高等數(shù)學(xué)第六版(上冊)第四至六章課后習(xí)題答案

高等數(shù)學(xué)第六版(上冊)第四至七章課后習(xí)題答案習(xí)題41求下列不定積分:(1)1dx;x2解1dxx2dx1x21C1C.(2)

x2xdx; 3

21

x31 2解x

xdxx2dx

x2312

Cx5

xC.x(3)1dx;x1

1 1x解 dxx

2dx

x212

C

xC.x23xdx; 7

1 73解x23xdxx3dx

x3713

C10

x33xC.1 dx;x2x1

1 53 1x2x解 dxx2x

2dx

x2C

C.2

2xx(6)(6)mxndx;m nm

1 n

m mn解mxndxxmdx

xmn1m

C

nmx

C.(7)5x3dx;4解5x3dx5x3dx5x4C.4(8)(x23x2)dx;解(x23x2)dxx2dx3xdx2dx1x33x22xC.3 22gh(9)2gh

(g是常數(shù));2hgdh 1 1 12hg2gh解 2gh2g

2dh2h2C2g

C.(10)(x2)2dx;3解(x2)2dx(x24x4)dxx2dx4xdx4dx1x32x24xC.3(11)(x21)2dx;解(x21)2dx(x42x21)dxx4dx2x2dxdx1x52x3xC.5 3x3(12)(x;x3 1 3解(

x

x31)dx(x2x

x3x2dxx2dxx2dxdx1x323 3

x22353

5x2xC.(13)

x)2

xdx;x(1x)2

12xx

1 1

1 43 25xx解 dxxx3x43x21

dx(x

22x2x2)dx2x23

x25

x2C.(14)

x213x43x21

dx;2 1 3解x21x2

dx(3x

)dxx21

arctanxC.(15)1x2dxx2

x211 1解1x2dx

1x

dx(11x2)dxxarctanxC.(16)(2ex3)dx;x解(2ex3)dx2exdx31dx2ex3ln|x|C.x x(17)(

3 1x211x2

2 )dx;解(32 )dx31dx21 dx3arctanx2arcsinxC.1x21x21x1x21x2ex

exxx e x e 解e(ex

1 12)dxex2x2C.3xexdx;xx

3xex解3

edx(3e)

dx C C.ln31(20)

23x523x23x52

dx;2x

(2)x3

5 2x解 3x dx[25(3

]dx2x

2C2xlnln3ln

(ln33

C.secx(secxtanx)dx;解secx(secxtanx)dx(sec2xsecxtanx)dxtanxsecxC.cos2xdx;2解cos2xdx1cosxdx1(1cosx)dx1(xsinx)C.2 2 2 2(23) 1 dx;1cos2x解1 dx1 dx1tanxC.1cos

2cos2x 2 cos2x dxcosxsinxcos2x

cos2xsin2x解cosxsinxdx

cosxsin

dx(cosxsinx)dxsinxcosxC. cos2x dx;cos2xsin2x 解 cos2x dx cos2xsin2x

cos2xsin2cos2xsin2x

dx

1sin

1xcos2

)dxcotxtanxC.x(26)(11)xx2

xdx;? 1

3 5

47 12解 ?1 ?2? x?

xdx(x4x

4)dx

x47

4C.一曲線通過點(e2,3),且在任一點處的切線的斜率等于該點橫坐標(biāo)的倒數(shù),求該曲線的方程.解設(shè)該曲線的方程為yf(x),則由題意得yf(x)1,x所以 y1dxln|x|C.x又因為曲線通過點(e2,3),所以有3213f(e2)ln|e2|C2C,C321.于是所求曲線的方程為yln|x|1.一物體由靜止開始運動,經(jīng)t3t2(m/s),問3秒后物體離開出發(fā)點的距離是多少？360m需要多少時間？解設(shè)位移函數(shù)為ss(t),則sv3t2,

s3t2dtt3C.因為當(dāng)t0時s0,所以C0.因此位移函數(shù)為st3.(1)3秒后物體離開出發(fā)點的距離是ss(3)3327.(2)由t3360,得物體走完360m所需的時間t33607.11s.

1e22

,ex

shx和

chx

exchx

的原函數(shù).ex證明 chxshx

exexexexe

exe

e2x.2 2因為(1e2x)e2x,2所以12

e2x

exchx

的原函數(shù).因為(exx

shx)ex

shx

chx

x(shxchx)ex

exe(2

exe2

)e2x,所以e

shx

chx

的原函數(shù).因為x x x x

xexe

exex 2x(echx)echxeshxx ex

(chxshx)e( 2

)e ,2所以e

chx

chx

的原函數(shù).習(xí)題42在下列各式等號右端的空白處填入適當(dāng)?shù)南禂?shù)使等式成立(例如dx1d(4x7:4(1)dxd(ax);

1d(ax).a(2)dx d(7x3);

