Page:Encyclopædia Britannica, Ninth Edition, v. 13.djvu/47

 1 N F INITESIMAL C A L C U L U S 37 vliere /Vac -b K ^ V 9, reduces to / f(z), by making e^^ 2 : and, accordingly, it can / J / y V O V be integrated whenever /(z) is a rational algebraic function of z. Again, if we take tan ^x = z, we get 2; 1 c 2 2dz sin a;= --, cos *=, -;, , dx=~ ~, and, tlic expression /^sin x, cos x)dx, / 9- i _ -2 o,/-. transforms into / ( -, , - ~- ) -^L . Consequently, whenever /(sin *, cos *) is a rational function, the integration of/^sin *, cos x)dx is reducible by the method of partial fractions. 119. Functions of this latter class are, however, usually more readily integrated by other processes. Thus, when /(sin *, cos a;) is a rational and integer function, its integration depends on that of the sum of a number of expressions of the form sin&quot; 1 * cos&quot;*^*. As a number of other forms are readily reducible to this type, it is proposed to devote a short space to its discussion here. In the first place it should be observed that whenever m or n is an odd integer, the expression sin &quot;* cos&quot;* dx can be immediately integrated. For, if we suppose n = 1r + 1, the integral transforms into by making z = sin *. Hence, as r is by hypothesis a positive integer, (1 -r J ) r can be expanded in a finite number of terms, and, the integral thus immediately obtained. Again, if m + n be an even negative integer, the expression can be readily integrated ; for, by assuming z = tan *, we get m+n_ cos n xdx=/z m i 1 +z&quot; I 2 dz. Again, fx. mJ r- This integral can be readily obtained by expansion. 120. AVhen neither of these methods is applicable it is usual to find the integral of sin m * cos&quot;*ffe by the method of successive re duction. The formula? of reduction can be easily obtained by the method of integration by parts ; thus /in &quot;* cos&quot; cdx= /cos&quot;&quot; 1 * sin&quot; ! *o ( sin *) = /cos&quot;- * f // s j n m+i r J J 7/1 + 1 cos&quot; 1 * sin &quot; +1 * n-1 /~. , = - / sm &quot;+-* cos&quot; -*a*. in + 1 m+ iy fsin m x(l - cos 2 *) cos&quot;- 2 *f?* (/ Substituting in the former equation, and transposing the latter integral to the other side of the equation, we get /, ,7 sin &quot;+ 1 * cos&quot; 1 * n- /* , sm m * cos&quot;** = h-- - /sm &quot;* cos&quot;&quot; -xdx. in + n m + nj Hence, when n is positive, the integral of sin m * cos&quot;*cZ* depends on that of sin &quot;* cos&quot;- 2 *^*. The corresponding formula? in which the degree of sin * is reduced can be immediately found. It should be noted that these formulae of reduction are perfectly general, and hold whether m and n be positive or negative, integer or fractional. Accordingly, changing the sign of m, our first equation may be written thus : /cos&quot;* 7 cos&quot;- 1 * n f cos&quot;- 2 * , dx= - -- / - -. dx . sin &quot;* (in- 1) sin &quot;- J * m-iy sin &quot;--* 121. These formula? of reduction, as well as many others, can be readily established by differentiation. For example, since (sin* cos&quot;*) = ??i sin- 1 * cos&quot;+ ] *- sin m + 1 * cos&quot;- 1 * dx = m sin- 1 * cos&quot; 1 *- (m + n) sin m + 1 * cos&quot;- 1 * , the integration of the expression sin m + 1 * cos&quot;- 1 *^* depends on that of sin*&quot;- 1 * cos&quot;- 1 *^* ; and similarly in other cases. It may be noted that the integral ( 118) /&quot; dx is at once reduced to the class here considered by making* - &amp;lt;x = /3tan 9, when it becomes To find Here - . n-l J tan&quot;- 1 * tan&quot;- 3 *. n~- n- 8 ^ Next, let us consider the integral r dx Let tan = -,- &amp;gt; find we get a cos x + b sin a- = ( 2 + & 2 )* sin (* + o). Hence, making a: + a z, the integral transforms into J sin&quot;z 122. In many applications the results depend on integrals of the form here discussed when taken between the limits and --. Such definite integrals are easily found when the indices m and n are positive integers. TT Commencing with the simple case of j 2 sin&quot;*a&quot;*, we have, since sin * cos&quot; l x vanishes for both limits, /&quot; 2 co$ n xdx = n * /&quot; 2&quot; cos&quot;- 2 * yo 5l yo - 2 *o&quot;*. By successive applications of this formula the definite integral in question can be always found when n is a positive integer; its form, however, depends on whether the index n is even or odd. (1) Suppose n even, and equal to 2r, then 2r 1 /* - = ~2^yo and, accordingly, by successive applications, we get r -f- ,,,, 1.3. 5 . . . (2r-l) TT_ y o &quot;&quot;2.4.6... Zr 2 (2) If n be odd, and equal to 2r + l, we get in like manner /j.4.1 7 2.4.6.. 2r 2 cos 2r+1 *a*=. It is evident that in all cases 9 /&quot; ~n~ / (3) In like manner, we have n ~ As in the former case, the value of this definite integral depends on whether the indices are odd or even. First suppose n odd, and equal 2r + 1, 2r Hence I 2 sin&quot; 1 * cos 2r + ] *c?* Jo 2?-(2r-2) ... 2 r 2. 4 . 6 ... 2? &quot;&quot;(m + !)(?+ 3). . . Xext let n be even, and equal to 2r, then Hence, as before, 1.3.5... (2r - 1 ) 1.3.5... (2r-l) .1.3.5... (2OT-1) w &quot;2.4.6 (2m + 2r) 2 When m and n are both fractional these definite integrals are