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166 CHAPTER 6 Transcendental Functions

SOLUTION
We have
1
(log4 9) · (log9 15) = · log9 16 = log4 16 = 2.
log9 4



6.4 Calculus with Logs and Exponentials to
Arbitrary Bases
6.4.1 DIFFERENTIATION AND INTEGRATION OF
loga x AND ax
We begin by noting these facts:
If a > 0 then
ax
dx
a = a x · ln a; equivalently, a dx = + C.
x
(i)
ln a
dx
1
d
(loga x) =
(ii)
x · ln a
dx
Math Note: As always, we can state these last formulas more generally as
du du
a = au · · ln a
dx dx
and
1 du 1
d
loga u = · · .
u dx ln a
dx
EXAMPLE 6.25
Calculate
d d d d
(log4 (x · cos x)).
(3cos x ),
(5x ), (log8 x),
dx dx dx dx
SOLUTION
We see that
dx
(5 ) = 5x · ln 5.
dx
For the second problem, we apply our general formulation with a = 3, u =
cos x to obtain
d cos x d
) = 3cos x · cos x · ln 3 = 3cos x · (’ sin x) · ln 3.
(3
dx dx
167
CHAPTER 6 Transcendental Functions

Similarly,
1
d
(log8 x) =
x · ln 8
dx
1
d d
(log4 (x · cos x)) = · (x · cos x)
(x · cos x) · ln 4 dx
dx
cos x + (x · (’ sin x))
= .
(x · cos x) · ln 4
EXAMPLE 6.26
Integrate

3cot x · (’ csc2 x) dx.

SOLUTION
For clarity we set •(x) = cot x, • (x) = ’ csc2 x. Then our integral becomes
1
3•(x) · • (x) dx = · 3•(x) · • (x) · ln 3 dx
ln 3
1
= · 3•(x) + C.
ln 3
Resubstituting the expression for •(x) now gives that
1
3cot x · (’ csc2 x) dx = · 3cot x + C.
ln 3

You Try It: Evaluate (log6 (x 3 )/x) dx.

You Try It: Calculate the integral
2
x · 3x dx.

Our new ideas about arbitrary exponents and bases now allow us to formulate a
general result about derivatives of powers:
For any real exponent a we have
da
x = a · x a’1 .
dx
EXAMPLE 6.27 √
’π, x 3, x e.
Calculate the derivative of x
168 CHAPTER 6 Transcendental Functions

SOLUTION
We have
d ’π
= ’π · x ’π ’1 ,
x
dx
d √3 √ √
x = 3 · x 3’1 ,
dx
de
x = e · x e’1 .
dx
2
You Try It: Calculate (d/dx)5sin x’x . Calculate (d/dx)x 4π .

6.4.2 GRAPHING OF LOGARITHMIC AND
EXPONENTIAL FUNCTIONS
If a > 0 and f (x) = loga x, x > 0, then



Y
1
f (x) =
x · ln a
FL
’1
f (x) = 2
x · ln a
AM

f (1) = 0.
Using this information, we can sketch the graph of f (x) = loga x.
If a > 1 then ln a > 0 so that f (x) > 0 and f (x) < 0. The graph of f is
TE



exhibited in Fig. 6.6.




Fig. 6.6

If 0 < a < 1 then ln a = ’ ln(1/a) < 0 so that f (x) < 0 and f (x) > 0. The
graph of f is sketched in Fig. 6.7.
Since g(x) = a x is the inverse function to f (x) = loga x, the graph of g is the
re¬‚ection in the line y = x of the graph of f (Figs 6.6 and 6.7). See Figs 6.8, 6.9.
169
CHAPTER 6 Transcendental Functions




Fig. 6.7




Fig. 6.8




Fig. 6.9


Figure 6.10 shows the graphs of loga x for several different values of a > 1.
Figure 6.11 shows the graphs of a x for several different values of a > 1.

You Try It: Sketch the graph of y = 4x and y = log4 x on the same set of axes.
170 CHAPTER 6 Transcendental Functions




Fig. 6.10




Fig. 6.11

6.4.3 LOGARITHMIC DIFFERENTIATION
We next show how to use the logarithm as an aid to differentiation. The key idea is
that if F is a function taking positive values then we can exploit the formula
F
[ln F ] = (—)
.
F
EXAMPLE 6.28
Calculate the derivative of the function
F (x) = (cos x)(sin x) , 0 < x < π.
171
CHAPTER 6 Transcendental Functions

SOLUTION
We take the natural logarithm of both sides:

ln F (x) = ln (cos x)(sin x) = (sin x) · (ln(cos x)). (†)
Now we calculate the derivative using the formula (—) preceding this example:
The derivative of the left side of (†) is
F (x)
.
F (x)
Using the product rule, we see that the derivative of the far right side of (†) is
’ sin x
(cos x) · (ln(cos x)) + (sin x) · .
cos x
We conclude that
’ sin x
F (x)
= (cos x) · (ln(cos x)) + (sin x) · .
cos x
F (x)
Thus
sin2 x
F (x) = (cos x) · (ln(cos x)) ’ · F (x)
cos x
sin2 x
= (cos x) · ln(cos x) ’ · (cos x)(sin x)
cos x

You Try It: Differentiate log9 | cos x|.