1d(7x3).7(3)xdx d(x2);解xdx

1d(x2).2(4)xdx d(5x2);解xdx(5)xdx

1 d(5x2).10d(1x2);解xdx12

d(1x2).(6)x3dx d(3x42);解x3dx

1 d(3x42).12e2xdx d(e2x);解e2x

1d(e2x).2e

x2dx

d

x2);x解e2dx2d(1e

x2).sin3xdx3 3

d(cos32

x);解sin xdx

d(cos

x).2dxx

3 2d(5ln|x|);解dx1x

d(5ln|x|).dxx

d(35ln|x|);解dx1

d(35ln|x|).

x 5dx

d(arctan3x);19x2解 dx 19x2 3

d(arctan3x).11x2

dx

d(1arctanx);1x1x2

(1)d(1arctanx).11x2

xdx

d(1x2).1x21x2

(1)d(

1x2).求下列不定積分(ab,,均為常數(shù)):(1)e5tdt;解e5tdt1e5xd5x1e5xC.5 5(2)(32x)3dx;解(32x)3dx1(32x)3d(32x)1(32x)4C.2 8(3)1dx;12x解1dx11d2x)1ln|12x|C.(4)

12x323323x

212x 2323x解323x

13

(23x)

1 13d(23x)32

2(23x)3C

1(23x)3C.222x(5)(sinaxeb)dx;x解 (sinaxeb)dx1xa

sin(ax)bebd(

)a

xxbcosaxbebC.xb(6)(6)sintdt;t解sintdt2sintdttan10xsec2xdx;

t2cos

tC. 解 tan10xsec2xdxtan10xdtanx1tan11x 11 dx ;xlnxlnlnx解dx 1 dlnx1dlnlnxln|lnlnx|C.xlnxlnln

lnxlnln

lnlnx1x21x2tan 1x21x2解tan

dxtan d d1x21x211x21x21x2sin1x2cos1x21x21x2 x 1x21x21x21x2 dx ;sinxcosx

ln|

|C.dx sec2x 1解sinxcosx

tanxdxtanxdtanxln|tanx|C.1 dx;exex1 ex

1 x x解exexdxe2x1dx1e2xde

arctan

C.xex2dx;解xex2dx1ex2d(x2)1ex2C.2 2xcos(x2)dx;解xcos(x2dx1cos(x2d(x2)1sin(x2)C.2 223x2x23x223x2解23x23x3

dx16

(23x2)

12d(23x2)

1(23x2)2C113 1

23x2C.(15)1x4dx;3x3 3

4 3 4解1x4dx41x4d

)ln|1x4

|C.(16)cos2(t)sin(t)dt;解cos2)sin(t)dt1cos2)d)1cos3)C. sinxdx;cos3x解sinxdxcos3xdcosx1cos2xC1sec2xC.cos3x 2 2sinxcosxdx;3sinxcosx3sinxcosx3sinxcosx解sinxcosxdx3sinxcosx3sinxcosx(sinxcos94x2(19)1x94x2

13d(sinxcosx)

(sinxcosx)3C.22294x294x294x2解x dx94x294x294x211 d(2x)11 d(94x2)1arcsin2x

94x94x21(2x)2391(2x)2394x2x3x(20)9x2dx;x3 1 x2 2

9 2 1 2 2解9x2dx29x2d(x

)2(19x2)d

)[x2

9ln(9x

)]C.(21)1 dx;2x21解1 dx1 dx1(1 1 )dx12x12x21 (2x2x2 2x1 2x112x1222x11222x1

d(2x1)1

d(2x1)1ln|22

2x1|122

22ln|22

2x1|C122

ln|

2x1|C.2x1(22) 1 dx(x1)(x2)解1 dx1(11)dx1(ln|x2|ln|x1|C1ln|x2|C.(x1)(x(23)cos3xdx;

3 x

x1 3

3 x13解cos3xdxcos2xdsinx(1sin2x)dsinxsinx1sin3xC.3(24)cos2(t)dt;解cos2(t)dt1[1cos2(t)]dt1t1sin2(t)C.2sin2xcos3xdx;

2 解sin2xcos3xdx1(sin5xsinx)dx1cos5x1cosxC.2cosxcosxdx;2

10 2解cosxcosxdx1(cos3xcos1x)dx1sin3xsin1xC.2 2 2 2 3 2 2sin5xsin7xdx;解sin5xsin7xdx1(cos12xcos2x)dx1sin12x1sin2xC.2 24 4tan3xsecxdx;解tan3xsecxdxtan2xsecxtanxdxtan2xdsecx(sec2x1)dsecx1sec3xsecxC.3(29)

102arccosx

11x2102arccosx11x2

dx10

2arccos

darccosx2

10

2arccos

d(2arccosx)

