課件ch6 26 2定積分計算法

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6

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ln

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2 5 例 0

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a21x例 x2

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g(01計算1

2x2xcosx 解原式 dx

1x dx1x

x2(1

)040

1

1

11

4

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f(x)dxa0

f(0 a

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f(x)例 設(shè)f(x)在[0,1]上連續(xù)，試

xf(sinx)dx

f(sinx)

xsin 1cos2

xf(sinx)

(t)f(sin(t))0 (x

t) (t)f(sint)dt f(sint)dt tf(sint) 2 xf(sinx)dx2

f(sinx) xsin

sin dx

dcos01cos2

2

cos2

201cos2

bba

uv

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xtan 0

3tanx0

|ln|cosx|0ln例計 1ln(1x) (2解

ln(1x)

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210 dln(11012 1

2ln3

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ln

ln(1x)ln(2x)15ln2

ln 例 求In

20

x 2 2 I0

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2

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cosx| n2

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(n 20

xsinn2x(n

0

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即In(n1)In2(n當(dāng)n為偶數(shù)時不妨設(shè)n 2m1 2m12m3 2m 2m12m331 2m12m331 2m 2m (2m1)!! 當(dāng)n為奇數(shù)時不妨設(shè)n2m

2m1

2m

I2m3

2m 2m

2m 2m

2m 2m 2m

2m 2m (2m 例解

求 2cos0t t 2

x In

sinntdt2

2sinntd0(2m1)!! n2m (2m n2m1例 0

cos6xdx

531642x2 例 設(shè)f(x)

dt，求0xfx 1xf(x)dx11f(x)dx01x

f(

1

sinx

x f(1)

sinx2dx1cos(x22

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e

td

sin2td 2I 2

e

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sin2x e2e

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dx(x

sin2x) ex

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xdx例

4ln(10

tanx (x4

t lntan(4t)dt4ln1

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dt

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4ln(1tant)d0

8sin(A+B)=sinAcosB+cosAsin