 1
ϩ ln( x ϩ 1 ϩ sx 2 ϩ 2x ϩ 4 )
s3 [ The resulting extra term Ϫ 3 ln(1͞s3 ) can be absorbed into the constant of integration.
2
Mathematica gives the answer ͩ 5
x
x2
ϩ ϩ
6
6
3 ͪ sx 2 ϩ 2x ϩ 4 Ϫ ͩ ͪ 3
1ϩx
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 3
sx 2 ϩ 2x ϩ 4 ͑2x 2 ϩ x ϩ 5͒ Ϫ 2 ln(sx 2 ϩ 2x ϩ 4 ϩ x ϩ 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 y x͑x 2 ϩ 5͒8 dx. SOLUTION Maple and Mathematica give the same answer:
1
18 x 18 ϩ 5 x 16 ϩ 50x 14 ϩ 1750 x 12 ϩ 4375x 10 ϩ 21875x 8 ϩ 218750 x 6 ϩ 156250x 4 ϩ 390625 x 2
2
3
3
2 5E08(pp 550559) 1/17/06 6:00 PM Page 551 SECTION 8.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS ❙❙❙❙ 551 It’s clear that both systems must have expanded ͑x 2 ϩ 5͒8 by the Binomial Theorem and
then integrated each term.
If we integrate by hand instead, using the substitution u x 2 ϩ 5, we get y x͑x  Derive and the TI89/92 also give this answer. 2 1
ϩ 5͒8 dx 18 ͑x 2 ϩ 5͒9 ϩ 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 dx. SOLUTION In Example 2 in Section 8.2 we found that 1 y sin x cos x dx Ϫ
5 1
3 2 cos 3x ϩ 2 cos 5x Ϫ 1 cos7x ϩ C
5
7 Derive and Maple report the answer
4
8
Ϫ 1 sin 4x cos 3x Ϫ 35 sin 2x cos 3x Ϫ 105 cos 3x
7 whereas Mathematica produces 10 5
1
3
1
Ϫ 64 cos x Ϫ 192 cos 3x ϩ 320 cos 5x Ϫ 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 5 0 EXAMPLE 8 If f ͑x͒ x ϩ 60 sin 4x cos 5x, ﬁnd the antiderivative F of f such that F͑0͒ 0. FIGURE 1 Graph F for 0 ഛ x ഛ 5. Where does F have extreme values and inﬂection points?
SOLUTION The antiderivative of f produced by Maple is 10
fª F 1
20
20
4
16
32
F͑x͒ 2 x 2 Ϫ 3 sin 3x cos 6x Ϫ 7 sin x cos 6x ϩ 7 cos 4x sin x ϩ 21 cos 2x sin x ϩ 21 sin x f 0 and we note that F͑0͒ 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 FЈ f in Figure 2 and observe that F has a local maximum
when x Ϸ 2.3 and a local minimum when x Ϸ 2.5. The graph of FЉ f Ј in Figure 2
shows that F has inﬂection points when x Ϸ 0.7, 1.3, 1.8, 2.4, 3.3, and 3.9. 5 _7 FIGURE 2  8.6 Exercises 1–4  Use the indicated entry in the Table of Integrals on the
Reference Pages to evaluate the integral. 1. y s7 Ϫ 2x 2
dx ; entry 33
x2 2. y 3x
dx ; entry 55
s3 Ϫ 2x 3. entry 71 4.
■ y sec ͑ x͒ dx ;
y e sin 3 d ; entry 98 3 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 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.
7.
9.
11. y 1 2x cosϪ1x dx 0 Ϫ3x ye 8. cos 4x dx dx
x 2 s4x 2 ϩ 9 y
y 6. 0 Ϫ1 10. t 2eϪt dt 12. tan ͑1͞z͒
dz
z2 y 1
dx
x 2 s4x 2 Ϫ 7 3 2 y 15. ye 17. y y s6 ϩ 4y Ϫ 4y 19. y sin x cos x ln͑sin x͒ dx 21.
23. x y 0 y ex
dx
3 Ϫ e 2x
5 y sec x dx y x sin͑x
yx y e ͑1 ϩ 2e 24. 2 2 CAS ͒ cos͑3x 2 ͒ dx ■ ■ ■ ■ ■ ■ 40. y sin 4x dx y tan5x dx 42. y x 5sx y 2 0 0 ■ ■ ■ 2 ■ ■ ■ 2 ϩ 1 dx ■ ■ ■ ■ x Ϫ 1 dx 2x Ϫ 1 dx 6 4 instead. Why do you think it was successful with this form of
the integrand? 2x dx CAS 44. Try to evaluate x 4eϪx dx
t y ͑1 ϩ ln x͒ s1 ϩ ͑x ln x͒ 2 2 ■ ■ dx 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. sin͑␣ t Ϫ 3͒ dt CAS
■ ■ 31. Find the volume of the solid obtained when the region under the curve y x s4 Ϫ x 2, 0 ഛ x ഛ 2, is rotated about the
yaxis.
32. The region under the curve y tan 2x from 0 to ͞4 is rotated about the xaxis. Find the volume of the resulting solid. 45–48  Use a CAS to ﬁnd an antiderivative F of f such
that F͑0͒ 0. Graph f and F and locate approximately the
xcoordinates of the extreme points and inﬂection points of F . 45. f ͑x͒ x2 Ϫ 1
x4 ϩ x2 ϩ 1 46. f ͑x͒ xeϪx sin x, 33. Verify Formula 53 in the Table of Integrals (a) by differentiation and (b) by using the substitution t a ϩ bu.
34. Verify Formula 31 (a) by differentiation and (b) by substituting Ϫ5 ഛ x ഛ 5
0ഛxഛ 47. f ͑x͒ sin x cos x,
4 u a sin . ■ y 2 x s2 sec 2 tan 2 ■ sec 4x dx x s4x Ϫ x dx
3 y sin
1 2 from human beings. Ask your CAS to evaluate y s9 Ϫ tan d x 4 dx
10
Ϫ2 y x s1 ϩ 2x dx ͒ 30. y sx y tan x If it doesn’t return an answer, ask it to try
x ye 29. 38. 43. Computer algebra systems sometimes need a helping hand dx x 28. 2x cos 2x dx 3 y 2 x s4 Ϫ 1 dx y se y x 2͑1 ϩ x 3 ͒4 dx x5
dx
ϩ s2 y 27. 36. ■ 26. y y sin x 39. sx dx s4 ϩ ͑ln x͒ 2
dx
x 25. ■ Ϫ1 y sin 22. 37. dx 2 x 2 cos 3x dx 20. 2 y x 2s5 Ϫ x 41. y 18. dy 35. s2y 2 Ϫ 3
dy
y2 16.
2  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. 3 14. sech͑e x ͒ dx 35–42 y csc ͑x͞2͒ dx 3 13. CAS 6 x Ϫx
x6 ϩ 1
3 48. f ͑x͒
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E08(pp 550559) 1/17/06 6:01 PM Page 553 DISCOVERY 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. 1 (i) y ͑x ϩ 2͒͑x ϩ 3͒ dx (iii) y ͑x ϩ 2͒͑x Ϫ 5͒ dx 1 (ii)
(iv) 1 y ͑x ϩ 1͒͑x ϩ 5͒ dx
y ͑x ϩ 2͒2 dx 1 (b) Based on the pattern of your responses in part (a), guess the value of the integral
1 y ͑x ϩ a͒͑x ϩ b͒ 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 2x dx (ii) y sin 3x cos 7x dx (iii) y sin 8x cos 3x 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 dx. 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
n x 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 ͑x͒ dx 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 s1 ϩ x 3 dx Ϫ1 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, b͔
into n subintervals of equal length ⌬x ͑b Ϫ a͒͞n, then we have y y b a n f ͑x͒ dx Ϸ ͚ f ͑x*͒ ⌬x
i i1 where x * is any point in the ith subinterval ͓x iϪ1, x i ͔. If x * is chosen to be the left endpoint
i
i
of the interval, then x * x iϪ1 and we have
i
0 x¸ ⁄ x™ x£ x¢ x y 1 f ͑x͒ dx Ϸ L n a (a) Left endpoint approximation ͚ f ͑x iϪ1 ͒ ⌬x i1 If f ͑x͒ ജ 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 ͑x͒ dx Ϸ Rn ͚ f ͑x ͒ ⌬x
i 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 iϪ1, 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 where
0 x
–¡ x™
– –
x£ –
x¢ (c) Midpoint approximation
FIGURE 1 x and f ͑x͒ dx Ϸ Mn ⌬x ͓ f ͑x1͒ ϩ f ͑x2 ͒ ϩ и и и ϩ f ͑xn ͔͒
⌬x bϪa
n xi 1 ͑x iϪ1 ϩ x i ͒ midpoint of ͓x iϪ1, x i ͔
2 5E08(pp 550559) 1/17/06 6:01 PM Page 555 SECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 555 Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2: y b a f ͑x͒ dx Ϸ 1
2 ͚ͫ
n ͬ n f ͑x iϪ1 ͒ ⌬x ϩ i1 ͚ f ͑x ͒ ⌬x
i i1 ⌬x
2 ͚ͫ ͬ n ͑ f ͑x iϪ1 ͒ ϩ f ͑x i ͒͒ i1 ⌬x
͓͑ f ͑x 0 ͒ ϩ f ͑x 1 ͒͒ ϩ ͑ f ͑x 1 ͒ ϩ f ͑x 2 ͒͒ ϩ и и и ϩ ͑ f ͑x nϪ1 ͒ ϩ f ͑x n ͔͒͒
2 ⌬x
͓ f ͑x 0 ͒ ϩ 2 f ͑x 1 ͒ ϩ 2 f ͑x 2 ͒ ϩ и и и ϩ 2 f ͑x nϪ1 ͒ ϩ f ͑x n ͔͒
2 y Trapezoidal Rule y b a f ͑x͒ dx Ϸ Tn ⌬x
͓ f ͑x0 ͒ ϩ 2 f ͑x1 ͒ ϩ 2 f ͑x2 ͒ ϩ и и и ϩ 2 f ͑xnϪ1 ͒ ϩ f ͑x n ͔͒
2 where ⌬x ͑b Ϫ a͒͞n and xi a ϩ i ⌬x.
