79. Supposing X to be always a function of x, and that it were required to integrate X l" x; we must write this expression under the form Sål" <=Xx=nski "x 1x, by the formula for integration by parts ; and then supposing SX: _X', we shall have by the same formula s špabc=X' /a-11 X *-==(n=15*3n. If we make sX':-x", we shall have -X", *, &c. Therefore SÅtx=X7"-n X (1+n(n-1) X"/^2- n(n-1) (-2) X" 1-3 *+&c.; an expression which depends upon the integratio of algebraic quantities only, and which will have a finite number of terms when n is a positive whole number. mti For example. Let X="2, and we shall have X= ( نر m+1' X' :=X"= 1+1 (m+1) n =X'= 2+1 . similarly Xmas &c. (m+1) (m+1)* Consequently Somil":=_ { n (n-1) m+1 (m +1) preg_" (n+1) (n+2) 1-5 *+&c. = )n x &c.} .(m+1)' The only case which does not come within the general formula is 11+1 that of m=-1, and then we have n+1 80. The foregoing formula applies equally to the case in which » is negative. But as then we obtain for the Auent an infinite geries, we shall explain another mode of integration. Xi Suppose the quantity were which being written under the (hr) form Xx. Ir= Xx gives s + lx) R1) flux lx (to which it is evidently equal, because flur eru -,) MASAT Aux (Xx). Now call : اس را-ls (n) = X' X" را ing flux (X.r) = X'.:, flur (X' x)=X":, flux (X" 2)=X., we shall have by continuing the process Xc -Xx S. (lx)(1-1) 7-tx (11-1) (1-2) (*-** (n-1)(1-2)(n-3)/31 "^1"
1 -&c., to a term of the form f (1-1)(n-2) (n-3)...2.1 LE the integration of which, if possible, will give that of the formula proposed. For example, let X=r”, we shall have X'=2" (m+1), X +1 (m+1)* <", X" = (m+1)*.x", &c. Therefore, I." (n-1) (m+1) (m + 1)"-? s -; therefore the proposed integral (11-1) (1-2)...1 It i is reduced to that of If we make mm+1 =u, this quantity will lux {1+ lx + 12 x + 1 x + + &c. &c.} u become lu a fluxion which has not yet been integrated. For this than by reason we cannot obtain the fluent of in any other way I" x a series, except in the case of m=-1; for then we find by the pre 1 ceding series, or indeed without its aid, S. **. 1 -12 1 81. Let it now be required to find the fluent of the exponential formula at X.č. I observe first that a' i la = flux (a"); therefore Sar= a"; and since, integrating by parts, s a* Xi = X fai la Så far, we have fat Xi =** X q* X_1 ya X. Let Å=X'., we shall la la • Because mista. (b)=(w)-a flux ls, of which the fluent is evidently iܫܐ as in the text. X' la 1 la 1 la la la Ра 13 a bave =f« x':="27" a sau i'; again let X'=x" i, and sa X X" we shall have fa'=a* X" for Å", &c. Hence fa Xi=_ax X-X a* X'+ a® X"_&c. till we arrive at an integral fai, 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 se Xi=eX-X'+eX"-e X"+eX"-&c. For example, let X=x", and we shall get X'=nx, X"=n(n-1) ***, X"=n(n-1) (12) ***, &e. Therefore, -1) fari+---+*(n+1) _" (n-1) (1-2)... -1) (, bahs +&c} la la (la) (la) and consequently also sex'i=r{z-now' +7 (921) 34-en (0–1) (0-2) s**+&c.} . . 83. To find the integral of content as the preceding rules do not apply, I reduce a into a series, and obtain +&c.)= tola+sep 0+&c. Therefore c a. sortzC+l+xla +3. B +$ +&c. and S + $ Let e=2, and we shall have the fluxion of a transcen tal quantity which is equal to the infinite series C+l. lz + 1 2.3 flux lz -=zUz-Sż Uz, it clear that file Iz + &c. C+* + } S + +&c. 2.3 2.3.4 z llz- - an integral which likewise depends upon that of the li transcendental quantity s 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 a + and it will be easily integrated. Thus let j==* ; by the above series y=: {1 + mx Ls + ma ma ľ? I m3 m3 13 +&e}==+mvé le + **** P*+&c. of 2 which the fluent may be found by that of time (79), and we obtain fom =x (1 + &c.) + 2.3 mt m2 x2 m2 z C + which in the particular case of x=1, becomes the converging serine moms 1. + 22 44 These examples are sufficient to enable us to integrate such sorts of quantities by means of series. +&c. oi o 4 ano Examples on Logarithmic and Exponential Fluxions. 1. Required the fluent of : Pa? 3. ? : Lx? 4. (x2+2x+1)irst 5 - a ri? 6. et a 7. Required the fluent of ? a' x 8. ? N (1 +.aor) ON THE INTEGRATION OF FLUXIONS CONTAINING SINES, COSINES, &c. n 1 ) n+1 1 1 85. Since cos x=flux sin x, and i sin == flux cos 2, it is evident that sicos x= sin s, and fë sin r=-cos r; as also that Sy cos ny,= S ný cos ny = sin ny, and similarly that sy sin ny 1 Sny sý -- cos ny. Again it is clear that få cos z (sin z)"=S (sin 2)" flux sin 23(sin z)+1 , and thats (sin z) cos" z=- 17(coszyn+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 will become cos (a +1)y+ cos (a-1) y. 2 (a+1) 2 (a--1) 86. A similar process will serve for å sin x sin ar, or i cos o cos ax, &c. With equal facility we may integrate r sin r sin ax cos bx, &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 i sin ’x, i sin 3x, * cos *x, &c.; but it is more simple to integrate them in the following manner. 87. The formula ; sin" =: sin . sin"-1 4. Consequently, integrating by parts, Së sin* == sin-Si sin =-;{ flux (sin x). x : jë sin z -} =-cos sin–13+(n-1) fi sinu? - cos? - =sir--++(n-1) /sin 8-(n-1) si sın" ; and transposing, fi sin* - = 1 *+ na? fi sin- s. By the same process we have fi sin"-27= cos e sine s+maafisins x COS X 1 n 1 3 n-2 n -2 |