高等數(shù)學(xué)第四章習(xí)題詳細(xì)解答

1、習(xí)題 4.1 1.求以下不定積分：解1x x 2d 5 x51d 5 x5xx12x71 0 x33 2 C1C22d5 d 2 x271x2dx12xx2dxx12x1x3d x2222C；xx113x2dx2x2dxx2d x2x4x32x5C3 2235axexdx=aexdx=ae xC=aexCalnae 1ln4 co t2xdx=csc2x1 dx=cotxxC2 3x3x5 2xdx =252xdx2 x5ln2 3x35 32ln6 sin21xdx=41xdx=4csc2x dxcotxCx 22 cossin22nxmdxxmdx11xm1CnnmxmnC. 7 nnna

2、rctan xmn812x2dx=x2 1x2dx 1121dxx2 1x22 x1x2xx2x9 x22dxx24x4dxx2dx4x dx4dx1x32 x24 xC. 310 x21 2dxx42x21 dxx4dx2x2dxdx1x52x3xC. 5311 3 x4x3x21dx3 x2x11dxx3arctanxC. 21212 1x22dxx211dx 1112dxxarctanxC. x1x2xx13 2ex3dx2exdx31dx2 ex3 ln|x |C. xxx11cot2xdxx11dx2 cscx1 d14 22=arcsinxcotxxCdxex21tanxdxtan

3、xCsecxC. 15 ex 1ex dxexx1x2x2C. 16 secx secxtanx dx2 secxdxsecx17 112xdx2121cotxC. xdxcossin2（18）cos2xxdxcosx-sincosxsinxdxsinxcosxcos2x-sin2xdxcosxsinx2. 解答以下各題：解1設(shè)所求曲線方程為y=fx,由題設(shè)有 f x=2x-2,f x 2x2 dxx22xC又曲線過(guò)點(diǎn) 1,0,故 f1=0 代入上式有1-2+ C=0 得 C=1,所以 ,所求曲線方程為f x x22x1 . x . 2由題意有 sinxf x ,即f x cosx , 故f

4、sinx , 所以f dxsinx xsinx xcosxC . 3由題意有f sinx,就f x sinx xcosxC 1于是f x dxcosxCd xsinxC xC . 其中C C 為任意常數(shù) ,取 1 2C 1C 20,得f x 的一個(gè)原函數(shù)為sin x . 留意此題答案不唯獨(dú).如如取C 11,C 20得f x 的一個(gè)原函數(shù)為sin x4由QP1 1000 3Pln 3得Q P1 P 1000 ln 33d x1 1000 ln 3 3Pdx1 1000 3PC.將 P=0 時(shí),Q=1000 代入上式得C=0 所以需求量與價(jià)格的函數(shù)關(guān)系是Q P1 1000 3P. （ 5）設(shè)位移函

5、數(shù)為s st, 就 sv 3 t 2, s3 t2dtt3C. 3. 由于當(dāng) t 0 時(shí), s 0, 所以 C 0. 因此位移函數(shù)為s t 3. 3 27. s. 1在 3 秒后物體離開(kāi)動(dòng)身點(diǎn)的距離是s s3 32由 t 3 360, 得物體走完360m 所需的時(shí)間t3360.7 11證明chexsh xexexexexexexe2x. xex22由于所以1e21e2 e 2x, 2x是chex的原函數(shù) . 2xsh x由于所以 eexshxe x shx exch x exsh xch x exexexexexe2x, 22xshx 是e ch xxsh x的原函數(shù) . 由于所以 eexch

6、 xee x ch xexshx exch xsh x exx e2exexex2 ex, 2xch x 是chxsh x的原函數(shù) . x習(xí)題 4.2 1.在以下各式等號(hào)右端的空白處填入適當(dāng)?shù)南禂?shù),使等式成立：解1、1 ；（2）、a1 ；（3）、-2；（4）、21 ；（5）-1；（ 6）、-1；（ 7）、52 ；（8）21 ；2（9）、3 ；（10）、 -1. 22.求以下不定積分：解 1e5 td t1e 5t5d1e 5 td 5 15 e txCx1xC.555（2） 原式1 1sinxdx1xdxdcosxtansin2x2 coscos2xcos31dxx=1d12x1ln|12|C