You Try It: Differentiate 3sin(3x) . Differentiate x sin 3x .
EXAMPLE 6.29
Calculate the derivative of F (x) = x 2 · (sin x) · 5x .

SOLUTION
We have

[ln F (x)] = [ln(x 2 · (sin x) · 5x )]
= [(2 · ln x) + ln(sin x) + (x · ln 5)]
2 cos x
=+ + ln 5.
sin x
x
Using formula (—), we conclude that
2 cos x
F (x)
=+ + ln 5
sin x
F (x) x
172 CHAPTER 6 Transcendental Functions

hence
2 cos x
F (x) = + + ln 5 · [x 2 · (sin x) · 5x ].
sin x
x

You Try It: Calculate (d/dx)[(ln x)ln x ].




6.5 Exponential Growth and Decay
Many processes of nature and many mathematical applications involve logarithmic
and exponential functions. For example, if we examine a population of bacteria,
we notice that the rate at which the population grows is proportional to the number
of bacteria present. To see that this makes good sense, suppose that a bacterium
reproduces itself every 4 hours. If we begin with 5 thousand bacteria then
after 4 hours there are 10 thousand bacteria
after 8 hours there are 20 thousand bacteria
after 12 hours there are 40 thousand bacteria
after 16 hours there are 80 thousand bacteria . . .
etc.
The point is that each new generation of bacteria also reproduces, and the older
generations reproduce as well.Asketch (Fig. 6.12) of the bacteria population against
time shows that the growth is certainly not linear”indeed the shape of the curve
appears to be of exponential form.




Fig. 6.12
173
CHAPTER 6 Transcendental Functions

Notice that when the number of bacteria is large, then different generations of
bacteria will be reproducing at different times. So, averaging out, it makes sense
to hypothesize that the growth of the bacteria population varies continuously as
in Fig. 6.13. Here we are using a standard device of mathematical analysis: even
though the number of bacteria is always an integer, we represent the graph of the
population of bacteria by a smooth curve. This enables us to apply the tools of
calculus to the problem.




Fig. 6.13


6.5.1 A DIFFERENTIAL EQUATION
If B(t) represents the number of bacteria present in a given population at time t,
then the preceding discussion suggests that
dB
= K · B(t),
dt
where K is a constant of proportionality. This equation expresses quantitatively
the assertion that the rate of change of B(t) (that is to say, the quantity dB/dt) is
proportional to B(t). To solve this equation, we rewrite it as
1 dB
· = K.
B(t) dt
We integrate both sides with respect to the variable t:
1 dB
· dt = K dt.
B(t) dt
The left side is
ln |B(t)| + C
and the right side is
Kt + C,
174 CHAPTER 6 Transcendental Functions

where C and C are constants of integration. We thus obtain
ln |B(t)| = Kt + D,
where we have amalgamated the two constants into a single constant D. Exponen-
tiating both sides gives
|B(t)| = eKt+D
or
B(t) = eD · eKt = P · eKt . ()
Notice that we have omitted the absolute value signs since the number of bacteria is
always positive. Also we have renamed the constant eD with the simpler symbol P .
Equation ( ) will be our key to solving exponential growth and decay problems.
We motivated our calculation by discussing bacteria, but in fact the calculation
applies to any function which grows at a rate proportional to the size of the function.
Next we turn to some examples.

6.5.2 BACTERIAL GROWTH
EXAMPLE 6.30
A population of bacteria tends to double every four hours. If there are 5000
bacteria at 9:00 a.m., then how many will there be at noon?

