INFINITESIMAL CALCULUS 35 fa* (cos fix + i sin px}dx = *""( flg + * ain ftr) J a + ip Ex, 4. Here dx Let cx + b = z, and it transforms into the preceding integral ; licnoc Ex. 5. u= i Let (a + cx-}*- x~, then (a + cx 2 )* c-z z and the integral transforms into Ex. 6. Let x = a sin -0 + )3 cos-0, and we get /(a-x)(x-p) -2 sin 1 f x ~& . V a-jB = /"l!f /<#-_a? J x V a; 4 - /3- c. 7. Here If we make a; 2 = - in the former, and x"=y in the latter, they each reduce to the preceding example. Ex. 8. / dx J a + b cos x Here u C J (a dx (a + b) cos 2 -^- + (a - b) sin 2 -^ where 3 = tan . This integral is a circular or a logarithmic function, according as a >, or, < b. (1) Let a > b, and suppose b = a cos a, then we have -l C ^ = 2 tanV tan -- tan ^ . a / ,.2 <* , ~2 ,,,-,,2 sin a V 2 2 , let = J cos a, then cos f- sin tan _L_ loc, 2 _J 2 6 sin a a a a 1 cos sin tan 2 22 a -e log (3) If a = b, the value of the integral is . a 2 114. The substitution of an imaginary expression for a constant in an integral is often useful in evaluating integrals. For example, if in the equation /,>az
- a *dx =
a we substitute 1 a + ip for a, it becomes 1 We shall throughout represent the imaginary symbol V- 1 by i, according to the usual notation. Hence, equating the real and also the imaginary parts, we get c "*( g COS 0* + 0i omitting the arbitrary constants. These results can be easily veri fied. 115. The method of integration ly parts depends on the follow ing equation, which is deduced immediately from the relation d(uv] = udv + vdu : Jvdv *= uv -fvdu . Hence the determination of the former integral is reduced to that of the latter, and vice versa. Ex. 1. To find / tan ~ l x dx . Hero /tan - *x dx = x tan - *x - f^- = x tan - U- - i log (1 + x-) . j y n- as 8 Ex. 2. Next, to find Jx n log x dx .
- lo
n+l Ex. 3. Again, to find /log (a; + Jx- + a 2 )dx . /i , / f xdx og (x + Vx 2 + a-)dx = x log (x + V.t 2 + u 1 ) - /r J/x* + a? = x log (a; + Va; 2 + a 2 ) - lx- + a- . Ex. 4. /(lg x) n dx . -= x(log a:)" - ?w. (log x)"- 1 + n(n - 1 /(log a;)" - 2 c?a; . Accordingly, by successive applications of this formula, the in tegral can be found whenever n is a positive integer. If n be a negative integer, the integral finally depends on /" , a form y ig * to be subsequently considered. This is at once reduced to the preceding by making s = a^+ 1 . Re. 6. This is immediately reducible to Ex. 4 by making c ax = z. It can also be deduced directly, since fx m e ax dx . cible to Ex nee /^rtfc.*5rl * y Ex.1. 116. In general, if -. + -- . - . . . +(-!)" -r-v i*-* r/.r^ 2 aai-3 v /7>. we have This result is readily proved by successive applications of the method of integration by parts, or can at once be verified by differentiation. As an example, let us consider the integral dx n+l where F(.r.) represents a rational integer algebraic function of a 1 , of the degree n. Let M=*F(*), an consequently we have v = -! - , then
