Calculus: Analytic Geometry and Calculus, with Vectors

8.4 Integration by trigonometric and other substitutions 473

we look at itand generate the idea that we should try setting x = a sec 8.
Then x' = a sec 0 tan B, so

(8.474)

• 1 ('sec0d8
  f
  J=

1 J (sin 8)-2 cos 8 d8

a2

J 1 - a sec 0 tan B do

J (a2 t^1

8

_ 1 (sin 8)-i

a'- -1 + c

x

a2

a2 + C.

x2-

It is not implied that we know in advance that the integrand should be

written in terms of sines and cosines, but trying this possibility and seeing

that the power formula can be applied is all a part of the game.

Our last problem has a lengthy solution. We can bravely start to

evaluate the integral in

(8.48) J2=j x2-a2dx

by setting x = a sec 8. Then x' = a sec 0 tan 0 and

J2 = f a2 tan2 0 a sec 0 tan 8 do

so

(8.481) J2 = a2f tang 8 sec 0 do.

It happens that this integral is an elementary function, but there is no

simple direct way of discovering the facts. Authors and teachers, as

well as students, can work for hours on this problem unless they remember

how to solve it or (and this much better) have a guiding principle. The

principle is the following. If we want to learn something about an

integral and other methods fail to be helpful, we try integration by parts.

By this we mean that we try to determine two functions u and v of the

variable of integration in such a way that the integral we are studying

will be the left member of the formula

(8.482) fu(8)v'(8) d8 = u(8)v(8) - fv(8)u'(6) d8

and, moreover, the "parts" on the right side enable us to make progress

with our problem. Assuming that u and v have continuous derivatives

over intervals where we use the formula,differentiation shows that the

formula is correct. The trick is to use it effectively. Naturally, we

want the functions u, v, u', v' to be as simple aspossible, and it is quite

easy to discover that the best way to convertJ2 into the left member of

(8.482) is to set

(8.483)

{ 1'( )=

aa' tan 0,

2 sec 0, vv(0) = sec

8.tan 8