Calculus: Analytic Geometry and Calculus, with Vectors

8.3 Inverse trigonometric functions 465

which involves completing a square. It may be time to renew the idea that
normal persons who have forgotten (or never remembered) needed integral
formulas either work them out to satisfy their vanities and preserve their
mathematical powers or stoop to hunting them in a book containing a table of
integrals.
2 Show that
v 1
o y2 + x2dx = 4y
when y > 0 and that
(° v^1 log a

Ji lie y2 + xsdx} dy=

a 4

when a > 0.

3 Supposing that -1 < x < 1 and

(1) T (x) = 2' " cos (n cos ' x),

find formulas for T',(x) and T;; (x).Then multiply TT(x) by n2, multiply 2 (x)

by -x, multiply T, (x) by (1 - xs), and add the results to discover that

(2) (1 -x2)dx2-xLy+n2y =0

when y = T.W. Remark: This problem involves much more than meets the

eye. When n is a positive integer, T. is the Tchebycheff polynomial of degree

n and (2) is the Tchebycheff differential equation of order n. One who wishes

a simple and brief discussion of these astonishing things can find it in the author's

textbook "Differential Equations," 2d ed, McGraw-Hill Book Company, Inc.,

New York, 1960.

4 Supposing that -1 < x < 1 and u = (sin-' x)', calculate the first two

derivatives of u and show that

(1 - x2)

d'u- xddu

= 2.

5 The eye of a man is at a point E that is, as in Figure 8.391, x feet from a

vertical wall bearing a picture the bottom and top of which are a and b feet above

Figure 8.391

eye level. The angle 0 which the picture subtends at the eye is surely small

when x is small and when x is large. Derive the formulas

0 =tan 1b- tan-'

a, dO (b - a)(ab - x2)

x x dx

• (a2 x2) (b2 + x2)