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sin 6x 1 1 1 2 1 + C; (c) sin5 x sin7 x + sin9 x + C; 39.12 (a) cos3 x + C; (b) x + 3 2 12 5 7 9 (d) (g) 1 8 3 1 1 5x + 2 sin 2x + sin 4x sin3 2x + C; (e) 2 8 6 16 sin 4x sin3 2x sin 8x 5 x + 8 8 3 64 + C; ( f ) 2 tan x x + C; 2
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3 1 1 3 1 tan5 x tan3 x + tan x x + C; (h) sec3 x tan x + sec x tan x + ln |sec x + tan x| + C; 5 3 4 8 8 3 1 1 1 1 1 5 (i) tan5 x + tan3 x + C; ( j) sec5 x sec3 x + C; (k) sec3 x tan x sec x tan x + ln |sec x + tan x| + C; 5 3 5 3 4 8 8 1 1 1 1 sin 5x sin 13x (2 cos 2 x cos 4 x) + C; (n) (6 sin 2x sin 12x) + C; (o) + + C. (l) cos 4x + C; (m) 8 8 24 2 5 13
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39.14 (a) 0; (b) . 39.15 (a) x 2 1 sec 1 x + C; (b) 2 sin 1 1 9 x x 4 x 2 + C; (c) 2 2 1 + x 2 + ln 1 + x2 1 + C; |x|
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x 3x 1 x2 9 x + C; ( f ) + C; tan 1 + 2 + C; (g) x 54 3 x +9 4 4 x2 1 4 16 9x 2 1 1 4 16 9x 2 (h) ln + C = ln + K; (i) [sin 1 x x(1 2x 2 ) 1 x 2 ] + C; |3x| |x| 4 4 8 1 x 3 2(x 3) 1 + C. tan 1 + 2 ( j) (sin 1 ex ex (1 2e2x ) 1 e2x ) + C; (k) 8 16 2 x 6x + 13 (d) 2 x 2 + C; (e)
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1 17 + ln ( 17 + 4). 4
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39.18 Same answer as to Problem 39.17 (because the two arcs are mirror images in the line y = x). 39.19 ln (2 + 3). 39.20 9 4 = 6 .
40.6 (a) x3 1 3 x 3 1 + C; (b) 3 ln |x + 3| 2 ln |x + 2| C; (c) ln + ln |x + 2| + ln |x 2| + C; 6 x+3 3 4 4 3 19 3 3 5 (d) ln |x 1| 9 ln |x 2| + ln |x 3| + C; (e) ln |x 1| ln |x + 1| + ln |x 3| + C; 2 2 4 8 8 5 x x2 9 1 13 7 1 1 ( f ) ln |x| + ln |x + 3| ln |x + 2| + ln |x 1| + C; (g) ln 2 + + C; + C; (h) 6 ln 6 6 6 6 10 x+1 x x 4 4 1 x 2 1 x 1 1 + + C; ( j) ln + C; (i) ln 4 x+2 x 2 9 x+3 3x+3 x2 1 5 1 x2 (k) + C; + 2x + 256 ln |x 4| + 19 ln |x + 1| + + C; (l) ln 2 2 25 x+1 10 x +5 1 x 9 13 1 1 (m) ln |x 1| + ln (x 2 + 4x + 5) tan 1 (x + 2) + C; (n) tan 1 x tan 1 + C; 10 20 10 3 6 2 1 1 1 1 x2 + [ln |x| 41 ln (x 2 + 9)] + C; (p) ln |x| ln (x 2 + 1) + (o) + C; 2 9 2 2 x2 + 1 x 1 1 x 4 3 1 ln |x 1| ln (x 2 + 4) + tan 1 + (q) + C; 25 50 100 2 10 x 2 + 4 2 x 1 3 2x + 1 1 x+1 + C; tan 1 (r) ln |x| ln (x 2 + x + 1) + C; (s) ln 2+x+1 2 9 3x x+2 3 1 11 (t) ln |x| + ln |x + 3| 3 ln |x + 2| + C; (u) x ln (1 + ex ) + C. 3 3 1 2 (3 ln 2 + 3); (b) ( 3 3 ln 2). 81 27 40.9 2 1 tan x + C. 2
40.8 (a)
40.10 (a) ln (1 sin x) + C; (b) ln |sin x 1| + C; (c) sin x 1. 3 ( 4 x 1)3 x 1 1 + 3x 1 + C; + 4 x 1 ln (1 + 4 x 1) + C; (c) ln 40.11 (a) ln 1 x 2/3 + C; (b) 4 2 3 2 1 + 3x + 1 4 (d) 2 x 3 3 x + 6 6 x ln (1 + 6 x) + C; (e) ( x 2) x + 1 + C; 3 1 + ex 1 + C = 2 1 + ex + 2 ln ( 1 + ex 1) x + C; ( f ) 2 1 + ex + ln x +1 1+e 1 x+1 3 2/3 2x 1/3 1 + ln x 1/3 + 1 ln 1/3 (g) x 3 tan 1 + C. 2 2 x +1 3
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