And these again will ultimately be reduced to one of the form

which have been already found, and there remains now only the integral

We may here give the integrals of one or two remarkable functions belong ing to this class.

de

=

de

4

a

tan.-1

-

a

b

log. (a + b tan. §) + a2 + b2°

a+b cos.

tan.-1 (tan. §)

may be obtained very simply by an algebraic

de

f a + b com # = √ 2+ b (cos-sin' }) cos. A S a+b(cos. 2

On Integration by Series.

When the integral of a proposed function cannot be exactly determined, we must have recourse to approximations. Thus in order to find

fx de

where X is a function of 2, we must develope X in a series according to as cending or descending powers of x, and then multiplying each term by de integrate them in succession. For example, we know that

Let P be the integral of X dø a function of r, and C the arbitrary constant which we must add in order to render the result perfectly general, we have

fx dx = P + C

So long as this calculation is altogether abstract, C may have any value whatever; but when we wish to apply this integral to the solution of some given problem the constant C ceases to be arbitrary and must answer certain condi

But since the required area P+C commences when &= AM=2, A ought to be=0 when we make = a in P + C, or

Q+C=0

Q being the value which the function of z represented by P assumes when x= a, hence we find

C=- Q

whence the area A = P -- Q.

It only now remains to substitute b for x, and we shall have the area included within the prescribed limits. We shall have several examples in what follows

CHAPTER IV.

APPLICATION OF THE INTEGRAL CALCULUS TO FINDING THE LENGTHS AND AREAS OF CURVES, And the surFACES AND VOLUMES OF SOLIDS OF REVOLUTION.

I. THE RECTIFICATION OF CURVES.

We have seen in p. 747 of the differential calculus, that if s represent the are of a curve,

?

= vậy? + dr

dr2

and we shall now apply this formula to a few examples.

(1.) To find the length of the arc of the common parabola. The equation is y2 = 4mx, where 4m is the parameter.

.. ydy = 2mdx .. =

dy 2m

dx y

y

=

4m

√y2 + 4m2 + m log (y + √y2 + 4m2) + C

If we suppose the arc to be measured from the vertex; then when y=0,=