7、2212x24 ln4x dxln4xdlnx ln5xCx55 1xde1d1 x1xC41x24 1dx24 exd2xexx6 x4x14d222x2x2217 x24 2Cx24C3xCx2Ctan10 x2 secxdx =tan2xdtanxtan11xC118xe2 xdx =1ex2dx21ex2C229 dxx221dx21d3x21arcsin2493x313x3210 221x2d1tan 1x2xdx=tan1x2d1xsin12 xcos1x2|=cos1x2dcos1x2lncos12 xC11x1xln | ln lnxln1lnxdxlnx1xdlnx1xdln

8、 lnxlnln lnln ln12xx2dx 3=x22333 dxdx3xdx3x22C11 dx23312d x231x23ln3x21322x22x2Cx212dxx1x1dx1x12x11dx1lnx233x114 sinx dx xdcosx1cos2xC3 coscos3x215 23x2dx924x2dx39xx2dx94x4112 x2d2x3914x2d94x2383arcsin2x394x2C3416 93 x2dx=x39x 9x dxxdx92 x19dxx231dx2C3 xxx2923=x23arctanxC2317 2dx1dx1212d2x1d2x1x22 x

9、22x12x1=212ln2x1C2x118cos4xdx1cos2x2dx12cos2xcos22xdx241cos2xcos22xdxxsin2x1cos4x dx424423xsin2xsin4xC4419 2 sin tdt11cos2tdt1dt1cos2t 2t2241t1 sincos2 4tC2202 costsintdt=1cos2tdcost13 costC321lncotxdxlncotxxdx1lncotxxcsc2xdx1lncotxxdcotxsin 2 x2sinxcos2cot2cot1lncotxdlncotx1lncotx 2C2422 sinxcosxdx

10、 =2sinxcos2xdx4cos2xdcosx4cos3xC222223223 sin 5 cos7xdx =1sin12xsin2xdx1cos12x1cos2xC224424 3sinxcosxdx3sin1cosxdsinxcosx 2 sinxcosx 23sinxcosxx25cos3xdxcos2xcosxdx1sin2xdsinx sinxsin3xC326 arcsin2 x dxarcsinxdarcsinxarcsin2xCx1xCx21x27ln tanxxdxln tanx1x2 secx xCcosxsintanln tanx dln tan 1ln tan 22

11、281lnxdxx1x1lnxdx1x2d lnx lnx2ln lnln29x2 dx2aa2x2a2x2dxa2ad xx2a2x2da222x利用教材 5.2 例 16 及公式 20可得 : 原式 =a2arcsinxa2arcsinx1xa2x2Ca2arcsinx1xa2x2C. a2a22a230令xtan , t t ,2 2,就dx2 sect t . 所以d x3 113 12 sect td tcos t tsintCx22 tantsec txtan ,sint1xx2原式1xx2C. 31arccos 10 xdx10arccosxdarccosx10arccosxC1

12、x2ln1032 令xsin , t t ,2 2,就d xcost t,1d x2 x1cos td t11 d tt2 sectdt 21cos t1 cos t2ttantCarcsinx11x2C.2x33 axd x ax2 a1x2d xad xx21d x22 xax2 a2a2aarcsin x aa2x2C. （34）設(shè)exsin t,就exdxcostdt,所以原式2cos2tdt 1cos2 tdtt1sin2tC2tcostsintCarcsinexex12 exC.（35）x34x2dx 2sint344sin2t2costdt32sin3 cos 2tdt5 cos

13、tCC3213 cost132sint 1cos2 t2 costdt32cos2t4 costdcost3544x2314x25C. 111tdttln1tC2xln12x. 35t1ttdt36 1dxx令2x21習(xí)題 4.3 1、求以下不定積分解：（1）xsin2xdx1xdcos2xxcos2x1cos2xdxx222xcos2x1sin2xC24（2）xex dxxdexxexexdxxexexC（3）x2lnxdxlnxdx3x3lnxx3dlnxx3lnxx2dx33333x3lnxx3C39（4）令x,t就x2t,dx2tdt,于是exdx=2tetdt2tdet2 tet2t

14、 edt=t 2 tet 2 eC2 exx1C（5）x2cosxdxx2dsinxx2sinxsinxdx2x2sinx2xsinxdxx2sinx2xdcosxx2sinx2xcosx2cosxdxx2sinx2xcosx2sinxC（6）由于exsin2xdxsin2xdexexsin2xexdsin2x exsin2x2cos2xdexexsin2x2excos2x2exdcos2exsin2x2 excos2x4exsin2xdx于是exsin2xdxexsin2x52excos2xC（7）x2arctanxdxarctanxdx3x3arctanxx3darctanxxdx333x

