课件ch6 26 2定积分计算法

bf(x)dxF(b)Fa

x

d1

arccos(ln

6

2ln

2

e

1e1e21

ln

11e2

(uex d arcsin 111 注注

0

e2

3ex3ex

e3e3ex0

(uex 1

(v

v23u则3u2v2 2

vd3

9

2)d

9

dx

sin2x|cosx|

xcosxdx

sin2xcosx

0

xdsinx

xdsinxsin25

2 5 例 0

(a 0

xaasin a(1sint)a2

acostda

2cos2t

a

2sintcos2t 2a 0

dta3(t2

sin2t)

a21x例 x2

(a1（令x t (1)dt4 tt4(at)2(at)2 1

2a

31]32a3a

[4a

31]

3]1[a23]3afx)，gx)分别为[ll] f(x)dx g(x)dx

g( 证 f(x)dxf(x)dxf( 0

l(xt

f(t)d(tll

f(0l f(t)dt l

f(l0

f(t)dt

f( 0

g(x)dx g(x)dx g(

2(xt

g(t)d(tl

g(0l g(t)d0

g(0l g(t)d0

g(0

g(01计算1

2x2xcosx 解原式 dx

1x dx1x

x2(1

)040

1

1

11

4

1(1x21x2dx4定理fx是以T为周期的连续函数，a

f(x)dxa0

f(0 a

f(x)dxa

f(x)dx0

f(x)dxTa

f(a f(T

(txT)0

f(tT)dt0

f(t)d f(x)dxa

f(0注

f(x)dx

f(x)T T2

f(x)0

TT0

f(x)例 设f(x)在[0,1]上连续，试

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

bba aa cos2|3| 3xdtan0

xtan 0

3tanx0

|ln|cosx|0ln例计 1ln(1x) (2解

ln(1x)

1ln(1 (2ln(1

210 dln(11012 1

2ln3

21 211

ln

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

ln 例 求In

20

x 2 2 I0

2dx

2

xdx

cosx| n2

In2sinnxdx2sin xd(cos cosxsinn1x|0

(n 20

xsinn2x(n

0

xdx(n

2sin0

即In(n1)In2(n当n为偶数时不妨设n 2m1 2m12m3 2m 2m12m331 2m12m331 2m 2m (2m1)!! 当n为奇数时不妨设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 设f(x)

dt，求0xfx 1xf(x)dx11f(x)dx01x

f(

1

sinx

x f(1)

sinx2dx1cos(x22

1cos1 例 sin2x2例 sin2x2e xee I

(xt2

e

td

sin2td 2I 2

e

2sin2xdx

sin2x e2e

e 2 2221

dx(x

sin2x) ex

I 2

xdx例

4ln(10

tanx (x4

t lntan(4t)dt4ln1

1tant4 1tant4

dt

4ln2dt

4ln(1tant)d 1tant

4ln(1tant)d0

8sin(A+B)=sinAcosB+cosAsin