and.. the integral will at length depend upon that of

cording as m is odd or even.

The formula (2) making n negative, and integrating for the cosine gives

fde sin."

1

=

cos."

And the integral will then be reduced to de sin." or to

If both m and n are negative, the integral becomes

de

sin. cos." A

de

de sin."

cos.

(4)

Multiply both numerator and denominator by sin.2 ◊ + cos.2 @

de

↓ sin." 8 cos." 6 = ↓ sin."-2 cos."

And by continuing the process the first of these fractions will be freed from the sine, and the second from the cosine, and the integral will be reduced to finding that of

If m and n be equal as in sin. x cos, x = sin. 2x, making 2x = z, the fraction becomes

We have now seen that in integrating the formula

'de sin.± é cos.± 6, we ultimately reduce it to one of the following forms.

Now we may find the first form of these by making m and n = 0 in the formulæ (1), (2), (3), (4), of last page, which will then give

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 belonging to this class.

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

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

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

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