0 x¸ ⁄ x™ x£ x¢ x The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates
the case f ͑x͒ ജ 0. The area of the trapezoid that lies above the ith subinterval is FIGURE 2 ⌬x Trapezoidal approximation y= ͩ f ͑x iϪ1 ͒ ϩ f ͑x i ͒
2 ͪ ⌬x
͓ f ͑x iϪ1 ͒ ϩ f ͑x i ͔͒
2 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 5 to
approximate the integral x12 ͑1͞x͒ dx.
SOLUTION (a) With n 5, a 1, and b 2, we have ⌬x ͑2 Ϫ 1͒͞5 0.2, and so the Trapezoidal Rule gives y
1 2 FIGURE 3 2 1 1
0.2
dx Ϸ T5
͓ f ͑1͒ ϩ 2 f ͑1.2͒ ϩ 2 f ͑1.4͒ ϩ 2 f ͑1.6͒ ϩ 2 f ͑1.8͒ ϩ f ͑2͔͒
x
2 ͩ 0.1 1
2
2
2
2
1
ϩ
ϩ
ϩ
ϩ
ϩ
1
1.2
1.4
1.6
1.8
2 ͪ Ϸ 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
2 1
y1 x dx Ϸ ⌬x ͓ f ͑1.1͒ ϩ f ͑1.3͒ ϩ f ͑1.5͒ ϩ f ͑1.7͒ ϩ f ͑1.9͔͒ 1
x
1 FIGURE 4 2 1
5 ͩ 1
1
1
1
1
ϩ
ϩ
ϩ
ϩ
1.1
1.3
1.5
1.7
1.9 Ϸ 0.691908
This approximation is illustrated in Figure 4. ͪ 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 ͑x͒ dx approximation ϩ error 1
2
dx ln x]1 ln 2 0.693147 . . .
x 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 Ϸ Ϫ0.002488 and EM Ϸ 0.001239 In general, we have
b ET y f ͑x͒ dx Ϫ Tn
a Module 5.1/5.2/8.7 allows you to compare approximation methods. Approximations to y 2 and b EM y f ͑x͒ dx Ϫ Mn
a 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 EL ER ET EM 5
10
20 Corresponding errors Ln 5
10
20 1 Ϫ0.052488
Ϫ0.025624
Ϫ0.012656 0.047512
0.024376
0.012344 Ϫ0.002488
Ϫ0.000624
Ϫ0.000156 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 B D
x i1 xi
– 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 ͑2n͒2 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 Љ͑x͒ measures how
much the graph is curved. [Recall that f Љ͑x͒ measures how fast the slope of y f ͑x͒
changes.] P A ❙❙❙❙ Խ xi Խ 3 Error Bounds Suppose f Љ͑x͒ ഛ K for a ഛ x ഛ b. If ET and EM are the errors
in the Trapezoidal and Midpoint Rules, then C
R Խ Խ ET ഛ P K͑b Ϫ a͒3
12n 2 Խ Խ EM ഛ and K͑b Ϫ a͒3
24n 2 B Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1. If
f ͑x͒ 1͞x, then f Ј͑x͒ Ϫ1͞x 2 and f Љ͑x͒ 2͞x 3. Since 1 ഛ x ഛ 2, we have 1͞x ഛ 1, so Q
A Խ f Љ͑x͒ Խ D FIGURE 5 2
2
ഛ 3 2
x3
1 Therefore, taking K 2, a 1, b 2, and n 5 in the error estimate (3), we see that  K can be any number larger than all the
values of f Љ͑x͒ , but smaller values of K
give better error bounds. Խ Ϳ Ϳ Խ ԽE Խ ഛ
T 2͑2 Ϫ 1͒3
1
Ϸ 0.006667
12͑5͒2
150 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 ͑1͞x͒ dx are accurate to within 0.0001? Խ Խ SOLUTION We saw in the preceding calculation that f Љ͑x͒ ഛ 2 for 1 ഛ x ഛ 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 2͑1͒3
Ͻ 0.0001
12n 2
Solving the inequality for n, we get
n2 Ͼ  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 nϾ 2
12͑0.0001͒
1
Ϸ 40.8
s0.0006 Thus, n 41 will ensure the desired accuracy. 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
2͑1͒3
Ͻ 0.0001
24n 2
nϾ which gives
y 1
Ϸ 29
s0.0012 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 1, and n 10, the Midpoint Rule gives y 1 0 2 e x dx Ϸ ⌬x ͓ f ͑0.05͒ ϩ f ͑0.15͒ ϩ и и и ϩ f ͑0.85͒ ϩ f ͑0.95͔͒
0.1͓e 0.0025 ϩ e 0.0225 ϩ e 0.0625 ϩ e 0.1225 ϩ e 0.2025 ϩ e 0.3025
ϩ e 0.4225 ϩ e 0.5625 ϩ e 0.7225 ϩ e 0.9025͔ 0 1 Ϸ 1.460393 x Figure 6 illustrates this approximation.
2
2
2
(b) Since f ͑x͒ e x , we have f Ј͑x͒ 2xe x and f Љ͑x͒ ͑2 ϩ 4x 2 ͒e x . Also, since
0 ഛ x ഛ 1, we have x 2 ഛ 1 and so FIGURE 6 2 0 ഛ f Љ͑x͒ ͑2 ϩ 4x 2 ͒e x ഛ 6e
Taking K 6e, a 0, b 1, and n 10 in the error estimate (3), we see that an upper
bound for the error is
6e͑1͒3
e
Ϸ 0.007
2
24͑10͒
400  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. 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, b͔ into n subintervals of
equal length h ⌬x ͑b Ϫ a͒͞n, but this time we assume that n is an even number. Then
on each consecutive pair of intervals we approximate the curve y f ͑x͒ ജ 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 , Piϩ1 , and Piϩ2 .
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 Ϫh, 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 ϩ Bx ϩ C and so the area under the parabola from x Ϫh to
x h is
 Here we have used Theorem 5.5.6. Notice
that Ax 2 ϩ C is even and Bx is odd. y h Ϫh h ͑Ax 2 ϩ Bx ϩ C͒ dx 2 y ͑Ax 2 ϩ C͒ dx
0 ͫ
ͩ ͬ
ͪ x3
2 A
ϩ Cx
3
2 A h 0 h3
h
ϩ Ch ͑2Ah 2 ϩ 6C ͒
3
3 But, since the parabola passes through P0͑Ϫh, y0 ͒, P1͑0, y1 ͒, and P2͑h, y2 ͒, we have
y0 A͑Ϫh͒2 ϩ B͑Ϫh͒ ϩ C Ah 2 Ϫ Bh ϩ C
y1 C
y2 Ah 2 ϩ Bh ϩ C
and therefore y0 ϩ 4y1 ϩ y2 2Ah 2 ϩ 6C Thus, we can rewrite the area under the parabola as
h
͑y0 ϩ 4y1 ϩ y2 ͒
3
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 ϩ 4y1 ϩ y2 ͒
3
Similarly, the area under the parabola through P2 , P3 , and P4 from x x 2 to x x 4 is
h
͑y2 ϩ 4y3 ϩ y4 ͒
3
If we compute the areas under all the parabolas in this manner and add the results, we get y b a f ͑x͒ dx Ϸ
h
h
h
͑y0 ϩ 4y1 ϩ y2 ͒ ϩ ͑y2 ϩ 4y3 ϩ y4 ͒ ϩ и и и ϩ ͑ynϪ2 ϩ 4ynϪ1 ϩ yn ͒
3
3
3
h
͑y0 ϩ 4y1 ϩ 2y2 ϩ 4y3 ϩ 2y4 ϩ и и и ϩ 2ynϪ2 ϩ 4ynϪ1 ϩ yn ͒
3 Although we have derived this approximation for the case in which f ͑x͒ ജ 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 ͑x͒ dx Ϸ Sn ⌬x
͓ f ͑x 0 ͒ ϩ 4 f ͑x 1 ͒ ϩ 2 f ͑x 2 ͒ ϩ 4 f ͑x 3 ͒ ϩ и и и
3
ϩ 2 f ͑xnϪ2 ͒ ϩ 4 f ͑xnϪ1 ͒ ϩ f ͑xn ͔͒ where n is even and ⌬x ͑b Ϫ a͒͞n.
