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1
[HRPTER
7
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
I I I I
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.
f xn dx
xn+l
+
= --
n
+
1
c
(n # - 1)
f e xdx =e x + c
Jsin x dx =
-cos x
+C
Jsec x dx = tan x + C
2
J sec x tan x dx =
J ~ dx =
In IxI + C
f a xdx
-
ax
=
sin x
J csc x dx =
-cot x
Jesc x cot x dx =
+C
c
J cos x dx =
2
sec x
+
-
In a
+C
+C
-esc x + C
Jsinh x dx = cosh x + C
J cosh x dx =
J tan x dx =
J cot x dx =
In Isin x I + C
f . . fn2 1-
dx
fx
1
2
+ a2
In I sec x I
dx
=
+C
_!_ tan_, (~) + C
a
a
x2
sinh x
=
+C
·-(X) + C
sm
I
-;;
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 7 . 1. Then we learn
meth~ds that are special to particular classes of functions such as trigonometric func-
tions 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 7 .5 we discuss a strategy for integration.
1111
7.1 Integration b~ 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
p arts.
The Product Rule states that if f and g are differentiable functions , then
d
dx [f(x)g(x)]
=
f(x)g'(x) + g(x)f'(x)
476
CHAPTER 7 TECHNIQUES OF INTEGRATION
In the notation for indefinite integrals this equation becomes
J[f(x)g'(x) + g(x)f'(x)] dx
=
Jf(x)g'(x) dx + J g(x)f'(x) dx =
or
f(x)g(x)
f(x)g(x)
We can rearrange this equation as
Jf(x)g'(x) dx
rn
=
f(x)g(x) -
Jg(x)f'(x) dx
Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u = f(x) and v = g(x). Then the differentials are
du = f'(x) dx and dv = g'(x) dx, so, by the Substitution Rule, the formula for integration
by parts becomes
[
[1]
EXAMPLE 1 Find
Ju dv
=
uv -
J
Jv du
f x sin x dx.
SOLUTION USING FORMULA l Suppose we choose f(x) = x and g'(x) = sin x. Then f'(x) = 1
and g(x) = -cos x. (For g we can choose any antiderivative of g' .) Thus, using Formula
1, we have
J x sin x dx = f(x)g(x)
J g(x)f'(x) dx
-
= x( -cos x) -
J (-cos x) dx
f
-
-X COS X
+
=
-X COS X
+ sin X + C
COS X
dx
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
1111 It is helpful to use the pattern:
u
=
D
dv
=
D
du
=
D
v
=
D
Then
u=x
dv =sin x dx
~
v = -cos x
du
and so
u
dx
dv
u
v
v
du
Jx sin x dx = J; ~~ = : '0~ - J'0~ dx
+f
dx
=
-X COS X
=
-X COS X
COS X
+ sin X + C
SECTION 7.1 INTEGRATION BYPARTS
477
NOTE " 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 f x sin x dx and expressed it in terms
of the simpler integral f cos x dx. If we had chosen u = sin x and dv = x dx, then
du = cos x dx and v = x 2/2, so integration by parts gives
f
x sin x dx
=
x2
(sin x) 2
1
- 2 Jx
2
cos x dx
Although this is true, f 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(x) to
be a function that becomes simpler when differentiated (or at least not more complicated)
as long as dv = g'(x) dx can be readily integrated to give v.
EXAMPLE 2 Evaluate
f ln x dx.
SOLUTION Here we don't have much choice for u and dv. Let
Then
u
=
ln x
du
=
-dx
dv
1
=
dx
v=x
X
Integrating by parts, we get
f ln
X
dx
= X
ln
X -
f
Jdx
I 1 dx as I dx.
=
xln x-
II II Check the answer by differentiating it.
=
xln x- x
II II It's customary to write
~
X
+C
Integration by parts is effective in this example because the derivative of the function
f(x) = ln xis simpler than f.
EXAMPLE 3 Find
f t e dt.
2
1
SOLUTION Notice that t 2 becomes simpler when differentiated (whereas e 1 is unchanged
when differentiated or integrated), so we choose
u
Then
du
= t2
2tdt
=
dv
=
e 1 dt
v
=
e1
2
Jte dt
Integration by parts gives
[I]
Jt e dt
2
1
=
t 2 e1
-
1
The integral that we obtained, f te 1 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
478
CHAPTER 7 TECHNIQUES OF INTEGRATION
dv = e 1 dt. Then du = dt, v = e 1, and
Jte dt
1
1
=
te
=
te 1
Je dt
1
-
-
e1
+C
Putting this in Equation 3, we get
Jt e dt = t e
2 1
2 1
t 2e 1
=
2
= t e
EXAMPLE 4 Evaluate
II II An easier method, using complex numbers,
is given in Exercise 48 in Appendix G.
f
1
Jte dt
1
...,....
2
-
2(te 1
-
2te + 2e + Ct
e1
-
1
+ C)
1
where Ct
=
-2C
e x 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 = ex dx and v = -cos X, so integration by
parts gives
[!]
f
e x sin
X
dx
= -ex COS X+
f
e x COS
X
dx
f
The integral that we have obtained, ex 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
rn
f
e x COS
dx
X
=
e x Sin
X -
f
e x sin
X
dx
At first glance, it appears as if we have accomplished nothing because we have arrived at
II II Figure 1 illustrates Example 4 by
showing the graphs of f(x) = e x sin x and
F(x) = !ex(sin x - cos x). As a visual check
on our work, notice that f(x) = 0 when F has
a maximum or minimum.
12
f ex sin x dx, which is where we started. However, if we put Equation 5 into Equation 4
we get
f ex sin X dx =
-ex COS
X
+
e x sin
X -
f ex sin X dx
This can be regarded as an equation to be solved for the unknown integral. Adding
ex sin x dx to both sides, we obtain
f
2
-3 1
:¥
<r
I I
f
e x Sin
X
dx
=
-ex COS
X
+
e x sin
X
16
Dividing by 2 and adding the constant of integration, we get
-4
FIGURE 1
Jex sin x dx = 4ex(sin x -
cos x)
+
C
If we combine the formula for integration by parts with Part 2 of the Fundamental
Theorem of Calculus, we can evaluate definite integrals by parts. Evaluating both sides of
Formula 1 between a and b, assuming f' and g' are continuous, and using the Fundamental
SECTION 7.1 INTEGRATION BY PARTS
479
Theorem, we obtain
[!]
J: f(x)g'(x) dx
[_
EXAMPLES Calculate
f(x)g(x)
=
J: - J: g(x)f'(x) dx _j
Jo' tan- x dx.
1
SOLUTION Let
u = tan- 1x
dv
dx
Then
du
=
1
=
dx
v=x
+ x2
So Formula 6 gives
JoI' tan- 1xdx =
xtan- x ]I0
1
JoI'
-
X
~ dx
i
= 1 · tan - l 1 - 0 · tan_, 0 II II Since tan- 1x;;. 0 for x;;. 0, the integral in
Example 5 can be interpreted as the area of the
region shown in Figure 2.
I'
7T
=
4 - Jo
X
l
o 1
+
x2
dx
X
1
+ x2
dx
To evaluate this integral we use the substitution t = 1 + x 2 (since u has another meaning
in this example). Then dt = 2xdx, so x dx = dt/2. When x = 0, t = 1; when x = 1,
t = 2; so
y
y = tan- 1x
iI
0
X
X
--2
1 +X
I I]
dt= 2ln
I
dx= 2I f 2t
t
I
2
1
= ~ (ln 2 - ln 1) = ~ ln 2
FIGURE 2
I' _,
Jo tan x dx
Therefore
=
7T
I'
4 - Jo
x
+
1
x 2 dx =
ln 2
7T
4 - - 2-
EXAMPLE 6 Prove the reduction formula
II II Equation 7 is called a reduction formula
because the exponent n has been reduced to
n - 1 and n- 2.
[I]
f
1
sinnx dx = --;cos
x sinn- lx
n-1s
+ --n-
sinn- 2x dx
where n ;::: 2 is an integer.
SOLUTION Let
Then
u = sinn- Ix
dv =sin x dx
du = (n - 1) sinn- 2x cos x dx
v
=
-cosx
so integration by parts gives
J sinnx dx =
- cos x sinn- Ix
+
(n - 1)
J sinn- x cos x dx
2
2
CHAPTER 7 TECHNIQUES OF INTEGRATION
480
Since cos 2x
=
1 - sin 2x, we have
Jsinnx dx
=
-cos x sinn- lx
+ (n - 1) Jsinn-zx dx - (n - 1) Jsinnx 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
or
Jsinnx dx
=
-cos x sinn- Jx
f sin
=
1
---;;cos
x sin n-Ix
11
X
dx
+ (n -
1)
Jsinn-zx dx
n-1s
+ --n-
sinn- zx dx
The reduction formula (7) is useful because by using it repeatedly we could eventually
express f sinnx dx in terms off sin x dx (if n is odd) or f (sin x) 0 dx = f dx (if n is even).
1111 7.1 Exercises
II II
1-2 1111 Evaluate the integral using integration by parts with the
indicated choices of u and dv.
f ln
2. f (} sec 0
1.
X
dx;
X
2
3-32
3.
1111
U =
d(};
u
dv
= X
dx
1.
Jx 2 sin
9.
f 1n(2x +
=
(},
dv
=
4.
Jxe- x dx
6.
f
dx
8.
f x 2 cos mx dx
1) dx
10.
7rX
arctan 4t dt
12.
Jsin- x dx
1
COS - X
dx
t 3e 1 dt
15.
Je 29 sin 3(}d(}
16.
f
e - 9 cos 2(}d(}
17.
f y sinh y dy
18.
f y cosh ay dy
20.
J
21.fln:dx
22.
f'lT/2 x csc 2x dx
'lT/4
26.
J
f,/3 arctan(l/x) dx
28.
29.
f cos(ln x) dx
30.
31.
r
32.
x 4(1n x) 2 dx
1
0
(x 2
r Ji
+
x 5xdx
1
r
o
r3
J4+f2
dr
4 + r
J: e s sin(t -
33.
Jsin Jx dx
34.
r
35.
J.,fir
36.
J x 5ex 2 dx
..fi12
0 3 cos( 0 2 ) d(}
s) ds
e.fidx
~ 37-40 1111 Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the function and its
antiderivative (take C = 0).
37.
f
39.
J(2x + 3)e xdx
X COS 'TrX
dx
38.
Jx 1
ln x dx
40.
f
2
l)e - x dx
ln t dt
1
0
33-36 1111 First make a substitution and then use integration by parts
to evaluate the integral.
f p 5 lnp dp
f
fo'lT tsin 3tdt
24.
Jcos x ln(sin x) dx
1
14.
X
- y2- dy
o e Y
27.
t sin 2t dt
J(ln x) 2 dx
l
r /2
sec 2 (J d(}
13.
19.
25.
Evaluate the integral.
f x cos Sx dx
f
X,
r
0
s. f re' 12 dr
11.
ln
23.
3 2
x 3ex dx
SECTION 7.1 INTEGRATION BY PARTS
41. (a) Use the reduction formula in Example 6 to show that
.
f sm
2
x dx
=
-
x
2
sin 2x
4
-
---
+
481
~ S3-S4 1111 Use a graph to find approximate x-coordinates of the
points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves.
C
y = (x - 2) 2
53. y = x sin x,
(b) Use part (a) and the reduction formula to evaluate
sin 4x dx.
54. y =arctan 3x,
f
y = x/2
42. (a) Prove the reduction formula
f
1 cos"- 1x sin x
cos"x dx = n
f
SS-58
1111 Use the method of cylindrical shells to find the volume
generated by rotating the region bounded by the given curves about
the specified axis.
n- 1 J cos"- 2x dx
+n
•
(b) Use part (a) to evaluate cos 2x dx.
(c) Use parts (a) and (b) to evaluate cos 4x dx.
f
43. (a) Use the reduction formula in Example 6 to show that
2. 4. 6 . . . . . 2n
(
+ 1)
3 . 5 . 7 . . . . . 2n
1 · 3 · 5 · · · · · (2n - 1) 7T
=
v(t)
-
2 · 4 · 6 · · · · · 2n
45.
J (ln x)"dx =
46.
Jx"e xdx =
x(ln x)"- n
x"ex- n
J (ln x)"-
1
about the x-axis
- gt -
=
m- rt
m
Ve ln -
2
where g is the acceleration due to gravity and t is not too
large. If g = 9.8 m/s 2, m = 30,000 kg, r = 160 kg/s, and
Ve = 3000 m/s, find the height of the rocket one minute
after liftoff.
Use integration by parts to prove the reduction formula.
1111
about they-axis
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
44. Prove that, for even powers of sine,
45-48
about the y-axis
1;
59. Find the average value of f(x) = x 2 ln x on the interval [1 , 3].
where n ;::::,: 2 is an integer.
(b) Use part (a) to evaluate f 07r12 sin 3x dx and f 07r12 sin 5x dx.
(c) Use part (a) to show that, for odd powers of sine,
i
56. y = eX, y = e-x, :X= 1;
58. y = e x, x = 0, y = 1r;
7r/2
n - 1 l7r/2
sin"- 2x dx
sin"x dx = - - o
n
o
7r/2 . 2
sm "xdx
o
~ x ~
57. y =e-x, y = 0, x = -1 , x = 0; about x = 1
i
[ Tr/ 2 . 2n+ I dxJo
sm
x
-
55. y = cos(7Tx/2), y = 0, 0
dx
61. A particle that moves along a straight line has velocity
Jx"- e xdx
v(t) = t 2e_, meters per second after t seconds. How far will
1
it travel during the first t seconds?
47.
f (x
2
62. If f(O) = g(O) = 0 and f" and g" are continuous, show that
+ a 2 )n dx
x(x2 + a 2)"
2n + 1
48.
f
sec"x dx
=
+
2na 2
2n + 1
tan x sec"- 2x
n-1
(n =I= -
J sec"- x dx
(n =I= 1)
n - 2
n-1
+ ---
2
D
=
+
f(a)g'(a) - f'(a)g(a)
J:f"(x)g(x) dx
63. Suppose that f(l) = 2, !(4) = 7, f'(l) = 5, !'(4) = 3, and
f (ln x) dx.
Use Exercise 46 to find f x e xdx.
51-52
foaf(x)g"(x) dx
f" ~s continuous. Find the value off~ xf"(x) dx.
64. (a) Use integration by parts to show that
Jf(x) dx =
3
49. Use Exercise 45 to find
50.
J(x 2 + a 2)n- l dx
xf(x) -
Jxf'(x) dx
4
1111
Find the area of the region bounded by the given curves.
51. y = xe- 0 .4X,
52. y = 5 ln X,
y = 0,
y=
X
ln
x = 5
(b) If f and g are inverse functions and f' is continuous, prove
that
/
f
b
a
X
f(x) dx
=
bf(b) - af(a) -
i f(b)
JW
g( y) dy
[Hint: Use part (a) and make the substitution y = f(x).]
482 II
CHAPTER 7 TECHNIQUES OF INTEGRATION
(c) In the case where f and g are positive functions and
b > a > 0, draw a diagram to give a geometric interpretation of part (b).
(d) Use part (b) to evaluate ln x dx .
(c) Use parts (a) and (b) to show that
2n
2n
g
65. We arrived at Formula 6.3.2, V =
J: 2 7Tx f( x ) dx, by using
and deduce that limn--->oo lzn+1/l zn = 1.
(d) Use part (c) and Exercises 43 and 44 to show that
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 one-to-one and therefore has an inverse function g. Use the figure to show that
v
=
7Tb 2d -
7Ta
2
c -
2 2 4 4 6 6
2n
2n
7T
lim-·-·-·-·-·-·····---·---=1 3 3 5 5 7
2n - 1 2n + 1
2
n --->oo
r
This formula is usually written as an infinite product:
7r[g(y)] 2 dy
J:
x= b
66. Let In =
f
12
7T
0
b
(a) Show that l zn+Z ~ l zn+ J ~ l zn·
(b) Use Exercise 44 to show that
lzn
llll 7.2
2n
4
6
6
--,--,
I
I
I
+1
+2
----
2n
4
ornEJ
EJJ
X
sinnx dx .
l zn+Z
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. x=g(y)
a
2
7T
-=-·-·-·-·-·-·
2
1 3 3 5 5 7
Make the substitution y = f( x) and then use integration by
27Tx f( x ) dx .
parts on the resulting integral to prove that V =
0
~ l zn+1 ~
1
fzn
+1
+2
I
I
I
I
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
Jcos x dx .
3
SOLUTION Simply substituting u
= cos x isn't helpful, since then du = -sin x dx. In order
to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of
sine would require an extra cos x factor. Thus, here we can separate one cosine factor
and convert the remaining cos 2x factor to an expression involving sine using the identity
sin 2x + cos 2x = 1:
cos 3x
=
cos 2x · cos x
=
(1 - sin 2x) cos x
We can then evaluate the integral by substituting u = sin x, so du = cos x dx and
Jcos x dx = Jcos x ·
= J(1 - u
2
3
=
GOS
2
)
x dx =
du = u -
sin x - ~ sin 3x + C
J(1 -
sin 2x) cos x dx
~ u3 + C
SECTION 7.2 TRIGONOMETRIC INTEGRALS
483
In general, we try to write an integrand involving powers of sine and cosine in a form
where we have only one sine factor (and the remainder of the expression in terms of
cosine) or only one cosine factor (and the remainder of the expression in terms of sine).
The identity sin 2x + cos 2x = 1 enables us to convert back and forth between even powers
of sine and cosine.
EXAMPLE 2 Find
Jsin x cos x dx
2
5
SOLUTION We could convert cos 2x to 1 - sin 2x, but we would be left with an expression in
terms of sin x with no extra cos x factor. Instead, we separate a single sine factor and
rewrite the remaining sin 4x factor in terms of cos x:
sin 5x cos 2x
1111 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?
(sin 2x) 2 cos 2x sin x
=
' " "
,
/
;o ' "'
(1 - cos 2x)2 cos 2x sin x
Substituting u = cos x, we have du = -sin x dx and so
Jsin x cos x dx J(sin x) cos x sin x dx
2
5
2
=
0.2
-7T I ,
=
' 17T
=
J(1 -
=
J (1
2
cos 2x) 2 cos 2x sin x dx
- u 2 ) 2 u 2 ( -du) = -
us
u3
=- (
- 0.2
FIGURE 1
=
2
J (u
2
-
2u 4
+
u 6 ) du
u7)
3- 2 5+7 +C
-tcos 3x + ~cos 5x- ~ cos 7x + C
In the preceding examples, an odd power of sine or cosine enabled us to separate a
single factor and convert the remaining even power. If the integrand contains even powers
of both sine and cosine, this strategy fails. In this case, we can take advantage of the following half-angle identities (see Equations 17b and 17a in Appendix D):
sin 2x
1111 Example 3 shows that the area of the region
shown in Figure 2 is 7T/2.
~( 1 - cos 2x)
=
and
cos 2x
=
~ (1 +cos 2x)
EXAMPLE 3 Evaluate fo7T sin 2x dx.
SOLUTION If we write sin 2x = 1 - cos 2x, the integral is no simpler to evaluate. Using the
half-angle formula for sin 2x, however, we have
1.5
y = sin 2 x
x
fo7T sin 2 dx =
=
0
-0.5
~
fo7T (1 - cos 2x) dx =
H
1T -
~ sin
27T) -
[Hx -
H0 -
~sin 2x)];
~ sin 0) = ~ 1T
7T
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 7 .1.
FIGURE 2
EXAMPLE 4 Find
f sin x
4
dx.
SOLUTION We could evaluate this integral using the reduction formula for
f sinnx dx
(Equation 7.1.7) together with Example 1 (as in Exercise 41 in Section 7.1), but a better
484
.!
CH APTER 1 TECHNIQUESOF INTEGRATION
method is to write sin 4x
=
(sin 2x? and use a half-angle formula:
Jsin x dx J(sin x) dx
4
2
=
2
c-
~
f
=
~ J (1 - 2 cos 2x + cos 2 2x ) dx
~os 2x )' dx
Since cos 2 2x occurs, we must use another half-angle formula
cos 2 2x
=
i(l
+ cos 4x )
This gives
Jsin x dx
4
=
~ J[1 - 2 cos 2x +
=
~ JG-
=
H~ x -
2 cos 2x
i(l
+ cos 4x )] dx
+ i cos 4 x ) dx
+ i sin 4x ) +
sin 2x
C
To summarize, we list guidelines to follow when evaluating integrals of the form
Jsinmx cosnx dx, where m ~ 0 and n ~ 0 are integers.