102arccos2ln10

C.nx(30)nxx(1x)

dx;nx解 nxx(1x)

dx2arctanxd(1x)

x2arctan

xdarctan

x(arctan

x)2C.1x2(31)dx 1x2(arcsinx)21(arcsinx)21x2

1(arcsinx)

darcsinx

1arcsin

C.1lnxdx;(xlnx)2解1lnxdx1 d(xlnx)1C.(xlnx)2lntanxdx;cosxsinx

(xlnx)2

xlnx解lntanxdxlntanxsec2xdxlntanxdtanxcosxsin

tan

tanxlntanxdlntanx1(lntanx)2C.2a2a2x22

dx(a>0);x2 令xasina2xa2x2

a2sin2acost

acostdta

sin

tdta

1cos2

dt,1a2

2t242

sin2tCa222

arcsinxxa2xa2x2

C.x2(35)x2x解dx 令xsect1 secttantdtdttCarccos1C.x21x sectx21或 dx 1 dx1 d1arccos1C.x2111x21x2111x211x2(x21)3(36)(x21)3(x21)3(tan2t1)3x21解dx 令xtant1 dtantcostdt(x21)3(tan2t1)3x21(37)

dx;x2x29x299sec2tx299sec2t93sect解 dx x

d(3sect)3tan

tdt3(

1cos2

3tant3tC

x293arccos3C.x2x(38)dx 12x2x解dx 令2xt1tdt(11)dttln(1t)C2x

2x

2x)C.11x2(39)1x21

1t

1t解dx 令xsint1 costdt(11 )dt(11sec2t)dt1x2111x211x2

1cost 2 2ttantCt2

sint1cost

Carcsinx

x C.1x2(40)1x2x解dx 令xsint1 costdt1costsintcostsintdt1x2x sint1x2

2 sintcost1dt11 d(sintcost)1t1ln|sintcost|C1x22 2sint1x21arcsinx1

x|C.2 2習(xí)題43求下列不定積分:xsinxdx;解xsinxdxxdcosxxcosxcosxdxxcosxsinxC.lnxdx;解lnxdxxlnxxdlnxxlnxdxxlnxxC.arcsinxdx;解arcsinxdxxarcsinxxdarcsinx1x2xarcsinx1x21x2xarcsin1x2xexdx;

C.解xexdxxdexxexexdxxexexCex(x1)C.x2lnxdx;解x2lnxdx1lnxdx31x3lnx1x3dlnx3 3 31x3lnx1x2dx1x3lnx1x3C.3 3 3 9excosxdx;解因為excosxdxexdsinxexsinxsinxdexexsinxexsinxdxexsinxexdcosxexsinxexcosxcosxdexexsinxexcosxexcosxdx,所以 excosxdx1(exsinxexcosx)C1ex(sinxcosx)C.2 2e2xsinxdx;2解因為e2xsinxdx2e2xdcosx2e2xcosx2cosxde2x2 2 2 22e2xcosx4e2xcosxdx2e2xcosx8e2xdsinx2 2 2 22e2xcosx8e2xsinx8sinxde2x2 2 22e2xcosx8e2xsinx16e2xsinxdx,2 2 2所以 e2xsinxdx2e2x(cosx4sinx)C.2 17 2 2xcosxdx;2解xcosxdx2xdsinx2xsinx2sinxdx2xsinx4cosxC.2 2 2 2 2 2x2arctanxdx;解x2arctanxdx1arctanxdx31x3arctanx1x31dx3 3 3 1x221x3arctanx1x dx21x3arctanx1(11)dx223 61x2 3

6 1x21x3arctanx1x21ln(1x2)C.3 6 6xtan2xdx2解xtan2xdxx(sec2x1)dxxsec2xdxxdx1x2xdtanx21x2xtanxtanxdx1x2xtanxln|cosx|C.2 2x2cosxdx;解x2cosxdxx2dsinxx2sinxsinx2xdxx2sinx2xdcosx解

x2sinx2xcosx2cosxdxx2sinx2xcosx2sinxC.te2tdt;te2tdt1tde2t1te2t1e2tdt2 2 21te2t1e2tC1e2t(t1)C.2 4 2 2ln2xdx;x解ln2xdxxln2xx2lnx1dxxln2x2lnxdxxxln2x2xlnx2x1dxxln2x2xlnx2xC.x解

xsinxcosxdx;xsinxcosxdx1xsin2xdx1xdcos2x1xcos2x1cos2xdx2 4 4 41xcos2x1sin2xC.4 8x2cos2xdx;2解x2cos2xdx1x2(1cosx)dx1x31x2dsinx1x31x2sinxxsinxdx2 2 6 2 6 21x31x2sinxxdcosx1x31x2sinxxcosxcosxdx6 2 6 21x31x2sinxxcosxsinxC.6 2解

xln(x1)dx;xln(x1ln(x21x2ln(x1x21dx2 2 2 x11x2ln(x1)1(x11)dx2 2 x11x2ln(x1)1x21x1ln(x1)C.2 4 2 217.(x21)sin2xdx;解(x21)sin2xdx1(x21)dcos2x1(x21)cos2x1cos2x2xdx2 2 21(x21)cos2x1xdsin2x2 21(x21)cos2x1xsin2x1sin2xdx2 2 21(x21)cos2x1xsin2x1cos2xC.18.

