微积分课后习题答案五

1、微积分(中国商业出版社 经管类)课后习题答案习题五(A)1 .求函数 f(x),使 f (x) (x 2)(3 x),且 f(1) 0 .解：f (x)x2 5x 6f(x)236x Cf(1) 01 35223f (x)-x x6x3262 . 一曲线y f(x)过点(0, 2),且其上任意点的斜率为lx 3ex,求f (x).1解：f (x) -x 3e1 2 x -f (x) x2 3ex C 4f(0) 23 C 2 C 1f (x) 1x2 3ex 1 42 .一3.已知f(x)的一个原函数为ex ,求f(x)dx.2、,2解：f (x) (ex )2xex_ x2_f (x)dxf

2、(x)C2xeC4. 一质点作直线运动，如果已知其速度为 3t2 sint ,初始位移为so 2 ,求s和t的函 dt数关系.S(t) t3 cost CS(0) 21 C 2 C 1S(t) t3 cost 15.设 In f(x)求 f(x).解：In f (x)1 x2In f (x) arctan x C1arctan x C1arctan x / 小f (x) e 1 sin 2x4 Ce (C 0)6 .求函数f(x),使f (x), 1e2 /-x、e (x e ). -dx 5 且 f(0) 0.1 x2f (x) In xCarcsin x - e2x 5x C21f (0)

3、 0 0 - 0 C 0f(x)In x 1 arcsin x1e2x 25x7 .求下列函数的不定积分(1)dx(2)dt. a(t 1)（3）m xndxx2 1丁”xdx x 1(6)sin 2x , dxsin x cosx(8).21 cos x , dx1 cos2xcos2x ,dxsin x cosx（9）cos2x , 22 dxsin xcos x（11）cos 2x 1_._2 _2sin xcos一 dx x(13)2 8x 38x5xdx(10)cos2 sin2 x dx2e2x 1(12)edxex 1x 1 x 1(14)dx10x(15)(16)(ex 2x)

4、(1 3x)dx(17)1 x '1dx(18 )(x2 1 1)、. 1 x25x ,dx(19)(21)解：(1)(2 )=(3 )=(4 )=(5 )=(6 )=(7 )=(8 )=(9 )=(10)=(11)=(12)=八1,八一2 2 dx2tan x Cxx -e 1 dx e x C3.x x 122-dxx (1 x )21 cos x(20)22- dx1 cos x sin x42(22 )-一1dx1 x2cos x(x2x2)dx32 2x2352 2x25d(t 1)1(t 1)212(t 1)2 Cann mmxmdx x m C m n,m 0n mnxm

5、dx In x Cm ndx x Cm 0dx2arctan xx2(x21)2xx21dxx3 x32arctanx_._2_2sin x cos xsinx cosx2sinxcosx , dx(sin x、2 cosx), dxsin x cosx(sin x cosx)dx sin x cosx C2_._2(cosxsin x)dxcos x sin x , dxsin x cosxsin x cosx12- cos xdx 1tanx2彳 21 cos x .2-dx_2_._2cos x sin_._2 _2 sin x cos-dxxsin12- dx cos xcot x t

6、an xcosx 11 cos2x dxcosxcos2x1 dx1sinx 22 cos x_._2 sin x2cos x_._2 _2 sin xcos x_._2 sin x , dxcos x(13) =2dxdx2x3In(14) =2dxdx-5 ln51 2 5ln2(15)dx(16)6x2x(3e)xdx2xln2(3e)xlln36xln6(17)xdx 2 .1一dx 2arcsinx2 x(18),5 dx,1 x21nx5arcsinx(19)dxarcsinx(20)21 cos x111xdx 一1 dx tan xC- 22coscos2xx222x(x21)

7、 11111 117 dx- 2 dx ln1 xarctan xx2(1x2)x x21 1xC2 x(22)x4 1 (x2 1)1 x22dxx21 x2dx2x2 arctan x8.用换元积分法计算下列各题(1)x 4 dxx 2(2)(3x2)8dx(3)2x e dx 4xe(4)dx2-cos 2x 一3(5)x2.4 x3dx(6)dxx2 2x 5dxx e(8)dxx e(9)dxcos2 x tan x 1(10)dxx(1 - lnx)(11)dxx11 ln2 x(12)xe2dx.9 ex(13)sin xcosxdx sin x(14)dxx dx1 2x2(1