SOLUTION
To answer this question, let B(t) be the number of bacteria at time t. For con-
venience, let t = 0 correspond to 9:00 a.m. and suppose that time is measured
in hours. Thus noon corresponds to t = 3.
Equation ( ) guarantees that
B(t) = P · eKt
for some undetermined constants P and K. We also know that
5000 = B(0) = P · eK·0 = P .
We see that P = 5000 and B(t) = 5000 · eKt . We still need to solve for K.
Since the population tends to double in four hours, there will be 10,000
bacteria at time t = 4; hence
10000 = B(4) = 5000 · eK·4 .
We divide by 5000 to obtain
2 = eK·4 .
175
CHAPTER 6 Transcendental Functions

Taking the natural logarithm of both sides yields
ln 2 = ln(eK·4 ) = 4K.
We conclude that K = [ln 2]/4. As a result,
B(t) = 5000 · e([ln 2]/4)t .
We simplify this equation by noting that
e([ln 2]/4)t = (eln 2 )t/4 = 2t/4 .
In conclusion,
B(t) = 5000 · 2t/4 .
The number of bacteria at noon (time t = 3) is then given by
B(3) = 5000 · 23/4 ≈ 8409.
It is important to realize that population growth problems cannot be described
using just arithmetic. Exponential growth is nonlinear, and advanced analytical
ideas (such as calculus) must be used to understand it.
EXAMPLE 6.31
Suppose that a certain petri dish contains 6000 bacteria at 9:00 p.m. and
10000 bacteria at 11:00 p.m. How many of the bacteria were there at
7:00 p.m.?

SOLUTION
We know that
B(t) = P · eKt .
The algebra is always simpler if we take one of the times in the initial data to
correspond to t = 0. So let us say that 9:00 p.m. is t = 0. Then 11:00 p.m. is
t = 2 and 7:00 p.m. is t = ’2. The initial data then tell us that
6000 = P · eK·0 (—)
10000 = P · eK·2 . (——)
From equation (—) we may immediately conclude that P = 6000. Substituting
this into (——) gives
10000 = 6000 · (eK )2 .
We conclude that

5
=√ .
eK
3
176 CHAPTER 6 Transcendental Functions

As a result,
√ t
5
B(t) = 6000 · √ .
3
At time t = ’2 (7:00 p.m.) the number of bacteria was therefore
√ ’2
5 3
B(’2) = 6000 · √ = · 6000 = 3600.
5
3
You Try It: A petri dish has 5000 bacteria at 1:00 p.m. on a certain day and 8000
bacteria at 5:00 p.m. that same day. How many bacteria were there at noon?

6.5.3 RADIOACTIVE DECAY
Another natural phenomenon which ¬ts into our theoretical framework is radio-
active decay. Radioactive material, such as C14 (radioactive carbon), has a half life.
Saying that the half life of a material is h years means that if A grams of material is
present at time t then A/2 grams will be present at time t + h. In other words, half
of the material decays every h years. But this is another way of saying that the rate
at which the radioactive material vanishes is proportional to the amount present.
So equation ( ) will apply to problems about radioactive decay.
EXAMPLE 6.32
Five grams of a certain radioactive isotope decay to three grams in 100
years. After how many more years will there be just one gram?
SOLUTION
First note that the answer is not “we lose two grams every hundred years
so ¦.” The rate of decay depends on the amount of material present. That is
the key.
Instead, we let R(t) denote the amount of radioactive material at time t.
Equation ( ) guarantees that R has the form
R(t) = P · eKt .
Letting t = 0 denote the time at which there are 5 grams of isotope, and
measuring time in years, we have
R(0) = 5 R(100) = 3.
and
From the ¬rst piece of information we learn that
5 = P · eK·0 = P .
Hence P = 5 and
R(t) = 5 · eKt = 5 · (eK )t .
177
CHAPTER 6 Transcendental Functions

The second piece of information yields
3 = R(100) = 5 · (eK )100 .
We conclude that
3
(eK )100 =
5
or
1/100
3
=
K
e .
5
Thus the formula for the amount of isotope present at time t is
t/100
3
R(t) = 5 · .
5
Thus we have complete information about the function R, and we can answer
the original question.
There will be 1 gram of material present when
t/100
3
1 = R(t) = 5 ·
5
or
t/100
1 3
= .
5 5
We solve for t by taking the natural logarithm of both sides:
t/100
3 t
ln(1/5) = ln = · ln(3/5).
5 100
We conclude that there is 1 gram of radioactive material remaining when
ln(1/5)
t = 100 · ≈ 315.066.
ln(3/5)
So at time t = 315.066, or after 215.066 more years, there will be 1 gram of
the isotope remaining.
You Try It: Our analysis of exponential growth and decay is derived from the
hypothesis that the rate of growth is proportional to the amount of matter present.
Suppose instead that we are studying a system in which the rate of decay is propor-
tional to the square of the amount of matter. Let M(t) denote the amount of matter
at time t. Then our physical law is expressed as
dM
= C · M 2.
dt
178 CHAPTER 6 Transcendental Functions