2ln3x2

2 4dx;ln3x 3 1 1 3 1 3 1 3 1 2解 x2dxln

xd x x

xxd

xlnx

x3x2

xdx1ln3x3ln2xd11ln3x3ln2x31dln2xx x x x x1ln3x3ln2x61ln1ln3x3ln2x6lnxd1x x x2 x x x1ln3x3ln2x6lnx61dxx x x x21ln3x3ln2x6lnx6C.x x x xe3xdx;3x xt3x xt 2t 2 解e dx 3tedt3tde3t2et6tetdt3t2et6tdet3t2et6tet6etdt3t2et6tet6etC33ex(3x223x2)C.coslnxdx;解因為xcoslnxdxxcoslnxxsinlnx1dxxxxcoslnxsinlnxdxxcoslnxxsinlnxxcoslnx1dxx所以解

xcoslnxxsinlnxcoslnxdx,coslnxdxx(coslnxsinlnx)C.2(arcsinx)2dx;1x2(arcsinx)2dxx(arcsinx)2x2arcsinx1x21x2x(arcsinx)21x21x2x(arcsinx)1x2

arcsinx2dx11x2

x(arcsinx)2exsin2xdx.

arcsinx2xC.解exsin2xdx1ex(1cos2x)dx1ex1excos2xdx,2 2 2而 excos2xdxcos2xdexexcos2x2exsin2xdxexcos2x2sin2xdexexcos2x2exsin2x4excos2xdx,excos2xdx1ex(cos2x2sin2x)C,5所以 exsin2xdx1ex1ex(cos2x2sin2x)C.2 10習(xí)題44求下列不定積分:x31.x3dx;x3x解x3dx

x327x3

dx

(x3)(x23x9)27x3 dx (x23x9)dx27 x31x33x29x27ln|x3|C.3 22. 2x3 dx;x23x10解2x3 dx1 d(x23x10)ln|x23x10|C.x23x10 x23x10x5x483.

x3

dx;x5x48解x3x

dx(x2

x

x2x8x3xdx1x31x2x8dx4dx3dx3 2

x

x11x31x2x8ln|x|4ln|x1|3ln|x1|C.3 24. 3dx;x31解3dx(1x2)dx(112x131 )dxx31

x

x2x

x

2x2x1

2x2x1ln|x1|11 d(x2x31 d(x1)32x2x13

2(x

1)2( )2 22 2x2x2x1

|x

3arctan2x1C.35. xdx (x1)(x2)(x3)3解xdx 1(413)dx(x1)(x2)(x3)2 x2 x1x31(ln|x2|3ln|x3|ln|x1|)C.2x216. (x1)2(xdx;x21解 2

dx[11

11

12]dx(x(x

2x1

2x1

(x1)1ln|x1|1ln|x1|1C2 2 x11ln|x227. 1 dx;x(x21)

1x

C.解1 dx(1x)dxln|x|1x2)C.x(x2x1x2 2 dx ;(x21)(x2x)解dx (11x111)dx(x21)(x2x) x2x212x1ln|x|1ln|x1|1x1dx2 2x21ln|x|1ln|x1|12xdx11dx2 4x21 2x21ln|x|1ln|x1|1ln(x21)1arctanxC.2 4 2 dx ;(x21)(x2x1)解dx (x1x)dx(x21)(x2xx2x1x2112x111 dx1ln(x22x2x12x2x1 21ln|x2x1|1ln(x211 dx2 2 2x2x11ln|x2x1|1ln(x21)3arctan2x1C.2 2 3 310. 1dx;x41解 1dx1 dxx41 (x22x1)(x22x2x1

2x1 4x2

22x

dx dx4 2x22x11(2x2)2 1(2x4 2x22x122 2dx22 2dx4 x2

2x

4 x2

2x12[d(x2

2xd(x2

2x1)]1(dx dx )8 x22x22x22x1x22x1

2x

x222

2x

4 x222

2x

x2

2x1ln|8

| 4

2x

arctan(4

2x1)C.x22(x2x1)2dx;x22 x1解(x2x1)2dx(x2x1)2dxx2x1dx12x1 dx31 dx1 dx2(x2x1)2

2(x2x1)2

x2x111 31 dx1 dx,2x2x因為

2(x2x1)2

x2x11 dx21 d(2x1)2arctan(2x1),x2x

31(2x1)2 3 3 33而 1 dx1 dx3(x2x1)23由遞推公式

[(x

1)2(2

)2]22dx 1 [ x (2n3)dx ],(x2a2

2a2(n

(x2a2)n1

(x2a2)n1得 1 dx1 dx3(x2x3

x

[(x

1)2(2

)2]223331 ( 2dx )12x122arctan2x1,3332( )22

x2x

x2x

3x2x13x22

1 1

2x1

2x1

2x1所以 (x2x1)2dx2x2x12x2x1

arctan3

arctan C3 3 3 x1 x2x1

arctan2x1C3 312.

