Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1
8.3 Inverse trigonometric junctions 467

10 Letting N be a positive integer (which we shall not call "arbitrary"),
manufacture and solve N problems of the nature of the preceding one.
11 Prove that


fol1 + x2dx = 4' Jo01 -I-x2dx = 2


1

12 Remark: We should not be so busy that we never have time to look at our
formulas and see that we can learn things from them. Supposing for simplicity
that x > 0, we can look at the formula


(1) tan1x=


1
0 1+t2dt
and realize that each member is equal to the area of the shaded region in Figure
8.393. Any information we can obtain about tan-' x or the area or the integral

therefore provides information about all three. While there are much more
complicated ways of getting the information, we can quickly learn some sig-
nificant facts by starting with the formula
1 ton+2
(2) 1

+

t2 = 1 - t2 +t4 - t6 .+ +- 14n -1 + 12

which can be obtained by long division and can be checked by multiplication
by (1 + 0). Integrating this over the interval 0 < t 5 x gives the formula

x3 x6 x7 x4n+1
(3) tan-' x = x - 3 + 5 - 7 -{- ... -{-4n {- 1 -R,(x),

where
f x ton+2
(4) R, (x) =Jo 1 -t2dt.

Suppose now that 0 5 x 5 1. Then 1 5 1 + t2 5 2 and therefore
x t4n+2 x
(5)
fo 2

dt5 Rt(x) 5 fo t4n+2dt

so

(6)

x4n+3 x4n+3
2(4n + 3)

<
Rn(x) 5 4n + 3

This gives us an excellent estimate of R,,(x) which implies the more crude estimate

(7) 0 5 Rn(x) 54n -l- 3^1
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