Strategy for Evaluating
Jsinmx cos"x dx
(a) If the power of cosine is odd (n = 2k + 1), save one cosine factor and use
cos 2x = 1 - sin 2x to express the remaining factors in terms of sine:
J sinmx cos
Then substitute u
2
k+
1
x dx
=
J sinmx (cos x)kcos x dx
=
J sinmx (1
2
- sin 2x)kcos x dx
sin x.
=
(b) If the power of sine is odd (m = 2k + 1), save one sine factor and use
sin 2x = 1 - cos 2x to express the remaining factors in terms of cosine:
J sin
2
k+
1
x COS 11X dx
=
J (sin x)kcOS
=
J(1
2
11
X
sin x dx
- cos 2x)kcOS 11X 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 half-angle identities
sin 2x
=
iO -
cos 2x
cos 2x)
=
It is sometimes helpful to use the identity
sin x cos x
=
i sin 2x
i(l
+ cos 2x )
SECTION 7.2 TRIGONOMETRIC INTEGRALS
485
We can use a similar strategy to evaluate integrals of the form I tanmx secnx dx. Since
2
2
= sec x, we can separate a sec x factor and convert the remaining (even)
power of secant to an expression involving tangent using the identity sec 2x = 1 + tan 2x.
Or, since (d/ dx) sec x = sec x tan x, we can separate a sec x tan x factor and convert the
remaining (even) power of tangent to secant.
(d/dx) tan x
EXAMPLE S Evaluate
Jtan x sec x dx.
6
4
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 + tan 2x. We can then evaluate the integral
by substituting u = tan x with du = sec 2x dx:
Jtan x sec x dx Jtan x sec x sec x dx
4
6
2
6
=
2
=
Jtan x (1 + tan x) sec x dx
=
Ju (1 + u
2
6
6
u7
2
)du =
2
J(u
6
+
8
u )du
u9
=-+-+C
7
9
=
EXAMPLE 6 Find
~ tan 7x +
t tan x + C
9
Jtan e sec ede.
5
7
e
SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with
a sec 5 e factor, which isn't easily converted to tangent. However, if we separate a
sec e tan efactor, we can convert the remaining power of tangent to an expression
involving only secant using the identity tan 2 e = sec 2 e - 1. We can then evaluate the
integral by substituting u = sec e, so du = sec e tan e de:
f tan
5
e sec 7 e de
=
f tan
=
J (sec e -
=
J (u
4
e sec 6 e sec e tan e de
2
2
-
1? sec 6 e sec e tan e de
1) 2 u 6 du
utt
u9
11
9
=
J (u
10
-
2u 8
+
u 6 ) du
u7
=--2-+-+C
=
7
fi sec e- §sec e + ~ sec e + C
11
9
7
The preceding examples demonstrate strategies for evaluating integrals of the form
I tanmx secnx dx for two cases, which we summarize here.
486
CHAPTER 7 TECHNIQUES OF INTEGRATION
Jtan"'x sec"x dx
Strategy for Evaluating
(a) If the power of secant is even (n = 2k, k ~ 2), save a factor of sec 2x and use
sec 2x = 1 + tan 2x to express the remaining factors in terms of tan x:
Jtanmx sec kx dx Jtanmx (sec x)k-l sec x dx
2
=
Then substitute u
2
=
2
Jtanmx (1 + tan x)k-l sec x dx
2
2
tan x.
=
(b) If the power of tangent is odd (m = 2k + 1), save a factor of sec x tan x and
use tan 2x = sec 2x - 1 to express the remaining factors in terms of sec x:
Jtan
2
k+
1
Then substitute u
x sec"x dx
=
J(tan x)ksec"- x sec x
=
J(sec x -
2
1
2
tan x dx
l)k sec"- 1x sec x tan x dx
sec x.
=
For other cases, the guidelines are not as clear-cut. 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):
J tan x dx =
ln I sec x I + C
We will also need the indefinite integral of secant:
Jsec x dx =
OJ
ln I sec x + tan x I + C
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x + tan x:
f
sec x dx
=
sec x + tan x
dx
sec x +tan x
f
sec x + sec tan
f sec +tan
sec x
2
=
x
x
x
x
dx
If we substitute u = sec x + tan x, then du = (sec x tan x + sec 2x) dx, so the integral
becomes (1/u) du = ln I u I + C. Thus, we have
J
Jsec x dx =
ln I sec x + tan x I + C
SECTION 7.2 TRIGONOMETRIC INTEGRALS
EXAMPLE 7 Find
487
I tan x dx.
3
SOLUTION Here only tan x occurs, so we use tan 2x = sec 2x - 1 to rewrite a tan 2x factor in
terms of sec 2x:
I tan x dx = I tan x tan x dx
2
3
I tan x (sec x - 1) dx
= I tan x sec x dx - I tan x dx
2
=
2
tan 2x
=---lnlsecxi+C
2
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
I sec x dx.
3
SOLUTION Here we integrate by parts with
u =sec x
du = sec x tan x dx
dv = sec 2x dx
v = tanx
I sec x dx = sec x tan x - I sec x tan x dx
= sec x tan x - I sec x (sec x - 1) dx
=sec x tan x- I sec x dx + I sec x dx
3
Then
2
2
3
Using Formula 1 and solving for the required integral, we get
I sec x dx = Hsec x tan x + ln I sec x + tan xI) + C
3
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 8. Integrals of
the form f cot' x csc x dx can be found by similar methods because of the identity
1 + cot 2x = csc 2x.
Finally, we can make use of another set of trigonometric identities:
11
11
[1] To evaluate the integrals (a) f sin mx cos nx dx, (b)
(c) f cos mx cos nx dx, use the corresponding identity:
II II These product identities are discussed in
Appendix D.
f sin mx sin nx dx, or
(a) sin A cos B = Hsin(A - B) + sin(A + B)]
(b) sin A sin B = Hcos(A - B) - cos(A + B)]
(c) cos A cos B = Hcos(A -B) + cos(A +B)]
488
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
EXAMPLE 9 Evaluate
J 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:
Jsin 4x cos 5x dx JHsin( -x) + sin 9x] dx
=
IW 1.2
~ J (-sin x + sin 9x) dx
=
Hcos x -
4cos 9x) + C
Exercises
1-47 1111 Evaluate the integral.
1.
=
Jsin 3x cos 2x dx
2.
Jsin 6x cos 3x dx
31.
Jtan x dx
32.
Jtan (ay) dy
33.
J tan 3()4 d()
34.
Jtan 2x sec x dx
35.
f1T/2
cot 2x dx
1T/6
36.
f1T/2
cot 3x dx
1T/4
37.
Jcot\~ esc\~ da
38.
Jcsc 4x cot 6x dx
39.
Jesc x dx
40.
f1T/3
csc 3x dx
1T/6
41.
Jsin 5x sin 2x dx
42.
J sin 3x cos x dx
43.
f
44 •
J cos x + sin x
45.
J1 -
46.
f .dx
5
cos ()
3. r7T/4 sin 5x cos 3x dx
4.
7T/2
S.
Jcos x sin 4x dx
5
6.
0
11.
f (1 +
cos 0) 2 d()
13.
i 1T/4 sin 4x cos 2x dx
15.
J
0
sin 3x Jcos x dx
dx
0
Jsin 3(mx) dx
8. i1T/ 2 sin 2(20) d()
7. i1T/ 2 cos20 d()
9. Jo.,. sin 4(3t) dt
i1T/2 cos 5x
0
10. Jo.,. cos 6() d()
12.
14.
~ 16.
Jx cos x dx
2
i1T/2 sin 2x cos 2x dx
0
Jcos () cos (sin 0) d()
17.
Jcos 2x tan 3x dx
18.
Jcot
19.
J1 -
20.
Jcos 2x sin 2x dx
COS X
5
()
Jsec 2x tan x dx
22.
i1T/ 2 sec 4(t/2) dt
23.
Jtan 2x dx
24.
Jtan 4x dx
25.
Jsec t dt
26. i1T/ 4 sec 40 tan 40 d()
27. i7T/ 3 tan 5x sec 4x dx
0
29.
Jtan3x sec x dx
/
28.
0
COS X -
1
Jt sec 2(t 2) tan 4(t 2) dt
48. If f0.,.14 tan 6x sec x dx
=
I, express the value of
f
8
14
.,. tan x sec x dx in terms of I.
0
~ 49-52 1111 Evaluate the indefinite integral. Illustrate, and check that
your answer is reasonable, by graphing both the integrand and its
antiderivative (taking C = 0).
49.
Jsin x dx
SO.
Jsin 4x cos x dx
51.
Jsin 3x sin 6x dx
52.
Jsec 4 f dx
5
0
J
tan 3(2x) sec 5 (2x) dx
30. i7T/3 tan 5x sec6x dx
0
dx
sin 2x
sin 40 d()
21.
6
cos 7 () cos 5() d()
tan 2x dx
sec 2x
5
47.
sin x dx
6
4
53. Find the average value of the function f(x)
the interval [- 1r, 1r].
=
sin 2x cos 3x on
\
S4. Evaluate f sin x cos x dx by four methods: (a) the substitution
u = cos x, (b) the substitution u = sin x, (c) the identity
1111
Find the area of the region bounded by the given curves.
SS. y = sin x,
y = sin 3x,
S6. y = sin x,
y
x = 0,
= 2 sin 2x,
x = Tr/2
~ S7-S8
1111 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.
S7.
Jof27r cos 3x dx
2
58. So sin 27rX
COS 57rX
65-67
dx
1111
Prove the formula, where m and n are positive integers.
6S. s:'lT sin mx cos nx dx
=
0
66. J'lT sin mx sin nx dx
=
if m =F- n
{O
7r if m = n
- ?T
59-62 1111 Find the volume obtained by rotating the region bounded
by the given curves about the specified axis.
67. J'lT cos mx cos nx dx
=
- ?T
S9. y = sin x, x = Tr/2, x
=
Tr,
y
= 0;
489
where t is the time in seconds. Voltmeters read the RMS (rootmean-square) voltage, which is the square root of the average
value of [E(t)] 2 over one cycle.
(a) Calculate the RMS voltage of household current.
(b) Many electric stoves require an RMS voltage of 220 V.
Find the corresponding amplitude A needed for the voltage
E(t) = A sin(1207rt).
x = Tr/2
x = 0,
II
64. Household electricity is supplied in the form of alternating
current that vanes from 155 v to -155 v with a frequency
of 60 cycles per second (Hz). The voltage is thus given by
the equation
E(t) = 155 sin(1207rt)
sin 2x = 2 sin x cos x, and (d) integration by parts. Explain the
different appearances of the answers.
SS-S6
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
{O7r
if m =F- n
if m = n
about the x-axis
2
60. y = tan x, y = 0, x = 0, x = Tr/4; about the x-axis
61. y = cosx, y = 0, x = 0, x = Tr/2;
62. y = cos x, y = 0, x
=
0, x = 7r/ 2;
68. A finite Fourier series is given by the sum
abouty = -1
about y
N
J(x)
1
=
=
~ an sin nx
n= L
= at
v(t) = sin wt cos wt, Find its position functions
j(O) = 0.
l!ll 7.3
=
+ a2 sin 2x + · · · + aN sin Nx
Show that the mth coefficient am is given by the formula
63. A particle moves on a straight line with velocity function
2
sin x
f(t) if
am
= -1
7r
J'lT J(x)
s~n
mx dx
- ?T
Trigonometric Substitution
I
In finding the area of a circle or an ellipse, an integral of the form ,Ja 2 - x 2 dx arises,
where a > 0. If it wete x)a 2 - x 2 dx, the substitution u = a 2 - x 2 would be effective
but, as it stands, ,Ja 2 - x 2 dx is more difficult. If we change the variable from x to eby
the SUbstitution X = a Sin 8, then the identity 1 - Sin 2 8 = COS 2 8 allOWS US tO get rid Of the
root sign because
I
I
)a 2
-
x2
=
)a 2
-
a 2 sin 2 8
=
)a 2 (1- sin 2 8)
=
)a 2 cos 2 8 = ajcos ej
Notice the difference between the substitution u = a 2 - x 2 (in which the new variable is
a function of the old one) and the substitution x = a sin e(the old variable is a function of
the new one).
In general we can make a substitution of the form x = g(t) by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that g has an inverse function; that is, g is one-to-one. In this case, if we replace u by x and x by t in the Substitution
Rule (Equation 5.5.4), we obtain
Jf(x) dx Jf(g(t)){j'(t) dt
=
This kind of substitution is called inverse substitution.
490
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
We can make the inverse substitution x = a sin e provided that it defines a one-to-one
function. This can be accomplished by restricting e to lie in the interval [- 7TI 2, 7TI 2].
In the following table we list trigonometric substitutions that are effective for the given
radical expressions because of the specified trigonometric identities. In each case the restriction on e is imposed to ensure that the function that defines the substitution is one-to-one.
(These are the same intervals used in Appendix Din defining the inverse functions.)
Table of Trigonometric Substitutions
Expression
.Ja2 -
Substitution
7T
7T
2
2
7T
7T
2
2
x2
x =a sin 8,
--~ 8~-
+ xz
x =a tan 8,
--< 8<-
~
x =a sec 8,
0
.Ja2
EXAMPLE 1 Evaluate
f
~
7T
8< -
2
or
Identity
1 - sin 2 8 = cos 2 8
0
7T ~
1
37T
8< -
+ tan 2 8 = sec 2 8
sec 2 8 - 1
2
=
tan 2 8
~
x 2 dx.
SOLUTION Let x = 3 sin e, where -7r12 ~ e ~ 7TI2. Then dx = 3 cos e de and
.J9 -
x2 =
.J9 -
9 sin 2e =
(Note that cos__e ~ 0 because -7r12 ~
gives
.J9 cos 2e
= 31 cos
e1 =
3 cos
e
e ~ 7TI2.) Thus, the Inverse Substitution Rule
= f cos e 3 cos e de
f~
x2
dx
9 sin2e
2
cos e
--de=
f sin 2e f coeede
= f (csc 2e- 1) de
3
=
=-cote-
~ 9-x 2
e+ c
Since this is an indefinite integral, we must return to the original variable x. This can be
done either by using trigonometric identities to express cot e in terms of sin e = xl3 or
by drawing a diagram, as in Figure 1, where e is interpreted as an angle of a right triangle.
Since sin e = xl 3, we label the opposite side and the hypotenuse as having lengths x and 3.
x 2 , so we
Then the Pythagorean Theorem gives the length of the adjacent side as
can simply read the value of cot e from the figure:
.J9 -
FIGURE 1
sin 8 =
cote=
3X
X
(Although e > 0 in the diagram, this expression for cote is valid even when e < 0.)
Since sin e = xl3, we have e = sin- 1(xl3) and so
2 dx = f -'-../_9_-_x_
x
2
.J9- x2 -sin- ~(~)
x
3
+C
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
491
EXAMPLE 2 Find the area enclosed by the ellipse
x2
2
-+L
a2
b2 =
1
SOLUTION Solving the equation of the ellipse for y, we get
y
y2
b2
(O,b )
1
x2
a2 _ x2
a2
a2
y
or
=
b
±-)a2 - x 2
a
(a, O)
x
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
FIGURE 2
x2
2
-+ y
a2
b2
= 1
-)a2 - x 2
a
=
-41 A
and so
=
i a -b
o a
v1a 2
o~x~a
-
x 2 dx
To evaluate this integral we substitute X = a sin e. Then dx =a cos e de. To change the
limits of integration we note that when X = 0, sin e = 0, SO e = 0; when X = a,
sine= 1, so e = 'TT/2. Also
) a 2 - x 2 = ) a 2 - a 2 sin 2 e = )a 2 cos 2 e = aJ cos ej =a cos e
since 0 ~
e ~ 7TI 2. Therefore
A = 4 -b i a .Ja 2
a o
-
x 2 dx = 4 -b iTr/2 a cos e · a cos e de
a o
[Tr/2
[Tr/2 1
2
= 4ab Jo cos e de= 4ab Jo 2 (1
= 2ab [ e
=
+ 21 sin 2e]Tr/2
0
=
+ cos 2e) de
2 + 0- 0 )
2ab ( 7T
7Tab
We have shown that the area of an ellipse with semiaxes a and b is 7Tab. In particular,
taking a = b = r, we have proved the famous formula that the area of a circle with
tadius r is 7Tr 2.
NOTE c 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 F.md
J x2)x12 + 4 dx.
SOLUTION Let x = 2 tan e, -'TT/2
<
e
)X2+4 = ) 4(tan 2 e +
<
'TT/2. Then dx = 2 sec 2 e de and
1) =
) 4 sec 2 e = 2J sec eJ = 2 sec
Thus, we have
f x )x
----=dx= = 2
2
+4 -
f 4 tan2 sece. e2 sec e - _!_4 f _se_c_e
tan e
2
2
de
-
2
de
e
492 II
CHAPTER 7 TECHNIQUES OF INTEGRATION
To evaluate this trigonometric integral we put everything in terms of sin 8 and cos 8:
sec8
tan 2 8
1
cos 2 8
cos 8 • sin 2 8
=
Therefore, making the substitution u
f x\/ xdx +
~X
2
sin 8, we have
=
1
4
2
J cos 8
4
=
__!_ (-__!_) + c =
sin 2 8 d
-
X
f
EXAMPLE 4 Find
x
+
4
+
1
4 sin 8
- --
+
c
c
2
dx
~
~
2...;x-
--;}2
.Jx + 4/x and so
We use Figure 3 to determine that esc 8 =
tan0=2
4
=
u
esc 8
J du
1
8
=
4
FIGURE 3
cos 8
sin 2 8
=
,fx' + 4
-
+C
4x
4
f~
dx.
+4
X
SOLUTION It would be possible to use the trigonometric substitution x = 2 tan 8 here (as in
Example 3). But the direct substitution u = x 2 + 4 is simpler, because then du = 2x dx
and
[
x
1 -. Ix 2 +
4 dx
=
1 J du
2 JU
JU + C = JX2+4 + C
=
NOTE " 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 S Evaluate
f .Jx
dx
2
_
a 2 , where a
> 0.
= a sec 8, where 0 < 0 < 7T/2 or 7T < 8 < 37T/2. Then
a sec 8 tan 8 d 8 and
SOLUTION l We let x
dx
=
.Jx 2 - a 2 = .Ja 2 (sec 2 8- 1)
)a 2tan 28
=
altan 81 =a tan 8
=
Therefore
f )x
dx
-a 2
--;===-=
~
2
=
J a sec 8 tan 8 d8
atan8
J sec 8 d 8 =
The triangle in Figure 4 gives tan 8 = )x 2
a
FIGURE 4
sec 0 = .:£
a
f ---;:::=dx==)x
a
2 -
=
2
ln
-
~~ +
a
=
ln I x
+
+ tan 81 ~ C
ln I sec 8
a 2/a, so we have
.Jx2 a
.Jx 2 -
a21+ C
a2 1
-
ln a
+C
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
Writing C 1
dx
f
1__ ?
_?
=
ln Ix
+ -Jx 2
SOLUTION 2 For x > 0 the hyperbolic substitution x
identity cosh 2 y - sinh 2 y = 1, we have
)x 2
=
=
+
a2 j
-
C1
a cosh t can also be used. Using the
a 2 = -ja 2 (cosh 2 t - 1) = )a 2 sinh 2 t =a sinh t
-
a sinh t dt, we obtain
fJ
dx
--r====-
Since cosh t
493
C - ln a, we have
=
rn
Since dx
1111
=
x2
x/a, we have t
=
J a sinh t dt = Jdt = t + c
-
a2
=
cosh- 1(x/a) and
a sinh t
J NJx_ a' ~cosh-'(:) + C
[I]
Although Formulas 1 and 2 look quite different, they are actually equivalent by
Formula 3.9.4.
NOTE o 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
i
3J3/ 2
o
x3
(
4x 2 + 9
) 3; 2
th.