EXAMPLE 4 Use Simpson’s Rule with n 10 to approximate x12 ͑1͞x͒ dx. SOLUTION Putting f ͑x͒ 1͞x, n 10, and ⌬x 0.1 in Simpson’s Rule, we obtain y 2 1 1
dx Ϸ S10
x
⌬x
͓ f ͑1͒ ϩ 4 f ͑1.1͒ ϩ 2 f ͑1.2͒ ϩ 4 f ͑1.3͒ ϩ и и и ϩ 2 f ͑1.8͒ ϩ 4 f ͑1.9͒ ϩ f ͑2͔͒
3 0.1
3 ͩ 1
4
2
4
2
4
2
4
2
4
1
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
ϩ
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2 ͪ Ϸ 0.693150
Notice that, in Example 4, Simpson’s Rule gives us a much better approximation
͑S10 Ϸ 0.693150͒ to the true value of the integral ͑ln 2 Ϸ 0.693147. . .͒ than does the
Trapezoidal Rule ͑T10 Ϸ 0.693771͒ or the Midpoint Rule ͑M10 Ϸ 0.692835͒. It turns out
(see Exercise 48) that the approximations in Simpson’s Rule are weighted averages of
those in the Trapezoidal and Midpoint Rules:
S2n 1 Tn ϩ 2 Mn
3
3 (Recall that ET and EM usually have opposite signs and Խ EM Խ is about half the size of Խ ET Խ.) 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 dx, the integral of y with respect to x.
EXAMPLE 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. D͑t͒ is the data
throughput, measured in megabits per second ͑Mb͞s͒. 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 SECTION 8.7 APPROXIMATE INTEGRATION ❙❙❙❙ 561 SOLUTION Because we want the units to be consistent and D͑t͒ is measured in megabits
per second, we convert the units for t from hours to seconds. If we let A͑t͒ be the
amount of data (in megabits) transmitted by time t, where t is measured in seconds, then
AЈ͑t͒ D͑t͒. So, by the Net Change Theorem (see Section 5.4), the total amount of data
transmitted by noon (when t 12 ϫ 60 2 43,200) is A͑43,200͒ y 43,200 0 D͑t͒ dt We estimate the values of D͑t͒ at hourly intervals from the graph and compile them in
the table.
t ͑hours͒ t ͑seconds͒ D͑t͒ t ͑hours͒ t ͑seconds͒ D͑t͒ 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 12 and ⌬t 3600 to estimate the integral: y 43,200 0 A͑t͒ dt Ϸ
Ϸ ⌬t
͓D͑0͒ ϩ 4D͑3600͒ ϩ 2D͑7200͒ ϩ и и и ϩ 4D͑39,600͒ ϩ D͑43,200͔͒
3
3600
͓3.2 ϩ 4͑2.7͒ ϩ 2͑1.9͒ ϩ 4͑1.7͒ ϩ 2͑1.3͒ ϩ 4͑1.0͒
3
ϩ 2͑1.1͒ ϩ 4͑1.3͒ ϩ 2͑2.8͒ ϩ 4͑5.7͒ ϩ 2͑7.1͒ ϩ 4͑7.7͒ ϩ 7.9͔ 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 . Խf ͑4͒ Խ ͑x͒ ഛ K for a ഛ x ഛ b.
If ES is the error involved in using Simpson’s Rule, then
4 Error Bound for Simpson’s Rule Suppose that ԽE Խ ഛ
S K͑b Ϫ a͒5
180n 4 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 ͑1͞x͒ dx is accurate to within 0.0001?
SOLUTION If f ͑x͒ 1͞x, then f ͑4͒͑x͒ 24͞x 5. Since x ജ 1, we have 1͞x ഛ 1 and so Խf
 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. Խ ͑4͒ ͑x͒ Ϳ Ϳ 24
ഛ 24
x5 Therefore, we can take K 24 in (4). Thus, for an error less than 0.0001 we should
choose n so that
24͑1͒5
Ͻ 0.0001
180n 4
n4 Ͼ This gives nϾ or 24
180͑0.0001͒
1
Ϸ 6.04
4
s0.00075 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.)
EXAMPLE 7 2 (a) Use Simpson’s Rule with n 10 to approximate the integral x01 e x dx.
(b) Estimate the error involved in this approximation.
SOLUTION (a) If n 10, then ⌬x 0.1 and Simpson’s Rule gives
 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 y 1 0 e x dx Ϸ
2 ⌬x
͓ f ͑0͒ ϩ 4 f ͑0.1͒ ϩ 2 f ͑0.2͒ ϩ и и и ϩ 2 f ͑0.8͒ ϩ 4 f ͑0.9͒ ϩ f ͑1͔͒
3
0.1 0
͓e ϩ 4e 0.01 ϩ 2e 0.04 ϩ 4e 0.09 ϩ 2e 0.16 ϩ 4e 0.25 ϩ 2e 0.36
3
ϩ 4e 0.49 ϩ 2e 0.64 ϩ 4e 0.81 ϩ e 1 ͔ Ϸ 1.462681
2 (b) The fourth derivative of f ͑x͒ e x is y=e ≈ f ͑4͒͑x͒ ͑12 ϩ 48x 2 ϩ 16x 4 ͒e x 2 and so, since 0 ഛ x ഛ 1, we have
0 ഛ f ͑4͒͑x͒ ഛ ͑12 ϩ 48 ϩ 16͒e 1 76e
0 FIGURE 10 1 x Therefore, putting K 76e, a 0, b 1, and n 10 in (4), we see that the error is at
most
76e͑1͒5
Ϸ 0.000115
180͑10͒4
(Compare this with Example 3.) Thus, correct to three decimal places, we have y 1 0 2 e x dx Ϸ 1.463 5E08(pp 560569) 1/17/06 5:57 PM Page 563 ❙❙❙❙ SECTION 8.7 APPROXIMATE INTEGRATION  8.7
1. Let I Exercises
x04 f ͑x͒ dx, where (Round your answers to six decimal places.) Compare your results
to the actual value to determine the error in each approximation. 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. y ■ ■ ■ ■ y 1 0 ■ eϪsx dx, n 6 ■ ■ ■ ■ ■ f y 9. y 11. y 13. y 15. y 17. y 1 2 3 4 x 2. The left, right, Trapezoidal, and Midpoint Rule approximations were used to estimate x02 f ͑x͒ dx, 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 ͑x͒ dx lie? ■ y 2 4
s1 ϩ x 2 dx, 0 ln x
dx,
1ϩx 2 1 1͞2 0
2 1
dy,
1 ϩ y5 3 0
■ ■ ■ ■ y 18.
■ y 16. n6 y 14. n8 y 12. n4 y 10. n 10 cos x
dx,
x 5 1 8. sin͑e t͞2 ͒ dt, n 8 e 1͞x dx, 1 n8 y ■ 1͞2 0
3 0
4 0
4 0 6 4
4 2
■ sin͑x 2 ͒ dx, n 4 dt
,
1 ϩ t2 ϩ t4 n6 s1 ϩ sx dx, n 8
sx sin x dx, n 8
ln͑x 3 ϩ 2͒ dx,
ex
dx,
x
■ n 10 n 10
■ ■ ■ 19. (a) Find the approximations T10 and M10 for the integral x02 eϪx 2 dx.
(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? 1 y=ƒ 2 1
20. (a) Find the approximations T8 and M8 for x0 cos͑x 2 ͒ dx. 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 cos͑x ͒ dx 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 dx 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? 2
; 4. Draw the graph of f ͑x͒ sin͑x ͞2͒ in the viewing rectangle ͓0, 1͔ by ͓0, 0.5͔ and let I x01 f ͑x͒ dx.
(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 ? 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. 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.) 1 0 x 2 sin x dx, n 8 ■ 7. 0 0 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 ͑4͒͑x͒ by hand. But computer algebra systems have no Խ Խ 5E08(pp 560569) 564 ❙❙❙❙ 1/17/06 5:57 PM Page 564 CHAPTER 8 TECHNIQUES OF INTEGRATION problem computing f ͑4͒ 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 ͑x͒ dx, where
f ͑x͒ e cos x.
(a) Use a graph to get a good upper bound for f Љ͑x͒ .
(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 ͑4͒͑x͒ .
(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? Խ Խ Խ CAS Use Simpson’s Rule to estimate the area of the pool. 6.2 Խ x  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? 0 ■ ■ y 26.
■ ■ ■ ■ 2 0 ■ e x dx ■ ■ ■ ■ ■  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? 27–28 27.