Here C is a (negative) constant of proportionality. We apply the method of
“separation of variables” described earlier in the section. Thus
dM/dt
=C
M2
so that
dM/dt
dt = C dt.
M2
Evaluating the integrals, we ¬nd that
1
’ = Ct + D.
M
We have combined the constants from the two integrations. In summary,
1
M(t) = ’ .
Ct + D


Y
For the problem to be realistic, we will require that C < 0 (so that M > 0 for
FL
large values of t) and we see that the population decays like the reciprocal of a
linear function when t becomes large.
AM

Re-calculate Example 6.32 using this new law of exponential decay.

6.5.4 COMPOUND INTEREST
TE



Yet a third illustration of exponential growth is in the compounding of interest. If
principal P is put in the bank at p percent simple interest per year then after one
year the account has
p
P · 1+
100
dollars. [Here we assume, of course, that all interest is reinvested in the account.]
But if the interest is compounded n times during the year then the year is divided
into n equal pieces and at each time interval of length 1/n an interest payment of
percent p/n is added to the account. Each time this fraction of the interest is added
to the account, the money in the account is multiplied by
p/n
1+ .
100
Since this is done n times during the year, the result at the end of the year is that
the account holds
pn
P · 1+ (—)
100n
179
CHAPTER 6 Transcendental Functions

dollars at the end of the year. Similarly, at the end of t years, the money accumulated
will be
p nt
P · 1+ .
100n
Let us set
n · 100
k=
p
and rewrite (—) as
k p/100
kp/100
1 1
P · 1+ =P · 1+ .
k k
It is useful to know the behavior of the account if the number of times the inter-
est is compounded per year becomes arbitrarily large (this is called continuous
compounding of interest). Continuous compounding corresponds to calculating the
limit of the last formula as k (or, equivalently, n), tends to in¬nity.
We know from the discussion in Subsection 6.2.3 that the expression (1 + 1/k)k
tends to e. Therefore the size of the account after one year of continuous
compounding of interest is
P · ep/100 .
After t years of continuous compounding of interest the total money is
P · ept/100 . ()
EXAMPLE 6.33
If $6000 is placed in a savings account with 5% annual interest compounded
continuously, then how large is the account after four and one half years?

SOLUTION
If M(t) is the amount of money in the account at time t, then the preceding
discussion guarantees that
M(t) = 6000 · e5t/100 .
After four and one half years the size of the account is therefore
M(9/2) = 6000 · e5·(9/2)/100 ≈ $7513.94.
EXAMPLE 6.34
A wealthy woman wishes to set up an endowment for her nephew. She
wants the endowment to pay the young man $100,000 in cash on the day of
his twenty-¬rst birthday.The endowment is set up on the day of the nephew™s
180 CHAPTER 6 Transcendental Functions

birth and is locked in at 11% interest compounded continuously. How much
principal should be put into the account to yield the desired payoff?
SOLUTION
Let P be the initial principal deposited in the account on the day of the
nephew™s birth. Using our compound interest equation ( ), we have
100000 = P · e11·21/100 ,
expressing the fact that after 21 years at 11% interest compounded continuously
we want the value of the account to be $100,000.
Solving for P gives
P = 100000 · e’0.11·21 = 100000 · e’2.31 = 9926.13.
The aunt needs to endow the fund with an initial $9926.13.

You Try It: Suppose that we want a certain endowment to pay $50,000 in cash
ten years from now. The endowment will be set up today with $5,000 principal and
locked in at a ¬xed interest rate. What interest rate (compounded continuously) is
needed to guarantee the desired payoff?



6.6 Inverse Trigonometric Functions
6.6.1 INTRODUCTORY REMARKS
Figure 6.14 shows the graphs of each of the six trigonometric functions. Notice that
each graph has the property that some horizontal line intersects the graph at least
twice. Therefore none of these functions is invertible. Another way of seeing this
point is that each of the trigonometric functions is 2π -periodic (that is, the function
repeats itself every 2π units: f (x + 2π ) = f (x)), hence each of these functions is
not one-to-one.
If we want to discuss inverses for the trigonometric functions, then we must
restrict their domains (this concept was introduced in Subsection 1.8.5). In this
section we learn the standard methods for performing this restriction operation
with the trigonometric functions.