+ 9) 312 = ()4x 2 + 9 ) 3 so trigonometric substitution
is appropriate. Although )4x + 9 is not quite one of the expressions in the table of
SOLUTION First we note that (4x 2
2
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 = ~ tan (},
which gives dx = ~ sec 2 8 d(} and
)4x 2 + 9 = )9 tan 2 8 + 9 = 3 sec (}
When x = 0, tan (} = 0, so (} = 0; when x = 3-/3/2, tan (} = -/3, so (} = 'TT/3.
1
3../3/ 2
o
3
x
- - 2- - . ,3 dx
(4x + 9) 12
=
=
=
J TT/3
27
8 tan
3
27 sec 3 (}
o
3 JTT/ 3
16
3 (}
o
-2
3
tan 8
sec (}
3 J TT/ 3
- - d ( } = 16
1. [TT/3 1 - cos
16
Jo
sec 2 (} d(}
cos
2 (}
2
(}
o
3
sin
cos
(}
--d
(}
2
(}
.
sm (} d(}
Now we substitute u = cos (}so that du = -sin(} d8. When (} = 0, u = 1; when
(} = 'TT/3 , u =
i.
494 I
CHAPTER 7 TECHNIQUES OF INTEGRATION
Therefore
f 3J3/2
Jo
3
x
3
(4x2+9) 3/2 dx= -T6
I
J
1 1/ 2
ft [ u + -;;
=
EXAMPLE 7 Evaluate
2
1- u
3 f 1/ 2
u 2 du=T6Jr (1-u -2 )du
Jl / 2
I
=
ft [G + 2) -
(1 + 1)] =
fz
f . . /3- 2xx
_ dx .
x
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 - 2x - x 2 = 3 - (x 2 + 2x)
=
=
4- (x + If
This suggests that we make the substitution u = x
f )3- 2x- x 2 dx
X
We now substitute u = 2 sin (),giving du =
II II Figure 5 shows the graphs of the integrand
in Example 7 and its indefinite integral (with
C = 0). Which is which?
3 + 1 - (x 2 + 2x + 1)
x
f --;::.====dx
)3 - 2x - x
2
=
+ 1. Then du =
f
U-
dx and x = u - 1, so
1
.J4"=U2 du
2 cos () d() and .J4"=U2 =
=
J 2 sin () 2 cos ()
.
1
2 cos (),so
2 cos() d()
3
-4
2
=
J (2 sin () -
=
-2 cos () - () +
=
-.J4"=U2- sin- 1 (~)
=
-
-5
FIGURE S
lllJ 7.3
2.
3.
JX ~
dx;
X -9
2
Jx
3
.)9 - x 2 dx;
x3
c
+C
)3 - 2x - x 2 - sin- 1( -X+2
1)
+C
Exercises
1-3 1111 Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle.
1.
1) d()
f .Jx2 + 9
dx .,
x
x
x=
=
=
3 sec ()
3 sin ()
4-30
1111
Evaluate the integral.
4 [ 2J3
x3
• Jo .)16 - x 2 dx
1
f .Jt2=1 dt
1
7. f x2~ dx
s.
3 tan()
2
../2 t 3
dx
9.
J .Jx2 + 16
6. Jo2 x 3.Jx2 + 4 dx
f
10. f
8.
.
.Jx2 -
a2
dx
X
t5
r-:-
dt
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION
13.
JJ1 - 4x 2dx
J Jx x2 - 9 dx
3
15.
J (a 2 -
17.
J Jx 2x- 7 dx
11.
19.
x2
x2 )3/2 dx
JJ 1:
x 2 dx
f ~dt
25- t
1
22. Jo
f J5 + 4x- x 2 dx
1
25. f
dx'
J 9x 2 + 6x- 8
24
ffi
p
0
Q
x2
26.
J J4X=X2
4x- x dx
J (x 2 + dx
2x + 2) 2
28
J (5 -
Jx J1=X4 dx
30. i ?T/2
dx
4x - x 2)512
R
X
36. Evaluate the integral
dx
f x4 Jx 2- 2
JX2+l dx
-
f Jt 2 - dt6t + 13
•
·
y
dx
x 2J 16x 2 - 9
J [(ax) 2 ~ b 2]3/2
20.
23.
29.
J
18.
0
•
J u~
du
•
16
•
21. r /3 x 3J4=9x2
4 - 9x 2 dx
27
equation x 2 + y 2 = r 2. Then A is the sum of the area of the
triangle POQ and the area of the region PQR in the figure.]
1
12. Jo xJx 2 + 4 dx
14
495
Graph the integrand and its indefinite integral on the same
screen and check that your answer is reasonable.
ffi
37. Use a graph to approximate the roots of the equation
x 2J4 - x 2 = 2 - x. Then approximate the area bounded by
the curve y = x 2J4 - x 2 and the line y = 2 - x.
38. A charged rod of length L produces an electric field at point
P(a, b) given by
o
cos t
dt
J1 + sin 2t
31. (a) Use trigonometric substitution to show that
dx
f ~ fx 2 +a ~
=
ln(x
+ Jx 2 + a 2) + C
E(P)
f
L- a
=
- a
A.b
47Tt:o ( X 2 + b 2)3/2
dx
where A. is the charge density per unit length on the rod and e 0
is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field E(P).
y
• P (a, b)
(b) Use the hyperbolic substitution x
a sinh t to show that
=
0
f ~I
dx
r 2
+
_
n2 -
. _1 ( x )
smh
--;;
+
L
X
C
These formulas are connected by Formula 3.9.3.
39. Find the area of the crescent-shaped region (called a lune)
bounded by arcs of circles with radii r and R. (See the figure.)
32. Evaluate
x2
f (x 2 + a 2)3/2 dx
(a) by trigonometric substitution.
(b) by the hyperbolic substitution x
33. Find the average value of f(x)
=
=
a sinh t.
~/x, 1 ::s.: x ::s.: 7.
34. Find the area of the region bounded by the hyperbola
9x 2 - 4y 2 = 36 and the line x = 3.
~ r 2 0 for the area of a sector of a circle
with radius rand central angle 0. [Hint: Assume 0 < 0 < 1r/2
and place the center of the circle at the origin so it has the
35. Prove the formula A
=
40. A water storage tank has the shape of a cylinder with diameter
10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the
total capacity is being used?
41. A torus is generated by rotating the circle x 2 + (y - R) 2 = r 2
about the x-axis. Find the volume enclosed by the torus.
CHAPTER 7 TECHNIQUES OF INTEGRATION
496
7.4 Integration of Rational Functions b~ Partial Fractions
111·
In this section we show how to integrate any rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called partial fractions, that we already know
how to integrate. To illustrate the method, observe that by taking the fractions 2/(x - 1)
and 1/(x + 2) to a common denominator we obtain
2
1
X -
2(x + 2) - (x - 1)
(x- 1)(x + 2)
1
x+2
x+5
x + x- 2
2
If we now reverse the procedure, we see how to integrate the function on the right side of
this equation:
dx- f (-2- - _1_) dx
f _X x+ +_5_
X- 2
X- 1
X+ 2
2
= 2lnlx- 11 -lnlx + 21 + C
To see how the method of partial fractions works in general, let's consider a rational
function
f(x)
=
P(x)
Q(x)
where P and Q are polynomials. It's possible to express f as a sum of simpler fractions
provided that the degree of Pis less than the degree of Q. Such a rational function is called
proper. Recall that if
P(x) = anxn + an- !Xn- i + ... + aix + ao
where an =I= 0, then the degree of P is n and we write deg(P) = n.
If f is improper, that is, deg(P) ~ deg(Q), then we must take the preliminary step
of dividing Q into P (by long division) until a remainder R(x) is obtained such that
deg(R) < deg(Q). The division statement is
A~
rn
~~
M=QW=~+QW
where Sand Rare also polynomials.
As the following example illustrates, sometimes this preliminary step is all that is
required.
EXAMPLE 1 Find
x2 + x
X-
I)x
3
f
x3
+x
X -
1
dx.
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
+2
+X
x3- xz
x 2 +x
x 2 -x
2x
2x-2
2
f
3
_x_+_x dx =
x-1
f
2
(x 2 + x + 2 + - - ) dx
x-1
x3
x2
=-+-+2x+ 2lnlx-1I+C
3
2
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
II 497
The next step is to factor the denominator Q(x) as far as possible. It can be shown that
any polynomial Q can be factored as a product of linear factors (of the form ax + b) and
irreducible quadratic factors (of the form ax 2 + bx + c, where b 2 - 4ac < 0). For instance, if Q(x) = x 4 - 16, we could factor it as
Q(x) = (x 2
-
4)(x 2
+ 4) = (x - 2)(x + 2)(x 2 + 4)
The third step is to express the proper rational function R(x)/ Q(x) (from Equation 1) as
a sum of partial fractions of the form
A
(ax + b)i
or
Ax+ B
(ax + bx + c) j
2
A theorem in algebra guarantees that it is always possible to do this. We explain the details
for the four cases that occur.
CASE I " The denominator Q(x ) is a product of distinct linear factors.
This means that we can write
Q(x)
=
(a1x + bi)(azx + hz) · · · (a kx + 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 A 1, A 2 , ••• , Ak such that
R(x)
Q(x)
[!]
Az
- -A1- - +
+ ... + - -Ak- alx + b1
azx + hz
akx + bk
These constants can be determined as in the following example.
EXAMPLE 2 Evaluate
x z + 2x- 1 dx.
2x
J 2x3 + 3x z -
SOLUTION Since the degree of the numerator is less than the degree of the denominator,
we don't need to divide. We factor the denominator as
2x 3
+ 3x 2
-
2x
=
x(2x 2 + 3x - 2)
=
x(2x - 1)(x + 2)
Since the denominator has three distinct linear factors, the partial fraction decomposition
of the integrand (2) has the form
[I]
II II Another method for finding A, B, and C
is given in the note after this example.
x 2 + 2x- 1
A
B
C
------=-+
+-x(2x - 1)(x + 2)
x
2x - 1
x +2
To determine the values of A, B, and C, we multiply both sides of this equation by the
product of the denominators, x(2x - 1)(x + 2), obtaining
[!]
x 2 + 2x - 1
=
A(2x - 1)(x + 2) + Bx(x + 2) + Cx(2x - 1)
Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get
[I]
x 2 + 2x - 1 = (2A + B + 2C)x 2 + (3A + 2B - C)x - 2A
498
CHAPTER 7 TECHNIQUES OF INTEGRATION
II II Figure 1 shows the graphs of the integrand
in Example 2 and its indefinite integral (with
K = 0). Which is which?
2
-3 1== /
~<;:
1
1
);> c
The polynomials in Equation 5 are identical, so their coefficients must be equal. The
coefficient of x 2 on the right side, 2A + B + 2C, must equal the coefficient of x 2 on the
left side-namely, 1. Likewise, the coefficients of x are equal and the constant terms are
equal. This gives the following system of equations for A, B, and C:
2A + B + 2C
=
1
3A + 2B- C
=
2
=
-1
I3
-2A
-2
FIGURE 1
f
1111 We could check our work by taking the terms
to a common denominator and adding them.
t B = t, and C = --fo, and so
Solving, we get A=
=
f (2 --_;- + S
=
~ ln Ix I +
2
x + 2x - 1
2x 3 + 3x 2 - 2x dx
1 1
1
1
2x - 1 -
fa ln l2x
1
1
lO x + 2
- 1I -
)
dx
fa ln Ix + 21
In integrating the middle term we have made the mental substitution u
gives du = 2 dx and dx = du/2.
=
+K
2x - 1, which
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 - 2A = - 1, or A = ~. Likewise,
x =~gives 5B/4 = i and x = -2 gives lOC = -1, soB=~ and C = -io. (You may object
that Equation 3 is not valid for x = 0, tor -2, so why should Equation 4 be valid for those
values? In fact, Equation 4 is true for all values of x, even x = 0, ~, and -2. See Exercise 67
for the reason.)
EXAMPLE 3 Find
fx
2
dx
, , where a =P 0.
-a
SOLUTION The method of partial fractions gives
1
1
A
(x - a)(x + a)
x - a
----- = --
x2- a2
+ -Bx + a
and therefore
A(x + a) + B(x - a)
=
1
Using the method of the preceding note, we put x = a in this equation and get
A(2a) = 1, so A = 1/(2a). If we put x = -a, we get B( -2a) = 1, soB = -1/(2a).
Thus
__
1 )dx
x- a
x +a
f _ d x _ 2a_I J(-1
x2
-
n2
-
1
(ln Ix - a I - ln Ix + a I) + C
2a
= -
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
Since ln x - ln y
1111
499
ln(x/y), we can write the integral as
=
f __dx__
[!]
x2
-
=
a2
1-x_-_a I + C
_1_ ln
2a
x +a
See Exercises 53-54 for ways of using Formula 6.
CASE II
a
Q(x) is a product of linear factors, some of which are repeated.
Suppose the first linear factor (a1x + bt) is repeated r times; that is, (a1x + btY occurs in
the factorization of Q(x). Then instead of the single term At/(atx + bt) in Equation 2, we
would use
A2
Ar
- -At- +
+ ... + - -alx + b1
(a1 x + b1?
(a1x + b1Y
[I]
By way of illustration, we could write
x3
+
x
-
1
A
B
x
C
D
x - 1
(x - 1)Z
----=
- + -2+ - - +
3
x 2(x - 1)
x
E
+-------,--3
(x - 1)
but we prefer to work out in detail a simpler example.
f x x3 - 2xx2 -+ x4x++ 1 dx.
4
EXAMPLE 4 Find
2
-
1
SO lUTION The first step is to divide. The result of long division is
x 4 - 2x 2 + 4 x + 1
x3 - x2 - x + 1
= x + 1 + x3 _
4x
x2 - x
The second step is to factor the denominator Q(x) = x 3 - x 2
we know that x - 1 is a factor and we obtain
x3 - x2
-
x + 1 = (x - 1)(x 2
=
-
+
1
x + 1. Since Q(l) = 0,
-
1) = (x - 1)(x - 1)(x
+ 1)
(x - 1) 2 (x + 1)
Since the linear factor x - 1 occurs twice, the partial fraction decomposition is
~
A
B
(x - 1)Z(x + 1)
x - 1
(x - 1)Z
-----=--+
C
Multiplying by the least common denominator, (x - 1)Z(x
[!]
II II Another method for finding the coefficients:
Put x = 1 in (8): B = 2.
Put x = - 1: C = - 1.
Put x = 0: A = B + C = 1.
4x
= A(x -
1)(x
= (A +
C)x 2
+
1)
+ B(x +
1)
+-x + 1
+ 1), we get
+ C(x -
1) 2
+ (B - 2C)x + (-A + B + C)
Now we equate coefficients:
A
+ C= 0
B- 2C
=
4
-A+ B + C
=
0
500
CHAPTER 7 TECHNIQUES OF INTEGRATION
Solving, we obtain A= 1, B = 2, and C = -1, so
dx = f [x + 1 + _1_ +
2
f _x_4x _-_2_x_x 22_+_4_x_+_1
- x + 1
x - 1
(x - 1) 2
3
-
_1_] dx
x + 1
x2
2
= - + x + In Ix - 1 I - -x--- - ln Ix + 1 I + K
1
2
2
2 + l n Ix-1 I +K
=x- + x - - 2
x-1
x+1
CASE Ill c Q(x) contains irreducible quadratic factors, none of which is repeated.
If Q(x) has the factor ax 2 + bx + c, where b 2 - 4ac < 0, then, in addition to the partial
fractions in Equations 2 and 7, the expression for R(x)/Q(x) will have a term of the form
Ax+ B
2
ax + bx + c
[!]
where A and B are constants to be determined. For instance, the function given by
f(x) = x/[(x - 2)(x 2 + 1)(x 2 + 4)] has a partial fraction decomposition of the form
x
(x - 2)(x 2 + 1)(x 2 + 4)
A
x - 2
-------- = --
+ Bx2 +
C
x + 1
Dx + E
+-x2 + 4
The term given in (9) can be integrated by completing the square and using the formula
~
)+ C
f x 2dx+a 2 =_!_tan-~(
a
~ c _j
ern
EXAMPLE S Evaluate
f
2x 2 - x
x
3
+4
+ 4x
dx.
SOLUTION Since x 3 + 4x = x(x 2 + 4) can't be factored further, we write
2x 2 - x + 4
x(x2 + 4)
A
x
- - - - - =-
Bx + C
+ -2- x +4
Multiplying by x(x 2 + 4), we have
2x 2 - x + 4 = A(x 2 + 4) + (Bx + C)x
= (A + B)x 2 + Cx + 4A
Equating coefficients, we obtain
A+B=2
C= -1
4A
=
4
Thus A = 1, B = 1, and C = -1 and so
dx = f (_!_ + _x_-_1 ) dx
f _2x_x2 -_x_+_4
+ 4x
x
x2 + 4
3
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
501
In order to integrate the second term we split it into two parts:
dx
f _x_-_I_
x2 + 4
J
=
x
dx x2 + 4
J __
I _ dx
2
x + 4
We make the substitution u = x 2 + 4 in the first of these integrals so that du = 2x dx.
We evaluate the second integral by means of Formula I 0 with a = 2:
dx
f _2_x_2x(x_-_x_+_4_
+ 4)
=
2
x
=
EXAMPLE 6 Evaluate
f
J_!._ dx + J
J__
I _ dx
2
x
dx x2 + 4
x + 4
ln lxl + t ln(x 2 + 4)- t tan- 1(x/2) + K
4x 2- 3x + 2
4x2 - 4x + 3 dx.
SOlUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain
4x 2 - 3x + 2
4x2 - 4x + 3
----- =
I
X -
I
+ -2 --4x - 4x + 3
Notice that the quadratic 4x 2 - 4x + 3 is irreducible because its discriminant is
b 2 - 4ac = -32 < 0. This means it can't be factored, so we don't need to use the
partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x 2
-
4x + 3
=
(2x - I)Z + 2
This suggests that we make the substitution u = 2x - I. Then, du = 2 dx and
x = (u + I)/2, so
J-:-;-:-=-!-:-:-~-dx ~
f (1+
= x +
=
1
2
4x'
~ ~: + 3) dx
f tcu + I) -
x + 4If
u 2 +2
U
u 2 +2
I
If
du = x + 4
du - 41 f
u - I
du
u 2 +2
I
du
u 2 +2
=
·
tI a n
x +
81 ln(u 2 + 2) - -I · - )2
4
=
x + s ln(4x 2 - 4x + 3)1
4
-1(-)2u )
+
c
I ( I)
)2 tan- 1 2x)2
+ C
NOTE " Example 6 illustrates the general procedure for integrating a partial fraction of
the form
Ax+ B
ax
2
+ bx + c
where b 2 - 4ac < 0
502
CHAPTER 7 TECHNIQUES OF INTEGRATION
We complete the square in the denominator and then make a substitution that brings the
integral into the form
+ D du
f Cuu +a
2
C
=
2
fu
u
du + D
+a 2
2
fu
1
2
+a
~
du
Then the first integral is a logarithm and the second is expressed in terms of tan _,.
CASE IV
c
Q(x ) contains a repeated irreducible quadratic factor.
If Q(x) has the factor (ax 2 + bx + cY, where b 2 - 4ac < 0, then instead of the single
partial fraction (9), the sum
A, x + B,
--,--------+
ax2 + bx + c
[TI]
A2x + B2
(ax 2 + bx + cf
Arx + Br
+ ... + ----,-----(ax 2 + bx + cY
occurs in the partial fraction decomposition of R(x)/ Q(x). Each of the terms in (11) can be
integrated by first completing the square.
II II It would be extremely tedious to work out by
hand the numerical values of the coefficients in
Example 7. Most computer algebra systems,
however, can find the numerical values very
quickly. For instance, the Maple command
convert(f, parfrac, x)
or the Mathematica command
Apart[ f)
B=
L c=
x3 + x2 + 1
x (x - 1)(x 2 + x + l)(x 2
+
D = - 1,
A
B
x - 1
=-+--+
E =lf, F =-L G=H =t
1 =-t
+ 1) 3
SOLUTION
x3 + x2 + 1
x (x - l)(x 2 + x + 1)(x 2
gives the following values:
A= - 1,
EXAMPLE 7 Write out the form of the partial fraction decomposition of the function
X
] =~
1) 3
Cx + D
x2 + x + 1
f 1 - x(+ +2x1
2
EXAMPLE 8 Evaluate
X X
-
+
x3
Gx + H
(x 2 + 1f
+
Ix + J
+-(x 2 + 1) 3
dx.
