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Now here is an apparent inconsistency; for it would appear from (61) and (62) that

sin-1 * +606-1 = Sande - Sen det = 0; (63) which result is not correct ; because sin- + cos->= : we

must therefore have recourse to the accurate process of definite integration; and let us take both integrals between the same limits; say, between the limits 0 and x; then by (11), Art. 5, po dx fir-x70_N

ren (64)

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sin

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Sin

therefore by addition
sin-+ cosa =

(66) which is a correct result; (61) and (62) therefore are not simultaneously true in the forms of indefinite integrals.

1. dx 32.) Integration of

(2 ax — x2)
since
Since d.versin-1 = _dx

"versih ā (2ax-x2)1'

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36.] The relations existing between some of these last integrals deserve consideration; for taking the definite integral of (69) between the limits x and 0, we have * dx

1x +(a? + x2),
= logas

(73) Jo (a? + x2) but the left-hand member may be put under the following form, whereby its integral is determined by equation (61): viz.

and the identity of (73) and (74) may thus be proved :

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replacing sin (2 V-1) in terms of its exponential value as given
by equation (30), Art. 61, of Vol. I;
i, e = x + (a? + x2).

a
sin-10V-1 - Loox + (a2 + x2).

(75) a which shews that the two results though different in form are identical in value. Similarly might

fx dx

might 22 I 2 be put in the form 1 fr d (x v 1)

-, and be integrated according to -11. a? — (xv1)2' Article 15, and the result shewn to be equal to

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(aza + b2 +c) whereby the integral is reduced to one or other of the forms (76) or (77). . dx

dix Ex. 1.

c(1–2x+3co) •

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39.] The integration of (a? — x2) dx, and of (a +x2) dx.

In these cases it is convenient to employ the method of integration by parts given by the formula, Ju dv = wv Svdu.

(78) To determine /(a? – x2)+dx.

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Then, adding and subtracting a? in the numerator of the last quantity, and writing the fraction in two parts, and cancelling in the second part (a? — 204)), which occurs in both numerator and denominator, we have

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