■ y 4 1
■ ■ y 28. sx dx
■ ■ ■ 2 Ϫ1 ■ ■ f ͑x͒ 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 (m͞s) t (s) v (m͞s) 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 a͑t͒ of a car measured in ft͞s2 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 dx
■ x (b) If it is known that Ϫ4 ഛ f Љ͑x͒ ഛ 1 for all x, estimate the
error involved in the approximation in part (a). 25–26 x 3 dx f ͑x͒ 0.0
0.4
0.8
1.2
1.6 1 1 4.8 value of the integral x03.2 f ͑x͒ dx. Ϫ1 y 5.6 5.0 4.8 31. (a) Use the Midpoint Rule and the given data to estimate the 24. Repeat Exercise 23 for the integral y s4 Ϫ x 3 dx. 25. 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 6 t y 34. Water leaked from a tank at a rate of r͑t͒ 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 4 x 30. The widths (in meters) of a kidneyshaped swimming pool were measured at 2meter intervals as indicated in the ﬁgure. 2 0 2 4 6 t 5E08(pp 560569) 1/17/06 5:57 PM Page 565 SECTION 8.7 APPROXIMATE INTEGRATION t P t P 0:00
0:30
1:00
1:30
2:00
2:30
3:00 1814
1735
1686
1646
1637
1609
1604 3:30
4:00
4:30
5:00
5:30
6:00 1611
1621
1666
1745
1886
2052 565 3
39. The region bounded by the curves y s1 ϩ x 3, y 0, x 0, 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.) ❙❙❙❙ 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 ͱ T4 L
t y ͞2 0 dx
s1 Ϫ k 2 sin 2x where k sin( 1 0 ) and t is the acceleration due to gravity.
2
If L 1 m and 0 42Њ, use Simpson’s Rule with n 10 to
ﬁnd the period. 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
I͑ ͒ N 2 sin 2k͞k 2, where k ͑ Nd sin ͒͞ and d is the
distance between adjacent slits. A heliumneon laser with
wavelength 632.8 ϫ 10Ϫ9 m is emitting a narrow band of
light, given by Ϫ10Ϫ6 Ͻ Ͻ 10Ϫ6, through a grating with
10,000 slits spaced 10Ϫ4 m apart. Use the Midpoint Rule with
10
n 10 to estimate the total light intensity xϪ10 I͑ ͒ d emerging from the grating. 0.4 Ϫ6 0 2 4 6 8 t (hours) Ϫ6 37. If the region shown in the ﬁgure is rotated about the yaxis to form a solid, use Simpson’s Rule with n 8 to estimate the
volume of the solid. x020 cos͑ x͒ dx. Compare your result to the actual value. Can
you explain the discrepancy? 43. Sketch the graph of a continuous function on ͓0, 2͔ for which the Trapezoidal Rule with n 2 is more accurate than the
Midpoint Rule. y
4 44. Sketch the graph of a continuous function on ͓0, 2͔ for which 2 0 42. Use the Trapezoidal Rule with n 10 to approximate the right endpoint approximation with n 2 is more accurate
than Simpson’s Rule. 2 4 6 10 x 8 45. If f is a positive function and f Љ͑x͒ Ͻ 0 for a ഛ x ഛ b, show that
b 38. The table shows values of a force function f ͑x͒ where x is measured in meters and f ͑x͒ in newtons. Use Simpson’s Rule
to estimate the work done by the force in moving an object a
distance of 18 m. Tn Ͻ y f ͑x͒ dx Ͻ Mn
a 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 ͑x͒ dx. x 0 3 6 9 12 15 18 47. Show that 2 ͑Tn ϩ Mn ͒ T2n. f ͑x͒ 9.8 9.1 8.5 8.0 7.7 7.5 7.4 48. Show that 3 Tn ϩ 3 Mn S2n. 1 1 2 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 ͑x͒ dx we dealt with a function f deﬁned on a ﬁnite interval ͓a, b͔ 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, b͔. 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 1͞x 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 A͑t͒ y t 1 1
1
2 dx Ϫ
x
x ͬ t 1Ϫ 1 1
t Notice that A͑t͒ Ͻ 1 no matter how large t is chosen.
y y= 1
≈
area=1  x=1
0 FIGURE 1 t 1 We also observe that ͩ ͪ lim A͑t͒ lim 1 Ϫ tlϱ 1
t tlϱ 1
t x 1 The area of the shaded region approaches 1 as t l ϱ (see Figure 2), so we say that the area
of the inﬁnite region S is equal to 1 and we write y ϱ 1 y y y
1
area= 2 0 1 2 x y area= 4
5 area= 2
3
0 t 1
1
dx lim y 2 dx 1
tlϱ 1 x
x2 1 3 x 0 1 area=1
5 x 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 ͑x͒ dx exists for every number t ജ a, then y ϱ a t f ͑x͒ dx lim y f ͑x͒ dx
tlϱ a provided this limit exists (as a ﬁnite number).
(b) If xtb f ͑x͒ dx exists for every number t ഛ b, then y b Ϫϱ f ͑x͒ dx lim t lϪϱ y t b f ͑x͒ dx provided this limit exists (as a ﬁnite number).
b
The improper integrals xaϱ f ͑x͒ dx and xϪϱ f ͑x͒ dx are called convergent if the
corresponding limit exists and divergent if the limit does not exist.
a
(c) If both xaϱ f ͑x͒ dx and xϪϱ f ͑x͒ dx are convergent, then we deﬁne y ϱ Ϫϱ f ͑x͒ dx y a Ϫϱ ϱ f ͑x͒ dx ϩ y f ͑x͒ dx
a 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 ͑x͒ ജ 0 and the integral xaϱ f ͑x͒ dx
is convergent, then we deﬁne the area of the region S ͕͑x, y͒ x ജ a, 0 ഛ y ഛ f ͑x͖͒ in
Figure 3 to be Խ ϱ A͑S͒ y f ͑x͒ dx
a y y=ƒ S FIGURE 3 0 a x This is appropriate because xaϱ f ͑x͒ dx is the limit as t l ϱ of the area under the graph of
f from a to t.
EXAMPLE 1 Determine whether the integral x1ϱ ͑1͞x͒ dx is convergent or divergent. SOLUTION According to part (a) of Deﬁnition 1, we have y ϱ 1 1
t 1
dx lim y
dx lim ln x
tlϱ 1 x
tlϱ
x Խ Խ] t 1 lim ͑ln t Ϫ ln 1͒ lim ln t ϱ
tlϱ tlϱ The limit does not exist as a ﬁnite number and so the improper integral x1ϱ ͑1͞x͒ dx 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
≈ y ϱ 1 ﬁnite area
0 x 1 FIGURE 4 y= 1
x y 1
dx diverges
x ϱ 1 Geometrically, this says that although the curves y 1͞x 2 and y 1͞x look very similar
for x Ͼ 0, the region under y 1͞x 2 to the right of x 1 (the shaded region in Figure 4)
has ﬁnite area whereas the corresponding region under y 1͞x (in Figure 5) has inﬁnite
area. Note that both 1͞x 2 and 1͞x approach 0 as x l ϱ but 1͞x 2 approaches 0 faster than
1͞x. The values of 1͞x don’t decrease fast enough for its integral to have a ﬁnite value.
EXAMPLE 2 Evaluate y 1
dx converges
x2 y 0 Ϫϱ xe x dx. SOLUTION Using part (b) of Deﬁnition 1, we have y 0 Ϫϱ inﬁnite area xe x dx lim t lϪϱ y t 0 xe x dx We integrate by parts with u x, dv e x dx so that du dx, v e x :
0 1 x y 0 t FIGURE 5 0 0 xe x dx xe x t Ϫ y e x dx
t Ϫte t Ϫ 1 ϩ e t
We know that e t l 0 as t l Ϫϱ, and by l’Hospital’s Rule we have
lim te t lim t lϪϱ t lϪϱ t
1
lim
t lϪϱ ϪeϪt
eϪt lim ͑Ϫe t ͒ 0
t lϪϱ Therefore y 0 Ϫϱ xe x dx lim ͑Ϫte t Ϫ 1 ϩ e t ͒
t lϪϱ Ϫ0 Ϫ 1 ϩ 0 Ϫ1
EXAMPLE 3 Evaluate y ϱ Ϫϱ 1
dx.
1 ϩ x2 SOLUTION It’s convenient to choose a 0 in Deﬁnition 1(c): y ϱ Ϫϱ 1
1
1
0
ϱ
2 dx y
2 dx ϩ y
2 dx
Ϫϱ 1 ϩ x
0 1 ϩ x
1ϩx We must now evaluate the integrals on the right side separately: y ϱ 0 1
t
dx
lim
lim Ϫ1
2 dx t l ϱ y
2 t l ϱ tan x
0 1 ϩ x
1ϩx t
0 lim ͑tan Ϫ1t Ϫ tanϪ1 0͒ lim tanϪ1t
tlϱ tlϱ
2 5E08(pp 560569) 1/17/06 5:58 PM Page 569 SECTION 8.8 IMPROPER INTEGRALS y 0 Ϫϱ 1
0
dx
dx lim y
lim tanϪ1x
t lϪϱ t 1 ϩ x 2
t lϪϱ
1 ϩ x2 ❙❙❙❙ 569 0
t lim ͑tan Ϫ1 0 Ϫ tan Ϫ1t͒
t lϪϱ ͩ ͪ 0Ϫ Ϫ
2
2 Since both of these integrals are convergent, the given integral is convergent and
1
y=
1+≈ y y area=π
0 FIGURE 6 ϱ Ϫϱ x 1
dx
ϩ
1 ϩ x2
2
2 Since 1͑͞1 ϩ x 2 ͒ Ͼ 0, the given improper integral can be interpreted as the area of
the inﬁnite region that lies under the curve y 1͑͞1 ϩ x 2 ͒ and above the xaxis (see
Figure 6).