)2
2
Ex + F
x2 + 1
SOLUTION The form of the partial fraction decomposition is
1 - x + 2x 2
x (x
Multiplying by x(x 2
-x 3
+ 2x 2
-
x
2
-
x3
+ 1f
A
=-+
X
Bx + C
x2 + 1
Dx + E
+-2
(x
+ 1?
+ 1?, we have
+ 1 = A(x 2 +
1f
+ (Bx + C) x (x 2 + 1) + (Dx + E)x
= A(x 4 + 2x 2 + 1) + B(x 4 + x 2 ) + C(x 3 + x) + Dx 2 + Ex
=(A
+ B)x 4 + Cx 3 + (2A + B + D)x 2 + (C + E)x
+A
If we equate coefficients, we get the system
A+B=O
2A + B + D
C= - 1
which has the solution A
=
1, B
=
- 1, C
=
=
-1 , D
2
=
C+E
1, and E
=
-1
= 0. Thus
A=1
SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
f
2
3
_1_-_x_+_2x_ _- _x_ dx
x (x 2 + 1) 2
f (_!_ _
=
1111 In the second and fourth terms we made the
mental substitution u = x 2 + 1.
dx
Jx
f ~-
ln Ix I -
=
503
x + 1 +
x
) dx
x2 + 1
(x 2 + 1)2
x
=
1111
2
x
+ 1 dx-
~ ln(x 2 +
Jx dx+ 1 + J (x x +dx1)2
2
1) - tan- 1x -
2
2
1
(
X
2
. + K
+1
We note that sometimes partial fractions can be avoided when integrating a rational function. For instance, although the integral
x
2
+1
f x (x2 + 3) dx
could be evaluated by the method of Case III, it's much easier to observe that if
u = x (x 2 + 3) = x 3 + 3x, then du = (3x 2 + 3) dx and so
f
1111
x2 + 1 ·
x (x 2 + 3) dx
=
t ln Ix 3 +
3x I + C
Rationalizing Substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions. In particular, when an integrand contains an expression of the form
v9{X5, then the substitution u = v9{X5 may be effective. Other instances appear in the
exercises.
rx-+4
f x dx.
rx-+4. Then u
EXAMPLE 9 Evaluate
SOLUTION Let u =
2
= x + 4, sox = u 2
-
4 and dx = 2u du.
Therefore
f JX+4 d x-- J u 2 u_
X
4
2u du
u2
2 J u 2 - 4 du
=
~ 2 f (1 + u' ~ 4 ) du
We can evaluate this integral either by factoring u 2
partial fractions or by using Formula 6 with a = 2:
-
4 as (u - 2)(u + 2) and using
J ../x x+ 4 dx ~ 2 Jdu + 8 J u' du- 4
=
2u
u _2l + C
+ 8 · - -1 l n I -
=
2)
x + 4 + 2ln rx+4-2
JX+4 + 2 l +c
2·2
u+2
l
504
CHAPTER 7 TECHNIQUES OF INTEGRATION
1111
l!ll 7.4 Exercises
1-6 1111 Write out the form of the partial fraction decomposition
of the function (as in Example 7). Do not determine the numerical
values of the coefficients.
1. (a) (x +
JX
•
+
(b) x 3 + 2x 2
1)
+x
lS.
1
X -
37.
x2
2
3x - 4
(b) (x - 1)(x2 + x + 1)
x3
2
f ___!!!_-x
X
(b) x 3 + x
4. (a) x 2 + 4x + 3
2x + 1
(b) (x + 1) 3(x 2 + 4) 2
x4
5. (a) x4 _ 1
t4 + t2 + 1
(b) (t2 + 1)(t2 + 4)2
x4
6. (a) (x 3 + x)(x 2 - x + 3)
1
(b) x 6 - x 3
1
-dx
1
32. (I
3-
4
f (x
2
34.
-
7.
39.
9J
x-9
• (x + 5)(x 11.
3
f
2
J-dr
r+4
f
l
1
X-
i
ax
dx
- bx
14.
f (x + a)(x + b) dx
16.
i
1
X -
i
2
2x + 3
(X+ 1)2 dx
4y 2
2
-
7y - 12
f y(y + 2)(y I
3)
dy
19.
J (x+5 )!(x-1 . dx
21.
J 5x 2 + 3x -
2 dx
x 3 + 2x 2
18.
f , ., dx
10
2S. f (x - 1)(x2 + 9) dx
x 3 + x 2 + 2x + 1
27.
f (x 2 + 1)(x 2 + 2)
X+ 4
29.
f + 2x + 5 dx
dx
x
2
l
x3
0
X
Jx
2
2
4
i
44. J
dx
dx
l
f :Vx 2x + 1 dx
45.
J JX ~
,r dx
3
1/ 3
[Hint: Substitute u
r
1, r dx
4
2x
e
dx
+ 3e x + 2
=
1
----;-c dx
0
43.
1
f x _ ~lx + 2 dx
1
+
yX
JX
x2 + x
dx
$.]
[Hmt:
. Substitute
.
u ·-= 12C.
....;x. ]
2x
48
f sin2x + sin
d
COS X
•
x
x
J (x-
·x·2
22.
f ______!!!_ 1
J ln(x 2 -
SO.
x + 2) dx
J x tan- x dx
1
Sl. Use a graph of f(x) = 1/(x 2 - 2x - 3) to decide whether
j(x) dx is positive or negative. Use the· graph to 'give a rough
J;
estimate of the value of the integral and then use partial
fractions to find the exact _value.
~
S2. Graph both y
x3
., dx
-
2
-
2(
x
2
2
-
4
1/(x 3 - 2x 2) and an antiderivative on the
S3-S4 1111 Evaluate the integral by completing the square and using
Formula 6.
53.
f
dx
X2 -
S4.
LX
J
2x + 1
dx
4x 2 + 12x- 7
)
x
2
=
same screen.
f
x
2x- 1
28. f (x-1 ) x +1 dx
2x + + 1
30.
f + 5x + 4 dx
x3
49.
., dx
x +6
3
dx
x + 3x
x2
1111 Use integration by parts, together with the techniques of
this section, to evaluate the integral.
~
s 2(s -
J (x
49-SO
+3 2x - 1 dx
20.
26.
+1
1
42.
X -
f
47. f e
4x- 10
dx
-X- 6
-X
24.
2
J x(x2 + f' dx
1
X
x2
23.
+ 3x +
o x2
38.
JX
46.
(t + 4)1(t - 1) dt
12.
1S.
17.
10.
1
fx
0
dx
2)
3
2x + 5x d
x 4 + 5x 2 + 4 x
(I
Jo
40.
(16
- 2 --dx
13.
l
8.
3
36
·
1
dx
f xJX+l
41. J9
r2
J x ~ 6 dx
X
39-48 1111 Make a substitution to express the integrand as a rational
function and then evaluate the integral.
Evaluate the integral.
1111
dx
13
3
J-x--dx
+1
x4
x-3
dx
+ 2x + 4) 2
3
7-38
x
Jo x2 + 4x +
x 2 + 2x
x 3 + 3x 2 + 4 dx
33 J s
2x
3)(3x
X- 1
2. (a) x 3 + x
3. (a) x 2 +
31.
SS. The German mathematician Karl Weierstrass (1815-1897)
noticed that the substitution t
=
tan(x/2) will convert any
SECTION 7.5 STRATEGY FOR INTEGRATION :1 505
rational function of sin x and cos x into an ordinary rational
function of t.
(a) If t = tan(x/2), -7r < x < 1r, sketch a right triangle or use
trigonometric identities to show that
(b) Show that
1 - t2
1+t2
=
and
sinx
2t
=
1+t2
58.
J
+
7712
7T/ 3
1
sin x - cos x
dx
[ill] 65. (a) Use a computer algebra system to find the partial fraction
1
s~
f 3 sin x -
59.
f 2 sin x + sin 2x dx
4 cos x
~
1
60-61 1111 Find the area of the region under the given curve from
a to b. "
60. y
· 61. y
=
X2 -
1 + 8'
6x
+
X
1
= --,
X -
1
a
=
a= 5, b
2, b
=
=
decomposition of the function
f (X )
4x 3 - 27x 2 + 5x - 32
30x5 - 13x 4 + 50x 3 - 286x 2 - 299x- 70
= --------------------------------
f
(b) Use part (a) to find f(x) dx (by hand) and compare with
the result of using the CAS to integrate f directly.
Comment on any discrepancy.
[ill] 66. (a) Find the partial fraction decomposition of the function
f(x)
=
100x 6
-
12x 5 - 7x 3 - 13x 2 + 8
80x 5 + 116x 4 - 80x 3 + 41x 2
-
20x
+4
(b) Use part (a) to find f f(x) 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 off f(x) dx.
10
3
67. Suppose that F, G, and Q are polynomials and
62. Find tQ.e volume of the resulting solid if the region under the
curve y = 1/(x 2 + 3x + 2) from x = 0 to x
about (a) the x-axis and (b) they-axis.
=
1 is rotated
63. One method of slowing the growth of an insect population
without using pesticides is to introduce into the population a
number of sterile males that mate with fertile females but produce no offspring. If P represents the number of female insects
in a population, S the number of sterile males introduced each
generation, and r the population's natural growth rate, then the
1111
rrn
dP
subtracting the same quantity. Use this factorization to evaluate
4
+ 1) dx.
S6-S9 1111 Use the substitution in Exercise 55 to transform the integrand into a rational function of t and then evaluate the integral.
dx
5 sin x
1)P - S]
f 1/(x
2
d x = - -2 dt
1+t
J3 _
P+S
f P[(r -
64. Factor x 4 + 1 as a difference of squares by first adding and
(c) Show that
56.
t =
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.)
cos(f) ~ ~ and sin(f) ~ '
cosx
female population is related to time t by
F(x)
Q(x)
--
G(x)
Q(x)
--
for all x except when Q(x) = 0. Prove that F(x)
all x. [Hint: Use continuity.]
68. Iff is a quadratic function such that f(O)
f(x)
f x2(x + 1)3
=
=
G(x) for
1 and
d
x
is a rational function, find the value of f'(O).
7.5 Strateg~ 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 7.1, and partial
fractions in Exercises 7 .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.
506 ... CHAPTER 7 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.
f xndx = -nxn+1+-1
3. f exdx =e x
1.
(n =I= -1)
2.
J ~ dx =
4.
ax
f axdx =Ina
ln I x I
S.
Jsin x dx =
-cos x
6.
J cos x dx =
sin x
7.
Jsec x dx =
tan x
8.
Jcsc x dx =
-cot x
9.
Jsec x tan x dx =
2
sec x
2
10.
J esc x cot x dx =
-esc x
11.
Jsec x dx =
ln I sec x + tan x I
12.
J esc x dx =
ln I esc x - cot x I
13.
J tan x dx =
ln I sec x I
14.
J cot x dx =
ln I sin x I
15.
Jsinh x dx =cosh x
16.
Jcosh x dx =sinh x
17.
f
a
18.
a
f .Ja2dx- x2 -_ sm. _ 1 (~)
*19
f
~x -al
*20.
•
dx
x2
x
2
_ -tan
1
_1( -x)
-
+
n2
dx
1
=-ln
-a
2a
x + a
-
a
2
f
dx
.Jx2 ± a 2
=
ln x + .Jx 2 ± a 2
I
1
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 four-step 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:
J.JX (1 + .JX) dx = J(.JX + x) dx
tan ()
Jsin ()
d() =
- cos d()
f sec ()
2
- -2
cos ()
=
()
Jsin () cos () d() = ~ Jsin 2() d()
SECTION 7.5 STRATEGY FOR INTEGRATION
I
J(sin x + cos xf dx J(sin x + 2 sin x cos x + cos x) dx
2
=
2
J(1 + 2 sin x cos x) dx
=
2. Look for an Obvious Substitution Try to find some function u = g(x) in the integrand whose differential du = g'(x) dx also occurs, apart from a constant factor.
For instance, in the integral
f
x2
~
1 dx
we notice that if u = x 2 - 1, then du = 2x dx. Therefore, we use the substitution u = x 2 - 1 instead of the method of partial fractions.
3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the
solution, then we take a look at the form of the integrand f(x).
(a) Trigonometric functions. If f(x) is a product of powers of sin x and cos x ,
of tan x and sec x, or of cot x and esc x, then we use the substitutions recommended in Section 7 .2.
(b) Rational functions. Iff is a rational function, we use the procedure of Section 7.4 involving partial fractions.
(c) Integration by parts. If f(x) is a product of a power of x (or a polynomial)
and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 7.1. If you look at the functions in Exercises 7.1, 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 .)±x2 ± a 2 occurs, we use a trigonometric substitution according to the
table in Section 7.3.
(ii) If -\1ax + b occurs, we use the rationalizing substitution u = -\1ax + b.
More generally, this sometimes works for -i9{X5.
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 7.1, we see that it works on tan- 1x, sin- 1x, and ln x,
and these are all inverse functions.
(c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the
denominator or using trigonometric identities) may be useful in transforming
the integral into an easier form. These manipulations may be more substantial
than in Step 1 and may involve some ingenuity. Here is an example:
f 1 - dx
----=
COS X
=
f
1
1-
•
COS X
f 1 +sm.
COS X
2
x
1
1
+ COS X
+ COS X
dx =
f(
dx
csc 2x
=
f 1+
1-
COS X )
+ -.
-2-
sm x
COS X
2
COS X
dx
dx
507
508
1111
CHAPTER 7 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, tan 2x sec x dx is a challenging integral, but if we make use of the
identity tan 2x = sec 2x - 1, we can write
I
Jtan x sec x dx Jsec x dx - Jsec x dx
2
3
=
I
and if sec 3x dx has previously been evaluated (see Example 8 in Section 7.2),
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
f
tan 3x
- 3- dx
COS X
In Step 1 we rewrite the integral:
f
3
tan x dx =
- -3
COS X
Jtan x sec x dx
3
3
I
The integral is now of the form tanmx secnx dx with m odd, so we can use the advice in
Section 7 .2.
Alternatively, if in Step 1 we had written
f
3
tan x
--dx=
cos 3x
J-sin- x- - d
1 x = J-sin-xdx
3
3
3
3
cos 6x
cos x cos x
then we could have continued as follows with the substitution u
f
3
sin x
- -6
dx
=
J1 -
2
cos x
6
COS X
sin x dx
=
COS X
=
EXAMPLE 2
J1 -
f
uz - 1
.
TJ
6
r.
=
u
cos x :
2
(-
du)
U
du
=
J u(
4
-
u - 6 ) du
f e JX dx
According to Step 3(d)(ii) we substitute u =
fx,. Then x = u 2 , so dx = 2u du and
f e JX dx 2 f ue udu
=
The integrand is now a product of u and the transcendental function e u so it can be integrated by parts.
SECTION 7.5 STRATEGY FOR INTEGRATION
EXAMPLE 3
fx
3
1111
509
x5 + 1
dx
- 3x 2 - lOx
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 7.4, remembering that the first step is to divide.
EXAMPLE 4
f x-ylnx
~
Here Step 2 is all that is needed. We substitute u
du = dx/ x, which occurs in the integral.
EXAMPLE S
=
ln x because its differential is
f \j!h
~ dx
Although the rationalizing substitution
~
\j!h
u=
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 ~. we have
-x
+X
f~
--dx=
1
1111
J 1-x
)1=X2
dx
1
=
f )1=X2 dx -
=
sin- 1x
f
X
)1=X2
-
dx
+ )1=X2 + C
Can We Integrate Rll 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 ex 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 (xa), 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
J
f(x)
=
~
x2
-
1
x + 2x- 1
3
+ ln(cosh x) - xe sin2x
is an elementary function.
Iff is an elementary function, then f' is an elementary
function but f(x) dx need not
2
be an elementary function. Consider f(x) = e x • Since f is continuous, its integral exists,
and if we define the function F by
J
F(x)
=
Joexet 2dt
510
CHAPTER 7 TECHNIQUES OF INTEGRATION
1111
then we know from Part 1 of the Fundamental Theorem of Calculus that
F'(x)
ex
=
2
2
Thus, f(x) = ex 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 f ex dx in terms of the functions we know. (In Chapter 11, however, we will see how
2
to express f ex dx as an infinite series.) The same can be said of the following integrals:
ex
J-;dx
Jsin(x
Jy0+ldx
f -lndx x
2
)
dx
1
Jcos(ex) dx
sin x
f --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.
Uli 1.S
1-80
Exercises
Evaluate the integral.
1111
J sin x + sec x dx
1•
3.
5.
r
0
tanx
earctan y
J
-1-2 dy
6.
Jx esc x
4
8.
r
10.
f
+y
-1
2.
7.
r
r ln rdr
9.
f
X-
x 2 - 4x
1
+
5
dx
•
x
+x +
dx
30.
f
f J2X=1
2x + 3 dx
32.
33.
JJ3 -
2x- x 2 dx
34• J?T/2 1 + 4 cotx dx
f
38.
i?T/ 4 tan 50 sec 30 dO
39.
J 1- x 2 +x J1=X2 dx
40.
f
J J1 + In x dx
37.
xln x
J1=X2
dx
4-cotx
cos 20 tan 20 dO
14.
0
?T/ 4
Jsin 4x cos 3x dx
J (1 - dxx2)3/ 2
J1=X2
1
36.
13.
0
2
ix - 4xl dx
_ x dx
1
2
2
f~
35.
X
3w- 1 dw
w+2
1
x 8 sin x dx
i?T/4
0
0
1
J4y 2 - 4y- 3
[ 7.
Jx sin 2x dx
18.
J- -e2r
4 - t dt
41.
J0 tan20 dO
42.
Jx 2 tan- x dx
19.
f ex+exdx
20.
Je¥ dx
43.
f ex~dx
44.
JJ[+eX dx
21.
Jt e- 21 dt
22.
Jx sin- x dx
45.
Jx 5e -x dx
46.
f -1 +d
ex
x
23.
J
24.
Jln(x 2 -
47.
J
48.
Jx
3
1
0
(1
+ Jx) 8 dx
1
+e
1
1) dx
dx
31.
Jsin x cos(cos x) dx
16. i J2/2
3x2- 2
3
x - 2x- 8
r
12.
dx
J
29.
cot x dx
X2
•
Jsin J(ii dt
2
4
26
28.
0
x- 1
dx
o x - 4x- 5
dx
Jcot x ln(sin x) dx
Jsin 30 cos 50 dO
X
3x2- 2
2
x - 2x- 8
27.
11.
15. f /2
J
3
tan 0 dO
4.f n
dx
3 - x4
(t-2t 3 )2 dt
fl
25
3
x +a dx
x 2 +a 2
1
1 - ex
4
x
-a
4
dx
dy
SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS
49
·
f xJ4~ + 1 dx
SO.
J x2~
dx
4x + 1
1
52
J
51. f
53.
2
xJ4x + 1
dx
54.
J (x +
56.
J~
xln x
dx
cdx
58.
Jx 2 ln(1
eX dx
60.
f~dx
X+ X
dx
62.
J (x +
64.
J .,./ 3
66.
r
Jx 2 sinh mx dx
SS. f
1
x+4+4JX+l
57.
J x~x +
59.
J eX3
61.
J X Jo +
63.
J JXeJX dx
1
•
dx
x(x 4 + 1)
dx
x4
65.
sin x) 2 dx
+ x) dx
67.
J~
x+1+
l lJ 7.6
JXdx
x
1)'o dx
~(tan x) dx
.,.;4 sm x cos x
2
u3 + 1
du
u3 - u2
Jt
I
68.
dt
f1+
1
2ex- e-x
511
dx
e2x
69.
f1
71.
J
73.
f (x -
75.
J sin x sin 2x sin 3x dx
76.
J (x
77.
f1+
JXx3 dx
18•
J sec x
79.
J x sin x cos x dx
80
J
x3
16
arctan,Jt
J 3
1111
+e x dx
x
+ 4x 2 + 3
x4
f 1 +Jtiff dt
74, f
dx
72.