6.6.2 INVERSE SINE AND COSINE
Consider the sine function with domain restricted to the interval [’π/2, π/2]
(Fig. 6.15). We use the notation Sin x to denote this restricted function. Observe
that
d
Sin x = cos x > 0
dx
181
CHAPTER 6 Transcendental Functions

3




2
sec x


1

cos x sin x


_3 _2 _1 1 2 3



_1

csc x

_2
tan x cot x


_3

Fig. 6.14




y




y = Sin x
1




F/2 x




Fig. 6.15
182 CHAPTER 6 Transcendental Functions

on the interval (’π/2, π/2). At the endpoints of the interval, and only there, the
function Sin x takes the values ’1 and +1. Therefore Sin x is increasing on its
entire domain. So it is one-to-one. Furthermore the Sine function assumes every
value in the interval [’1, 1]. Thus Sin : [’π/2, π/2] ’ [’1, 1] is one-to-one and
onto; therefore f (x) = Sin x is an invertible function.
We can obtain the graph of Sin’1 x by the principle of re¬‚ection in the line
y = x (Fig. 6.16). The function Sin’1 : [’1, 1] ’ [’π/2, π/2] is increasing,
one-to-one, and onto.
y

F/2
_1
y = Sin x



x
1



_ F/2




Fig. 6.16

The study of the inverse of cosine involves similar considerations, but we must
select a different domain for our function. We de¬ne Cos x to be the cosine function
restricted to the interval [0, π]. Then, as Fig. 6.17 shows, g(x) = Cos x is a one-to-
one function. It takes on all the values in the interval [’1, 1]. Thus Cos : [0, π] ’
[’1, 1] is one-to-one and onto; therefore it possesses an inverse.
We re¬‚ect the graph of Cos x in the line y = x to obtain the graph of the function
Cos’1 . The result is shown in Fig. 6.18.


EXAMPLE 6.35
Calculate

√ √
3 2
Sin’1 Sin’1 0, Sin’1 ’
, ,
2 2
√ √
3 2
Cos’1 Cos’1 0, Cos’1
’ , .
2 2
183
CHAPTER 6 Transcendental Functions

y




y = Cos x
1




F/2 F x




Fig. 6.17

y


F



F/2 _1
y = Cos x




x
1




Fig. 6.18

SOLUTION
We have

3 π
Sin’1 = ,
2 3
Sin’1 0 = 0,
184 CHAPTER 6 Transcendental Functions

2 π
Sin’1 ’ =’ .
2 4

Notice that even though the sine function takes the value 3/2 at many different
values of the variable x, the function Sine takes this value only at x = π/3.
Similar comments apply to the other two examples.
We also have

3 5π
Cos’1 ’ = ,
2 6
π
Cos’1 0 = ,
2

2 π
Cos’1 =.
2 4

We calculate the derivative of f (t) = Sin’1 t by using the usual trick for inverse
functions. The result is
1 1
d
(Sin’1 (x)) = =√ .
1 ’ sin2 (Sin’1 x) 1 ’ x2
dx

The derivative of the function Cos’1 t is calculated much like that of Sin’1 t. We
¬nd that
1
d
(Cos’1 (x)) = ’ √ .
1’x
dx 2

EXAMPLE 6.36
Calculate the following derivatives:
1
d d d
Sin’1 x Sin’1 (x 2 + x) Sin’1
, , √.

dx dx dx x
x= 2/2 x=’ 3
x=1/3

SOLUTION
We have

1
d
Sin’1 x √ =√ = 2,

x= 2/2
1 ’ x2
dx x= 2/2

1 15
d
Sin’1 x 2 + x = · (2x + 1) =√ ,
1 ’ (x 2 + x)2
dx 65
x=1/3 x=1/3

1 1 1
d
Sin’1 (1/x) = ·’ = ’√ .
√ x2 √
1 ’ (1/x)2
dx 6
x=’ 3 x=’ 3
185
CHAPTER 6 Transcendental Functions

You Try It: Calculate (d/dx)Cos’1 [x 2 + x]. Also calculate (d/dx)Sin’1 —
[ln x ’ x 3 ].
EXAMPLE 6.37
Calculate each of the following derivatives:

d d d
Cos’1 x Cos’1 (ln x) Cos’1 ( x)
, , .

dx dx dx
x=1/2 x= e x=1/2

SOLUTION
We have
1 2
d
Cos’1 x =’ √ = ’√ ,
1 ’ x2
dx 3
x=1/2 x=1/2

’2
1 1
d
Cos’1 (ln x) =’ · = ’√ ,

1 ’ (ln x)2 √
dx x 3e
x= e x= e

√ 1 1 ’1/2
d
Cos’1 ( x) =’ · = ’1.
√ x
1 ’ ( x)2 2
dx x=1/2 x=1/2

You Try It: Calculate (d/dx) ln[Cos’1 x] and (d/dx) exp[Sin’1 x].