8、5)arctan x ,Tdx1 ex(16)(17)(19)(21)(23)(25)(27)(29)(31)(33)(35)(37)(39)解：(1 arctanxdx1 x22(18)(x 1)ex 2x 4dx(20 )1nx dxx . 2 ln xsinx1 sin2xdxdx一. -72(sinx 2cosx)dx2sin x 3cos xdx,1 x (J x)3(22 )(24 )(26 )(28 )sin x cosx . dx1 sin 2xln tanx ,dxsin xcosxxx (1 In x)dxdx72 7xlnx(ln x 1)ds.2x 1.2x1xdxx4

9、 2x2 5x(30)dxxx2 1(32 ) arcsinxdx1 x(1 x)dxJxsin . x cos v xcosxcos2 xdxarctan、一 x(34dx.x(1 x)(36 ) sin 2xcos3 xdx(cosx sin x) cos2xdx(38)sin xcosx；41 sin xdx1_._4 sin xdx(40)tan3 xdx311) = x 2 6dx . x 2 -6d(x 2) -(x 2)2 12(x 2)2 C.x 2. x 23(2) =1 (3x 2)8d (3x 2) (3x 2)9 C 3272x2x/_ 1 d e1. e c(3)=5

10、 arctanC2 3e2x 22 3、31 tan 2x - C23d 2x (4) =1 322 °cos 2x 一3(5)=33 d(x )31d(4x)"4x3Cd(x 1)(x 1)2d(ex)d(ex)d (tan x_ 1arctan 42xarctane11n 2x e""x e1)tanx 12(tan x11)2d(1 ln x)1 ln xd (In x)*1ln2 xd e29 e2sin xd (sin x)ln 1Inxarcsinlne2 c2arcsin C3一一 2sin xd(2 sin2 x)2 sin2 x1ln

11、2 sin 22xd(1 2x2)- 1 2x212,1 2x22d(x2 1)x x、e (1 e )dx1 , arctan dx1.arctan xd (arctan x)x、d(e )x、d(e )x Ie (1ex22x31 x34 2d(x2xd(x3)令 x3t 1 13 d(t)31 tarctan - d x4)1t 一二3(tln(1x2)32； 八一(arctan x) 2 C 3d(1x, e )xelnarctan1 xarctan-215(t1)312(t1)31 x2e22x 431 td(t)1)3d(t)(t1)3d(t)1 / 3 5(x1)31(x3 1)

12、3 C(6)=(8)=(9)=(10 )=(11)=(12 )=(13 )=(14 )=(15 )=(16 )=(17 )=(18 )=(19 )=(20 )In xd(ln x)令 ln ttd(t)(21)(22 )(23 )(24 )(25 )(26 )(27 )(28 )(29 )(30 )(31)(32 )3344t2 C (arcsin ' x)2 C3312 t 2d (t)d(2 t) (2 t) d(x2 1) (x2 1)2 4d(2 t) 2 ,2 t,2 t31312 ko2大k_-(2t)24(2t)2C-(2ln x)24(2ln x)2C3 3d (cos

13、 x), cosx - arcsin C,2 cos2 x 2d (sin x cosx)1 小-2 (sin x cosx) C(sin x cosx)d (tan x 2)1 cln tanx .d(tan x) tan xy (tanx 2) 1 C (tanx 2)212 -ln tanxd(ln tan x) (ln tanx) Cd (tanx) 1 d( 3tan x) 122tan(.3tanx) C1 3tan2x 3 1 ( 3tanx)2. 32x 1 2x 1 dY 12222 cdx (2x 1)(2x 1) C 24 33331-(2x 1)2 (2x 1)2 C6

14、，d(x 1)3 令 A 1 t 上百dt.x 1 (. x 1)3 t t32 1-？ dt 2arctant C 2arctan ,1 x C1 t21x2 1 _arctan C42xlnxxln x xln x 八 x 八e (1 ln x)dx ld e e C x C3/22 .1221 31 / 25 小x(x x 1)dx x dx . x 1d(x 1) x (x 1)2 C233d(ln x)d(t)1T2令 lnx t 2-1ln x(ln2 1)t(t2 1)1d(t2)d(t2 1)2tT1 t2ln 72t21-ln ln x C ln ln x 1ln(ln2 x

15、 1) C2 ln2x 12令 arcsin 4 x t ,贝U dt 2sintcostdt2 、-tdtt22 2sintcostdtsin2t cos2t(33)d( x)sin x cos, xd(tan . x)tan , x21n tan JXC(34)arctan. x2 d ,一x1 ( x)22 arctan . xd arctan x (arctan. x )2C(35)d (sin x) 1 ,2 sin x 八2lnC2 sin x 2.2/ 2 sin x(36)3 .2sin xcosxcos xdx-42 cosxd cosx-cos5 x C 5(37), x