1
2)(x2 + 4) dx
dx
e -e -x
X
2
2
-
bx) sin 2x dx
cos 2x dx
sin x + sec x
•
sin x cos x
sin 4 x + cos 4x dx
2
2
81. The functions y = ex andy = x 2 e x don't have elementary
antiderivatives, but y
2
(2x 2 + 1)e x dx.
f
=
2
2
(2x + 1)e x does. Evaluate
Integration Using Tables and Computer Hlgebra Sqstems
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 ex or the other functions described at the end of Section 7.5.
1111
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 y-axis. Find the volume of the resulting solid.
SOLUTION Using the method of cylindrical shells, we see that the volume is
V
J27Tx arctan x dx
1
=
0
512 ,1'1 CHAPTER 1 TECHNIQUES OF INTEGRATION
II II 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:
f
u tan- 1u du
u2 + 1
u
tan- 1u - - + C
2
2
=
Thus, the volume is
x + 1 tan- x - -x]
1
2
V
27T J[I x tan- 1x dx
=
=
1
27T [
2
10
= 1r[(x + 1) tan- x 2
1
7T[2( 7T/4) - 1]
=
=
x]~
47T 2
2
0
1
= 7T(2 tan- 1 - 1)
7T
-
x2
EXAMPLE 2 Use the Table of Integrals to find
J .J5 -
dx.
4x 2
SOlUTION If we look at the section of the table entitled Forms involving )a 2
that the closest entry is number 34:
f .Ja
2
u 2, we see
2
-r=:=u==::c du
2
-
u
-
=
2
.!!_ .Ja 2
2
-
u2
-
+ _a_ sin - t (.!!_) + C
a
2
This is not exactly what we have, but we will be able to use it if we first make the substitution u = 2x:
f
2
x
.J5 - 4x2 dx =
Then we use Formula 34 with a 2
=
J
(u/2)
2
du
~
5 (so a=
1
2
=
8
J
u
2
~ du
.j5):
J )5- 4x' dx- g J ~ du = 8 -2~ + 2sin-' ,/5) + C
x2
_ _!_
=
-
u2
8x.J5 -
1
u
(
5 sin - 1 (
+ l6
4x 2
EXAMPLE 3 Use the Table of Integrals to find
Jx
3
5
u
2x) + c
J5
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:
Jx
85.
f u"
COS
=
sin x dx
=
-
x 3 cos x
+ 3 Jx 2 cos x dx
f
u du
u" sin u - n
3
J u"-
1
sin u du
We now need to evaluate x 2 cos x dx. We can use the reduction formula in entry 85
with n = 2, followed by entry 82:
Jx
2
cos x dx
Jx sin x dx
=
x 2 sin x - 2
=
x 2 sin x- 2(sin x- x cos x)
+K
SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS
1111
513
Combining these calculations, we get
I x sin x dx = - x cos x + 3x 2sin x + 6x cos x 3
3
6 sin x
+C
where C = 3K.
EXAMPLE 4 Use the Table of Integrals to find
I x)x
2
+ 2x + 4 dx.
SOLUTION Since the table gives forms involving .ja 2 + x 2, .ja 2 - x 2, and
not ..}ax 2 + bx + c, we first complete the square:
x 2 + 2x + 4
If we make the substitution u = x
pattern ..}a 2 + u 2:
I x .j
x2
=
(x
.Jx 2 -
a 2, but
+ 1) 2 + 3
+ 1 (sox= u - 1), the integrand will involve the
+ 2x + 4 dx
=
=
I (u - 1) .fu2+3 du
I u .fu2+3 du - I .fu2+3 du
The first integral is evaluated using the substitution t = u 2 + 3:
I
21.
J.Ja
2
+ u 2 du
a
=
~ .Ja 2 + u 2
u.fu2+3 du
=~I .jt dt = ~ . h3/2 = i(u2 + 3)3/2
For the second integral we use Formula 21 with a=
-/3:
I .fu2+3 du = -2 .fu2+3 + ~ ln(u + .fu2+3)
2
u
+ -ln(u + .Ja 2 + u 2 ) + C
2
Thus
I x .Jx
2
+ 2x + 4 dx
= i(x 2 + 2x
1111
+ 4) 312 -
~
..)x 2 + 2x + 4
2
-
~ ln(x + 1 + ..)x 2 + 2x + 4) + C
Computer Rlgebra Sqstems
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 answer.
To begin, let's see what happens when we ask a machine to integrate the relatively
simple function y = 1/(3x - 2). Using the substitution u = 3x - 2, an easy calculation
by hand gives
I 3x ~ 2 dx
=
i In l3x -
21 + C
514
I
CHAPTER 7 TECHNIQUES OF INTEGRATION
whereas Derive, Mathematica, and Maple all return the answer
~ ln(3x - 2)
The first thing to notice is that computer algebra systems omit the constant of integration. In other words, they produce a particular antiderivative, not the most general one.
Therefore, when making use of a machine integration, we might have to add a constant.
Second, the absolute value signs are omitted in the machine answer. That is fine if our
problem is concerned only with values of x greater than ~ . But if we are interested in other
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 S Use a computer algebra system to find
Jx)x
2
+ 2x + 4 dx.
SOLUTION Maple responds with the answer
~ (x 2 + 2x + 4) 312 - ~(2x + 2).jx 2 + 2x + 4 - ~2 arcsinh J3
(1 + x)
3
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
1111 This is Equation 3.9.3.
=
ln(x +
.jXT+l)
Thus
. J3 + x)
arcsmh3(1
=
)3
ln [ 3(1
+ x)1
+ .j3(1 + x? + 1 ]
=
ln
~
=
ln
J3
1
[1 + x + .J(1 + x) 2 + 3]
+ ln( x + 1 + .Jx 2 + 2x + 4)
The resulting extra term - ~ ln( 1/)3) can be absorbed into the constant of integration.
Mathematica gives the answer
2
X x ) .Jx 2 + 2x + 4 - -3 arc sinh ( 1 +X)
5
(- + - + 6
6
3
2
J3
Mathematica combined the first two terms of Example 4 (and the Maple result) into a
single term by factoring.
Derive gives the answer
i.Jx 2
+ 2x + 4 (2x 2 + x + 5) - ~ ln(.Jx 2 + 2x + 4 + x + 1)
The first term is like the first term in the Mathematica answer, and the second term is
identical to the last term in Example 4.
EXAMPLE 6 Use a CAS to evaluate
Jx(x
2
+ 5) 8dx.
SOLUTION Maple and Mathematica give the same answer:
fg xt8 +
~
x16
+ 50xt4 + 17;o x12 + 4375xl0 + 21875x8 + 2183750 x6 + 156250x4 + 3902625 x2
SECTION 7.6 INTEGRATION USING TABlES AND COMPUTER AlGEBRA SYSTEMS UR 515
It's clear that both systems must have expanded (x 2 + 5) 8 by the Binomial Theorem and
then integrated each term.
If we integrate by hand instead, using the substitution u = x 2 + 5, we get
Jx(x
1111 Derive and the Tl-89/92 also give this answer.
2
+ 5) 8 dx
=
fg(x 2 + 5) 9 + C
For most purposes, this is a more convenient form of the answer.
EXAMPLE 7 Use a CAS to find
J sin 5x cos 2x dx.
SOLUTION In Example 2 in Section 7.2 we found that
Jsin x cos x dx = -i cos x + ~ cos x -
m
2
5
9cos7x +
5
3
C
Derive and Maple report the answer
1·4
- 7 Slll X
3
COS X -
4·2
35
3
Slll X COS X -
8
3
105 COS X
whereas Mathematica produces
10
5
I
-64 cos x - 192 cos 3x
3
320
cos 5x -
I
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
0
+
5
EXAMPLE 8 If f(x) = x
FIGURE 1
Graph F for 0
~
x
~
+ 60 sin 4x cos 5x, find the antiderivative F off such that F(O) = 0.
5. Where does F have extreme values and inflection points?
SOLUTION The antiderivative off produced by Maple is
w
~----------------~
F(x) = ~ x 2
0 IL < I
/
\ ' ·V
\
!
~ sin 3x cos 6x - ~sin x cos 6x
+ ~ cos 4x sin x + ~ cos 2x sin x + ~sin x
and we note that F(O) = 0. This expression could probably be simplified, but there's no
need to do so because a computer algebra system can graph this version ofF as easily as
any other version. A graph of F is shown in Figure 1. To locate the extreme values of F,
we graph its derivative F' =fin Figure 2 and observe that F has a local maximum
when x = 2.3 and a local minimum when x = 2.5. The graph ofF"= f' in Figure 2
shows that F has inflection points when x = 0.7, 1.3, 1.8, 2.4, 3.3, and 3.9.
I5
-7
FIGURE 2
llll 7.6
-
Exercises
1-4 1111 Use the indicated entry in the Table of Integrals on the
Reference Pages to evaluate the integral.
2·
J J3 3x- 2x dx;
entry 55
3.
J sec
3
4.
Je
sin 3(} d(} ;
26
(
7TX) dx;
entry 71
entry 98
516 J CHAPTER 7 TECHNIQUES OF INTEGRATION
5-30 1111 Use the Table of Integrals on the Reference Pages to
evaluate the integral.
5.
1.
9
•
11.
Sol
2x COS -
IX
dx
6.
J e- xcos 4x dx
3
8.
3
rl
Jx 2~dx
36.
Jx (1 + x
37.
Jsin 3x cos 2x dx
38.
Jtan x
39.
JxJ1+2xdx
40.
Jsin x dx
41.
Jtan x dx
42.
Jx #+1 dx
J
csc 3(x/2) dx
12. fa',. x 2cos 3x dx
t 2e- t dt
15.
J exsech(ex) dx
dz
17. f yJ6 + 4y- 4y 2 dy
14.
J sin- JX dx
16.
Jx sin(x
18.
J
2
2
3 4
)
dx
sec 4x dx
2
2
xs
~
)
cos(3x 2 ) dx
J
22.
23.
J sec 5x dx
24.
25.
J J4 + (ln x)2 dx
26 fi x 4 e- x dx
2
0
x4dx
f2x~ dx
x J4x 2 - x 4 dx
3
instead. Why do you think it was successful with this form of
the integrand?
Jsin 2x dx
6
[ill] 44. Try to evaluate
J(1 + ln x) J1 + (x ln x)2 dx
28.
Je
30•
J sec 0 tan 0 dO
1
sin(at - 3) dt
2
J9-
2
tan 2 0
31. Find the volume of the solid obtained when the region under
=
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.
[ill] 45-48 1111 Use a CAS to find an antiderivative F off such
that F(O) = 0. Graph f and F and locate approximately the
x-coordinates of the extreme points and inflection points of F.
xJ4 - x 2, 0:::; x:::; 2, is rotated about the
45. f(x)
x2
=
X
tan 2x from 0 to 7T /4 is rotated
about the x-axis. Find the volume of the resulting solid.
32. The region under the curve y
=
33. Verify Formula 53 in the Table of Integrals (a) by differentiation
and (b) by using the substitution t
=
a
dx
If it doesn't return an answer, ask it to try
• Jo
27. f~dx
the curve y
y-axis.
f 2x J4x"-=-I
20. f ex(l + 2ex)
2
5
[ill] 43. Computer algebra systems sometimes need a helping hand
from human beings. Ask your CAS to evaluate
~ dx
J--dx
3- ex
J
5
4
1
dx
Jsin x cos x ln(sin x) dx
ex
29.
35.
dx
J tan3(1/z)
~
21.
dx
2
f x\/4X2 + 9
13.
19.
1
f x2J4x2- 7
[ill] 35-42 1111 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.
+ bu.
34. Verify Formula 31 (a) by differentiation and (b) by substituting
u =a sin 0.
46. f(x)
=
4
1
-
+X
2
+1
xe-x sin x,
4
6
47. f(x)
=
sin x cos x,
48. f(x)
=
x3 - x
- -x6
+1
-5 :::; x:::; 5
0 :::; x :::;
7T
DISCOVERY PROJECT PATIERNS IN INTEGRALS '!~I 517
ill!
DISCOVERY PROJECT
cmJ 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, I
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.
. f (x + 2)(1
1
(iii) f '
(1)
f (x +
. f (x +1 2)2 dx
(1v)
(ii)
-- dx
_, dx
_1_
,
dx
(b) Based on the pattern of your responses in part (a), guess the value of the integral
f
1
(x
+ a)(x + b)
dx
if a =tf 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)
Jsin x cos 2x dx
(ii)
Jsin 3x cos 7x dx
(iii)
Jsin 8x cos 3x dx
(b) Based on the pattern of your responses in part (a), guess the value of the integral
Jsin 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)
JIn x dx
Jx ln x dx
3
(ii)
(v)
Jx ln x dx
Jx ln x dx
(iii)
Jx
2
In x dx
7
(b) Based on the pattern of your responses in part (a), guess the value of
f xninxdx
(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)
f xe x dx
(ii)
f x ex dx
(iv)
J x 4ex dx
(v)
J x 5ex dx
2
(iii)
f x ex dx
3
(b) Based on the pattern of your responses in part (a), guess the value off x 6exdx. 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
f xne x dx
when n is a positive integer.
(d) Use mathematical induction to prove the conjecture you made in part (c).
518
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
llll 7.7
Hpproximate 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 f(x) 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 7 .5). For
example, it is impossible to evaluate the following integrals exactly:
J:
fl
Jo ex dx
rl~dx
2
The second situation arises when the function is determined from a scientific experiment through instrument readings or collected data. There may be no formula for the function (see Example 5).
In both cases we need to find approximate values of definite integrals. We already know
one such method. Recall that the definite integral is defined as a limit of Riemann sums,
so any Riemann sum could be used as an approximation to the integral: If we divide [a, b]
into n subintervals of equal length ~x = (b - a)/n, then we have
y
b
f
a
n
f(x) dx
= i~ j(x'[) ~X
where x! is any point in the ith subinterval [xi- !, xJ If xt is chosen to be the left endpoint
of the interval, then xt = Xi- I and we have
0
I
Xo
XI
X2
X3
X4
X
(a) Left endpoint approximation
b
OJ
f
a
f(x) dx
= Ln =
n
i~ j(Xi- I) ~X
If f(x) ;;:::: 0, then the integral represents an area and (1) represents an approximation of this
area by the rectangles shown in Figure l(a). If we choose x! to be the right endpoint, then
y
x ! = xi and we have
b
rn
0I
Xo
XI
x2
X3
X4
X
(b) Right endpoint approximation
y
fa f(x) dx = Rn =
n
i~ f(xi) ~X
[See Figure l(b).] The approximations Ln 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
the subinterval [xi- 1, xJ Figure l(c) shows the midpoint approximation Mn, which appears
to be better than either Ln or Rn.
Midpoint Rule
J: f(x) dx = Mn
where
0
XI
:x2 :x3 :x4
(c) Midpoint approximation
FIGURE 1
X
and
~x
=
=
~x [f(x,) + j(x2) + · · · + f(xn)]
b- a
n
xi = 4(xi- l +xi) =midpoint of [xi- !, xi]
SECTION 7.7 APPROXIMATE INTEGRATION
1111
519
Another approximation, called the Trapezoidal Rule, results from averaging the approximations in Equations 1 and 2:
n
1f(x) dx = 21 [ i~f(xi-d
~x + i~nf(xi) ~x ]
b
~X
=
2
=
2
~X
[(f(xo)
[f(xo)
=
2~X [ i~n (J(xi- 1) + f(xi)) ]
+ f(x1)) + (J(xl) + f(x2)) + · · · + (j(Xn- 1) + f(xn))]
+ 2f(x1) + 2f(x2) + · · · + 2j(Xn- 1) + f(xn)]
y
Trapezoidal Rule
b
f
~X
= Tn =2- [f(xo) + 2f(xt) +, 2f(x2) + · · · + 2j(Xn- 1) + f(xn)]
f(x) dx
a
where ~x = (b - a)/nand Xi = a
0I
x0
X1
X2
X3
X4
X
+ i ~x.
The reason for the name Trapezoidal Rule can be seen from Figure 2, which illustrates
the case f(x) ~ 0. The area of the trapezoid that lies above the ith subinterval is
FIGURE 2
Trapezoidal approximation
f(x,~ ,)2+ f(x ,)) = ~x [f(x;~J)
L\x (
+ f(x;)]
and if we add the areas of all these trapezoids, we get the right side of the Trapezoidal
Rule.
1
y=~
EXAMPLE 1 Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n
approximate the integral 12 (1/x) dx.
f
=
5 to
SOLUTION
(a) With n = 5, a = 1, and b = 2, we have
zoidal Rule gives
2
2
1
1
X
J-
dx
= T5 =
0.2
-
2
[j(l)
~x
= (2 - 1)/5 = 0.2, and so the Trape-
+ 2/(1.2) + 2/(1.4) + 2/(1.6) + 2/(1.8) + /(2)]
1)
1
2
2
2
2
=0.1 ( - + - + - + - + - + 1
1.2
1.4
1.6
1.8
2
FIGURE 3
= 0.695635
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
J
2
l
1
- dx
X
= ~x [/(1.1) + /(1.3) + /(1.5) + /(1.7) + /(1.9)]
u
1(1
=
2
FIGURE 4
5
1
1
+ 1.3 + 1.5 +
= 0.691908
This approximation is illustrated in Figure 4.
u1
1)
+ 1.9
520 1m CHAPTER 7 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,
1
]2
= lnx = ln2 = 0.693147 ...
J2-dx
1
r
f(x) dx
=
approximation
+ error
1
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
Er = - 0.002488
and
EM = 0.001239
In general, we have
Er
Module 5.1/5.2/7.7 allows you to compare approximati on methods.
. .
Approx1mat10ns to
f1-1 dx
I
X
Corresponding errors
=
s:
f(x) dx - Tn
and
EM =
s:
f( x) dx - Mn
The following tables show the results of calculations similar to those in Example 1, but
for n = 5, 10, and 20 and for the left and right endpoint approximations as well as the
Trapezoidal and Midpoint Rules.
n
L"
R"
T"
M,
5
10
20
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
-0.052488
-0.025624
- 0.012656
0.047512
0.024376
0.012344
-0.002488
-0.000624
- 0.000156
0.001239
0.000312
0.000078
We can make several observations from these tables:
1. In all of the methods we get more accurate approximations when we increase the
value of n. (But very large values of n result in so many arithmetic operations that
we have to beware of accumulated round-off error.)
2. 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.
II II It turns out that these observations are true
in most cases.
3. The Trapezoidal and Midpoint Rules are much more accurate than the endpoint
approximations.
4. 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.
S. 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
SECTION 7.7 APPROXIMATE INTEGRATION
c
A
X;
521
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 (2n) 2 = 4n 2 • The fact that the estimates depend on the size of the
second derivative is not surprising if you look at Figure 5, because f"(x) measures how
much the graph is curved. [Recall that f"(x) measures how fast the slope of y = f(x)
changes.]
D
X;-!
Ill!
rn
Error lnl41 Suppose I f"(x) I ~ K for a ~
in the Trapezoidal and Midpoint Rules, then
X;
c
1Eri~K(b-a)3
IEMI~K(b-a)3
and
12n 2
~ b. If Er and EM are the errors
X
24n 2
Let's apply this error estimate to the Trapezoidal Rule approximation in Example 1. If
f(x) = 1/x, then f'(x) = -1/x 2 and f"(x) = 2/x 3 . Since 1 ~ x ~ 2, we have 1/x ~ 1, so
A
If"( x) I ~ I~3 ~
D
,;;
~3 ~ 2
FIGURE S
Therefore, taking K = 2, a = 1, b = 2, and n = 5 in the error estimate (3), we see that
1111 K can be any number larger than all the
values of I f"(x) 1. but smaller values of K
give better error bounds.
3
I Erl
~
2(2 - 1)
12(5)Z
=
1_
150
_
= 0.006667
Comparing this error estimate of 0.006667 with the actual error of about 0.002488, we see
that it can happen that the actual error is substantially less than the upper bound for the
error given by (3).