EXAMPLE 4 For what values of p is the integral y ϱ 1 convergent? 1
dx
xp SOLUTION We know from Example 1 that if p 1, then the integral is divergent, so let’s
assume that p 1. Then
ϱ 1
t
y1 x p dx tlim y1 x Ϫp dx
lϱ lim tlϱ lim tlϱ xϪpϩ1
Ϫp ϩ 1 ͫ ͬ xt x1 ͬ 1
1
pϪ1 Ϫ 1
1Ϫp t If p Ͼ 1, then p Ϫ 1 Ͼ 0, so as t l ϱ, t pϪ1 l ϱ and 1͞t pϪ1 l 0. Therefore y ϱ 1 1
1
dx
xp
pϪ1 if p Ͼ 1 and so the integral converges. But if p Ͻ 1, then p Ϫ 1 Ͻ 0 and so
1
t 1Ϫp l ϱ
t pϪ1 as t l ϱ and the integral diverges.
We summarize the result of Example 4 for future reference: 2 y ϱ 1 1
dx is convergent if p Ͼ 1 and divergent if p ഛ 1.
xp Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function deﬁned on a ﬁnite interval ͓a, b͒ 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 x=b t A͑t͒ y f ͑x͒ dx
a 0 a x t b If it happens that A͑t͒ approaches a deﬁnite number A as t l bϪ, then we say that the
area of the region S is A and we write FIGURE 7 y b a t f ͑x͒ dx lim ya f ͑x͒ dx
Ϫ
tlb 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
 Parts (b) and (c) of Deﬁnition 3 are illustrated
in Figures 8 and 9 for the case where f ͑x͒ ജ 0
and f has vertical asymptotes at a and c,
respectively. Definition of an Improper Integral of Type 2 (a) If f is continuous on ͓a, b͒ and is discontinuous at b, then y b a t f ͑x͒ dx lim ya f ͑x͒ dx
Ϫ
tlb if this limit exists (as a ﬁnite number).
(b) If f is continuous on ͑a, b͔ and is discontinuous at a, then y y b a b f ͑x͒ dx lim y f ͑x͒ dx
ϩ
tla t if this limit exists (as a ﬁnite number).
0 a t b The improper integral xab f ͑x͒ dx is called convergent if the corresponding limit
exists and divergent if the limit does not exist. x (c) If f has a discontinuity at c, where a Ͻ c Ͻ b, and both xac f ͑x͒ dx and
xcb f ͑x͒ dx are convergent, then we deﬁne FIGURE 8
y y b a EXAMPLE 5 Find
0 a b x c y 5 2 b a c 1
dx.
sx Ϫ 2 SOLUTION We note ﬁrst that the given integral is improper because f ͑x͒ 1͞sx Ϫ 2
has the vertical asymptote x 2. Since the inﬁnite discontinuity occurs at the left endpoint of ͓2, 5͔, we use part (b) of Deﬁnition 3: FIGURE 9 y 5 2 y y= c f ͑x͒ dx y f ͑x͒ dx ϩ y f ͑x͒ dx 1
œ„„„„
x2 dx
5
dx
lim y
t l2ϩ t sx Ϫ 2
sx Ϫ 2
lim 2sx Ϫ 2
ϩ
t l2 5 t lim 2(s3 Ϫ st Ϫ 2 )
ϩ
t l2 2s3 3
area=2œ„
0 1 FIGURE 10 2 3 4 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 SECTION 8.8 IMPROPER INTEGRALS y EXAMPLE 6 Determine whether /2 0 ❙❙❙❙ 571 sec x dx converges or diverges. SOLUTION Note that the given integral is improper because lim x l͑ /2͒Ϫ sec x ϱ. Using part (a) of Deﬁnition 3 and Formula 14 from the Table of Integrals, we have y /2 0 sec x dx
lim t l ͑ /2͒Ϫ y t 0 sec x dx Խ lim ln sec x ϩ tan x t l ͑ /2͒Ϫ Խ] t 0 lim ͓ln͑sec t ϩ tan t͒ Ϫ ln 1͔ t l ͑ /2͒Ϫ ϱ
because sec t l ϱ and tan t l ϱ as t l ͑͞2͒Ϫ. Thus, the given improper integral is
divergent.
EXAMPLE 7 Evaluate y dx
if possible.
xϪ1 3 0 SOLUTION Observe that the line x 1 is a vertical asymptote of the integrand. Since it
occurs in the middle of the interval ͓0, 3͔, we must use part (c) of Deﬁnition 3 with
c 1:
3
dx
1
dx
3
dx
y0 x Ϫ 1 y0 x Ϫ 1 ϩ y1 x Ϫ 1 y where 1 0 dx
t
dx
lim y
lim ln x Ϫ 1
t l1Ϫ 0 x Ϫ 1
t l1Ϫ
xϪ1 Խ Խ Խ Խ lim (ln t 1 Ϫ ln Ϫ1
Ϫ
t l1 Խ] t 0 Խ) lim ln͑1 Ϫ t͒ Ϫϱ
Ϫ
t l1 because 1 Ϫ t l 0 ϩ as t l 1Ϫ. Thus, x01 dx͑͞x Ϫ 1͒ is divergent. This implies that
x03 dx͑͞x Ϫ 1͒ is divergent. [We do not need to evaluate x13 dx͑͞x Ϫ 1͒.]  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:
WARNING ■ y 3 0 dx
ln x Ϫ 1
xϪ1 Խ Խ] 3
0 ln 2 Ϫ 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 ͑x͒ dx you must decide, by looking
at the function f on ͓a, b͔, whether it is an ordinary deﬁnite integral or an improper
integral.
EXAMPLE 8 Evaluate y 1 0 ln x dx. SOLUTION We know that the function f ͑x͒ ln x has a vertical asymptote at 0 since lim x l 0ϩ ln x Ϫϱ. Thus, the given integral is improper and we have y 1 0 1 ln x dx lim y ln x dx
ϩ
tl0 t 5E08(pp 570579) 572 ❙❙❙❙ 1/17/06 5:42 PM Page 572 CHAPTER 8 TECHNIQUES OF INTEGRATION Now we integrate by parts with u ln x, dv dx, du dx͞x, and v x : y 1 t 1 1 ln x dx x ln x t Ϫ y dx
t 1 ln 1 Ϫ t ln t Ϫ ͑1 Ϫ t͒
Ϫt ln t Ϫ 1 ϩ t
To ﬁnd the limit of the ﬁrst term we use l’Hospital’s Rule:
lim t ln t lim
ϩ t l 0ϩ tl0 ln t
1͞t y lim
ϩ
tl0 0 1 1͞t
Ϫ1͞t 2 lim ͑Ϫt͒ 0
ϩ x tl0 area=1 y Therefore 1 0 ln x dx lim ͑Ϫt ln t Ϫ 1 ϩ t͒
ϩ
tl0 Ϫ0 Ϫ 1 ϩ 0 Ϫ1 y=ln x 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 Comparison 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 f ͑x͒ ജ t͑x͒ ജ 0 for x ജ a.
(a) If xaϱ f ͑x͒ dx is convergent, then xaϱ t͑x͒ dx is convergent.
(b) If xaϱ t͑x͒ dx is divergent, then xaϱ f ͑x͒ dx 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 ͑x͒ is ﬁnite, then so is the area under the bottom curve
y t͑x͒. And if the area under y t͑x͒ is inﬁnite, then so is the area under y f ͑x͒.
[Note that the reverse is not necessarily true: If xaϱ t͑x͒ dx is convergent, xaϱ f ͑x͒ dx may
or may not be convergent, and if xaϱ f ͑x͒ dx is divergent, xaϱ t͑x͒ dx may or may not be
divergent.]
EXAMPLE 9 Show that y ϱ 0 2 e Ϫx dx is convergent.
2 SOLUTION We can’t evaluate the integral directly because the antiderivative of eϪx is not an elementary function (as explained in Section 8.5). We write y ϱ 0 2 1 2 ϱ 2 eϪx dx y eϪx dx ϩ y eϪx dx
0 1 5E08(pp 570579) 1/17/06 5:42 PM Page 573 SECTION 8.8 IMPROPER INTEGRALS y y=e y ϱ 1 t eϪx dx lim y eϪx dx
tlϱ 1 lim ͑eϪ1 Ϫ eϪt ͒ eϪ1 x 1 tlϱ FIGURE 13 2 Thus, taking f ͑x͒ eϪx and t͑x͒ eϪx in the Comparison Theorem, we see that
2
2
x1ϱ eϪx dx is convergent. It follows that x0ϱ eϪx dx is convergent.