dx ;3sin2x解 dx 3sin2x

14cos2x

dx

12334tan2x3233

dtanx1 14tan2x(

32

dtanx

arctan2tanxC.13. 1 dx;3cosx1 1 dx

xd()2解3cosxdx2 x x x1cos2 cos2(1sec2)dtanx

2 2 2tanx222tan2x22

arctan

2C.2令utanx 或 1 dx 3cosx

132u1u

2 du1u2

tanx1 du1arctanuC1arctan 2C.222u2(2)2 222214. 1 dx;2sinx1 dx

xd()2解2sinxdx x x x x x22sincos sin2(csc2cot)2 2 2 2 2d(cotx)

d(cot

x1)2 22 cot2xcotx1

(cotx1)2(

3)223

2arctan

22cotx123

22 2C.令utanx 或 1 dx 2sinx

122u1u

2 du1u21 du1 du3u2u13

(u

1)2( )22 22arctan2u1C3 3

3

2tanx23

C. dx ;1sinxcosxdx 1

xd(tan)2 x解1sinx

cos

2

cos2x(1tan2

x)2

1

xln|tan2|C.2令utanx 或 dx 1sinxcosx1

12u1u2

2 du1u21u2 1u21duln|u1|Cln|tanx1|C.u1 2 dx ;2sinxcosx5令utanx 解 dx 2sinxcosx

4u1u

11u1u

21u5

du

1 du3u22u23tanx111 du1arctan1C1arctan 2 C.5553(u1)2(5553

5)2 53令utanx 或 dx 2sinxcosx5

4u1u

11u1u

2 du1u251 du11 du522u5

3(u

1)2( )23 31arctan3u1C5 5

1arctan5

3tanx125

C.13x117.13x1令3x1令3x113x1

13u1u

2du3(u11)du1u3u23u3ln|1u|C33(x1)233x13ln(13x1)C.2 2(x)(x)31x1

dx;(x)31x1(x)31x1

x)2

x1]dx12

x223

3x2xC.x19.x1xx1 解 x11dx令x1uu12udu2(x1 u1

2u

)du2(1u2

2u2ln|u1|)C(x4x4x20.x4x

x14ln(

x11)C.解 dxx4x

令xu

1 u2u

3du4(u11)du2u24u4ln|1u|C1u2x4x4ln(14x)C.21.

1xdx;1xx解令1x1x

1u2u,則x 1u2

dx

4u(1u2)

du,1x1x dxu1u24u du2(1x1xx 1u

(1u2)

u2

1u21x1xln|u1|1x1xln|

1x1x

1x|2arctan1x

C.3(x23(x2(x4x1解令3 u,則xx1

u3,u3

dx

6u2(u32

,代入得3(x3(x2(x4

32

du

3uC2

3 x13 C.2 x1總習(xí)題四求下列不定積分(其中a,b為常數(shù)):dx ;exex1dx e1

1 x

ex1解exexe2x1dxe2x

ln|2 ex

|C. x dx;(1x)3 解 x dx (1x)3x2

1(x

dx

1(x

dx

1x

12

1x)

C.3.a6x6dx(a>0);x2 1 1 3 1 x3a3解a6x6dx3(a32(x32d(x

) ln| |C.6a3 x3a31cosxdx;xsinx解1cosxdx1 d(xsinx)ln|xsinx|C.xsinxlnlnxdxx

xsinx解lnlnxdxlnlnxdlnxlnxlnlnxlnx11dxlnxlnlnxlnxC.xsinxcosxdx;1sin4x

lnxx 解 sinxcosxdx 1sin4xtan4xdx;

sinx1sin4x

dsinx2

11(sin2x)2

d(sin2x)1arctansin2xC.2解tan4

4sinx xdxcos2xdtanxsinx

xsin2

xdtanx tan4xtan2x1

dtanx(tan2x11 )dtanxtan2x11tan3xtanxarctantanxc1tan3xtanxxc.3 3sinxsin2xsin3xdx;2解sinxsin2xsin3xdx1(cos3xcosx)sin3xdx21cos3xsin3xdx1cosxsin3xdx2 21cos3xd(cos3x)1(sin4xsin2x)dx6 41cos23x1cos4x1cos2xC.12 16 8 dx ;x(x64)dx 1 1 x5 1 1 6解x(x64)4xx64)dx4ln|x|24ln(x

4)C.10.解

axdx(a0);axa2x2a2x2a2x2 axdxa2x2a2x2a2x2 a2x2a2x2ax)x)