EXAMPLE 2 How large should we take n in order to guarantee that the Trapezoidal and
Midpoint Rule approximations for 12 (1/x) dx are accurate to within 0.0001?
f
SOLUTION We saw in the preceding calculation that I f"(x) I ~ 2 for 1 ~ x ~ 2, so we can
take K = 2, a = 1, and b = 2 in (3). Accuracy to within 0.0001 means that the size of
the error should be less than 0.0001. Therefore, we choose n so that
3
2(1) < 0.0001
12n 2
Solving the inequality for n, we get
nz >
1111 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.
1
or
Thus, n
=
41 will ensure the desired accuracy.
522 .... CHAPTER 7 TECHNIQUES OF INTEGRATION
For the same accuracy with the Midpoint Rule we choose n so that
2(1)3
24n 2 < 0.0001
n>
which gives
y
1
Jo.oo1 2
= 29
EXAMPLE 3
(a) Use the Midpoint Rule with n = 10 to approximate the integral f~ ex dx.
(b) Give an upper bound for the error involved in this approximation.
y=
2
SOLUTION
ex2
(a) Since a= 0, b = 1, and n = 10, the Midpoint Rule gives
Jol I ex 2dx = Llx [/(0.05) + /(0.15) + · · · + /(0.85) + /(0.95)]
=
0
1
0 .1[e o.oo25 + e o.o225 + e o.o625 + e o.J225 + e o.2o25
+ e 0.4225 + e o.5625 + e o.n25 + e o.9025J
+ e o.3025
= 1.460393
X
FIGURE 6
Figure 6 illustrates this approximation.
2
2
2
(b) Since f(x) = ex , we have f'(x) = 2xe x and f"(x) = (2 + 4x 2 )ex . Also, since
2
0 : : : ; x ::::::; 1, we have x ::::::; 1 and so
2
0 :::S f"(x) = (2 + 4x 2 )ex :::S 6e
II II Error estimates are upper bounds for
the error. They give theoretical, worst-case
scenarios. The actual error in this case turns
out to be about 0.0023.
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
6e(1)3
e
24(10) 2 = 400 = 0.007
1111
Simpson's Rule
Another rule for approximate integration results from using parabolas instead of straight
line segments to approximate a curve. As before, we divide [a, b] into n subintervals of
equal length h = Llx = (b - a)/n, but this time we assume that n is an even number. Then
on each consecutive pair of intervals we approximate the curve y = f(x) ~ 0 by a parabola
as shown in Figure 7. If Yi = f(xi), then Pi(xi, yi) is the point on the curve lying above Xi.
A typical parabola passes through three consecutive points Pi, Pi+ I, and Pi+2 ·
y
y
PI
0
a=x 0
FIGURE 7
XI
x2
x3
x4
Xs
x6=b
X
-h
FIGURE 8
0
h
X
SECTION 7.7 APPROXIMATE INTEGRATION
1111
523
To simplify our calculations, we first consider the case where xo = -h, x1 = 0, and
X2 = h. (See Figure 8.) We know that the equation of the parabola through Po, P1, and P2
is of the form y = Ax 2 + Bx + C and so the area under the parabola from x = - h to
x =his
J~h (Ax 2 + Bx +
Iill Here we have used Theorem 5.5.7. Notice
that Ax 2 + Cis even and Bx is odd.
C) dx
=
2 foh (Ax
2 + C) dx
3
~ 2[A ~
=
h
2( A 3
+
Cx
J:
3
+ Ch )
=
3h (2Ah + 6C)
2
But, since the parabola passes through Po( -h, yo), P1(0, Y1), and P2(h, y2), we have
Yo= A( -h) 2 + B( -h) + C
=
Ah 2 - Bh + C
Y1 = C
Y2
=
Ah 2 + Bh + C
Yo + 4yl + Y2
and therefore
=
2Ah 2 + 6C
Thus, we can rewrite the area under the parabola as
h
3 (yo + 4yl + y2)
Now, by shifting this parabola horizontally we do not change the area under it. This means
that the area under the parabola through Po, P1 , and P2 from x = xo to x = x2 in Figure 7
is still
h
3 (yo + 4yl + y2)
Similarly, the area under the parabola through P2, P 3 , and P4 from x
=
x2 to x
=
X4 is
h
3 (y2 + 4y3 + Y4)
If we compute the areas under all the parabolas in this manner and add the results, we get
b
fa
f(x) dx
h
h
= -(yo
+ 4yl + Y2) + 3
3
h
(y2 + 4y3 + Y4) + · · · + - (Yn-2 + 4Yn- l + Yn)
3
h
=-(yo + 4yl + 2y2 + 4y3 + 2y4 + · · · + 2Yn-2 + 4Yn-l + Yn)
3
Although we have derived this approximation for the case in which f(x) ~ 0, it is a reasonable approximation for any continuous function f and is called Simpson's Rule after
the English mathematician Thomas Simpson (1710-1761). Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, ... , 4, 2, 4, 1.
524
Ill
CHAPTER 7 TECHNIQUES OF INTEGRATION
Simpson's Rule
1111 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
best-selling calculus textbook. entitled A New
Treatise of Fluxions.
~X
b
f f(x) dx = Sn
=-
3
a
+ 4f(xt) + 2f(x2) + 4f(x3) + · · ·
[f(xo)
+ 2j(Xn- 2) + 4j(Xn-l ) + J(xn)]
where n is even and
~x =
(b - a)/n.
EXAMPLE 4 Use Simpson's Rule with n
=
SOLUTION Putting f(x) = 1/x, n = 10, and
2
10 to approximate
~x
J (1/x) dx.
2
1
= 0.1 in Simpson's Rule, we obtain
1
f -dx = Sw
IX
~X
=
3
[!(1)
+ 4f(l.l) + 2!(1.2) + 4/(1.3) + ... + 2/(1.8) + 4!(1.9) + !(2)]
0.1 ( 1
2
4
2
4
1)
- + 4- + ++++ -2 + -4 + -2 + -4 + 3
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
=-
= 0.693150
Notice that, in Example 4, Simpson's Rule gives us a much better approximation
(S 10 = 0.693150) to the true value of the integral (ln 2 = 0.693147 ... ) than does the
Trapezoidal Rule (Tw = 0.693771) or the Midpoint Rule (M 10 = 0.692835). It turns out
(see Exercise 48) that the approximations in Simpson's Rule are weighted averages of
those in the Trapezoidal and Midpoint Rules:
S2n
=
t Tn + ~ Mn
(Recall that ET and EM usually have opposite signs and IEM I is about half the size of IET1.)
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 y dx, the integral of y with respect to x.
J:
EXAMPLE S Figure 9 shows data traffic on the link from the United States to SWITCH,
the Swiss academic and research network, on February 10, 1998. D(t) is the data
throughput, measured in megabits per second (Mb/s). Use Simpson's Rule to estimate
the total amount of data transmitted on the link up to noon on that day.
D
8
6
4
2
FIGURE 9
0
3
6
9
12
15
18
21
24 t (hours)
SECTION 7.7 APPROX IMATE INTEGRATION
81 525
SOLUTION Because we want the units to be consistent and D(t) is measured in megabits
per second, we convert the units fort from hours to seconds. If we let A(t) be the
amount of data (in megabits) transmitted by timet, where tis measured in seconds, then
A'(t) = D(t). So, by the Net Change Theorem (see Section 5.4), the total amount of data
transmitted by noon (when t = 12 X 60 2 = 43,200) is
A(43,200)
[ 43 200
=
Jo .
D(t) dt
We estimate the values of D(t) at hourly intervals from the graph and compile them in
the table.
t (hours)
t (seconds)
D(t)
t (hours)
t (seconds)
D(t)
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
43200
1
.
A(t) dt = -
0
~t
3
~t
= 3600 to estimate the integral:
[D(O) + 4D(3600) + 2D(7200) + · · · + 4D(39,600) + D(43,200)]
3600
= -3- [3.2 + 4(2.7) + 2(1.9) + 4(1.7) + 2(1.3) + 4(1.0)
+ 2(1.1) + 4(1.3) + 2(2.8) + 4(5.7) + 2(7.1) + 4(7.7) + 7.9]
=
143,880
Thus, the total amount of data transmitted up to noon is about 144,000 megabits, or
144 gigabits.
In Exercises 27 and 28 you are asked to demonstrate, in particular cases, that the error
in Simpson's Rule decreases by a factor of about 16 when n is doubled. That is consistent
with the appearance of n 4 in the denominator of the following error estimate for Simpson's
Rule. It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules, but
it uses the fourth derivative of f.
[!]
Error Bound for Simpson's Rule Suppose that I f (4l(x) I ~ K for a ~ x ~ b.
If Es is the error involved in using Simpson's Rule, then
IEs i~K(b-a)s
180n 4
526
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
EXAMPLE 6 How large should we taken in order to guarantee that the Simpson's Rule
approximation for
SOLUTION If f(x)
Jl2(1/x) dx is accurate to within 0.0001?
1/x, then j <4)(x)
=
24/x 5 • Since x ~ 1, we have 1/x ~ 1 and so
=
IJ <'l(x) I ~I~
1111 Many calculators and computer algebra systems have a built-in algorithm that computes an
approximation of a definite integral. Some of
these machines use Simpson's Rule; others use
more sophisticated techniques such as adaptive
numerical integration. This means that if a function fluctuates much more on a certain part of
the interval than it does elsewhere, then that
part gets divided into more subintervals. This
strategy reduces the number of calculations
required to achieve a prescribed accuracy.
Therefore, we can take K
choose n so that
=
1.;
24
24 in (4). Thus, for an error less than 0.0001 we should
5
24(1)
180n 4
< 0.0001
This gives
= 6.04
or
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
(a) Use Simpson's Rule with n = 10 to approximate the integral f~ ex dx.
(b) Estimate the error involved in this approximation.
2
SOLUTION
(a) If n = 10, then dx = 0.1 and Simpson's Rule gives
1111 Figure 10 illustrates the calculation in
Example 7. Notice that the parabolic arcs are
2
so close to the graph of y = ex that they are
practically indistinguishable from it.
i
t
0
dx
ex dx = - [f(O)
2
3
=
0.1 [eo
3
y ~
+ 4 eo.o1 +
2eo.04
+ 4 eo.o9 + 2eo.16 + 4 e o.zs + 2eo.36
+ 4 eoA9 + 2eo.64 + 4eo.s1 + e 1]
= 1.462681
I
y=ex2 J
(b) The fourth derivative of f(x)
/
-
+ 4f(0.1) + 2f(0.2) + · · · + 2f(0.8) + 4f(0.9) + f(l)]
FI GURE 10
f (4)(x)
ex is
=
(12
+ 48x 2 +
16x 4)ex
2
//
f-- _ _ ,..
0
2
=
and so, since 0
"""
~
x
~
1, we have
0 ~ f( 4)(x) ~ (12
1
X
+ 48 +
16)e 1
=
76e
Therefore, putting K = 76e, a = 0, b = 1, and n = 10 in (4), we see that the error is at
most
76e(l)s
180{10)4 = 0.000115
(Compare this with Example 3.) Thus, correct to three decimal places, we have
fl 2
Jo ex dx
= 1.463
SECTION 7.7 APPROXIMATE INTEGRATION IIU 527
mJ 7.7
Exercises
1. Let I= f~f(x) dx, where f is the function whose graph is
shown.
(a) Use the graph to find L 2, R 2, and M 2.
(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 Ln, Rn, Mn, Tn, and I in
increasing order.
y
3
/
f l/ v
2
11/
0
1---"
1/
v
v
(Round your answers to six decimal places.) Compare your results
to the actual value to determine the error in each approximation.
S. Jo7T x 2 sin x dx,
2
3
4
X
were used to estimate f02 f(x) 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
f02 f(x) dx lie?
y
Jo' e-.JX dx,
n
6
=
1111 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.
J
.v'1
9.
J
lnx
--dx,
1 +X
11.
2. The left, right, Trapezoidal, and Midpoint Rule approximations
6.
7-18
2
0
2
I
1
n= 8
+ x 2 dx, n
Jof 1/2 sin(e 12 )
1
13. fe' 1xdx,
1S.
dt,
i
3
0
8
10
n= 8
12.
1
+y
n
r
0
sin(x 2 ) dx,' n = 4
1
+
dt
t2 + t4 '
n=6
J.)1 + JX dx, n=8
14. JJX sin x dx, n=8
4
0
0
8
16.
X
1
•
f /2
0
4
n=4
- - -5 dy,
8.
10
=
Jscos x dx, n=
I
17.
n
=
=
6
18.
r
4
f
2
ln(x 3 + 2) dx,
ex
-dx,
X
n
=
n = 10
10
19. (a) Find the approximations T 10 and M 10 for the integral
s;
e-x2 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?
0
2
X
~ 3. Estimate f~ cos(x 2 ) 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?
~ 4. Draw the graph of f(x) = sin(x 2/2) in the viewing rectangle
[0, 1] by [0, 0.5] and let I= f~ f(x) dx.
(a) Use the graph to decide whether L 2, R 2, M 2, and T2 underestimate or overestimate I.
(b) For any value of n, list the numbers Ln, Rn, Mn, Tn, and I in
increasing order.
(c) Compute Ls, Rs, Ms, and T5 • From the graph, which do you
think gives the best estimate of I?
S-6 1111 Use (a) the Midpoint Rule and (b) Simpson's Rule to
approximate the given integral with the specified value of n.
20. (a) Find the approximations T8 and M8 for f~ cos(x 2 ) dx.
(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?
21. (a) Find the approximations T10 and S 10 for f~ e xdx 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?
22. How large should n be to guarantee that the Simpson's Rule
2
approximation to f~ ex dx is accurate to within 0.00001?
[ill] 23. The trouble with the error estimates is that it is often very difficult to compute four derivatives and obtain a good upper bound
K for I j<4l(x) I by hand. But computer algebra systems have no
528
CHAPTER 7 TECHNIQUES OF INTEGRATION
IIJl
problem computing j <4l 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= J:.,. f(x) dx, where
j(x)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
= e cosx.
Use a graph to get a good upper bound for I f"(x) I·
Use M 10 to approximate/.
Use part (a) to estimate the error in part (b).
Use the built-in numerical integration capability of your
CAS to approximate/.
How does the actual error compare with the error estimate
in part (c)?
Use a graph to get a good upper bound for I j <4l(x) 1.
Use Sw to approximate I .
Use part (f) to estimate the error in part (g).
How does the actual error compare with the error estimate
in part (h)?
How large should n be to guarantee that the size of the
error in using S,. is less than 0.0001?
[N] 24. Repeat Exercise 23 for the integral
f
1
.j4 - x 3 dx.
25-26 1111 Find the approximations L,., R,., T,., and M,. for n = 4, 8,
and 16. Then compute the corresponding errors EL, ER, Er, 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, whathappens to the errors when
n is doubled?
25.
1
Jo
Use Simpson's Rule to estimate the area of the pool.
3
J
2
26.
x dx
0
e-" dx
31. (a) Use the Midpoint Rule and the given data to estimate the
value of the integral sg· 2 f(x) dx.
X
f(x)
X
f(x)
0.0
0.4
0.8
1.2
1.6
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
(b) If it is known that - 4 ::::; f"(x) ::::; 1 for all x, estimate the
error involved in the approximation in part (a).
32. A radar gun was used to record the speed of a runner during
the first 5 seconds of a race (see the table). Use Simpson's
Rule to estimate the distance the runner covered during those
5 seconds.
t (s)
v (m/s)
t (s)
v (m/s)
0
0.5
1.0
0
4.67
7.34
8.86
9.73
L0.22
3.0
3.5
4.0
4.5
5.0
10.51
10.67
10.76
L0.81
10.81
1.~
27-28
Find the approximations T,., M,., and S,. for n = 6 and 12.
Then compute the corresponding errors Er, 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.
r
1111
f
28.
.[;dx
2.0
2.5
33. The graph of the acceleration a(t) of a car measured in ft/s 2 is
shown. Use Simpson's Rule to estimate the increase in the
velocity of the car during the 6-second time interval.
a
1
xe -" dx
/ ........
v
12
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.
4
0
I
\
J
1\
\
\
v
v
v
2
4
6 t
y
/v
I
1 /
0
'
v
v
1
2
34. Water leaked from a tank at a rate of r( t) liters per hour, where
the graph of r is as shown. Use Simpson's Rule to estimate the
total amount of water that leaked out during the first six hours.
'\
\
""'""'
3
4
r
4 "'
X
30. The widths (in meters) of a kidney-shaped swimming pool
were measured at 2-meter intervals as indicated in the figure.
2
""""'"'
!"-- t-~---.
.....
0
2
4
6 t
SECTION 7.7 APPROXIMATE INTEGRATION
35. The table (supplied by San Diego Gas and Electric) gives the
power consumption in megawatts in San Diego County from
midnight to 6:00A.M. on December 8, 1999. Use Simpson's
Rule to estimate the energy used during that time period. (Use
the fact that power is the derivative of energy.)
t
p
t
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
1111
529
= {/1 + x 3, y = 0, x = 0,
and x = 2 is rotated about the x-axis. Use Simpson's Rule with
n = 10 to estimate the volume of the resulting solid.
39. The region bounded by the curves y
[ill] 40. The figure shows a pendulum with length L that makes a maximum angle 00 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
(L
dx
[ 7T/ 2
T= 4\jg Jo ~sin 2x
where k = sin( k00 ) and g .is the acceleration due to gravity.
If L = 1m and 00 = 42°, 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 8:00A.M. Dis 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
0.4
I
I Oo
I
I
I
'
I
'-..... ......... ___ _LI ___ ,..,.. ,.,./
!.---'
7
(
h
/
V' r---..v
I
41. The intensity of light with wavelength A traveling through
a diffraction grating with N slits at an angle 0 is given by
l(O) = N 2 sin 2k/k 2 , where k = ( 1rNd sin 0)/ A and dis the
distance between adjacent slits. A helium-neon laser with
wavelength A = 632.8 X 10- 9 m is emitting a narrow band of
light, given by -10- 6 < 0 < 10- 6, through a grating with
10,000 slits spaced 10- 4 m apart. Use the Midpoint Rule with
n = 10 to estimate the total light intensity f~~~-· 1(0) dO emerging from the grating.
IV
0
4
2
8 t (hours)
6
37. If the region shown in the figure is rotated about the y-axis to
form a solid, use Simpson's Rule with n
volume of the solid.
=
8 to estimate the
42. Use the Trapezoidal Rule with n
=
10 to approximate
f020 cos( 1rx) dx. Compare your result to the actual value. Can
you explain the discrepancy?
43. Sketch the graph of a continuous function on [0, 2] for which
y,
4
2
v
/.,......
~
/
1\
\
1/
0
2
4
the Trapezoidal Rule with n
Midpoint Rule.
6
=
2 is more accurate than the
44. Sketch the graph of a continuous function on [0, 2] for which
the right endpoint approximation with n
than Simpson's Rule.
=
2 is more accurate
lQ X
8
45. Iff is a positive function and f"( x) < 0 for a : : : ; x : : : ; b, show
that
38. The table shows values of a force function f(x) where x is
measured in meters and f( x) in newtons. Use Simpson's Rule
to estimate the work done by the force in moving an object a
distance of 18 m.
Tn <
s:f(x) dx < Mn
46. Show that if f is a polynomial of degree 3 or lower, then
Simpson's Rule gives the exact value of f(x) dx.
J:
X
0
3
6
9
12
15
18
47. Show thad(Tn + Mn) = Tzn·
f(x )
9.8
9.1
8.5
8.0
7.7
7.5
7.4
48. Show that Tn + ~ Mn = Szn·
-
-
'
i
530
CHAPTER 7 TECHNIQUES OF INTEGRATION
1111
llll 7.8
Improper Integra Is
J:
In defining a definite integral f(x) dx we dealt with a function f defined on a finite interval [a , b] and we assumed that f does not have an infinite discontinuity (see Section 5.2).
In this section we extend the concept of a definite integral to the case where the interval is
infinite and also to the case where f has an infinite discontinuity in [a, b ]. In either case
the integral is called an improper integral. One of the most important applications of this
idea, probability distributions, will be studied in Section 8.5.