2 In Example 9 we showed that x0ϱ eϪx dx 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 s͞2. Table 1 illustrates the deﬁnition of an improper integral by showing how
2
the (computergenerated) values of x0t eϪx dx approach s͞2 as t becomes large. In fact,
2
these values converge quite quickly because eϪx l 0 very rapidly as x l ϱ. TABLE 1 x0t eϪx t
1
2
3
4
5
6 573 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 ജ 1 we have x 2 ജ x, so Ϫx 2 ഛ Ϫx
2
and therefore eϪx ഛ eϪx. (See Figure 13.) The integral of eϪx is easy to evaluate: _x 2 y=e_x 0 ❙❙❙❙ 2 dx 0.7468241328
0.8820813908
0.8862073483
0.8862269118
0.8862269255
0.8862269255 EXAMPLE 10 The integral because y ϱ 1 1 ϩ eϪx
dx is divergent by the Comparison Theorem
x
1 ϩ eϪx
1
Ͼ
x
x and x1ϱ ͑1͞x͒ dx 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 (a) y (c) y 1
2 0 0.8636306042
1.8276735512
2.5219648704
4.8245541204
7.1271392134
9.4297243064 Exercises 1. Explain why each of the following integrals is improper.
ϱ x1t ͓͑1 ϩ eϪx ͒͞x͔ dx 2
5
10
100
1000
10000  8.8 t 4 x 4eϪx dx (b) y x
dx
x 2 Ϫ 5x ϩ 6 (d) y ͞2 0
0 Ϫϱ sec x dx
1
dx
x2 ϩ 5 2. Which of the following integrals are improper? Why? 1
dx
2x Ϫ 1
ϱ
sin x
(c) y
2 dx
Ϫϱ 1 ϩ x
(a) y 2 1 (b) y (d) y 1 0
2 1 1
dx
2x Ϫ 1
ln͑x Ϫ 1͒ dx 5E08(pp 570579) 574 ❙❙❙❙ 1/17/06 5:44 PM Page 574 CHAPTER 8 TECHNIQUES OF INTEGRATION 3. Find the area under the curve y 1͞x 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 ജ 1. 1.1
0.9
; 4. (a) Graph the functions f ͑x͒ 1͞x and t͑x͒ 1͞x in the viewing rectangles ͓0, 10͔ by ͓0, 1͔ and ͓0, 100͔ by ͓0, 1͔.
(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 ജ 1, if it exists. 5–40  Determine whether each integral is convergent or
divergent. Evaluate those that are convergent. 5.
7. y 1
dx
͑3x ϩ 1͒2 ϱ 1 y 1
dw
s2 Ϫ w Ϫ1 Ϫϱ 9. y 11. y 13. y ϱ Ϫy͞2 6.
8. y Ϫϱ y 0
Ϫ1 10. y x
dx
1 ϩ x2 12. y xeϪx dx 14. y 15. y sin d 16. y 17. y xϩ1
dx
x 2 ϩ 2x 18. y 19. y se Ϫ5s ds 20. y 21. y ln x
dx
x 22. y 23. ϱ 2 Ϫϱ
ϱ 2 ϱ 1
ϱ 0
ϱ 1 y ϱ Ϫϱ y 27. y 31. ϱ Ϫϱ 25. 29. e ϱ 1
3 0 y 0 Ϫ1 y 3 Ϫ2 33. y 35. y 37. y 33 0 0
1 Ϫ1 x2
dx
9 ϩ x6 24. y ϱ 0
3 0 y 9 1 y 1 0 ͑x Ϫ 1͒ Ϫ1͞5 dx 34. y sec x dx 36. y 38. y 1 0
4 0
2 0 ; 44.
; 45.
; 46. ■  40.
■ ■ ■ y 1 0 ■ ln x
dx
sx ■ ■ ■ ■ Sketch the region and ﬁnd its area (if the area is ﬁnite). Խ
S ͕͑x, y͒ Խ x ജ Ϫ2, 0 ഛ y ഛ e ͖
S ͕͑x, y͒ Խ 0 ഛ y ഛ 2͑͞x ϩ 9͖͒
S ͕͑x, y͒ Խ x ജ 0, 0 ഛ y ഛ x͑͞x ϩ 9͖͒
S ͕͑x, y͒ Խ 0 ഛ x Ͻ ͞2, 0 ഛ y ഛ sec x͖
S {͑x, y͒ Խ Ϫ2 Ͻ x ഛ 0, 0 ഛ y ഛ 1͞sx ϩ 2 }
Ϫx/2 2 2 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 48. (a) If t͑x͒ 1͞(sx Ϫ 1), use your calculator or computer to make a table of approximate values of x2t t͑x͒ dx for t 5,
10, 100, 1000, and 10,000. Does it appear that x2ϱ t͑x͒ dx is
convergent or divergent?
(b) Use the Comparison Theorem with f ͑x͒ 1͞sx to show
that x2ϱ t͑x͒ dx is divergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 2 ഛ x ഛ 20. Use your graph to explain intuitively why
x2ϱ t͑x͒ dx is divergent. 49–54  Use the Comparison Theorem to determine whether the
integral is convergent or divergent.
ϱ cos 2x
dx
1 ϩ x2 50. y dx
x ϩ e 2x 52. y dx
x sin x 54. y 49. y 1
dx
xsx 51. y 1
dx
3
sx Ϫ 9 53. y dx
s1 Ϫ x 2 ϱ 2 ϩ e Ϫx
dx
x 55. The integral 1
dx
x2 ϩ x Ϫ 6
xϪ3
dx
2x Ϫ 3 ■ make a table of approximate values of x1t t͑x͒ dx for t 2,
5, 10, 100, 1000, and 10,000. Does it appear that x1ϱ t͑x͒ dx
is convergent?
(b) Use the Comparison Theorem with f ͑x͒ 1͞x 2 to show
that x1ϱ t͑x͒ dx is convergent.
(c) Illustrate part (b) by graphing f and t on the same screen
for 1 ഛ x ഛ 10. Use your graph to explain intuitively why
x1ϱ t͑x͒ dx is convergent. x arctan x
dx
͑1 ϩ x 2 ͒ 2 1
dy
4y Ϫ 1 ■ 2
2
; 47. (a) If t͑x͒ ͑sin x͒͞x , use your calculator or computer to eϪԽ x Խ dx ln x
dx
x3 ■ ■ re r͞3 dr ϱ 1 ; 43. x 2eϪx dx Ϫϱ y ex
dx
e Ϫ1 6 Ϫϱ ϱ 42. dz
z 2 ϩ 3z ϩ 2 ϱ z 2 ln z dz 41. S ͕͑x, y͒ x ഛ 1, 0 ഛ y ഛ e x ͖ dt cos 2␣ d␣ 2 0 41–46 3 0 28. x ϱ y ■ ͑2 Ϫ v 4͒ dv ϱ 0 1
dx
sx 32. ϱ Ϫϱ y 1
dx
x4 e Ϫϱ 26. 30. Ϫ2t Ϫϱ ln x
dx
x2 1
dx
x2 x
dx
2
͑x ϩ 2͒ 2 ϱ dy 4 1
dx
2x Ϫ 5 0 39. ■ 1
ϱ 1 /2 0
■ ■ ■ ■ ■ y ϱ 0 ■ 1
ϱ 1
1 0
■ x
dx
s1 ϩ x 6
eϪx
dx
sx
■ ■ ■ 1
dx
sx ͑1 ϩ x͒ is improper for two reasons: The interval ͓0, ϱ͒ 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:
ϱ
1
1
1
ϱ
1
y0 sx ͑1 ϩ x͒ dx y0 sx ͑1 ϩ x͒ dx ϩ y1 sx ͑1 ϩ x͒ dx ■ 5E08(pp 570579) 1/17/06 5:45 PM Page 575 SECTION 8.8 IMPROPER INTEGRALS 56. Evaluate y ϱ 2 1
dx
x sx 2 Ϫ 4 57–59  Find the values of p for which the integral converges and
evaluate the integral for those values of p. y 59. y ■ 1
dx
xp 1 0
1 58. y ϱ e 1
dx
x ͑ln x͒ p ■ ■ y͑s͒ y R s 2r
x ͑r͒ dr
sr 2 Ϫ s 2 If the actual density of stars in a cluster is x ͑r͒ 1 ͑R Ϫ r͒2,
2
ﬁnd the perceived density y͑s͒.
67. A manufacturer of lightbulbs wants to produce bulbs that last x p ln x dx 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
y͑s͒, where s is the observed planar distance from the center of
the cluster, and x ͑r͒ is the actual density, it can be shown that by the same method as in Exercise 55. 57. ❙❙❙❙ ■ ■ ■ ■ ■ ■ ■ ■ 60. (a) Evaluate the integral x0ϱ x neϪx dx for n 0, 1, 2, and 3. (b) Guess the value of x0ϱ x neϪx dx when n is an arbitrary positive integer.
(c) Prove your guess using mathematical induction. ϱ
61. (a) Show that xϪϱ x dx is divergent. about 700 hours but, of course, some bulbs burn out faster than
others. Let F͑t͒ be the fraction of the company’s bulbs that burn
out before t hours, so F͑t͒ 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 r͑t͒ FЈ͑t͒?
(c) What is the value of x0ϱ r͑t͒ dt ? Why?
68. As we will see in Section 10.4, a radioactive substance decays (b) Show that exponentially: The mass at time t is m͑t͒ m͑0͒e kt, where
m͑0͒ is the initial mass and k is a negative constant. The mean
life M of an atom in the substance is t lim y x dx 0 tlϱ ■ Ϫt This shows that we can’t deﬁne y ϱ Ϫϱ ϱ M Ϫk y te kt dt
0 t f ͑x͒ dx lim y f ͑x͒ dx
tlϱ Ϫt 62. The average speed of molecules in an ideal gas is
v 4
s ͩ ͪ
M
2RT 3͞2 y ϱ 0 2 v 3eϪMv ͑͞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
v ͱ For the radioactive carbon isotope, 14 C, used in radiocarbon
dating, the value of k is Ϫ0.000121. Find the mean life of a
14
C atom.
