Second criterion. If 1. Let du Here m = 2 n 1 x dx m n √ (a + x) = = (a + x)−x dx. 2, and the first criterion is satisfied. Let then a + x = z2; then x2 = (z2 — a)2 = z1 — 2 a z2 + a2; ... x dx = 2zdz-2az dz 2 (z az) dz; = ( z − 2 a z − ( a + (x − 2 a) (a + √ (a2 - x2). d x = x)1. 2 f *2 dz-2afd z · (a2 — x•)* x−* dx. 5 1 2 + 2 Here m n criterion is satisfied; hence we may assume - √ (a2 x2) a2 — x2 = x2x2; then (ax)= zx, and x2 = a2 (22 + 1)"'; du (a) da s (2 + 1) dz; (a2 — x2) хо = 15 a1 x3 3a2 2x2 } 15 a* x3 (a2 — xo) F. (2 * − 1) (1+x*)*. +29) 3x8 INTEGRATION BY SUCCESSIVE REDUCTION. 98. When neither of these criteria is satisfied, binomial differentials may still be modified so as to assume a simpler form, and to admit of integration by other means. If we can make x (a + bx")" dx depend on another integral of the same form, but with smaller values of m or p, it is evident that by successive repetitions of the process we shall finally arrive at an integral which can be determined by an elementary form. This is called the method of reduction: it is applicable to a great number of functions, and is very convenient in practice. I. Let it be required to integrate du = x dx x2−1 (a2 − x2)* + (n − 1) ƒ x2−2 (a2 — x2)* dx XR-1 = − x2-' (a2 — 2o) * + (n x2−2 (a2 — x2) d x (a2 — x2)* − 1) f x2-2 dx (a2 — x2)+ x dx ૨) Transposing the last term of the second member, and dividing by n, (n − 1) a2 By the successive application of this formula of reduction, the proposed integral will be reduced, according as n is even or odd, to These integrals may be continued at pleasure, and the method employed in the investigation of the formula of reduction (A) should be followed by the student in each of the preceding examples. The method, and not the formula, should be recollected. II. Let it be required to integrate du = (2 ax-x) x2-1 dx x dx (2ax - x2)+ (ax) dx - x* −1 (2 ax − x2) * + (n − 1) ƒ x2-2 (2 a x − x2) * dx. But fx-(2ax-x2)* dx = f x2-2 S x2-1dx น, (2ax-x2) x" - (2 a x − 2o) * + 2 a (n − 1) S x-1dx (2 a x - x2) therefore by transposition, and collecting like terms, we obtain IV. Integrate du = (a3- x2)dx, where n is an odd number. Here u = VOL. I. n = ƒ (aa — x*) * . dx = x (aa — xo)" + 1⁄2 ƒ 2 xoa (aa—xo)1 ̈'dx − 2 2 E = x (a2 − x2); + nf {a2 — (a2 — x2)} (a2 − x2)TM TM dx - 2 = x (a2 — − xo)2 + a3n ƒ (aa − x2) dx — nf (a2 — x2)1 dx; * • ' . (n+1) u = x (a2 — x2) * +na2ƒ x2), dx ƒ (aa — x3) d x = x (aa — xo) n+1 EXAMPLES. z (a* − 2*)* + % sin~ 2. x (a* 4 5 a2 + 3 65 ƒ (a2 − x2) * dx x2)* x = 6 x” (x2 — a2)1 = + (n + 1) √ x2+1 (x2 - a2) d x xTM+2 (x2 — a2)‡ + (n + 1) u − (n + 1) a2 (n + 1) a2 f; 2, or n+2 into n, then will -+ (n-1) S dx dx x2+2 (x2 — a2) a2) + - (n-1) a2u; whence, by transposition, collecting, and dividing by (n - 1) a3, |