1111
T~pe 1: Infinite Intervals
Consider the infinite region S that lies under the curve y = 1/x 2, above the x-axis, 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 I Module 6
I How To Calculate
I Start of Improper Integrals
A(t) =
J -x1 dx =
1
--
t
I
2
]I
X
1
= 1- t
I
Notice that A(t) < 1 no matter how large tis chosen.
y
0
FIGURE 1
X
We also observe that
limA(t) =lim
t-> OO
t ->OO
(1 -I_)
t
The area of the shaded region approaches 1 as t ~
of the infinite region S is equal to 1 and we write
f Iro -;z
1
dx = r->oo
lim
Y+
oo
= 1
(see Figure 2), so we say that the area
f _!__x 2 d x-- 1
1
y
I
yr \
yr \
area= 1
Oj
1
2
X
Ol
1
3
X
ol
I~
1
5
X
01
1
X
FIGURE 2
Using this example as a guide, we define the integral off (not necessarily a positive
function) over an infinite interval as the limit of integrals over finite intervals.
SECTION 7.8 IMPROPER INTEGRALS
OJ
1111
531
Definition of an Improper Integral of Type 1
(a) Iff~ f(x) dx exists for every number t ~a, then
{~ f(x) dx
~~11,! J: f(x) dx
=
provided this limit exists (as a finite number).
(b) If
ft f(x) dx exists for every number t ~ b, then
f~"' f(x) dx
!iJEoo
=
1
ib f(x) dx
provided this limit exists (as a finite number).
J:
The improper integrals f(x) dx and f~"' f(x) dx are called convergent if the
corresponding limit exists and divergent if the limit does not exist.
(c) If both f: f(x) dx and f~"' f(x) dx are convergent, then we define
f~"' f(x) dx = J~"' f(x) dx + I"' f(x) dx
In part (c) any real number a can be used (see Exercise 74).
Any of the improper integrals in Definition 1 can be interpreted as an area provided that
f is a positive function. For instance, in case (a) if f(x) ~ 0 and the integral f: f(x) dx
is convergent, then we define the area of the regionS= {(x, y) I x ~a, 0 ~ y ~ f(x)} in
Figure 3 to be
A(S)
=
I"' f(x) dx
y
FIGURE 3
01
a
X
This is appropriate because
J: f(x) dx is the limit as t ~ oo of the area under the graph of
f from a tot.
EXAMPLE 1 Determine whether the integral f~ (1/x) dx is convergent or divergent.
SOlUTION According to part (a) of Definition 1, we have
1 dx = lim
foo
I
X
1 ~00
II -1 dx = lim ln I x I]
I
X
= lim (ln t
1~00
1~00
-
I
1
ln 1) = lim ln t =
oo
1~00
The limit does not exist as a finite number and so the improper integral f~ (1/x) dx is
divergent.
532
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
y
Let's compare the result of Example 1 with the example given at the beginning of this
section:
1
y = xz
oo
f
I
finite area
0
X
FIGURE 4
oo
f
X
I
1
- dx diverges
X
Geometrically, this says that although the curves y = 1/x 2 and y = 1/x look very similar
for x > 0, the region under y = 1/x 2 to the right of x = 1 (the shaded region in Figure 4)
has finite area whereas the corresponding region under y = 1/x (in Figure 5) has infinite
area. Note that both 1/x 2 and 1/x approach 0 as x ~ oo but 1/x 2 approaches 0 faster than
1/x. The values of 1/x don't decrease fast enough for its integral to have a finite value.
EXAMPLE 2 Evaluate
y
1
2 dx converges
f~oo xex dx.
SOlUTION Using part (b) of Definition 1, we have
o
f
-00
infinite area
We integrate by parts with u
0
=X,
xex dx
.
=
ro
t~~oo Jr xex dx
dv = ex dx so that du = dx, v = ex:
X
io xex dx
xex]~ - io ex dx
=
FIGURE 5
- -tet- 1
We know that et ~ 0 as t ~
-oo,
+
et
and by !'Hospital's Rule we have
1
t
lim te t = lim --=-; = lim
t ~-oo
r ~-oo
e
- -_-t
-e
r ~-oo
= lim (-e r) = 0
t~-00
Therefore
o
f
xexdx
-00
.
t!:~oo (-tet- 1
=
+
et)
= -0- 1 + 0 = -1
EXAMPLE 3 Evaluate
oo
1
f-oo 1 +X
~
dx.
SOlUTION It's convenient to choose a = 0 in Definition 1(c):
f-oo 1 +1 x
oo - - - d x
2
=
JO
-oo
1
dx
1 + x2
+
i oo - -1- d x
o 1 + x2
We must now evaluate the integrals on the right side separately:
roo --1- -2 dx =lim [t
Jo 1 + x
r~ oo Jo
1
= lim (tan ( ~00
dx
+x
2
=lim tan- 1 x]~
r ~oo
1T
It -
tan- 1 0) = lim tan- 1t = (~ 00
2
SECTION 7.8 IMPROPER INTEGRALS ..;1 533
1_ dx =
fo __
1+
x2
-co
ro _dx_
lim
lim tan- 1 x]~
1 + x2
t -'>-OO Jt
t -'>-00
= lim (tan - 1 0- tan- 1t)
t -'>-00
=0-(-;)=;
Since both of these integrals are convergent, the given integral is convergent and
y
f
(X)
-co
0
FIGURE 6
X
1
~
---d
x=1 + x2
2
~
+ -=
2
~
Since 1/(1 + x 2 ) > 0, the given improper integral can be interpreted as the area of
the infinite region that lies under the curve y = 1/(1 + x 2 ) and above the x-axis (see
Figure 6).
EXAMPLE 4 For what values of p is the integral
(X)
f
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=- 1. Then
f1
(X)
1
-dx
=lim
Xp
ft x - p dx
t -'><X>
1
= lim _x_-_P_+
_1_] x=t
t -'>oo -p + 1 x= 1
=
If p > 1, then p - 1 > 0, so as t ~
1I. m 1- - [ -1- 1 ]
l -'>00 1 - p
tp- 1
oo,
1
tp- t ~ oo and 1/tp- t ~ 0. Therefore
1
f -xPd x = - (X)
1
p- 1
if p > 1
and so the integral converges. But if p < 1, then p - 1 < 0 and so
1
tp- 1 = ti -p~
00
as t ~
oo
and the integral diverges.
We summarize the result of Example 4 for future reference:
[!]
1111
(X)
f
I
-
1
Xp
dx
is convergent if p > 1 and divergent if p
~
1.
Tqpe 2: Discontinuous lntegrands
Suppose that f is a positive continuous function defined on a finite interval [a, b) but has
a vertical asymptote at b. LetS be the unbounded region under the graph off and above
the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a
534
1111
CHAPTER 7 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
x=b
0
a
t b
X
A(t)
=
J: f(x) dx
If it happens that A(t) approaches a definite number A as t
area of the region S is A and we write
J: f(x) dx
FIGURE 7
=
~
b- , then we say that the
~~~ J: f(x) dx
We use this equation to define an improper integral of Type 2 even when
tive function, no matter what type of discontinuity f has at b.
f
is not a posi-
II] Definition of an Improper Integral of Type 2
(a) Iff is continuous on [a, b) and is discontinuous at b, then
1111 Parts (b) and (c) of Definition 3 are illustrated
in Figures 8 and 9 for the case where f(x) ;;:;. 0
and f has vertical asymptotes at a and c,
respectively.
J: f(x) dx
=
~~~ J: f(x) dx
if this limit exists (as a finite number).
y
(b) Iff is continuous on (a, b] and is discontinuous at a, then
J: f(x) dx
=
~~~ Jb f(x) dx
if this limit exists (as a finite number).
0
b
a t
The improper integralS: f(x) dx is called convergent if the corresponding limit
exists and divergent if the limit does not exist.
X
(c) Iff has a discontinuity at c, where a< c < b, and both S; J(x) dx and
S: f(x) dx are convergent, then we define
FIGURE 8
y
J: f(x) dx J:f(x) dx + J: f(x) dx
=
EXAMPLE S Find
5
f
2
0
b
c
a
1
r--;:;
... ! X - 2
dx.
SOLUTION We note first that the given integral is improper because f(x) = 1/~
has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left end-
X
FIGURE 9
point of [2, 5], we use part (b) of Definition 3:
fs --==dx==- _
2
y
1
y=
.Jx - 2 .
~-
lim
2~ ]:
= lim
2()3 - ~)
t-->2+
=
FIGURE 10
2
3
4
5
~
=lim
t -->2+
0
rs_dx_
t-->2+ J r
2)3
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.
SECTION 7.8 IMPROPER INTEGRALS
[-rr/2
EXAMPLE 6 Determine whether Jo
•
sec x dx converges or diverges.
SOLUTION Note that the given integral is improper because lim x~(-rr/2)- sec x =
part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have
-rr/2
i0
sec x dx =
=
=
because sec t ----7
divergent.
oo
and tan t ----7
oo
535
1111
•
lim
it
lim
ln Isec
lim
[ln(sec t
t ~ (-rr/2) -
t ~ (-rr/2) -
t~ (-rr/2) -
0
oo.
Using
sec x dx
x + tan xI]~
+ tan t)
- ln 1]
00
as t ----7
(
7T/2t. Thus, the given improper integral is
EXAMPLE 7 Evaluate [ ~ if possible.
Jo x- 1
3
SOLUTION Observe that the line x = 1 is a vertical asymptote of the integrand. Since it
occurs in the middle of the interval [0, 3], we must use part (c) of Definition 3 with
c = 1:
[3 dx
[1 dx
J3 dx
Jo ~ = Jo x - 1 + 1 x - 1
1
where
dx= l i m
-1
t~l-
i0 X -
it
dx
]t
--=limlnlx-11
0
1
t~1 -
0 X -
= lim (ln 1t - 1 1 - ln I -1
t ~1 -
= lim ln(l - t) =
I)
-oo
t~l -
because 1 - t ----7 o+as t ----7 1- . Thus, f~ dx/(x - 1) is divergent. This implies that
sg dx/(x - 1) is divergent. [We do not need to evaluate f13dx/(x - 1).]
~
li!.llii
WARNING o 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:
3
dx
i - - = ln I
0 X -
1
]3
x - 1 I 0 = In 2 - In 1 = In 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 J:J(x) dx you must decide, by looking
at the function f on [a, b ], whether it is an ordinary definite integral or an improper
integral.
EXAMPLE 8 Evaluate
s; ln x dx.
SOLUTION We know that the function f(x) = ln x has a vertical asymptote at 0 since
lim x~ o +
In x =
-oo.
Thus, the given integral is improper and we have
1
[
Jo
1
In x dx = lim [ In x dx
t ~ o+ Jt
536
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
Now we integrate by parts with u = ln x, dv = dx, du = dx/x, and v = x:
f ln x dx
f dx
=
x ln x]; -
=
1ln 1 - t ln t - (1 - t)
=
-tln t - 1
+t
To find the limit of the first term we use !'Hospital's Rule:
.
. ln t
11mtnt=
1
1I m r-> o+ 1/t
, _. o+
lim_!/!___
-ljt2
=
t -> O+
= lim ( -t) = 0
t -> O+
area= 1
fl ln x dx
Therefore
Jo
=
lim ( -t ln t - 1
r-> o+
+ t)
= -0- 1 + 0 = -1
Figure 11 shows the geometric interpretation of this result. The area of the shaded region
above y = ln x and below the x-axis is 1.
FIGURE 11
1111
RComparison 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 g are continuous functions with
f(x) ?: g(x) ?: 0 for x ?: a.
J: f(x) dx is convergent, then J: g(x) dx is convergent.
(b) If J: g(x) dx is divergent, then J: f(x) dx is divergent.
(a) If
y
0
a
FIGURE 12
X
We omit the proof of the Comparison Theorem, but Figure 12 makes it seem plausible.
If the area under the top curve y = f(x) is finite, then so is the area under the bottom curve
y = g(x). And if the area under y = g(x) is infinite, then so is the area under y = f(x).
[Note that the reverse is not necessarily true: If g(x) dx is convergent, f(x) dx may
or may not be convergent, and if f(x) dx is divergent,
g(x) dx may or may not be
divergent.]
J:
J:
EXAMPLE 9 Show that
Jo
[ ""
J:
J:
2
e -x dx is convergent.
2
SOLUTION We can't evaluate the integral directly because the antiderivative of e-x is not
an elementary function (as explained in Section 7.5). We write
Jof"" e-x dx
2
=
Jof e-x dx + J""I e-x dx
I
2
2
SECTION 7.8 IMPROPER INTEGRALS
y
1111
537
and observe that the first integral on the right-hand side is just an ordinary definite integral. In the second integral we use the fact that for x ~ 1 we have x 2 ~ x, so -x 2 ~ -x
2
and therefore e-x ~ e-x. (See Figure 13.) The integral of e-x is easy to evaluate:
f
oo
e-x dx
=
I
0
lim
ft e-x dx
I
t-> OO
= lim (e- 1
X
e- 1 ) = e- 1
-
/---'>00
FIGURE 13
2
Thus, taking f(x) = e-x and g(x) = e-x in the Comparison Theorem, we see that
2
2
f~ e-x dx is convergent. It follows that e-x dx is convergent.
J;
J;
TABLE 1
t
f~ e-x2 dx
1
2
3
4
5
6
0.7468241328
0.8820813908
0.8862073483
0.8862269118
0.8862269255
0.8862269255
2
In Example 9 we showed that e-x 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 8.5; using the methods of multivariable calculus it can be shown that the exact
value is .f;/2. Table 1 illustrates the definition of an improper integral by showing how
2
the (computer-generated) values of J~ e-x dx approach /Tr/2 as t becomes large. In fact,
2
these values converge quite quickly because e-x ~ 0 very rapidly as x ~ oo.
I
I
J 1 +X e-x dx is divergent by the Comparison Theorem
00
I
EXAMPLE 10 The integral
because
I
1 + e-x
1
--->X
X
and J~ (1/x) 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
s; [(1
t
2
5
10
100
1000
10000
IUJ 7.8
4
x 4e-x dx
f
(c)
io x
(a)
1
X
2
~
0.8636306042
1.8276735512
2.5219648704
4.8245541204
7.1271392134
9.4297243064
Exercises
1. Explain why each of the following integrals is improper.
oo
+ e-x )/x] dx
2
-
5x
+6
dx
[71"/ 2
(b)
Jo
secxdx
(d)
Jo
_ 2_ 1_ dx
+5
-oo X
2. Which of the following integrals are improper? Why?
-1d x
f -2x1
sin
(c)
- dx
f1 +X
(a)
2
(b)
1
oo
-oo
X
2
(d)
i' -2x--1d x
o
f2ln(x I
1
1) dx
538
CHAPTER 7 TECHNIQUES OF INTEGRATION
1111
3
= 1/x from x = 1 to x = t
and evaluate it for t = 10, 100, and 1000. Then find the total
area under this curve for x ~ 1.
3. Find the area under the curve y
~ 4. (a) Graph the functions f(x) = 1/xu and g(x) = 1/x 0·9 in the
viewing rectangles [0, 10] by [0, 1] and.[O, 100] by [0, 1].
(b) Find the areas under the graphs of f and g from x = 1 to
x = t and evaluate fort = 10, 100, 10 4 , 10 6 , 10 10, and 10 20 .
(c) Find the total area under each curve for x ~ 1, if it exists.
S-40
Determine whether each integral is convergent or
divergent. Evaluate those that are convergent.
5
•
1111
"' 1
f (3x + 1)
1
2
dx
1
.7. s-1
dw
-ex> ~
6.
8.
so
_1_dx
-"' 2x- 5
i"' (
+ 2) 2 dx
2
21
"' X- dx
f-"' 1 +X
12.
13. _"' xe-x dx
f"'
14. _"' x 2e-x dx
f"'27T sin 0 d 0
16. Jo"' cos 2a da
11.
15.
2
"' X+
f +2
19. J
: se - ds
1
17.
X
I
2
dx
i"'
20.
f "'
6
f3
-dx
X2
Jx1 dx
27. Jo
o
26 •
X
1
1
30.
f ~X- 9
1
35. fo7T sec x dx
36.
3
-2 X
3
f1
- 1
1) - I/ S dx
_e_x_ dx
ex - 1
9
dx
=
{(x, y) I 0 ~ y ~ 2/(x 2
~ 44. S
=
{(x, y) I x ~ 0, 0 ~ y ~ x/(x 2 + 9)}
~ 45. S
=
{(x, y) I 0 ~ x
~ 46. S
=
{(x,y)l-2 < x ~
y
e-x12 }
+ 9)}
< n/2, 0 ~ y ~
1
dx
.J1 -
0, 0 ~ y ~ 1/$+2}
~ 48. (a) If g(x) = 1/(Jx - 1), use your calculator or computer to
make a table of approximate values of s~ g(x) dx fort= 5,
10, 100, 1000, and 10,000. Does it appear that 2"' g(x) dx is
convergent or divergent?
(b) Use the Comparison Theorem with f(x) = 1/JX to show
that 2"' g(x) dx is divergent.
(c) Illustrate part (b) by graphing f and g on the same screen
for 2 ~ x ~ 20. Use your graph to explain intuitively why
"' g(x) dx is divergent.
2
J"'
2
J
f"' ~
X+ e 2x
52.
f"' J1+X6
X
dx
1
1+x
53. i7TI2----$---
54.
1
cos x dx
1 + x2
I
1
X2 +X- 6
f 2 ~dx
Jo 2x- 3
x sm x
f"'
Jo
~ 1 dy
4
"' 2 + e-x
dx
I
X
SO.
49.
f e-x
Jx dx
0
55. The integral
x2
0
i
sec 2x}
~ 47. (a) If g(x) = (sin 2 x)/x 2 , use your calculator or computer to
make a table of approximate values of s; g(x) dx fort= 2,
5, 10, 100, 1000, and 10,000. Does it appear that 1"' g(x) dx
is convergent?
(b) Use the Comparison Theorem with f(x) = l/x 2 to show
that 1"' g(x) dx is convergent.
(c) Illustrate part (b) by graphing f and g on the same screen
for 1 ~ x ~ 10. Use your graph to explain intuitively why
"' g(x) dx is convergent.
1
o
I
fl
32. Jo
38.
~ 43. S
-2, 0
0 XyX
34 · fo 4y
X
~
{(x, y) I x
51 •
r:. dx
f 2dx
1
31. f 4dx
33. s: (x- 1
~
=
49-54 1111 Use the Comparison Theorem to determine whether the
integral is convergent or divergent.
-dx
3
28.
1
~
42. S
f
f"' x arctan x dx
Jo (1 + x2)2
3
S={(x,y)lx~ 1, O~y~e x }
f
3
ln x
I
Sketch the region and find its area (if the area is finite).
1111
f
+ 3z + 2
2
re r/ dr
"'
f-
0
f
dz
o z
41.
40.
f
dv
3
24.
I
•
)
x2
J-"'"' - dx
9 +X
ln x
37
4
22. J~"' e-l xldx
"'
25.
f29.
- v
J"' lnXx dx
I
23.
18.
X
55
21.
J~"' (2
f"'
2
41-46
J ~ dx
1 1
z ln z dz
f
10. J~~ e- dt
9. {"' e -y/ 2 dy
J02 2
X
X
0
39.
dx
1
Jx (1
+ x) dx
is improper for two reasons: The interval [0, oo) 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:
io"' Jx (11 + x) dx ilo Jx (11 + x) dx + f"' Jx (11 + x) dx
--==----
=
1
SECTION 7.8 IMPROPER INTEGRALS
56. Evaluate
oo
f
2
by the same method as in Exercise 55.
57-59 1111 Find the values of p for which the integral converges and
evaluate the integral for those values of p.
(1 1
59.
f.
58.
oo
e
539
a photograph. Suppose that in a spherical cluster of radius R
the density of stars depends only on the distance r from the
center of the cluster. If the perceived star density is given by
y(s), where sis the observed planar distance from the center of
the cluster, and x(r) is the actual density, it can be shown that
1
x.Jx2 _ 4 dx
57. . Jo -;p dx
1111
1
dx
X(ln X) P
1
y(s)
(R
2r
Js ~x(r)dr
=
If the actual density of stars in a cluster is x(r)
find the perceived density y(s).
=
hR -
r) 2,
67. A manufacturer of lightbulbs wants to produce bulbs that last
Jo x Pln x dx
60. (a) Evaluate the integral So"" x ne-x dx for n
= 0, 1, 2, and 3.