16、.、2,(cos sin x) (cosxsin x)dx(cosx.、2 ,，.、sinx) d(cosx sin x)(38)(39)(40)1 ,.、3-(cosx sinx) C321d(sin x)21 sin4 x12 sVdx sin x2arctan sin xd (cot x)2-sin x(cot21,3,x 1) d (cot x) 一 cot x cotx3,2(sec x 1)tan xdx tan xd tan xtan xdx1,.、2 ,(tan x) ln cosx9.求下列函数的不定积分(1)dxx(1 x7)(2)x2 . 1xdx(3)1 dx.1 2x

17、 3(4)dx(1 x). 1 - x(5)dx,x 3 x(6)x 1 dxx. x 2T=dx 1. 1 x2(8)1 exdx(9),x2x 2 dx2x 4(10)(1解：（1）x6,7- dxx (1 x )dx7 x7(1x7)1nx1 -ln 17x7t6 , dx2tdt(1 t2)t(2t)dt64(t 2tt2)dt2(7t71 3-t ) C 3(3)令 J1 2xt,1 t22dx tdt1(t)dtt 3(1t 3ln t 3 C1T2 3ln d1 2xx 1 t2,dx 2tdt(5 )(6 )(7 )(8 )(9 )2t2- dt(2 t ).t令浙t,6t5-

18、32-dtt3 t2dt2.1n 22. 2：j21 x2.2，-2ln <2 V1 xt6dx6t5dtt3 dt t 1(t2 t1)1 dtt 11 31 26(-t3t2322t3 3t2In t1)6t61n tt2 , dx2tdtt2 3-2 .2tdt t2 22 (11-2一)dt2tt2 242 arctan. 2-.x 22 七 x 2. 2 arctan ：1令(1 x2)32dt t3t2dtt2In t1) C?(12 x2)3Jt2 1t 3.,t3 3 dt(x22x(1(10)令x2 tt(1 t)1 x212(t ln214)3In(11n(t21 ,

19、dxtdt,t2 331n1 x2)31),dxdt2tdtt2 12(-.1ex11n 2 1ex,.1exdt,t2 32x3(1 t)dt(t13)231n t1(t 1)21(t 1)222111122(t 1)212(t 1)4(t 1)21112( x2 1)4(x2 1)222x2 12 T24(x2 1)210.设 F(x)sinxasin x bcosxdx,G(x)cosxa sinx bcosxdx求 aF(x) bG(x) ; aG(x) bF (x) ; F (x);G(x).解：aF(x) bG(x)asin x bcosx dx x Casin x bcosxaG

20、 (x)bF (x)a cosx bsinx dx a sinx bsin xd(asin x bcosx)asin x bcosxdx ln asin x bcosx CG(x)(a ln asin x bcosbx) CF(x)122( bln asinx bcosxa bax)11.用三角代换求下列不定积分(1)dxx2 1 x2(2)dx一(1-x2)3(3)x2dx.2h1 x(4)22.x adxx(5)dx x2 .(12、3 x )(6)x9810Tdx(1 x2) 2解：（1）令xsint贝U dx costdt(tcost.2."sin t costx sintc

21、ost嬴dtx sintdtdt.2.sin tcottcot(arcsinx) Cdxdt27 cos t则dxcostdt(t-)tantcostdt.2.sin tcost. 23dtsin tdt cost1-arcsin x 214 sin 2(arcsin x)C tan(arcsinx)CC(t 2)1 cos2t.1dt21t 21八1 sin 2t C41Carcsin x21x.1 x4令 x a sect ,贝 U dxa sectant , (0 t ) 21atant.asecttant dtasect2 .,2a tan tdt a (sec t 1) dt a(t

22、ant1) C, x2_2 _ _ a 八a a arccos Cs22,x aa(aa arccos) C x(5)令x sint ,则dx costdt(6)costsin2t cos3tcott tant令 x sint ,. 98.sin tcost101 x cos tdtdt1sin2tcos2tdt1 x2dx costdt.98sin1007 cos t(1、1 x2 Cdt1 cos2t)cos2t dttan98td tan工 tan99 9999127 cos t99,1 x2解：（1）(1 3x x2)dex (1 3xx2)exex(2x 3)dx12.用分部积分法计

