高數(shù)-5-12積分換復(fù)合函數(shù)求導(dǎo)法改變變量

分換小復(fù)合函數(shù)求導(dǎo)法（改變求導(dǎo)變量求導(dǎo)法——微分法中的“換元法積分法中的“換元法（改變積分變量積分法）一、第一類換元第一類換gxf(xx)且f(u)duF(uC則g(x)dxf((x))(x)dx（f((x))d((x)）(f(u)du)u(x)(F(u)C)u(x)——（湊微分法）g 湊成f[x)]第一類換定理若f(u)duF(u)C,且(x)可導(dǎo) f((x))(x)dxF((x))證∵dF(( d[F(u)]

(u( f(u)( f((x))(F(x))f(x))x)的一個(gè)原F((x))Cf(x))x)的不定例1求sin(2x解sinuducosuc,為套令u2x3則dusin(2x3)dx1sin(2x3)2dx1sin(2x3)d(2x2sin(2x

湊微分2

sin(2x3)d(2xu3

sin

基本公 2還

cosu2 s2xC;例2

1

dx2

2 1(x2ux 21u2du2arctanu還原2

arctanx2C例3

x(12ln

dx

d(lnx)12lnx 12ln

d(12ln uln

11du1ln|u|C 1ln|12lnx|C2例4tan

sin cos

dx

dcosxcosx1 lncosx1 1xx1xx

dx

dxxxx xxx

1 xd21 xd(1

234u33

C

34 x)2C43例 (14x2)(arctan2x d2 [1(2x)2](arctan2x d(arctan2x2(arctan2x 2(arctan2x例7

dx1 a2x1a2xa2 x2 x21a

dxaa

1arctanxC a2x21(xa2x21(x)2a同理,

dx arcsinxCa1例81ex11ex

1ex(1ex 1e dx

1e

d(1exxln(1ex)C例9

csccscxdx

1sin

dx

xcos

x 2

dtanx

x 2x2 2 ln|

x|Cln|cscxcotx|C2類似地可secxdxln|secxtanx|C解法二cscxdx

sin

dx sinxsin2 d(cosx)1cos2xucos

(1u)(1

1u2

2

(1u)(1 1 du1

d(1u)

1cos21 1 2 1 1cos11112

C12

C.解法

cscxdx cscxcscxcotxdxcscxcot xcscxcot cscxcot d(cscxcotx)cscxcotxlncscxcotxC但根據(jù)不定積分定義,它們之間只相差例10求

1cos

解法

dx

1cos1cosx1cosx1cosxdx

dx

d(sinsin2cotx

sin

sin2C

sin2解法

dx

d

sec2xd1costanxC2

2例11求cos3xcos2解cosAcosB1[cos(ABcos(A2cos3xcos2x1(cosxcos5x22cos3xcos2xdx1(cosxcos5x21sinx1sin5xC 一般可求:sinmxcos sinmxsincosmxcos32x32xx2解由于(32xx222-1(2-2x)32xx232x32xx232xx232xx32xx2-32xx32xx232xx32xx4(x2

d(32xx2)

d(x32xx2 arcsin32xx22類似地可計(jì)算

xx26x

例13(1)設(shè)f(x)dx(lnx)2C,則xf(x)dx 解由于f(x)[(lnx)2]2lnx1 f(x)22lnx故x

(x)dx2

1ln

dxx

lnx

2[lnx1(lnx)2]C2lnx(lnx)2C2(2)exfx)則x2f(lnx 解由題fx(ex)exf(lnx)elnx1xx2f(lnx)dxx21)dx1x2C 二、第二類換元定 設(shè)x(t)單調(diào)）可導(dǎo)且(t)f[(t)](t)dtF(t) f(x)dxF(1(x))其中1xxt ∵d

F

1(x))F(t)

(t

f[(t)]f(xf(x)dxF(1(x))C

第二類換元法求積分過(guò)程 x( f( g(tt1(xF(t) F(1(x))C注從運(yùn)算形式上看是第一類換元法的反向使用例14求a2x2dxa解令xasint,dxacostdta2x2dxacostacos a2cos2a2

a(1cos2t)dta

(t1sin2t)ata2xata2xa2xa ) 1(a2arcsinx

a2x2)a2x2例

(ax2a2txx2a2tx令xatantdxasec2 22t, 221x2a21x2a2

asec

asec2tdtsecx2ax2a2aaln(secttant)C aaln(x

x2x2a2)C例16

(a1x2a2令x1x2a2

則dxasecttanxx2a2xx2a2txat

xat

,2 xa dx

asecttant

asecttantdtx2a2

a|tant

atanx2x2sectdtln(secttant)C

x2aln x2axa dx

asecttantdt

secx2a2

x2ax2a2aln|secttant|

x2a2x2x2a2x2a2

C x1x2x1x2a2x2a2xa2

Clnx

C可使用三角代換去掉如下二次根 a2a2a2a2x2xx2

可令xasintta2或xa2x2a2txx2a2t或xacott; xtxasectxtx2a2xx2a2a當(dāng)分母的階>>分子的階時(shí),可考慮試用倒代換x1.t1x1t

1例

x(x7

17 t

t2t

d2t

dt

7 12t

11ln|12t7|C1ln|2x7|1ln|x|C 例x

x21dx

（分母的階>>分子的階131

dx

1t1t1t1t12

t1t

d(t2

ut2

u1u1duu 1(u1 u12)du1u32u12 1(11)32(11)12 Cx2 x2解法

x21dx

令xtanxdxsec2 1sec2 cos 1 dt dttan4tsec sin4

1sin2sin4

dsin sin4

dsint

sin2

dsinxx2t 1 C 3sin3(1x2(1x2 x

sin1x1xx例19

x

dx 3x u去根號(hào)u33x

1解原

u3x

u2du (u42u)du3

1(3x1)531(3x1)23C. u3x

u1

11u3另解原 1u3 1

1u2

2u

3

x3xx3x 令x1t6,dx6tx3x3x

6tt3t

t t6(t2t1

t

t t

t

t

)x 33x166x1x

x

C例21求 1ex1e1e

xln(u21),dx

u211exdxu 2(1 )u2

u2 2[11( 1 )]du 2u u

uu

)1e1e1e1e1ex注一般地，第一類換元法比第二類換元法用起來(lái)方例

22)txe

x

2etdt2

C2exC

u解法

xe x

u

e2udu euduu2euC2exC*積分表tanxdxln|cosx|*積分表cotxdxln|sin