6.6.3 THE INVERSE TANGENT FUNCTION
De¬ne the function Tan x to be the restriction of tan x to the interval (’π/2, π/2).
Observe that the tangent function is unde¬ned at the endpoints of this interval.
Since
d
Tan x = sec2 x
dx
we see that Tan x is increasing, hence it is one-to-one (Fig. 6.19). Also Tan takes
arbitrarily large positive values when x is near to, but less than, π/2. And Tan takes
negative values that are arbitrarily large in absolute value when x is near to, but
greater than, ’π/2. Therefore Tan takes all real values. Since Tan : (’π/2, π/2) ’
(’∞, ∞) is one-to-one and onto, the inverse function Tan’1 : (’∞, ∞) ’
(’π/2, π/2) exists. The graph of this inverse function is shown in Fig. 6.20. It
is obtained by the usual procedure of re¬‚ecting in the line y = x.

EXAMPLE 6.38
Calculate
√ √
Tan’1 1, Tan’1 1/ 3, Tan’1 (’ 3).
186 CHAPTER 6 Transcendental Functions

y


y = Tan x




x




Fig. 6.19

y




_1
y = Tan x



x




Fig. 6.20

SOLUTION
We have
π
Tan’1 1 = ,
4
√ π
Tan’1 1/ 3 = ,
6
√ π
Tan’1 (’ 3) = ’ .
3
187
CHAPTER 6 Transcendental Functions

As with the ¬rst two trigonometric functions, we note that the tangent func-
√ √
tion takes each of the values 1, 1/ 3, ’ 3 at many different points of its
domain. But Tan x takes each of these values at just one point of its domain.
The derivative of our new function may be calculated in the usual way. The
result is
1
d
Tan’1 t = .
1 + t2
dt
Next we calculate some derivatives:
EXAMPLE 6.39
Calculate the following derivatives:
d d d
Tan’1 x Tan’1 (x 3 ) Tan’1 (ex )
, √, .
dx dx dx
x= 2
x=1 x=0

SOLUTION
We have
1 1
d
Tan’1 x = =,
1 + x2 2
dx x=1 x=1
1 2
d
Tan’1 (x 3 ) = · 3x 2 =,
√ √
1 + (x 3 )2 3
dx x= 2 x= 2
1 1
d
Tan’1 (ex ) = · ex =.
1 + (e x )2 2
dx x=0 x=0

You Try It: Calculate (d/dx)Tan’1 [ln x + x 3 ] and (d/dx) ln[Tan’1 x].

6.6.4 INTEGRALS IN WHICH INVERSE
TRIGONOMETRIC FUNCTIONS ARISE
Our differentiation formulas for inverse trigonometric functions can be written in
reverse, as antidifferentiation formulas. We have
du
= Sin’1 u + C;

1 ’ u2
du
= ’Cos’1 u + C;

1 ’ u2
du
du = Tan’1 u + C.
1+u 2

The important lesson here is that, while the integrands involve only polynomials
and roots, the antiderivatives involve inverse trigonometric functions.
188 CHAPTER 6 Transcendental Functions

EXAMPLE 6.40
Evaluate the integral
sin x
dx.
1 + cos2 x

SOLUTION
For clarity we set •(x) = cos x, • (x) = ’ sin x. The integral becomes
• (x) dx
’ .
1 + • 2 (x)
By what we have just learned about Tan’1 , this last integral is equal to
’Tan’1 •(x) + C.
Resubstituting •(x) = cos x yields that
sin x
dx = ’Tan’1 (cos x) + C.


Y
1 + cos 2x
FL
x/(1 + x 4 ) dx.
You Try It: Calculate
EXAMPLE 6.41
AM

Calculate the integral
3x 2
√ dx.
TE



1 ’ x6
SOLUTION
For clarity we set •(x) = x 3 , • (x) = 3x 2 . The integral then becomes
• (x) dx
.
1 ’ • 2 (x)
We know that this last integral equals
Sin’1 •(x) + C.
Resubstituting the formula for • gives a ¬nal answer of
3x 2
dx = Sin’1 (x 3 ) + C.

