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7. ∫sin(x/2)cos(x/2)dx = (1/2)∫sinxdx = -(1/2)cosx + C
8. y' = 6e^x-3(secx)^2
9. y' = -2/(1-2x)
∫<1, 2>dx/(2x-1)^2 = (1/2)∫<1, 2>d(2x-1)/(2x-1)^2
= -(1/2)[1/(2x-1)]<1, 2> = 1/3
∫<0, 1>xe^xdx = ∫<0, 1>xde^x
= [xe^x]<0, 1> - ∫<0, 1>e^xdx
= e - [e^x]<0, 1> = 1
8. y' = 6e^x-3(secx)^2
9. y' = -2/(1-2x)
∫<1, 2>dx/(2x-1)^2 = (1/2)∫<1, 2>d(2x-1)/(2x-1)^2
= -(1/2)[1/(2x-1)]<1, 2> = 1/3
∫<0, 1>xe^xdx = ∫<0, 1>xde^x
= [xe^x]<0, 1> - ∫<0, 1>e^xdx
= e - [e^x]<0, 1> = 1
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后面还有两个小题
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2 (1) f(x) = cos(x+π/3-π/3)
= (1/2)cos(x+π/3)+(√3/2)sin(x+π/3),
代 cosu,sinu 级数展开式
(2) f(x) = 1/(4-x) = 1/[2-(x-2)]
= (1/2)/[1-(x-2)/2]= (1/2) ∑<n=0,∞>[(x-2)/2]^n
-1<(x-2)/2<1, 0<x<4
(3) f(x) = lgx = lg[1+(x-1)] = ln[1+(x-1)]/ln10
代 ln(1+u)级数展开式
(4)f(x) = 1/(x^2+4x+9) = 1/[5+(x+2)^2]
= (1/5)/[1+(1/5)(x+2)^2]
= (1/5)∑<n=0,∞>(-1)^n[(x+2)^2/5]^n
= ∑<n=0,∞>(-1)^n (x+2)^(2n)/5^(n+1)
-1<(1/5)(x+2)^2<1, -(√5+2)<x<√5-2
= (1/2)cos(x+π/3)+(√3/2)sin(x+π/3),
代 cosu,sinu 级数展开式
(2) f(x) = 1/(4-x) = 1/[2-(x-2)]
= (1/2)/[1-(x-2)/2]= (1/2) ∑<n=0,∞>[(x-2)/2]^n
-1<(x-2)/2<1, 0<x<4
(3) f(x) = lgx = lg[1+(x-1)] = ln[1+(x-1)]/ln10
代 ln(1+u)级数展开式
(4)f(x) = 1/(x^2+4x+9) = 1/[5+(x+2)^2]
= (1/5)/[1+(1/5)(x+2)^2]
= (1/5)∑<n=0,∞>(-1)^n[(x+2)^2/5]^n
= ∑<n=0,∞>(-1)^n (x+2)^(2n)/5^(n+1)
-1<(1/5)(x+2)^2<1, -(√5+2)<x<√5-2
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在下愚昧,表示看不懂阁下所言
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