Calculus: Analytic Geometry and Calculus, with Vectors

464 Trigonometric functions

The graph of y = csc x presents the same difficulty that the graph of

y = sec x presented. The best we can do is let D, be the interval

-7r/2 < x < a/2 with the center point

' Y=CsC_1 X x = 0 omitted. The inverse cosecant

X=cscy

is then the inverse of the function fi

-1 with domain D, for which fi(x) = cscx.

1F x The graph of y = csc-1 x, which coin-

cides with a part of the graph of

Figure 8.383 x = csc y, is shown in Figure 8.383.

If x > 1 or x < -1 and y = csc-1 x,

then csc y = x, -7r/2 < y < a/2, y 0, and hence csc y cot y > 0.

Again Theorem 8.33 implies that dy/dx exists, so

• csc y cot y

dy

and

(8.384) dcsc-1 x =

dy -1 -1

dx dx csc y cot y Icsc yl Icot yJ

Icsc yI csc2 y - 1 JxJ x2 - 1

Problems 8.39

(1)

(2)

1 Supposing that a > 0 and b2 < 4ac, show how the first of the formulas

f 1-}-usdutan-' u+c, f a2+u2du=titan'U+Ca

1 dx = 2 tan-'

tax + b

faxe + bx + c c

V 4 -. - b' 74=ac==-_h=1::ac - b2

can be used to obtain the other two. Remark: Everyone should know how the

last of these formulas and many similar ones are presented in standard integral

tables. Let a > 0 and let

X = axe + bx + c, q = b2 - 4ac.

Then (2) takes the form

1 2 2ax + b

(3) fX dx =

-v/-

tan 1 + c

when q < 0 and hence -q > 0. The derivation of (3) is based upon the identity

_ f b r ( b l (c _ b2 l1

(4) X-a[x2 I ax+ac]=a[\x2+ax+4

b2

a2/+a 4a2 /J

= a C\x + 2a/2 +l 4 2a b2\21= a L\x +2a/2+`"/ZJ