23、算下列积分(1)2、x , (1 3x x )e dx(2)x L xe dx(3)x/e (cosx sin x)dx(4)xcosxdxarcsin xdx(6)ln(4 x2)dx4xsinxcosxdx(8),2arctan . x ldx(9)ln(ln x), -dxx(10)2.2xsec xdx(11)arctanx ,2 dx x(12)2(arccosx)2dx(13)ln x , 9dxx2 4x 4(14)2x,.2 arctan xdx1 x2(15)2 3、,.6x (1 x )arctan xdx(16)ln tanx ,2 dx cos x(17)ln secx

24、 ?sin xdx(18)sin 2x?ln tan xdx(19)x2 ln 2 x ln x dx(20)x2 arctan . xdx(1 3x x2t cost 2 tdsint t cost 2(t sin t)ex ex(2x 3) 2 exdx-xde x 1 xe x e xdxC (1 x)e (1 x) ce3xxx xe cosxdx e sin xdx cosxde e sin xdxxe cosxxxxe sin xdx e sin xdx e cosx C(2 )(3 )(4 )(5 )(6 )(7 )(8 )(9 )(10)(11)(12)(13)sdsin x

25、xsin xsin xdx xsin x cosx Cxarcsinx-xxarcsin x 1 d(1 x ).1 x22 J /2xarcsin x 1 x C22ln(4 x )dx xln(4 x )2x25dxxln(4 x ) 24 x24x dx4 x2.2xxln(4 x ) 2x 4arctan C 21 xd cos2x xcosx cos2xdx xcos2x sin 2x C2xarctan x2 1x.2x 2 .x2 11 ( ,x2 1)2dx2,s arctan、x 1,2,xarctan , x 1In x3x21 CIn x tln(ln x)d(ln x)

26、 t In tdt t In t t Cx eInx.ln(lnx) In x C In x(ln(ln x) 1) C2 xd (tan x)2x tan x 2 tan xdx2xtanx 2ln cosx C,Z1 arctan x 11arctanxd ()2x x x 1 x2rctanx12、CIn x -ln(1 x ) Carctanxdxarccosx tdx sin tdtt2sin tdtt2cost 2 cost.tdt.、2sintdt) t cost 2tsint 2cost Cxarccosx 2arccosx-. 1 x2 2x CIn x , 2dx (x 2

27、)2,1In xd x 2Inx.x 2dx12In xdxIn x12ln(14)1TVarctanxdxarctanxdx1,2 arctan xdx1 xxarctan xx .5 dx1 x2arctan xd (arctan x)xarctan x12ln(11,2- arctan x C 2(15)arctan xdx31 x32arctan xx31dxx2x32arctan xx62x31dxx32arctan x2x2xdxx32arctan x1 3 -x 3,2ln x(16)In tan xd (tan x)令 tanxln tdt t ln t ttan x.ln t

28、an xtan x(17)In secxd cosxInsecxcosxcosx.cosx.cosx.tanxdxIn secx.cosxsin xdx cos.ln secx cos C(18)(19)1 1 x2 ex2C2In tan xd sinsin21x33x.ln tan x,2ln xln2 xln2ln2ln2.22sin x.ln tanx sintan xdx2ln x321n321nx 32ln x3lnxsin xcosxdxarctan - xdsin2x.ln tanx13,2-x ln 3x2 ln xdx3 ln xd xln xx3-lnx.x39In co

29、sx2ln x3x2dxx3 2lndxdxx2dx一 x arctan . x 3ln22x dx92 .x dx1 x 2. x(20)1 x3 arctan. x 1 365x21 3,1dx - x arctan-j x 1 x 3-6311-22-2x2x2x2“x1 xdx3 x3 arctan . x51 2x21531x2x2arctan 、 x C93313.计算下列有理函数的不定积分Exdx(x 1)(x 2)(x 3)(3 )x2 x 12(x 1)2(x 2)dx(4 )2x 32dxx2 2x 3(6 )x2 12dx4x2 5x 2(7 )(9 )2x 1 人(8)

30、 2dx(x 1)( xx 1)-23x 8x 1(10)3dx(x 1) (x 2)解:(1)13x11 3xdxIn x In 1 3x CIn1 3x(2 )111 dx1ln(x 1)(x 3)2(x 1)x 22(x 3)2(x 2)2C(3 )12(x 2)2dxIn x(4 )dx4(x 3)4(x 1)ln x(5 )1x2 1122 xdxx11x31arctan x21 In4(6 )(x 1)2dx4ln x 15ln x 2 C(7 )1 x2- 2(x2(8 )1)2( x 1)dx12 x2(x 1)dx41nx21arctanx 221nxx 2x2 x 1dx-