(b) Guess the value of So"" xne-x dx when n is an arbitrary positive integer.
(c) Prove your guess using mathematical induction.
61. (a) Show that J.:'"" x dx is divergent.
about 700 hours but, of course, some bulbs bum out faster than
others. Let F(t) be the fraction of the company's bulbs that bum
out before t hours, so F(t) always lies between 0 and 1.
(a) Make a rough sketch of what you think the graph ofF
might look like.
(b) What is the meaning of the derivative r(t) = F'(t)?
(c) What is the value of So"" r(t) dt? Why?
68. As we will see in Section 9 .4, a radioactive substance decays
(b) Show that
lim J t xdx
( ~CO
0
=
- t
exponentially: The mass at timet is m(t) = m(O)e kt, where
m(O) is the initial mass and k is a negative constant. The mean
life M of an atom in the substance is
This shows that we can't define
M
J~""f(x) dx = ~~~ t
t f( x) dx
62. The average speed of molecules in an ideal gas is
v
=
4
-J:;;.
( M
2RT
)3/2fo"" v 3e- Mvz/(2RT ) dv
where M is the molecular weight of the gas, R is the gas
constant, T is the gas temperature, and v is the molecular
speed. Show that
For the radioactive carbon isotope, 14 C, used in radiocarbon
dating, the value of k is -0.000121. Find the mean life of a
14
C atom.
t9. Determine how large the number a has to be so that
f.
{8RT
-y---;;M
63. We know from Example 1 that the region
9ft = {(x, y) I x ~ 1, 0 ~ y ~ 1/x} has infinite area. Show
that by rotating 9ft about the x-axis we obtain a solid with
finite volume.
2
deterrnine .the density of stars in a star cluster from the
observed (two-dimensional) density that can be analyzed from
2
2
the sum of Soe-x dx and S4"" e-x dx. Approximate the first integral by using Simpson's Rule with n = 8 and show that the
4
"" e - x dx, which is less than
second integral is smaller than S
4
0.0000001.
71. If f(t) is continuous for t
~
0, the Laplace transform off is
the function F defined by
F(s)
tion 6.4 to find the work required to propel a 1000-kg satellite
out of Earth's gravitational field.
66. Astronomers use a technique called stellar stereography to
1
x 2 + 1 dx < 0.001
70. Estimate the numerical value of So"" e-x dx by writing it as
64. Use the information and data in Exercises 29 and 30 of Sec-
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 4mv 5supplies the needed work.
oo
a
4
v=
-k fo"" te kt dt
=
=
fo"" f(t)e -st dt
and the domain ofF is the set consisting of all numbers s for
which the integral converges. Find the Laplace transforms of
the following functions .
(b) f(t) = e t
(c) f(t) = t
(a) f(t) = 1
~ f(t) ~ Me at fort ~ 0, where Manda are
constants, then the Laplace transform F(s) exists for s > a.
72. Show that if 0
73. Suppose that 0 ~ f(t) ~ Me at and 0 ~ f'(t) ~ Ke at fort ~ 0,
where f' is continuous. If the Laplace transform of f(t) is F(s)
540
CHAPTER 7 TECHNIQUES OF INTEGRATION
1111
77. Find the value of the constant C for which the integral
and the Laplace transform of f'(t) is G(s), show that
G(s)
s> a
sF(s) - f(O)
=
J:(k- x~2)dx
74. If f.:'oo f(x) dx is convergent and a and b are real numbers,
show that
1t(x) dx
converges. Evaluate the integral for this value of C.
00
t oo t(x) dx +
=
r oo t(x) dx + t oo t(x) dx
78. Find the value of the constant C for which the integral
r(x,:
75 • Show that Joroo x 2e -x 2d x = 2I Joroo e -x 2dx.
2
76. Show that fooo e-x dx = f~ .J -In y dy by interpreting the
WJ 7 Review
c
CONCEPT CHECH
1. State the rule for integration by parts. In practice, how do you
use it?
2. How do you evaluate f sinmx cosnx dx if m is odd? What if n is
odd? What if m and n are both even?
.Ja
2
x
-
a
.J
factor (not repeated)? What if the quadratic factor is repeated?
- - - - - - - - - - - - - c
2.
3.
x(x
(
XX
x+2
+4
(
2
xx-4
x2
(
4
i
oo
6.
f
0
I
)
A
B
can be put in the form 2 + - - .
)
A
can be put in the form-+ -
4
-
2
+4
X
x
X
-
.
1
x-4
X
- 2 --dx
X
x-2
A
B
C
) can be put in the form + - - + --.
-4
X
x+2
x-2
2
x2
XX
A
B
can be put in the form-- + - - .
+4
x2
4.
5.
+ 4)
x 2 -4
X
B
2 --.
+4
I
=
5. State the rules for approximating the definite integral
J:
f(x) dx
with the Midpoint Rule, the Trapezoidal Rule, and Simpson's
Rule. Which would you expect to give the best estimate? How
do you approximate the error for each rule?
(a) 1 oo f(x) dx
(b)
J~oo f(x) dx
7. Define the improper integral
(c)
J~oo f(x) dx
J:
f(x) dx for each of the following cases.
(a) f has an infinite discontinuity at a.
(b) f has an infinite discontinuity at b.
(c) f has an infinite discontinuity at c, where a < c < b.
8. State the Comparison Theorem for improper integrals.
TRUE-FALSE OUIZ
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.
1.
- - - - - - - - - - - - - -
6. Define the following improper integrals.
4. What is the form of the partial fraction expansion of a rational
function P(x)/ Q(x) if the degree of Pis less than the degree of
Q and Q(x) has only distinct linear factors? What if a linear
factor is repeated? What if Q(x) has an irreducible quadratic
2
c
2
occurs in an integral, what substitution might you try? What if a 2 + x 2 occurs? What if
2
2
occurs?
.Jx
~ 1 ) dx
converges. Evaluate the integral for this value of C.
integrals as areas.
3. If the expression
1 - 3x
c - - - - - - - - - - - - -
8. The Midpoint Rule is always more accurate than the
Trapezoidal Rule.
9. (a) Every elementary function has an elementary derivative.
(b) Every elementary function has an elementary antiderivative.
J
10. Iff is continuous on [0, oo) and 1oo f(x) dx is convergent, then
fooo f(x) dx is convergent.
11. Iff is a continuous, decreasing function on [1, oo) and
limx--+oo f(x) = 0 , then 1oo f(x) dx is convergent.
J
12. If J: f(x) dx and faoo g(x) dx are both convergent, then
J: [f(x)
+ g(x)] dx is convergent.
2 ln 15
13. If faoo f(x) dx and faoo g(x) dx are both divergent, then
faoo [f(x) + g(x)] dx is divergent.
1
.J2 dx is convergent.
7. Iff is continuous, then f.:'oo f(x) dx
14. If f(x) :::::; g(x) and fooo g(x) dx diverges, then fooo f(x) dx also
=
limr--+oo tr f(x) dx.
diverges.
CHAPTER 7 REVIEW
- - - - - - - - - - - - - - o
EXERCISES
Note: Additional practice in techniques of integration is provided in
Exercises 7.5.
1-40
3• i7T/2
o
J
cos 0
dO
1 + sin 0
tan x sec 3x dx
7
J sin(ln t) dt
1•
t
9.
11.
r
r JX2=1
X
1
dx
13.f~
X +X
2.
4
r
41. f oo
(1
2x
+ 2x) 2
dx
40. J7T/ 3
JtallO
7T/4 sin 28 d8
dt
6. f
1
2
y - 4y- 12
8 f
dx
• x 2Jl+X1
10. 11 Jarctan2x dx
o
12.
r
1 + x
Jx2 +
18.
Jx2 +
Jcsc 4 4x dx
25.
J
3x 3 - x 2 + 6x - 4
dx
(x 2 + 1)(x 2 + 2)
. 2x dx
27. i7T/2 cos 3x sm
x 5 sec x dx
ex~ dx
ex + 8
r
8x- 3
dx
46. it_1_dx
o 2 - 3x
r
r
+ 1
48. _1 x{jX4
dx
2 dx
ox -x-2
-1
50• J oo tan 2 x dx
I
X
J (x +x 31)1o dx
•
~
+
~
•
J exJ1 dx-
1
X- 1 dx
e -2x
32 • i7T/4 x sin3 x dx
o
cos x
33.
J (4 - xx22)3/2 dx
34.
J (arcsin x)2dx
35.
J Jx ~ x 3/2 dx
36.
J1 -
37.
J(cos x + sin x) 2cos 2x dx
38.
tan 8 d8
1+tan8
Jx (tan - x) 2dx
1
x
dx
52. f .. fx2 + 1
+ 2x + 2) dx
=
cos2x sin3x and use the graph to
[ill] 54. (a) How would you evaluate f x 5e- 2xdx by hand?
(Don't actually carry out the integration.)
(b) How would you evaluate f x 5e- 2xdx using tables?
(Don't actually do it.)
(c) Use a CAS to evaluate f x 5e- 2xdx.
(d) Graph the integrand and the indefinite integral on the same
screen.
26.f~
1 + ex
28 f
3
2
guess the value of the integral J;7T f(x) dx. Then evaluate the
integral to confirm your guess.
dt
sin 2t + cos 2t
Jln(x
~ 53. Graph the function f(x)
24. f ex COS X dx
30
1~ dx
~ 51-52 1111 Evaluate the indefinite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C = 0).
51.
x 3 + 3x 2
•
0
t - 1
y
44. 2 ~dy
y - 2
cJx
• -oo 4x 2 + 4x + 5
tan 0
20 f
22.
f -t22-+- d1 t
0
49 f oo
2 dx
Jx sec x tan x dx
23.
42.
sinx dx
- 1 1 + x2
17.
21 f
dx
• Jx 2 - 4x
Evaluate the integral or show that it is divergent.
43. f oo ____!!!___
2 x lnx
45. Jo4
J sec60
2 dO
31. ilnlO
dy
x +2
X+ 1
dx
9x 2 + 6x + 5
1111
1
dx
1 (2x + 1) 3
16•
o
xe
t/2
• 1 (2t + 1) 3
14.
5
1
fa
41-50
Jsin 20 cos 0 dO
f
39.
J: ye-o.6y dy
15.
29.
o - - - - - - - - - - - - - - -
47.
x 312 ln X dx
19. f
541
Evaluate the integral.
1111
1. r--x-dx
0 X + 10
5.
1111
55-58 1111 Use the Table of Integrals on the Reference Pages to
evaluate the integral.
55. f e x ~dx
57.
56.
J csc t dt
5
J.Jx 2 + x + 1 dx
59. Verify Formula 33 in the Table of Integrals (a) by differentiation 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 fooo xn dx is
convergent?
542
1111
CHAPTER 7 TECHNIQUES OF INTEGRATION
62. For what values of a is So"' eax cos x dx convergent? Evaluate the
integral for those values of a.
point to be 53 em. The circumference 7 em from each end is
45 em. Use Simpson's Rule to make your estimate.
63-64
1111 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.
JJI+X4 dx
1
64.
0
[7T/2 r:::::::-=
Jo v sm x dx
------- 28cm -------
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 = ex/x from x = 1 to x = 4.
67. The speedometer reading (v) on a car was observed at
1-rninute intervals and recorded in the chart. Use Simpson's
Rule to estimate the distance traveled by the car.
t (min)
t (min)
v (mi/h)
40
42
45
49
52
54
0
1
2
3
4
5
v (mi/h)
6
7
8
9
10
56
57
57
55
56
71. Use the Comparison Theorem to determine whether the integral
x3
- 5- - d x
1 x + 2
is convergent or divergent.
"'
f
72. Find the area of the region bounded by the hyperbola
y2
-
x2
=
1 and the line y
3.
=
73. Find the area bounded by the curves y
between x = 0 and x = 1r.
with six subintervals to estimate the increase in the bee population during the first 24 weeks.
y
=
1/(2
+ JX), y
1/(2 -
=
12000
;I\
4000
0
\
v
4
8
12
16
"-.. t-20
=
cos 2x
=
1.
= cos x, 0 ~ x ~ 1rj2, is rotated
about the x-axis. Find the volume of the resulting solid.
2
76. The region in Exercise 75 is rotated about they-axis. Find the
volume of the resulting solid.
77. Iff' is continuous on [0, oo) and limx->oo /(x)
=
=
0, show that
-j(O)
78. We can extend our definition of average value of a continuous
function to an infinite interval by defining the average value of
on the interval [a, oo) to be
f
1
lim-t - a
fl f(x) dx
a
(a) Find the average value of y = tan- 1x on the interval [0, oo).
(b) If f(x) ~ 0 and fa"' f(x) dx is divergent, show that the average value off on the interval [a, oo) is lim x_."' f(x), if this
limit exists.
(c) If Sa"' f(x) dx is convergent, what is the average value off
on the interval [a, oo)?
(d) Find the average value of y = sin x on the interval [0, oo).
I \
I 1
\
I
8000
JX), and x
75. The region under the curve y
1-> 00
r
cos x and y
74. Find the area of the region bounded by the curves
fo"' f'(x) dx
68. A population of honeybees increased at a rate of r(t) bees per
week, where the graph of r is as shown. Use Simpson's Rule
=
t
(weeks)
24
79. Use the substitution u
=
i"'
1/x to show that
ln x
---dx=O
+ x2
o 1
[ill] 69. (a) If f(x) = sin(sin x), use a graph to find an upper bound
for I / (4)(x)l.
(b) Use Simpson's Rule with n = 10 to approximate
So7T f(x) 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 em long. You use a
piece of string and measure the circumference at its widest
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
F=-q-
47reor2
where r is the distance between the charges and eo 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.
PLUS
Cover up the solution to the example and try it yourself first.
EXAMPLE
(a) Prove that iff is a continuous function, then
foa f(x) dx
=
foa f(a - x) dx
(b) Use part (a) to show that
TT/2
sinnx
7T
------dx=i o sinnx + cosnx
4
for all positive numbers n.
1111 The principles of problem solving are
discussed on page 80.
SOLUTION
(a) At first sight, the given equation may appear somewhat baffling. How is it possible
to connect the left side to the right side? Connections can often be made through one of
the principles of problem solving: introduce something extra. Here the extra ingredient is
a new variable. We often think of introducing a new variable when we use the Substitution Rule to integrate a specific function. But that technique is still useful in the present
circumstance in which we have a general function f.
Once we think of making a substitution, the form of the right side suggests that it
should be u =a-x. Then du = -dx. When x = 0, u =a; when x =a, u = 0. So
faaf(a - x) dx
- {
f(u) du
=
foa f(u) du
But this integral on the right side is just another way of writing
equation is proved.
J; f(x) dx. So the given
(b) If we let the given integral be I and apply part (a) with a
7T/2, we get
I=
1111 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.
0
=
TT/2
sinnx
dx
i 0 sinnx + cosnx
=
iTT/2
0
=
sinn( 7T/2 - x)
dx
sinn( 7T/2 - x) + cosn( 7T/2 - x)
A well-known trigonometric identity tells us that sin( 7T/2 - x)
cos( 7T/2 - x) = sin x, so we get
I=
=
cos x and
TT/2
COSnX
i 0 cosnx + sinnx dx
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
2I =
FIGURE 1
TT/2 sinnx + COSnX
iTT/2
7f
dx =
1 dx = i 0 sinnx + cosnx
0
2
Therefore, I= 7T/4.
543
ffi
1111 PROBLEMS
1. Three mathematics students have ordered a 14-inch pizza. Instead of slicing it in the traditional 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
2. Evaluate S- 1 - X
-X
dx.
The straightforward approach would be to start with partial fractions, but that would be brutal.
Try a substitution.
3. Evaluate S (-{/1 - x 1
1
-
0
14----
14 in-------+~
~1 -
x 3 ) dx.
4. A man initially standing at the point 0 walks along a pier pulling a rowboat by a rope of
length L. The man keeps the rope straight and taut. The path followed by the boat is a curve
called a tractrix and it has the property that the rope is always tangent to the curve (see the
figure).
(a) Show that if the path followed by the boat is the graph of the function y = f(x), then
FIGURE FOR PROBLEM 1
y
dy
f'(x)
(b) Determine the function y
=
= -
dx
-.JL 2
x2
-
= -----'--X
f(x).
5. A function f is defined by
(L,O)
0
f(x)
X
=
Jo'r cos t
0
cos(x - t) dt
~X~
27r
Find the minimum value of f.
FIGURE FOR PROBLEM 4
6. If n is a positive integer, prove that
S(In x)n dx
1
= ( -1)nn!
0
7. Show that
l
i
(1 - x 2 )n dx
22n (n!)2
=
--'---'--
(2n
0
+ 1)!
Hint: Start by showing that if In denotes the integral, then
h+I
ffi
=
2k + 2
_2_k_+_3 h
8. Suppose that f is a positive function such that f' is continuous.
(a) How is the graph of y = f(x) sin nx related to the graph of y
as n ___,. oo?
(b) Make a guess as to the value of the limit
lim
n ..... oo
fi f(x)
Jo
=
f(x)? What happens
sin nx dx
based on graphs of the integrand.
(c) Using integration by parts, confirm the guess that you made in part (b). [Use the fact that,
since f' is continuous, there is a constant M such that I f'(x) I ~ M for 0 ~ x ~ 1.]
9. If 0 < a < b, find lim
r ..... o
544
{i l
o
[bx
+ a(1
- x)J dx
}1/t.
~ 10. Graph f(x)
sin(e x) and use the graph to estimate the value oft such that J:+1 f(x) dx is a
maximum. Then find the exact value of t that maximizes this integral.
y
=
11. The circle with radius 1 shown in the figure touches the curve y
of the region that lies between the two curves.
=
l2x I twice. Find the area
12. A rocket is fired straight up, burning fuel at the constant rate of b kilograms per second. Let
v = v(t) be the velocity of the rocket at timet and suppose that the velocity u of the exhaust
gas is constant. Let M = M(t) be the mass of the rocket at timet and note that M decreases
as the fuel burns. If we neglect air resistance, it follows from Newton's Second Law that
dv
F=M--ub
X
FIGURE FOR PROBLEM 11
dt
where the force F
=
-Mg. Thus
dv
M - - ub = -Mg
rn
dt
Let M 1 be the mass of the rocket without fuel, M 2 the initial mass of the fuel, and
M 0 = M 1 + Mz. Then, until the fuel runs out at time t = Mzb, the mass is M = Mo - bt.
(a) Substitute M = M 0 - bt into Equation 1 and solve the resulting equation for v. Use the
initial condition v(O) = 0 to evaluate the constant.
(b) Determine the velocity of the rocket at timet = Mz/b. This is called the burnout velocity.
(c) Determine the height of the rocket y = y(t) at the burnout time.
(d) Find the height of the rocket at any timet.
t.
13. Use integration by parts to show that, for all x > 0,
0<
i
oo
0
sin t
2
dt<--ln(l + x + t)
ln(l + x)
~ 14. The Chebyshev polynomials Tn are defined by
Tn(x)
=
cos(n arccos x)
n
=
0, 1, 2, 3, ...
(a) What are the domain and range of these functions?
(b) We know that T0(x) = 1 and T1(x) = x. Express Tz explicitly as a quadratic polynomial
and T3 as a cubic polynomial.
(c) Show that, for n ~ 1,
Tn+J(x)
=
2xTn(X) - Tn- J(x)
(d) Use part (c) to show that Tn is a polynomial of degree n.
(e) Use parts (b) and (c) to express T4, T5 , T6 , and T1 explicitly as polynomials.
(f) What are the zeros of Tn? At what numbers does Tn have local maximum and minimum
values?
(g) Graph Tz, T3 , T4, and T5 on a common screen.
(h) Graph T5 , T6 , and T1 on a common screen.
(i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the
zeros of Tn+J? What about the x-coordinates of the maximum and minimum values?
(j) Based on your graphs in parts (g) and (h), what can you say about J~ 1 Tn(x) dx when n is
odd and when n is even?
(k) Use the substitution u = arccos x to evaluate the integral in part (j).
(1) The family of functions f(x) = cos(c arccos x) are defined even when cis not an integer
(but then f is not a polynomial). Describe how the graph off changes as c increases.
545
