高數(shù)上2012年10月18日條子題

b1p129- b1p129- b1p129- b1p129-

b1p144- b3章§3 b1p129-2x2 xdx 3解令2x3

1 du

xdx2x313 xdx 12x313 251u3 1u325 2x2 xdx 251u3du1u325 3512x35

513 b1p129-12ex2xxx

則

12x2 12ex2x

ex eu

eub1p129-e2xe2x

x解令

t,

te2t1e2t1e21e1e21e2

arcsinetarcsinexb1p129-xlnlnxxlnx解令ln

則du

lnln1xlnlnln1lnudlnu

1

u221lnlnx222b1p129-

uexexee2xexee2x u2

212

ex1exex1ex2b1p129-解法1 1sinx cos22 2cos

tanxsecxdxtanxsecx解法 1cosx 2cos2x dx 42

tan

x cos2x 解法

dtan1sinx 12sinx

1sec2 sinx

dtanx1sec222dtanx2

2tanx2

21tanx2解法

,2tan,sinx

,u

tan2du2

1

x

dx

2du u2

u21sinx 12u

u12

u2 u1

2duu2122

C

Cu12

1tanb1p129-求I

1x5解法I

x201x5

d1x5 1x5 511x535解法令u

1x5

du

I1

x5

d1

x5

1u125u33 1x55u33

2 1du

u2

u 151x5b1p129-

x5

1

x5 xx52

x5x5

ux53 uu23 uu2

1 2u 2u2

1

uu2

1 u u2

x52x52xx52 b1p129-I

lnx2lnxxx2 注意dlnx

lnx

xx2故24I24

lnx2lnxdlnx2lnxulnx2lnx2 2

1u244

lnx

lnx2b1p129-x3分項積分x3x3x

ux21x22dx1x22

1u11111u111111132 132

23 1udu23

11u2

c2

dx

11

3x2

c1c21x2x1x211x2x1x21x23

c3x3

dx

11

x221x231x23e2x3e2x3

解令

1ex,

exdx,e2x

exe2x3e2x3 531ex 5b1P134-dd b1p134-求I

xarctanx51x225意d 3 35dx35

x212

x212I

33 3

x2

1233

1

dx3x2

12

x2

1255xtant

dx

dtx212

sec3

tdt

1sin2

dsint3sint3

1

t 1 331x23總I x2x233arctanx

391x22 3引求Ix 2n求Ix1x意d

12nxdx

x2

n12

x2

n2I

12n

x2

n1212n

n 12n n1 x1 x11 xtant dx

sec2

dt

x2

n2

sec2n1

tdsint

cos2

tn1dsint1sin2tn1dsint.1u2n1du.b第3章§3例IxIx

x1x1解法 原解法改

1t2x.1tx.d 1t2 1t2

4t ,dxt2

4t dtt2I

1t21t2

t 4t dtt2

t2t21t2解法

I

x2x1x2x1

x2x2x2Jx2xx

1dxx2xx2x2J1

x2 根 P128例12,即x2J1 1lnx x21x2 x2x2

x2x2x2x2x1x2x2x2x2

C2總x2x2I 1 x2x2

x21 x2x2

1lnx

x21x2x2x2x2

b1p134-19x2

x1t233t233t21lntt2t21lntt21t9x21ln3x 9x211 9x2 b1p134-1x2ln11x2xln1xln1

1x21x21x21x2

1x21x2b1p144-I

J

x21

1

x2

2 2 1 xx2

dx1 x12 x2x2 2IJ2

1

x21

1

c2 x2 x2 2x2 2 1 xx2

dx1 x12 x2x2 2x2 2JI

ln222

c2總2 1arctan22

1 x2x2 2x2 2x2 2解法I

x4

x4 x2

x2 AxBx2 2x1CxDx2

2x2222x3ACx2BD 2222xA

C

B AC22B D 22B22 22AC BDA2,B1,C2,D x2 x22x x2 x2I

dx dx

x

dx1

22

222

x

1

2

x

2 2x2 2x2 x2 2422242arctan2x 224

2b1p144-2解法I

cos3

dx

cos3xcosxsinxcosxsinx coscos4

x2

1cos2x221cos2x1cos22 cos4xdx

dx

x

1cos8cos 4cos 8x

1sin2x1sin

1sin

cos3xsinx

ucosx

cos 12u2

u

du 2 2u 222

lnu2

1

1u284184

cos2x

1cos2

c22Ix2

1sin2x1sin

1sin818

cos2x

1cos2

4解法4I

cos3xcosxsinxJ

sin3xcosxsinxIJ1

1sin2xdxx1cos

88又cos3xsin3xcosxsinx

cos2xsin2xcos2xsin2xcosxsinxcosxsinx2cos2x1cosxsinxcosxsinx2

2cos2x2sin2x41sin2x,故,IJ

cos3xsin3xdx

2cos2x2sin2xcosxsinx 41sin2xtsin2x

2t 41t1t1lnt

c2 1sin2x1ln2sin

c2 I1x1cos2x1sin2x1ln2sin2x2 b1p144-I

xcos2

I

xcos2

xdcosxucosx

1u1

2

1u3 1cos7x2cos5x1

x b1p144-I

xcos4

1I1

2xcos2

8 8

2x1cos2xdx

dx

2xdsin2x8 161x1sin4x1

2x b1p144-解法I

11

J

IJ

1x

1xx2

3x1 22arctan2x2

I

1xx2x2 1x3 dxx2 lnx

1

1lnx3

332I

2arctan2x332lnx332

1

1lnx3

3C1333 2I1arctan2x33 21lnx

1

1lnx3

1fx

1arctan2 2 21lnx1

1lnx3

1解法

A

BxCx2x1兩邊同乘以x 得 ABxCx

1,x2 x2以x

代入兩

A3 31x3

3x1

BxCx2x1x兩邊同乘以 得x 1x3

3x1

BxCx2x1令x 得01BB 在

3x1

1xC x2x

兩邊代x0,得1313 213于 1

1 x2 1

3x1

3x2x11lnx1

1lnx2

x3 arctan32x13 解法

A

BxCx2x1兩邊同乘以x 得 ABxCx