79. Supposing X to be always a function of x, and that it were required to integrate X x; we must write this expression under the form ƒ'Xl" x=Xl* x—n ƒ Xx px, by the formula for integra x X', we shall have by the same formula ƒ ̃Ñ¿11 x=X' ¿a-' x—(n−1 ) ƒX1 à T. ƒ X" b−2 x=X" b122 x—(1−2 ) ƒ X _3 z, &c. Therefore ƒX ba x=X l^ »—n X1 l−−1 x+n (n−1) X" [~~2 x— n (n−1) (n—2) X-3x+&c.; an expression which depends upon the integration of algebraic quantities only, and which will have a finite number of terms when n is a positive whole number. For example. Let Xxx, and we shall have X= m+1 Consequently fax la x =. n (n+1) (n+2) Įn−3 x+&c. (m + 1 ) 3 The only case which does not come within the general formula is 80. The foregoing formula applies equally to the case in which s is negative. But as then we obtain for the fluent an infinite series, we shall explain another mode of integration. form Xx. flux lx (to which it is evidently equal, because flux lr. Suppose the quantity were ing flux (Xr) = X' x, flux (X' x)=X" x, flux (X" x)=XTM ¿, we shall have by continuing the process Xx S -Xx X' x X" * (lx)" (n−1) l-1 x (n−1) (n−2) l−x (n−1)(n−2)(n−3) lTM3 s -&c., to a term of the form 1 (n−1) (n−2) (n−3)...2.1° the integration of which, if possible, will give that of the formula proposed. For example, let X=r", we shall have X'=x" (m+1), X"= (m+1)2x", X = (m+1)3 x", &c. Therefore (n-2) (n-3) (n-4) S therefore the proposed integral xm x If we make TMTM+1 u, this quantity will become- a fluxion which has not yet been integrated. น For this lu a series, except in the case of m=— -1; for then we find by the pre 81. Let it now be required to find the fluent of the exponential formula a Xx. I observe first that aa x la = flux (a*); therefore a; and since, integrating by parts, fa* Xx = X fa2 — = *. (lx)—*—— (Lx )—* flux lx, of which the fluent is evidently integral fax, which at least will be the most simple transcendental integral of its kind, if it be not susceptible of an exact integration. 82. We may observe that if e is the number whose logarithm =1, we shall have fe Xx=e X-e2 X'+e* X"—e2 XTM+e® X1—&c. For example, let X=x", and we shall get X'=n*, X"=n(n−1)r, Xm=n (n−1)(n-2) &B Therefore, _n (n−1) (n−2) +3+&c } &c} and consequently also ants to (1) anh (m-1) (mm) gas +&c.} 83. To find the integral of 4*_*, as the preceding rules do not apply, I reduce a" into a series, and obtain == (1+xla+1+1^+&c.) == +la+Pa+&c. Therefore 2 84. When the preceding rules will not apply to the integration of an exponential quantity, we must reduce it to a series by the formula x2 la x3 13 a πρα a=1+xla+ 2 and it will be easily integrated. + + +&c. 2.3 2.3.4 Thus let y=xTM; by the above series = {1+mx le + 13 x + m3 3 12 = +&e } = x + mx_x lx + m2x2 à la x + &c. of 2.3 2 which the fluent may be found by that of alx (79), and we obtain which in the particular case of x=1, becomes the converging series These examples are sufficient to enable us to integrate such sorts of quantities by means of series. Examples on Logarithmic and Exponential Fluxions. 1. Required the fluent of x ON THE INTEGRATION OF FLUXIONS CONTAINING SINES, COSINES, &c. 85. Since x cos x=flux sin x, and - sin x flux cos x, it is evident that st cos x= sin r, and ƒr sin x=-cos r ; as also that Jy cos ny n sin z. = === Sny cos ny = n 1 n sin ny, and similarly that fy sin ny cos ny. Again it is clear that fz cos z (sin z)"=ƒ (sin x)" flux 1 (sin x)+1, and that f (x sin x) cos" z= n+1 (cos 2)+1 n+1 Similarly if we desire to integrate y sin y cos ay, we must make sin y cos ay sin (a+1) y-sin (a-1)y, and the integral x 86. A similar process will serve for x sin x sin ax, or x cos x cos ax, &c. With equal facility we may integrate r sin r sin ax cos ba, &c. if we first reduce these products to simple sines or cosines, by means of the values of sin a cos b, sin a sin b, &c. given in the article trigonometry. The same treatment will serve to integrate x sin x, x sin 3x, x cos x, &c.; but it is more simple to integrate them in the following manner. 87. The formula i sin" x=x sin x. sin"-1. Consequently, integrating by parts, så sin" x= sin^-1 x ƒx sin x-ƒ{ flux (sinTM1 x). få sin x} ——cos x sin11 x + (n−1) fr sin22 * cos2 x = COS X sin21 x+(n−1) ƒ'r sin *2 x—(n−1 )ƒx sın" ; and transposing, process we have fi sin"-2 = 1 |