Calculus: Analytic Geometry and Calculus, with Vectors

8.3 Inverse trigonometric functions 461

R being the closed interval -1 <- y 5 1. When we are called upon to
select a domain D, to which we can apply Theorem 8.33, the graph of
y = sin x can remind us of known properties of its derivative which show
us that if we are going to be able to apply Theorem 8.33, we must let
D, be the interval -7r/2 < x < a/2 or some other interval lying n7r
(n an integer) units to the right or left of it. Making the simplest choice,
we let fi be the function, defined only when -a/2 5 x <- 7r/2, for which
f,(x) = sin x. The inverse of fl is called the inverse sine. If y = fi(x),

so that y = sin x and -,7r/2 < x <- 7r/2 and -1 5 y < 1, we write

x = sin-' y. The graph of x = sin-' y coincides with the part of the

graph of y = sin x for which -,r/2 < x < Ir/2, and the graph is shown in

Figure 8.351. When we interchange x and y, which amounts to replacing

Figure 8.351 Figure 8.352

each point (x,y) by its "image" (y,x) in the line y = x, we find that the

graph of the equation y = sin-' x is the same as the part of the graph

of x = sin y for which -a/2 < y < 7r/2 and that the graph is shown in

Figure 8.352. When -1 < x < 1 and y = sin-' x, we have sin y = x,

-7r/2 < y < a/2, and cos y > 0. Since Theorem 8.33, with x and y

interchanged, implies that dy/dx exists, we can use the chain rule to

obtain cos y(dy/dx) = 1 and hence

(8.353) dx sin-' x=

dx cosy 1-sin y 1 z 1-xs^1

While we could not be expected to guess the result in this formula, we

can look at Figure 8.352 and see that it seems to be a very reasonable

answer. One who feels that inverse sines and their derivatives could

never be useful should be informed at once that (8.353) gives the formula

(8.354)^11

1 - x2

dx = sin-' x + c (Ixj < 1)

which is useful because the integral appears in important problems.