C.x)1(x)2解dxx)1(x)2

x2ln(

x1(

x)2)C2ln(

x1x)C.12.解

xcos2xdx;xcos2xdx1(xxcos2x)dx1x21xdsin2x2 4 41x21xsin2x1sin2xdx1x21xsin2x1cos2xC.4 4 4 4 4 813.eaxcosbxdx;解因為eaxcosbxdx1cosbxdeax1eaxcosbxbeaxsinbxdxa1ax b

aax 1

ab

b2 axae cosbxa2sin

e cosbx a a2

sinbxa2

cosbxdx,ax a2 1

b ax所以

cosbxdx (a2b2a

cosbx a2

sinbx)C 1a2b2

eax(acosbxbsinbx)C.1ex14.1exdx 令1dx 令1exu1ex

dln(uu

1)

1u2

du(

1u1

1u

)du.ln|

uu

|cln

c. ex ex1ex1x2x2解dx 令xsect1 secttantdtcostdtsintCx21x2 sec2tx21xx21x16.

dx ;(a2x2)5/2解 dx(a2x2)5/

令xasint 1 acostdt(acost)511 dt1(tan2ttanta413a4

cos4ttan3t1a4

a4tantC(a2x2(a2x2)3a4a2x23a4

C.1x217.1x2x41x2解dx 令xtant1 1x2x4cos3t

cos2

tan4tsectsin4tdtsin4tdsint(

1sin4

1sin2

)dsint

13sin3

1Csintx2)33x3

C.11x218.

x

xdx;解xsin

令xt2xdx tsint2tdt2tsintdt22t2dcost2t2cost2cost2tdt2t2cost4tdsint2t2cost4tsint4sintdt2t2cost4tsint4costC2xcos

x

x

x4cos

xC.19.解

ln(1x2)dx;ln(1x2)dxxln(1x2)x

2xdx1x2xln(1x2)2(1

11x

)dxxln(1x2)2x2arctanxC.sin2xcos3xdx;sin2x

sin2x

tanx解cos3xdxcosxdtanx(tanxtan2x1)dtanx1tan2x1ln(tan2x1)C.2 2arctanxdx;解arctan

xdxxarctan

xx1dx1xxx

x(11)dx1xxxarctanx

xarctan

xC

(x1cosxdxsinx

xxC.2cosx cot|解1cosxdx2dx cot|

xxcscd

2ln|cscx xC.sinxx3x23.(1x8)2dx;

2sin

xcosx2 2

2 2 2 23解 x dx1382 4

1 482dx4

11[x

arctanx88

]C.x)

x

421x提示:已知遞推公式dx 1 [ x (2n3)dx ].(x2a2x11

2a2(n

(x2a2)n1

(x2a2)n124.x83x42dx;x11 1 x

4令x4t1 t2解x83x42dx4x83x42dx1(13t2)dt1(141)dt

4t23t2dt4 t23t

4 t

t11tln|t2|1ln|t1|C41x44

ln

44x41x4x41

C.25. dx;16x4解dx1 dx1(1 1 )dx16x4 (4x2)(4x2) 8 4x2 4x21(184

ln|

2x2x

12

x)C21ln|2x|1arctanxC.32 2x 16 226.解

sinxdx;1sinx sinxdx 1sinx

sinx(1sin1sin2x

dx

sinxsin2dxcos2x(sinx11 )dxsecxxtanxC.27.

cos2xxsinxdx1cosx

cos2x解xsinxdxxsinxdx1x dx1sinxdx1cos

2cos2x2

2cos2x2

2cos2x2xdtanxtanxdx2 2xtanxtanxdxtanxdxxtanxC.2 2 2 228.e

sinxxcos3xsinxcos2x

dx;解e

sinxxcos3xsincos2x

dx

sin

cosxdxe

sin

tanxsecxdxxesinxdsinxesinxdsecxxdesinxsecxesinxsecxdesinxxesinxesinxdxsecxesinxsecxesinxcosxdxxesinxsecxesinxC.3xx3xx(x3x)

dx;3xx(x3xx(x3x)

令xt6

t

t2(t3t2

6t)

5dt6(1t

1t

)dt6ln

tt

Cln

x C.(6x1)6(6x1)6(1ex)2dx 1ext1

1 1 12解2ex)2

t2t1dt(t1t

)dtln(t1)lnt1Ctxln(1ex)e3xex

11e

C.31.e4xe2x1dxe3xex

exex

1 x x解e4xe2x1dxe2x1e2xdx1(exex)2d(ee )arctan(exex)Carctan(2shx)C.xex32. (ex2dx;xex x x 1解(ex2dx(ex2d(e

