(^242) Integrals
20 This problem is interesting because it shows how a basic formula involving
areas (a well-known formula which we have not yet proved) can be used to obtain
preliminary derivations of formulas involving trigonometric and inverse trigo-
nometric functions. Rigorous derivations will be given in Chapter 8. Supposing
that 0 < t < a, construct and look at an appropriate figure to derive the formula
(1)
fol

a2-x2dx=mot a2-t +Ta2sin 1a

in which the first term on the right is the area of a particular triangle and the

last term is the area of a circular sector having radius a and central angle 0,

where 6 = sin-' (t/a) and 0 < B < it/2. Anyone who is short on information

about areas of circular sectors is reminded that the area of a sector having central

angle 0 is, as it ought to be, the product of 6/27r and the area 7ra2 of the whole

circle. We can suddenly become interested in (1) if we realize that we have

theorems and rules that enable us to write formulas for the derivatives with

respect to t of everything in it except the last term and hence that we can obtain

a formula for the derivative of the last term. To capitalize this idea, put (1) in

the form

(2) sin 'a a2 C2 fo.Va2- x2 dx - t a2 - t2

and then differentiate and simplify results to obtain the formula

(3) d sin-' Q = - t2

a2

1

Remark: We invest a moment to look at the formula

(4) dsin-1 t =

at 1 - t2

to which (3) reduces when a = 1. At least in the case where 0 < t < 1 and

0 < 8 < 7r/2, trigonometry books emphasize the fact that the angle 0 of Figure

4.495 is "the angle whose sine is t" or "the inverse sine of t," so that 6 = sin-' t

t

i

Figure 4.495 Figure 4.496

and t = sin 0. The graph in Figure 4.496 shows how 0 and t are related. The

relation (4) is equivalent to the relation

(5)(l.o AtAe

1 - t2