69. Determine how large the number a has to be so that y ϱ a 1
dx Ͻ 0.001
x2 ϩ 1
2 70. Estimate the numerical value of x0ϱ eϪx dx by writing it as
2 8RT
M 63. We know from Example 1 that the region Խ ͕͑x, y͒ x ജ 1, 0 ഛ y ഛ 1͞x͖ has inﬁnite area. Show
that by rotating about the xaxis we obtain a solid with
ﬁnite volume.
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.
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 2 the sum of x04 eϪx dx and x4ϱ eϪx dx. Approximate the ﬁrst integral by using Simpson’s Rule with n 8 and show that the
second integral is smaller than x4ϱ eϪ4x dx, which is less than
0.0000001. 71. If f ͑t͒ is continuous for t ജ 0, the Laplace transform of f is the function F deﬁned by
ϱ F͑s͒ y f ͑t͒eϪst dt
0 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 ͑t͒ 1
(b) f ͑t͒ e t
(c) f ͑t͒ t
72. Show that if 0 ഛ f ͑t͒ ഛ Me at for t ജ 0, where M and a are constants, then the Laplace transform F͑s͒ exists for s Ͼ a. 73. Suppose that 0 ഛ f ͑t͒ ഛ Me at and 0 ഛ f Ј͑t͒ ഛ Ke at for t ജ 0, where f Ј is continuous. If the Laplace transform of f ͑t͒ is F͑s͒ 5E08(pp 570579) ❙❙❙❙ 576 1/17/06 5:46 PM Page 576 CHAPTER 8 TECHNIQUES OF INTEGRATION and the Laplace transform of f Ј͑t͒ is G͑s͒, show that
G͑s͒ sF͑s͒ Ϫ f ͑0͒ 77. Find the value of the constant C for which the integral sϾa y ϱ
Ϫϱ f ͑x͒ dx is convergent and a and b are real numbers,
show that 74. If x y a Ϫϱ ϱ f ͑x͒ dx ϩ y f ͑x͒ dx y 75. Show that x0ϱ x 2eϪx dx
2 76. Show that x0ϱ eϪx dx
2 1
2 x0ϱ eϪx 2 f ͑x͒ dx ϩ y f ͑x͒ dx 1
C
Ϫ
xϩ2
sx 2 ϩ 4 ͪ dx converges. Evaluate the integral for this value of C.
78. Find the value of the constant C for which the integral b dx. y ϱ 0 x01 sϪln y dy by interpreting the ͩ x
C
Ϫ
x2 ϩ 1
3x ϩ 1 ͪ dx converges. Evaluate the integral for this value of C. integrals as areas.  ͩ ϱ b Ϫϱ a ϱ 0 8 Review ■ CONCEPT CHECK 1. State the rule for integration by parts. In practice, how do you use it?
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? ■ b
5. State the rules for approximating the deﬁnite integral xa f ͑x͒ 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?
6. Deﬁne the following improper integrals. 3. If the expression sa 2 Ϫ x 2 occurs in an integral, what sub stitution might you try? What if sa 2 ϩ x 2 occurs? What if
sx 2 Ϫ a 2 occurs? 4. What is the form of the partial fraction expansion of a rational function P͑x͒͞Q͑x͒ if the degree of P is less than the degree of
Q and Q͑x͒ has only distinct linear factors? What if a linear
factor is repeated? What if Q͑x͒ has an irreducible quadratic
factor (not repeated)? What if the quadratic factor is repeated? ■ (a) ϱ a f ͑x͒ dx (b) y b Ϫϱ f ͑x͒ dx (c) y ϱ Ϫϱ f ͑x͒ dx b
7. Deﬁne the improper integral xa f ͑x͒ dx 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 Ͻ c Ͻ b.
8. State the Comparison Theorem for improper integrals. 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. y ■ 8. The Midpoint Rule is always more accurate than the Trapezoidal Rule. 1. x ͑x ϩ 4͒
B
A
can be put in the form
ϩ
.
x2 Ϫ 4
xϩ2
xϪ2 2. x ϩ4
B
C
A
can be put in the form ϩ
ϩ
.
x ͑x 2 Ϫ 4͒
x
xϩ2
xϪ2 10. If f is continuous on ͓0, ϱ͒ and x1ϱ f ͑x͒ dx is convergent, then 3. x ϩ4
B
A
can be put in the form 2 ϩ
.
x 2͑x Ϫ 4͒
x
xϪ4 11. If f is a continuous, decreasing function on ͓1, ϱ͒ and 4. A
x Ϫ4
B
can be put in the form ϩ 2
.
x ͑x 2 ϩ 4͒
x
x ϩ4 2 2 2 2 5. y 6. y 4 0
ϱ 1 x
dx 1 ln 15
2
x2 Ϫ 1
1
dx is convergent.
x s2 ϱ
t
7. If f is continuous, then xϪϱ f ͑x͒ dx lim t l ϱ xϪt f ͑x͒ dx. 9. (a) Every elementary function has an elementary derivative. (b) Every elementary function has an elementary antiderivative. x0ϱ f ͑x͒ dx is convergent. lim x l ϱ f ͑x͒ 0 , then x1ϱ f ͑x͒ dx is convergent. ϱ
12. If xa f ͑x͒ dx and xaϱ t͑x͒ dx are both convergent, then xaϱ ͓ f ͑x͒ ϩ t͑x͔͒ dx is convergent. 13. If xaϱ f ͑x͒ dx and xaϱ t͑x͒ dx are both divergent, then xaϱ ͓ f ͑x͒ ϩ t͑x͔͒ dx is divergent. 14. If f ͑x͒ ഛ t͑x͒ and x0ϱ t͑x͒ dx diverges, then x0ϱ f ͑x͒ dx also diverges. 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
dx
x ϩ 10 5 0 y ͞2 0 cos
d
1 ϩ sin
7 4. 3 5. y tan x sec x dx 7. sin͑ln t͒
dt
t 9. y
y 4 1 8. 10. sx 2 Ϫ 1
dx
x 2 11. y 13. yx 15. y sin 17. 6. x 3͞2 ln x dx 1 3 2 23. y csc 25. y 4x dx 3x 3 Ϫ x 2 ϩ 6x Ϫ 4
dx
͑x 2 ϩ 1͒͑x 2 ϩ 2͒
cos 3x sin 2x dx y 29. y 31. y 33. y ͑4 Ϫ x 35. y sx ϩ x 37. y ͑cos x ϩ sin x͒ 0 y x2 ϩ 2
dx
xϩ2 y sec 6
d
tan 2 Ϫ1 y
y
y x3
dx
͑x ϩ 1͒10 ye 26. y 1ϩe ■ ■ ■ y dx
x ln x 44. y ln x
dx
sx 46. y dx
x2 Ϫ x Ϫ 2 48. y 50. y ϱ 4 0
3 0 dx
4x ϩ 4x ϩ 5 ϱ 2 Ϫϱ ■ ■ ■ ■ ■ ■ Evaluate the integral or show that it is divergent.
42. 2 stan
d
sin 2 ■ 1
dx
͑2x ϩ 1͒3 1 ͞4 ■ ■ ■ ■ ■ ■ 1 0
6 2
1 0
1 Ϫ1
ϱ 1 t2 ϩ 1
dt
t2 Ϫ 1
y
dy
sy Ϫ 2
1
dx
2 Ϫ 3x
xϩ1
dx
3
sx 4
tanϪ1x
dx
x2 ■ ■ ■  Evaluate the indeﬁnite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C 0). y ln͑x ■ ■ 2 ϩ 2x ϩ 2͒ dx
■ ■ ■ 52.
■ ■ y sx x3
dx
ϩ1 2 ■ ■ ■ ■ ■ 2
3
; 53. Graph the function f ͑x͒ cos x sin x and use the graph to guess the value of the integral x02 f ͑x͒ dx. Then evaluate the
integral to conﬁrm your guess. CAS 54. (a) How would you evaluate x x 5eϪ2x dx by hand? (Don’t actually carry out the integration.)
(b) How would you evaluate x x 5eϪ2x dx using tables?
(Don’t actually do it.)
(c) Use a CAS to evaluate x x 5eϪ2x dx.
