INTEGKATION [LLUSTRATIVE EXAMPLE 1. Find C sin 2 x cos 5 xdx. Solution. f sin 2 x cos 5 xdx = f sin 2 x cos 4 x cosxdx = I sin 2 x (1 sin 2 x) 2 cosxdx = f (sin 2 x 2 sin 4 x + sin 6 x) cosxdx = J (sin x) 2 cos xdx 2 f (sin x) 4 cos xdx + f (sin x) 6 cos xdx Here v = sin x, du = cos:xdx, and n = 2, 4, and 6 respectively. ILLUSTRATIVE EXAMPLE 2. Find f cos 3 xdx. Solution. J cos 3 xdx = f cos 2 x cosxdx f (1 sin 2 x) cosxdx = I cos xdx I sin 2 x cos xdx
by 28, p. 2 By (4) sin 3 x = sin x --- 1- G. EXAMPLES . fsi J . fsin 2 x cosxdx = ^^ t/ 3 . sinx cosxdx = . f sin 3 60 cos60d0 = ^- + (7. J 24 fsi J . f / . f CO J si
+ C. COs3 ^ sin 4 x 3 . . cos 2 a = esc x - esc 3 x + C. 3 = sec a + cos a + O. . I cos 4 xsin 3 xdx = ^cos 5 x + |cos 7 x + C. . j sin 5 xdx = cosx + - cos 3 x -- + (7. J 36 . I cos 5 xdx = sinx -- sin 3 x - --- h G. J 35 . fsin^ cos 3 0d</ = ^ sinV ^ sinV + C.
- This was integrated by the power formula taking n = 1, u = sin x, dv = cos xdx. To illus-
trate how an answer may take on different forms when more than one method of integration is possible, let us take /i = l, v = cosJC, dv = sin xdx, and again integrate by the power formula. Then //* COS3J sin x cos xdx = - I (cos x) (- sin x dx) = -- - + C', a result which differs from the first one in the arbitrary constant only. For,
2
Hence, comparing the two answers, C=
+