1)xdex1x1dxx1 dexex1

ex

ex

ex(ex1)x(11)dexex1 ex ex1 xex1

lnexln(ex1)Cxexex1

lnex

1)C.33.ln2(x

1x2)dx;ln2(x

1x2)dxxln2(x

1x2)x[ln2(x

1x2)]dxxln2(x

1x2)2ln(x

1x2)x dx1x2xln21x2

1x2)2ln(x

1x21x211x21x21x2xln2(xxln2(x

1x2)1x2)

ln(xln(x

1x2)21x2)2

[ln(x

1x2)]dx1x2xln2(1x2

1x2)2

ln(x

1x2)2xC.1x234. lnx1x2(1x2)3/2解因為 1 dx令xtanx2

1sec3t

sec2tdtcostdtsintC

x C1x2所以 lnx dxlnxd( x )xlnxx 1x21x21x21x21x21x21x21x21x2

ln(x

1x2)C.11x2

arcsinxdx;1x22解 arcsinxdx令xsinttcos2tdt1(ttcos21x221t21tsin2t1t21tsin2t1sin2tdt4 4 4 4 41t21tsin2t1cos2tC4 4 81x21(arcsinx)21x arcsinx1x21x236.

41x2x31x2

2 4 1dx;x3arccos11x2

dxx

arccosx x dxx 1x2

arccosxd1x21x2x2

arccosx1x211x21x2

(x2arccosx)dx1x21x21x2(2xarccosxx1x21x21x1x2

arccosx2

arccosxdxx2dx1x1x2

arccosx1x32arccosxd1x1x2x2)3x2

arccosx1x3

arccosx2(1x2)dx1x1x2x2)3x2

arccosx1x32(1x2)3arccosx2x2x3C1x23 3 31x213

1x2(x21)arccosx1x(x26)C.9cotxdx;1sinx解 cotx1sinx

dx

1sinx(1sin

dsinx(

1sin

11sinx

)dsinxln|sinx|ln|1sinx|Cln|cscx1|C. dx ;sin3xcosx解 dxsin3xcos

1sinxcos

dcotx

cosxsinxcos2

dcotxcot

1cos2

dcotx

(1cot

cotx)dcotxln|cotx|1cot2xCln|tanx|2

12sin2

C1. dx (2cosx)sinx解令utanx,則2dx 1 2du

1u2du(2cosx)sinx

(2

1u1u

)2u1u

1u

(u23)u1

du11du1ln(u23)1ln|u|C3u23 3u 3 31ln|tan3x3tanx|C.3 2 2sinxcosxdx;sinxcosxsinxcosx

(sinxcosx)sinxcos

sin2xcosx cos2xsinx解sinxcosxdx

sinx2cos2

du2sin2x1dx2cos2x1dx sin2x2sin2x

dsinx

cos2x2cos2x

dcosx1(12

12sin2x

)dsinx12

12cos2x

)dcosx1sinx

ln|

2sinx1|1cosx

cosx1|C2 2 2sinx1 2 2 2cosx11(sinxcosx)2

2

sinx1|C2cosx1習(xí)題51利用定積分定義計算由拋物線y=x21,兩直線x=a、x=b(b>a)及橫軸所圍成的圖形的面積.:在區(qū)間[a,b]n1

abai(i1,2,,n1),把區(qū)間[a,b]分n成n個長度相等的小區(qū)間,各個小區(qū)間的長度為:

xi

ba(i1,2,,n).n第二步:在第i個小區(qū)間[xi1,xi](i1,2,,n)上取右端點i

abai,作和nn S f)x

[(abai)21]bani1

i ii1 n nbn

n[ai1

2a(ba)in

(ba)n2

i21](bn

[na

2a(ba)n

n(n2

(ba)n2

n(n1)(2n6

n](ba)[a

a(ba)(nn

(ba)2(n1)(2n6n2

1].第三步:令max{x1,x2,,xn}ba,取極限得所求面積nSbf(x)dxlimnf)x i0i1lim(ba)[an

a(ba)(nn

(ba)2(n1)(2n6n2

1](ba)[a2a(ba)1(ba)21]1(b3a3)ba.3 3利用定積分定義計算下列積分:bxdx(a<b);a1exdx.0(1)

abai(i1,2,,n1),則n

ba(i1,2,,n).在第i個小區(qū)n間上取右端點

aban

(i1,2,,n).于是bxdx

n

n(abai)baini1

i nin n(ba)2

lim[a(ba)n

(ba)2n(n]2n2

1(b2

a2).(2)xi(i1,2,,n1),則x1(i1,2,,n).i個小區(qū)間上取右端點i n i niixii

(i1,2,,n).于是1x ni

1 1 2 nedxlimenlim

(enenen)0 n

nnn1 1 1lim

1en[1(en)n

en

e1.nn

11en

n

1n(1en)利用定積分的幾何意義說明下列等式:(1)12xdx1;0(2)0

1x2dx;4sinxdx0 cosxdx22cosxdx. 02解(1)12xdx表示由直線y2x、x軸及直線x1所圍成的面積,顯然面積為1.11x21(2)1

1x2dx表示由曲線y x軸及y軸所圍成的四分之一圓的面積,即圓x2y211411

1x2dx112.0 4 4ysinx為奇函數(shù),在關(guān)于原點的對稱區(qū)間[,]x軸所夾的面積的代數(shù)和為零,即sinxdx2cosxdxycosxx軸上[ 2

,]一段所圍成的圖形的面積.因為cosx2為偶函數(shù),所以此圖形關(guān)于y軸對稱.因此圖形面積的一半為2cosxdx,即0 2cosxdx22cosxdx. 02水利工程中要計算攔水閘門所受的水壓力,已知閘門上水的壓強p(單位面積上的壓力大小)是水深h的函數(shù),且有p98h(kN/m2).若閘門高H3m,寬L2m,求水面與閘門頂相齊時閘門所受的水壓力P..