1 ’ x6
You Try It: Evaluate the integral
x dx
√ .
1 ’ x4
189
CHAPTER 6 Transcendental Functions

6.6.5 OTHER INVERSE TRIGONOMETRIC FUNCTIONS
The most important inverse trigonometric functions are Sin’1 , Cos’1 , and Tan’1 .
We say just a few words about the other three.
De¬ne Cot x to be the restriction of the cotangent function to the interval (0, π)
(Fig. 6.21). Then Cot is decreasing on that interval and takes on all real values.
Therefore the inverse
Cot’1 : (’∞, ∞) ’ (0, π)

y




y = Cot x




F x




Fig. 6.21

is well de¬ned. Look at Fig. 6.22 for the graph. It can be shown that

1
d
Cot’1 x = ’ .
1 + x2
dx

De¬ne Sec x to be the function sec x restricted to the set [0, π/2) ∪ (π/2, π]
(Fig. 6.23). Then Sec x is one-to-one. For these values of the variable x, the cosine
function takes all values in the interval [’1, 1] except for 0. Passing to the recip-
rocal, we see that secant takes all values greater than or equal to 1 and all values
less than or equal to ’1. The inverse function is

Sec’1 : (’∞, ’1] ∪ [1, ∞) ’ [0, π/2) ∪ (π/2, π]
190 CHAPTER 6 Transcendental Functions

y

F



_1
y = Cot x


x




Fig. 6.22

y
y = Sec x




1




F x




Fig. 6.23


(Fig. 6.24). It can be shown that
1
d
Sec’1 x = |x| > 1.
√ ,
|x| · x 2 ’ 1
dx
The function Csc x is de¬ned to be the restriction of Csc x to the set [’π/2, 0) ∪
(0, π/2]. The graph is exhibited in Fig. 6.25. Then Csc x is one-to-one. For these
values of the x variable, the sine function takes on all values in the interval [’1, 1]
except for 0. Therefore Csc takes on all values greater than or equal to 1 and all
values less than or equal to ’1; Csc’1 therefore has domain (’∞, ’1] ∪ [1, ∞)
and takes values in [’1, 0) ∪ (0, 1] (Fig. 6.26).
191
CHAPTER 6 Transcendental Functions

y

F

_1
y = Sec x




x
1




Fig. 6.24

y




_ F/2 F/2 x




Fig. 6.25


It is possible to show that
1
d
Csc’1 x = ’ |x| > 1.
√ ,
|x| · x 2’1
dx
√ √
’1 ’1
You Try It: What is Sec (’2/ 3)? What is Csc (’ 2)?
192 CHAPTER 6 Transcendental Functions

y




F/2



x


_ F/2




Fig. 6.26


Summary of Key Facts About the Inverse Trigonometric Functions
π π
Sin x = sin x, ’ ¤ x ¤ ; Cos x = cos x, 0 ¤ x ¤ π;
2 2
π π
Tan x = tan x, ’ < x < ; Cot x = cot x, 0 < x < π;
2 2
Sec x = sec x, x ∈ [0, π/2) ∪ (π/2, π]; Csc x = csc x, x ∈ [’π/2, 0) ∪ (0, π/2].
1 1
d d
Sin’1 x = √ Cos’1 x = ’ √
, ’1 < x < 1; , ’1 < x < 1;
1 ’ x2 1 ’ x2
dx dx
1 1
d d
Tan’1 x = Cot’1 x = ’
, ’∞ < x < ∞; , ’∞ < x < ∞;
1 + x2 1 + x2
dx dx
1 1
d d
Sec’1 x = Csc’1 x = ’
, |x| > 1; , |x| > 1;
√ √
|x| · x 2 ’ 1 |x| · x 2 ’ 1
dx dx
du du
= Sin’1 u + C; = ’Cos’1 u + C;
√ √
1 ’ u2 1 ’ u2
du du
du = Tan’1 u + C; du = ’Cot’1 u + C;
1+u 1+u
2 2

du du
= Sec’1 u + C; = ’Csc’1 u + C.
√ √
|u| · u 2’1 |u| · u 2’1


You Try It: What is the derivative of Sec’1 x 2 ?
193
CHAPTER 6 Transcendental Functions

6.6.6 AN EXAMPLE INVOLVING INVERSE
TRIGONOMETRIC FUNCTIONS
EXAMPLE 6.42
Hypatia is viewing a ten-foot-long tapestry that is hung lengthwise on a
wall. The bottom end of the tapestry is two feet above her eye level. At what
distance should she stand from the tapestry in order to obtain the most
favorable view?

SOLUTION
For the purposes of this problem, view A is considered more favorable than
view B if it provides a greater sweep for the eyes. In other words, form the
triangle with vertices (i) the eye of the viewer, (ii) the top of the tapestry, and
(iii) the bottom of the tapestry (Fig. 6.27). Angle ± is the angle at the eye of
the viewer. We want the viewer to choose her position so that the angle ± at the
eye of the viewer is maximized.




