Pre-Calc Homework Solutions 257

Pre-Calc Homework Solutions 257 - Section 6.3 31. Let u (3)...

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Unformatted text preview: Section 6.3 31. Let u (3) dx, so 9 257 28. Let w dx e 2 3 3x 9 3x 3x dx 9. Then dw 9 dw w 1 2 3x xn nx n 1 dv dx v cos x dx sin x (sin x)(nx n 1dx) n xn dv 1 du 2 w dw. 3 x n cos x dx 2 w ew dw 3 x n sin x x n sin x 2 (e ) w dw 3 sin x dx Let u du w dv dw v e dw ew w ew w ew (w ew dw ew 1)ew 33. Let u du 1)e 3x 9 w 32. Let u du xn nx n 1 sin x dx cos x ( cos x)(nx n 1 w ew dw dx v x n sin x dx (x n)( cos x) x ncos x dx) n xn dv 1 cos x dx e 3x 9 dx 2 w ew dw 3 2 (w 1)ew 3 2 ( 3x 9 3 xn nx n 1 e ax dx 1 ax e a 1 ax e (nx n 1 dx) a dx 1 a v C x ne ax dx (x n) e ax x ne ax a 29. Let w x e 7 x2 x 2. Then dw dx (x ) e 2 3 x2 2x dx. x dx 1 3 w w e dw. 2 n n 1 ax x e dx, a a 0 Use tabular integration with f (x) f(w) and its derivatives w3 3w2 6w 6 0 (+) () (+) () g(w) and its integrals ew ew ew ew w 3 and g(w) e w. 34. Let u du (ln x)n n(ln x) x n 1 dv dx v x dx (ln x)n dx (ln x)n(x) x(ln x) n n(ln x)n x x 1 dx dx f (y) dy. y f (y) dy n (ln x) n 1 35. (a) Let y f 1 (x). Then x (x) dx f (y), so dx (y)[f (y) dy] dv f ( y)dy f (y) f (y) dy f (y) dy Hence, f (b) Let u du 1 y dy y f ( y) f 1 ew v w 3 e w dw w3 ew (w 3 3w 2e w 3w 2 6w 6w e w 6)e w 6e w C C y f (y) dy (x)(x) x 7e x dx 2 1 3 w w e dw 2 1 3 (w 3w 2 6w 6)e w C 2 2 (x 6 3x 4 6x 2 6)ex C 2 1 dr, and so dr r Hence, f 1 (x) dx y f ( y) dy xf 1 (x) dx x f ( y) dy. 36. Let u du f 1 (x) dv v d f 1(x) dx dx 30. Let y ln r. Then dy r dy e y dy. f 1 (x) dx xf f 2 1 (x) d x f 1(x) dx dx Using the result of Exercise 13, we have: sin (ln r) dr (sin y)e dy 1 y e (sin y cos y) C 2 1 ln r e [sin (ln r) cos (ln r)] C 2 r [sin (ln r) cos (ln r)] C 2 y 37. (a) Using y 2 1 (x) sin 1 x and f (y) sin y, y 1 , we have: x sin x sin x sin 1 1 1 sin x dx x x x sin y dy cos y cos (sin C 1 x) C ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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