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

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

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MATH 1804 University Mathematics A Assignment 5 – Solution April 15, 2011 1. (Note that Z x 0 t sin(sin t ) dt 0 as x 0.) lim x 0 Z x 0 t sin(sin t ) dt x 3 = lim x 0 x sin(sin x ) 3 x 2 ( 0 0 , L’Hˆ opital’s Rule) = lim x 0 sin(sin x ) 3 x = lim x 0 cos(sin x ) · cos x 3 ( 0 0 , 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 s ds (1 - s 4 ) = 1 2 Z dt (1 - t 2 ) ( t = s 2 , dt = 2 s ds ) = 1 2 arcsin t + C = 1 2 arcsin( s 2 ) + C 1

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3. (a) Z x sec 2 x dx = Z x d tan x = x tan x - Z tan x dx = x tan x - Z sin x dx 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 bx dx thus, a 2 + b 2 a 2 Z e ax sin bx dx = 1 a e ax sin bx - b a 2 e ax cos bx + C Z e ax sin 2 x dx = a a 2 + b 2 e ax sin bx - b a 2 + b 2 e ax cos bx + C 0 . (c) Z sin(ln x ) dx = x sin(ln x ) - Z x d sin(ln x ) = x sin(ln x ) - Z cos(ln x ) dx = x sin(ln x ) - x cos(ln x ) + Z x d cos(ln x ) = x sin(ln x ) - x cos(ln x ) - Z sin(ln x ) dx thus, Z sin(ln x ) dx = x sin(ln x ) - x cos(ln x )
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Assignment5(1804)(Solution)(v2-6-4-11) - MATH 1804...

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