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

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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.

Y

SOLUTION FL

We observe that the integrand is a product. Let us use the integration by parts

formula by setting u(x) = x and dv = cos x dx. Then

u(x) = x du = u (x) dx = 1 dx = dx

AM

v(x) = sin x dv = v (x) dx = cos x dx

Of course we calculate v by anti-differentiation.

TE

According to the integration by parts formula,

x · cos x dx = u dv

=u·v’ v du