数学分析8.2换元积分法与分部积分法(讲义)

-1-1d-1d111-1sinx1-1-1d-1d111-1sinx1第章

不积换积法分积法(讲)一换积法定法设g(u)在[,βu=φ在a,b]上且α(x)β，x∈[a,b]记f(x)=g((x))’x[a,b].1、(第一元分若g(u)在αβ]数G(u)，f(x)在数，F(x)=G((x))+C∫∫g(φ(x))φ’g(u)du=G(u)+C=G((x))2第换积法若φ’(x)≠[a,b]题f(x)在数F(x)，[,β]数G(u)，且G(u)=F(φ∫g(u)du=g((x))φ∫φ证1、G(φ’(φφ’(x)=g(φφ’(x)=f(x)dx∫∫φ(x))’(x)dx=∫g(u)du=G(u)+C=G(φ(x))+C.(：g(φφ’(x)dx=∫g(φ(x))+C2φ’(x)≠0,∈[a,b]u=φ(x)有反数(x)=du

x(x)

,F(φ

-1

(u))=F’(x)·=f(x)·=g(φ(x))φ’(x)=g(φφφφ∴∫g(u)du=∫g(φφ’∫f(x)dx=F(x)+C=F(φ例∫解∫∫

∫令，cosx∫∫u

=∫=2x=1111x-=∫1∫+∫=∫=2x=1111x-=∫1∫+∫例∫

a

dxx

解∫

a

2

d1xaa1例∫

dx

-x2解∫

a

2-x

2

=∫

1-

a例∫

x

dx-

(a解∫

dxx-22a

∫

1xx

[∫∫2axx=

ln2a2ax例∫secxdx.解法1∫

dxcosx

=∫

cos

xxxxsincos2xxxxxsincos22222=(∫2

xxxxcos2xxxxsin2

∫

xxxx22xxxxcos22

1dx=∫lndx=∫66x311xx32x2222a22xx1dx=∫lndx=∫66x311xx32x2222a22xx2x

xxxxcossinsin222

2sin

cosx解2：∫

cosx1sinxx1-2x2x解3：∫

secx(secxsecx

du例∫u解令u=x

6

式=∫

=6∫dx=6∫(x-xx-1

x3x-1x

)dx=6∫(x=6[∫∫xdx+∫xdx+∫+x

+6x+C=6ln|u+33+66u例∫

2-x2

dx解令则2a式=∫

-at

∫∫(=[∫∫=aaa=(

1-

a

xxxx1xx2-a21x=∫∫tdt=∫2111+333xxxx1xx2-a21x=∫∫tdt=∫2111+3331axx例∫

dxx2-

解令0<t<,则2a式=∫

d(asect)(asect)a

=∫

secttant

=∫1+aa

1-

a

1+a

x-

1

x-

例∫

(x

2

dx)

2

解令,则2式=∫

[(atant)2]

2

1sect1ata3

=∫[∫∫34a34a3=

1ttantt2a2a2t)2a

a3(1

a

22

)

+

3

=(3x22例10：求∫

dx

x

x

-1解1：运用法

1∫122121∫12212式=∫

x

3

-

x2

=-∫

1)xx1-x

=1-x2x2x

+C解2：运用法令则式=∫

2tsec2

t-1

=∫

sect

dt=

1-

1sec2

=

1-

11x2x

x2

+C二分积法：定(分法若u(x)与v(x)可导分u则∫∫’(x)dx=u(x)v(x)-∫u：∫分积公)证由u(x)v(x))’=u’(x)，∫(u(x)v(x))dx=∫[u’’(x)]dx=∫’∫’(x)dx，即有∫’(x)dx=∫(u(x)v(x))dx-u’(x)v(x)dx=u(x)v(x)-∫u’例11：求∫解∵sinxdx=-cosx+C，∫xcosxdx=∫xdsinx=xsinx-∫例12：求∫解∵xd(arctanx)=∫

x11xx2

+1)+C2arctanxdx=xarctanx-∫2

31t33tt4t4t114t4t4t4t4t4t142-x2-x2-x-x2-x-x-x-x-x-x-x-x-x-x2-x-x2-x31t33tt4t4t114t4t4t4t4t4t142-x2-x2-x-x2-x-x-x-x-x-x-x-x-x-x2-x-x2-x-x22-x-x-x2-x-x-xaxax11axaxaxaxaaa111axaxaxaxaaa-bI=ecoxbx例13：求∫x解令t=lnx，则x=e∫x∫e∫etdt=∫tde4∵∫edt=e+C，∫=(te-e444

116

e∴原=x16例14：求∫xedx.解∫xe∫x∫edx=2∫xex.∵∫e+C∫xde=xe-edx=xe+e+C，∫edx+e)+C，式=-(xe-edxe+ee-2e