(d) Graph the integrand and the indeﬁnite integral on the same
screen. dx ■ sx ϩ 1 ■ ͞3 y ; 51–52 cos x dx x  ϱ ■ dt
sin 2t ϩ cos 2t x y 51. x 2 ϩ 8x Ϫ 3
dx
x 3 ϩ 3x 2 20. y 49. sin x
dx
1 ϩ x2 1 28. y sx Ϫ 1 dx 30. y e s1 Ϫ e 32. y dx 34. y ͑arcsin x͒ dx dx 36. y 1 ϩ tan d 38. y x ͑tan x 5 sec x dx ln 10 47. sarctan x
dx
1 ϩ x2 1 0 y ■ 40. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 3 27. 1 y 24. y Ϫ1 y y 45. dx
x 2s1 ϩ x 2 22. 21. dx
sx 2 Ϫ 4x ͞2 y 1
dy
y 2 Ϫ 4y Ϫ 12 y 43. dt
͑2t ϩ 1͒ 3 4 1 ■ 41. y 18. y 0 y xe 2x
dx
͑1 ϩ 2x͒ 2 1͞2 0 41–50 yeϪ0.6y dy 16. cos 5 d y x sec x tan x dx 4 5 0 14. xϩ1
dx
9x 2 ϩ 6x ϩ 5 19. y 12. dx
ϩx y ■ 2. 577 ■ 39. Evaluate the integral.  ❙❙❙❙ e xse x Ϫ 1
dx
ex ϩ 8
x2 3 55–58  Use the Table of Integrals on the Reference Pages to
evaluate the integral. dx ͒ 1 3͞2 2 cos 2x dx s1 Ϫ e 2x dx Ϫ2x 55. ye ͞4 x sin x
dx
cos 3 x 57. y sx 0 ■ 2 3͞2 x x ■ 2 ■ ■ ■ y csc t dt 58. ϩ x ϩ 1 dx
■ 5 56. y s1 ϩ 2 sin x dx ■ cot x ■ ■ ■ ■ 2 1 Ϫ tan Ϫ1 x͒2 dx 59. Verify Formula 33 in the Table of Integrals (a) by differentia tion and (b) by using a trigonometric substitution.
60. Verify Formula 62 in the Table of Integrals.
61. Is it possible to ﬁnd a number n such that x0ϱ x n dx 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 x0ϱ 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 s1 ϩ x 4 dx 0 64. y ͞2 0 ssin x dx
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?
66. Use Simpson’s Rule with n 6 to estimate the area under the curve y e x͞x from x 1 to x 4. 1minute intervals and recorded in the chart. Use Simpson’s
Rule to estimate the distance traveled by the car.
t (min) v (mi͞h) t (min) v (mi͞h) 0
1
2
3
4
5 40
42
45
49
52
54 6
7
8
9
10 56
57
57
55
56 x3
dx
1 x ϩ 2
is convergent or divergent. y ϱ 5 y 2 Ϫ x 2 1 and the line y 3.
73. Find the area bounded by the curves y cos x and y cos 2x between x 0 and x . 74. Find the area of the region bounded by the curves y 1͞(2 ϩ sx ), y 1͞(2 Ϫ sx ), and x 1. 75. The region under the curve y cos 2x, 0 ഛ x ഛ ͞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.
77. If f Ј is continuous on ͓0, ϱ͒ and lim x l ϱ f ͑x͒ 0, show that 68. A population of honeybees increased at a rate of r͑t͒ bees per 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 y ϱ 0 f Ј͑x͒ dx Ϫf ͑0͒ 78. We can extend our deﬁnition of average value of a continuous function to an inﬁnite interval by deﬁning the average value of
f on the interval ͓a, ϱ͒ to be
1
t
lim
y f ͑x͒ dx
tlϱ t Ϫ a a
(a) Find the average value of y tanϪ1x on the interval ͓0, ϱ͒.
(b) If f ͑x͒ ജ 0 and xaϱ f ͑x͒ dx is divergent, show that the average value of f on the interval ͓a, ϱ͒ is lim x l ϱ f ͑x͒, if this
limit exists.
(c) If xaϱ f ͑x͒ dx is convergent, what is the average value of f
on the interval ͓a, ϱ͒?
(d) Find the average value of y sin x on the interval ͓0, ϱ͒. 8000
4000 0 4 8 12 16 20 t
24
(weeks) 69. (a) If f ͑x͒ sin͑sin x͒, use a graph to ﬁnd an upper bound Խ integral 72. Find the area of the region bounded by the hyperbola 67. The speedometer reading (v) on a car was observed at CAS 71. Use the Comparison Theorem to determine whether the Խ for f ͑4͒͑x͒ .
(b) Use Simpson’s Rule with n 10 to approximate
x0 f ͑x͒ dx 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 1͞x to show that y ϱ 0 ln x
dx 0
1 ϩ x2 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 0 r 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 y a 0 a f ͑x͒ dx y f ͑a Ϫ x͒ dx
0 (b) Use part (a) to show that y ͞2 0 sin n x
dx
sin x ϩ cos n x
4
n 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 Ϫ x. Then du Ϫdx. When x 0, u a; when x a, u 0. So y a 0 0 a f ͑a Ϫ x͒ dx Ϫy f ͑u͒ du y f ͑u͒ du
a 0 But this integral on the right side is just another way of writing x0a f ͑x͒ dx. So the given
equation is proved.
(b) If we let the given integral be I and apply part (a) with a ͞2, we get
Iy ͞2 0  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. sin n x
͞2
sin n͑͞2 Ϫ x͒
dx y
dx
0
sin n x ϩ cos n x
sin n͑͞2 Ϫ x͒ ϩ cos n͑͞2 Ϫ x͒ A wellknown trigonometric identity tells us that sin͑͞2 Ϫ x͒ cos x and
cos͑͞2 Ϫ x͒ sin x, so we get
Iy ͞2 0 1 3
4 2 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 y ͞2 0 0 FIGURE 1 cos n x
dx
cos n x ϩ sin n x π
2 sin n x ϩ cos n x
͞2
1 dx
n
n dx y
0
sin x ϩ cos x
2 Therefore, I ͞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
dx.
x7 Ϫ x
The straightforward approach would be to start with partial fractions, but that would be brutal.
Try a substitution. 2. Evaluate y 1 3
7
3. Evaluate y (s1 Ϫ x 7 Ϫ s1 Ϫ x 3 ) dx. 0 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 ͑x͒, then FIGURE FOR PROBLEM 1 y pier f Ј͑x͒ dy
ϪsL 2 Ϫ x 2
dx
x (b) Determine the function y f ͑x͒.
L (x, y) 5. A function f is deﬁned by
(L, 0)
O FIGURE FOR PROBLEM 4 f ͑x͒ y cos t cos͑x Ϫ t͒ dt x 0 0 ഛ x ഛ 2 Find the minimum value of f .
6. If n is a positive integer, prove that y 1 0 ͑ln x͒n dx ͑Ϫ1͒n n! 7. Show that y 1 0 2 2n ͑n!͒2
͑2n ϩ 1͒! ͑1 Ϫ x 2 ͒n dx Hint: Start by showing that if In denotes the integral, then
Ikϩ1 2k ϩ 2
Ik
2k ϩ 3 ; 8. Suppose that f is a positive function such that f Ј is continuous. (a) How is the graph of y f ͑x͒ sin nx related to the graph of y f ͑x͒? What happens
as n l ϱ?
(b) Make a guess as to the value of the limit
lim nlϱ y 1 0 f ͑x͒ 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 Ј is continuous, there is a constant M such that f Ј͑x͒ ഛ M for 0 ഛ x ഛ 1.]
9. If 0 Ͻ a Ͻ b, ﬁnd lim
tl0 580 ͭy 1 0 ͮ Խ 1͞t ͓bx ϩ a͑1 Ϫ x͔͒ t dx . Խ 5E08(pp 580581) 1/17/06 5:39 PM Page 581 tϩ1
x
; 10. Graph f ͑x͒ sin͑e ͒ and use the graph to estimate the value of t such that xt f ͑x͒ dx is a maximum. Then ﬁnd the exact value of t that maximizes this integral.
y Խ Խ 11. The circle with radius 1 shown in the ﬁgure touches the curve y 2x 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 v͑t͒ be the velocity of the rocket at time t and suppose that the velocity u of the exhaust gas is constant. Let M M͑t͒ 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 FIGURE FOR PROBLEM 11 FM x dv
Ϫ ub
dt where the force F ϪMt. Thus
1 M dv
Ϫ ub ϪMt
dt Let M1 be the mass of the rocket without fuel, M2 the initial mass of the fuel, and
M0 M1 ϩ M2 . Then, until the fuel runs out at time t M2 b, the mass is M M0 Ϫ bt.
(a) Substitute M M0 Ϫ bt into Equation 1 and solve the resulting equation for v. Use the
initial condition v ͑0͒ 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 y͑t͒ 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 Ͼ 0, 0Ͻy ϱ 0 sin t
2
dt Ͻ
ln͑1 ϩ x ϩ t͒
ln͑1 ϩ x͒ ; 14. The Chebyshev polynomials Tn are deﬁned by
Tn͑x͒ cos͑n arccos x͒ n 0, 1, 2, 3, . . . (a) What are the domain and range of these functions?
(b) We know that T0͑x͒ 1 and T1͑x͒ x. Express T2 explicitly as a quadratic polynomial
and T3 as a cubic polynomial.
(c) Show that, for n ജ 1,
Tnϩ1͑x͒ 2x Tn͑x͒ Ϫ TnϪ1͑x͒
(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 Tnϩ1 ? What about the xcoordinates of the maximum and minimum values?
1
(j) Based on your graphs in parts (g) and (h), what can you say about xϪ1 Tn͑x͒ dx 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 ͑x͒ cos͑c arccos x͒ 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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This note was uploaded on 02/08/2010 for the course M 340L taught by Professor Lay during the Spring '10 term at École Normale Supérieure.
 Spring '10
 Lay
 Linear Algebra, Algebra

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