31、dx 1x2 x 1dx-2dxx2 x 1In x 11ln x2 x 1 2J3arctan 2x 1(9)11 2x 1 2 dx x xx 1 x 12x 12dx2x2 1dx(10)12ln1x2 11F-dx x2 11 21nx1 1n 141-arctan x C 2dx12In(B)1.填空题(1)设 f(lnx) 1 2x ,贝U f(x)=(2)设函数f(x)满足下列条件 f(0) 2, f( 2) 0 ;f(x)在x 1 , x 5处有极值；f(x)的导数是x的二次函数，则f(x)=.(3)若 xf(x)dx x2ex C ,则 -edx=.f(x)(4)设 f(x2

32、 1) 1nx,且 f (x) 1nx,则 (x)dx _ x2 23x(5)设 f(x) lnx ,则 e x e f (ex)dx2-e4x(6)?lnx)”._x ； f (ln x) 设f(x)的一个原函数为sin-x ,则xf (2x)dx . x(8)若 sin f(x)dx xsin f (x) cosf (x)dx ,贝U f (x)解：(1) f x x 2ex Cf lnx 1 2x 1 2elnxf x1 2exf x x 2ex C由已知可设f (X) ax3 bx1 2 cx d有 f x 3ax2 2bx C8a 4b 2cd 0c 15d 2f 1 3a 2b c

33、 0f 575a 10b c 03_ 2_f x x3 6x2 15x 2(3) ln2xf xdxC xf2xex2exx xexdx f(x)LxxIn 2(4) x2ln xf(x2 1)In2 xF xf(x)lnUx 1In(x) 1(x) 1(x)(x)dx1dx(1)dxx 2ln x 1(5)1 2xe214一 21ne2xTx e原式e3x. . dx4x x e e2xe2x2 e4xdx2x14,2 1n(6)2.f(ln x)原式d(f(ln x),f (In x)2,f(lnx)cos2x sin 2x4xf(x)dx一、 xcosxf (c)sinx原式xd(f (

34、2x)1 1xf (2x)21-f (2x)dx41 x.22xcos2x4x2sin2x1 sin 2x4 2xcos2xsin2x 0C4x(8) lnxsin g(x)dx xsin f (x)xcos f (x) f (x)dx2.选择题(1)设 f (sin x)cos1 2贝 U f(x)dxA. x lx3 C31 4一x12Cx C11x412c. xi 1x4212(2)设 F(x)f(x)g(x) f(x)(x)g2(x)A. tan xcot xarctan xarc cot x2f (x)dx sinxxf®H)dx1sin(2x2 21)1sin(2x2 4

35、1)C. Isinx Cx C112 (2x2 1) C 2(4)设f(x).y- dx sin2xg(x)?f(x),2.cot xdx ,f (x) , g(x)分别是(DA.f(x)In cosx ,g(x)tan xf(x)In cosx , g(x)cot xC.f(x)In sin x ,g(x)tan x. f(x)Insinx , g(x)cot x解:(1)f (sin x)cosx_._2 sin xf (x)f(x)f(x)dx根据1(-)首先排除C、D,再将选项AB分别代入原条件中，得(3) B2f(x) 2xsinxf . 2x2 1222；2x2 1 sin(2x2

36、 1)12-sin(2x2 1) C , 2原式 2 xsin(2x2 1)dx122.2 sin(2x2 1)d(2x2 1)4(4) D取 g(x) cotx 则f(x)2- dxf(x)d( cotx)sin x上式f(x)g(x) cotxdf(x)与条件比较，得df(x) cotx f(x) Insinx , g(x) cotx ,得 D3 .计算下列不定积分(1)入n -1 x21 xdx(2)ex(1 sinx)dx1 cosx(3 )dxx2x7e (1 e )(4)1snE<dx(5 )_x .xe cosxdx(6)x 1 dx- x 1(7 )dxcos x(8 )

37、axdx a2(9 )dx3 . 9 x2(10).1 xx(提示令x sin2t)(11)x3 8dx(12)arcsin x ,dxx2.1 x2(提示令 arcsin xt , x sint ,再用分部积分法)(13)2 .x(arctanx) dx(14 )arctan xearctan xdx 1 x2(15)(16)sin(ln x)x3dx (提示经过两次分部积分，又出现原积分形式，移项后便可得到所要结xln x ,2dx(x2 3)2(2 )(3 )(4 )1 x x 2 x . ex(1 2tg- tg2-)dx 222果)11x1x12 1xc角系：(1) lnd(ln) ln C21x1x41xex(sin cos)2 22-dx2 x cos 一21 x x x 12 x Xie tg e dx tg e dx2 2221 x . x x 1 x,. 2X1 . 2 x x .x. x _etg ee (t