<< . .

. 21
( : 41)



. . >>

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 = ’ .

<< . .

. 21
( : 41)



. . >>