10 ft



G
=
2 ft
O
x ft



Fig. 6.27


The ¬gure shows a mathematical model for the problem. The angle ± is the
angle θ less the angle ψ. Thus we have

± = θ ’ ψ = Cot’1 (x/12) ’ Cot’1 (x/2).

Notice that when the viewer is standing with her face against the wall then
θ = ψ = π/2 so that ± = 0. Also when the viewer is far from the tapestry then
θ ’ ± is quite small. So the maximum value for ± will occur for some ¬nite,
positive value of x. That value can be found by differentiating ± with respect
to x, setting the derivative equal to zero, and solving for x.
194 CHAPTER 6 Transcendental Functions


We leave it to you to perform the calculation and discover that 24 ft is the
optimal distance at which the viewer should stand.
You Try It: Redo the last example if the tapestry is 20 feet high and the bottom
of the tapestry is 6 inches above eye level.



Exercises
1. Simplify these logarithmic expressions.
a 2 · b’3
(a) ln 4
c ·d
log3 (a 2 b)
(b)
log2 (a 2 b)
(c) ln[e3x · z4 · w ’3 ]

(d) log10 [100w · 10]
2. Solve each of these equations for x.
(a) 3x · 5’x = 2x · e3
3x
= 10x · 102
(b)
5’x · 42x
(c) 23x · 34x · 45x = 6
5 2
= x ’x
(d)
2 ·3
23x · e5x
3. Calculate each of these derivatives.
d
ln[sin(x 2 )]
(a)
dx
x2
d
(b) ln
x’1
dx
d sin(ex )
(c) e
dx
d
(d) sin(ln x)
dx
4. Calculate each of these integrals.

e’x x 3 dx [Hint: Guess p(x) · e’x , p a polynomial.]
(a)
195
CHAPTER 6 Transcendental Functions

x 2 · ln2 x dx [Hint: Guess p(x) · ln2 x + q(x) ln x + r(x), p, q, r
(b)
polynomials.]
e ln x
(c) dx
x
1
2 ex
(d) dx
ex + 1
1
5. Use the technique of logarithmic differentiation to calculate the derivative
of each of the following functions.
x2 + 1
(a) x · 3
3
x ’x
sin x · (x 3 + x)
(b)
x 2 (x + 1)
(c) (x 2 + x 3 )4 · (x 2 + x)’3 · (x ’ 1)
x · cos x
(d)
ln x · ex
6. There are 5 grams of a certain radioactive substance present at noon on
January 10 and 3 grams present at noon on February 10. How much will be
present at noon on March 10?
7. A petri dish has 10,000 bacteria present at 10:00 a.m. and 15,000 present at
1:00 p.m. How many bacteria will there be at 2:00 p.m.?
8. A sum of $1000 is deposited on January 1, 2005 at 6% annual interest,
compounded continuously. All interest is re-invested. How much money
will be in the account on January 1, 2009?
9. Calculate these derivatives.
d
Sin’1 (x · ex )
(a)
dx
d x
Tan’1
(b)
x+1
dx
d
Tan’1 [ln(x 2 + x)]
(c)
dx
d
Sec’1 (tan x)
(d)
dx
10. Calculate each of these integrals.
2x
(a) x dx
1 + x4
196 CHAPTER 6 Transcendental Functions

3x 2

(b) dx
1 ’ x6
π/2 2 sin x cos x
(c) dx
1 ’ sin x 4
0

dx
(d)
5 + 2x 2
CHAPTER 7



Methods of
Integration
7.1 Integration by Parts
We learned in Section 4.5 that the integral of the sum of two functions is the sum
of the respective integrals. But what of the integral of a product? The following
reasoning is incorrect:

x 2 dx = x · x dx = x dx · x dx

because the left-hand side is x 3 /3 while the right-hand side is (x 2 /2) · (x 2 /2) =
x 4 /4.
The correct technique for handling the integral of a product is a bit more subtle,
and is called integration by parts. It is based on the product rule

(u · v) = u · v + u · v .

Integrating both sides of this equation, we have

(u · v) dx = u · v dx + u · v dx.

The Fundamental Theorem of Calculus tells us that the left-hand side is u · v. Thus

u·v = u · v dx + u · v dx

197
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
198 CHAPTER 7 Methods of Integration

or

u · v dx = u · v ’ v · u dx.

It is traditional to abbreviate u (x) dx = du and v (x) dx = dv. Thus the
integration by parts formula becomes

u dv = uv ’ v du.

Let us now learn how to use this simple new formula.
EXAMPLE 7.1
Calculate

x · cos x dx.



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