Hi(i1,2,,n1)將區(qū)間[0,H]分為n分個小區(qū)間,各n小區(qū)間的長為

H(i1,2,,n).n在第i個小區(qū)間[xi1,xi]上,閘門相應(yīng)部分所受的水壓力近似為Pi9.8xilxi.閘門所受的水壓力為Plim

n9.8x

n9.8Llim

HiH9.8LH2limn(n1)4.8LH2. i ni1

ni1n

n 2n將L2,H3代入上式得P88.2(千牛).證明定積分性質(zhì):bkf(x)dxkbf(x)dx;a abdxba.a ab 證明(1)kf(x)dxlim kf

nklim

f)x

bf(x)dx.0i1

i i 0i1

i i (2)

n nxlim

lim(ba)ba.lim0i1

i 0

i0估計下列各積分的值(1)4(x21)dx;15(2)44

(1sin

x)dx;13xarctanxdx;13ex2xdx.2解(1)因為當(dāng)1x4時,2x2117,所以2(41)4(x21)dx17(41),1即 64(x21)dx51.1(2)x5時11sin2x2,所以4 45 5 2 5 (44)44

(1sin

x)dx2(),4 45即 44

(1sin

x)dx2.3f(x)xarctanx在區(qū)間[1,3

3]上的最大值M與最小值m.f(x)arctanx

x1x

.因為當(dāng)

1x

時,f(x)0,所以函數(shù)f(x)xarctanx在區(qū)間3333633[13333633

3]上單調(diào)增加.于是33mf(1)133

arctan1

,Mf(

3)

3.3因此 63

31)

3xarctanxdx31331

31),33即 3xarctanxdx.339 1 33f(x)ex2x在區(qū)間[,2M.f(x)ex2x(2x1),駐點為x1.2比較f(0)1,f(2)e2,

f(12

)

14,m

14,Me

2.于是1

x2x 2e4(20)0

dx

(20),即 2e

2

x2x

1dxdx2e4.7.f(x)g(x)在[ab]上連續(xù),證明:(1)若在[a,b]上f(x)0,且bf(x)dx0,則在[a,b]上f(x)0;aa(2)若在[ab]上,f(x)0,f(x)?0,則bf(x)dx0a(3)若在[a,b]上,f(x)g(x),且bf(x)dxbg(x)dx,則在[ab]上f(x)g(x).a a證明(1)假如f(x)?0,則必有f(x)0.根據(jù)f(x)在[a,b]上的連續(xù)性,在[a,b]上存在一點x0,使f(x0)0,且f(x0)為f(x)在[a,b]上的最大值.再由連續(xù)性,存在[c,d][a,b],且x0[c,d],使當(dāng)x[c,d]時,

f(x)f(x0).于是2bf(x)dxcf(x)dxdf(x)dxbf(x)dxdf(x)dxf(x0)(dc)0.a a c d c 2這與條件bf(x)dx0相矛盾.因此在[a,b]上f(x)0.a(2)f(x)在[a,b]上連續(xù),所以在[a,b]上存在一點x0,使f(x0)0,且f(x0)為f(x)在[a,b]上的最大值.再由連續(xù)性,存在[c,d][a,b],且x0[c,d],使當(dāng)x[c,d]時,

f(x)f(x0).于是2b f(x)dxf(x)dxf(x0)(dc)b a c 2證法二因為f(x)0,所以bf(x)dx0.假如bf(x)dx0不成立.則只有bf(x)dx0,a a a根據(jù)結(jié)論(1),f(x)0,矛盾.因此bf(x)dx0.a(3)令F(x)g(x)f(x),則在[a,b]上F(x)0且bF(x)dxb[g(x)f(x)]dxbg(x)dxbf(x)dx0,a a a a由結(jié)論(1),在[ab]F(x)0,f(x)g(x).4.7題的結(jié)論,說明下列積分哪一個的值較大:x2dx還是x3dx？0 02x2dx還是2x3dx？1 12lnxdx還是2(lnx2dx？1 1xdx還是x)dx？0 01exdx還是x)dx？0 0解(1)因為當(dāng)0x1時,x2x3,所以1x2dx1x3dx.0 0又當(dāng)0x1時,x2x3,所以1x