Assignment5(1804)(Solution)(v2-6-4-11)

Assignment5(1804)(Solution)(v2-6-4-11) - MATH 1804...

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Unformatted text preview: MATH 1804 University Mathematics A Assignment 5 – Solution April 15, 2011 1. (Note that Z x t sin(sin t ) dt → 0 as x → 0.) lim x → Z x t sin(sin t ) dt x 3 = lim x → x sin(sin x ) 3 x 2 ( , L’Hˆ opital’s Rule) = lim x → sin(sin x ) 3 x = lim x → cos(sin x ) · cos x 3 ( , L’Hˆ opital’s Rule) = 1 3 2. (a) Z dx (1 + √ x ) 2 √ x = 2 Z du u 2 ( u = 1 + √ x, du = dx 2 √ x ) = 2 ·- 1 u + C =- 2 1 + √ x + C (b) Z cos(ln x ) x dx = Z cos u du ( u = ln x, du = dx x ) = sin u + C = sin(ln x ) + C (c) Z sds √ (1- s 4 ) = 1 2 Z dt √ (1- t 2 ) ( t = s 2 , dt = 2 sds ) = 1 2 arcsin t + C = 1 2 arcsin( s 2 ) + C 1 3. (a) Z x sec 2 xdx = Z xd tan x = x tan x- Z tan xdx = x tan x- Z sin xdx cos x = x tan x + Z d cos x cos x = x tan x + ln | cos x | + C (b) Z e ax sin bx dx = 1 a Z sin bx d ( e ax ) = 1 a e ax sin bx- 1 a Z e ax d sin bx = 1 a e ax sin bx- b a Z e ax cos bx dx = 1 a e ax sin bx- b a 2 Z cos bx d ( e ax ) = 1 a e ax sin bx- b a 2 e ax cos bx + b a 2 Z e ax d cos bx = 1 a e ax sin bx- b a 2 e ax cos bx- b 2 a 2 Z e ax sin bxdx thus, a 2 + b 2 a 2 Z e ax sin bxdx = 1 a e ax sin bx- b a 2 e ax cos bx + C Z e ax sin2 xdx = a a 2 + b 2 e ax sin bx- b a 2 + b 2 e ax cos bx + C ....
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This note was uploaded on 04/19/2011 for the course MATH 1804 taught by Professor Ng during the Spring '11 term at HKU.

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Assignment5(1804)(Solution)(v2-6-4-11) - MATH 1804...

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