15、3arctanx11x32dxx3arctanx1x3x2xdx33x331xx3arctanx1x2ln1x2C33（8）xcos2xdxx1cos2xdx1xxcos2x dxx21xcos22242x21xdsin2xx21xsin2x1sin2xdx44444xx21xsin2x1cos2xC448（9）1arcsinxdx2arcsinxdx2xarcsinx2xdarcsinx2xarcsinx1xdx2xarcsinx21xC1（10）x2e3x dx1x2d3 exx2e3x23 xexdxx2e3x2xde3x33339x2e3x2xe3x2e3xC3927（11）由于cos

16、lnx dxxcoslnxxdcoslnxxcoslnxsinlnxdxxcoslnxxsinlnxxdsinlnxxcoslnxxsinlnxcoslnxdx于是coslnx dxxcoslnx2xsinlnxC（12）fxxdxxdfxxfxfxdxxfxfxC習(xí)題 4.4 求以下不定積分 : 解 1. x3dx 3; 27327dxx3x23x39 1027dxxxx33dxx3xx3x23x9 dx27x13dxln|x2x|C. 1x33x29x27ln|x3|C. 32x23 x102. x22x310dxx2110d3 x3xx23xx8dx3. 5 xx3x48dxx2x1dx

17、xxx8dx4113 x1x2dxx3dx 132xxC. 13 x1x2x8 ln|x|4ln|x|13 ln|x|1324.解5. dx3113xx21d x3Cx31x2xdx1x2xx11dx1xd x1x2123162124lnx11lnx2x11arctan2x136331lnx2 111arctan2x31C6x2x3xdxx3 1x42x11x33 dxx1 x2 21ln|x|23ln|x3|ln|x1 |C. 26. dx1x11111x12 dxx21x12x122x1 dx1ln|x|11ln|x|1x11C227. 1ln|x2|1x11C. 2x 11dx11x2

18、dxln|x|1ln1x2C. x2xx283x12dxx3 x11 dxx72x41x23x2x7x12dx4x11dx7ln|x 2| 4ln|x 1| C9. x2dx2x11x1111dxx1 1dx 1C1 xx2x212xln|x|1ln|x|11x1dx22x21ln|x|1ln|x|11x2x1dx1x11dx24222ln|x|1ln|x|11lnx211arctanxC. 24210. x3xxx1dxxx21 dx1dx121 x2x12x2. 1lnx11lnx21 1arctanxC24211. x11dxx22x1x22x1 dx41 x22x11dxx22xx1d

19、x424222x1212x212dx21 2x212dx22224x22x4x22x1x2dx2dx22x1dx22x1 1x2dxx8x22x1x22x1422x2ln|x22x1|2a r c t a n x1 2a r c t a n x1 C. 8x22x1441112. 6 x2x11 x24dx4x21x11 2dx2x1x 1x4lnx2lnx1x11C2lnxx13.3dx2x412xdx41x3dtanxsincostan21tan2x132dtanx213arctan2tanxC. 43214.1sinxd xxsin作變換ttan x 2,就dx122dt,sinx12t

20、2,1sinxxd xtsint2 t122dt=112 t2 t2122dtC11t4 txt2dtC122dtt2 1tt1t=12arctant=-1x12d tt1tan215. 2dxxcos作變換ttan x 2就有dx122dtcosx1t22t1t2dt2dxx211t22231t2dt2112dttcos3t321t32arctant3C2arctan1tanx 2C33316sin5xdxsin4xdcosx 1cos 2x 2dcosxcos 4xcos4xcos 4x 12x1xdcosxcos2cos4cosx2x313xCcoscos17. sinxcotxxd x

21、1costanxt ,就2sinx12t2,cosx1t2,cotx12t2,x2arctan ,dx2t1ttt2就sincotxx1d xu2 t21t2t21122d tdu1td t111 d tt1d t2 t1xcosxt2 t221t1t218. 2dx令1lnt1 2tC1lntanx1tanxC2du22222tanx 212sinxcos54 u1u251u23 u22u1u21u2或19. 1u12152du1arctan3 u51C1arctan3tanx 21C. C. 3555332 sinxdxx5令utanx14 u21u25122du2cos1uu1u2C.

22、3x1 3 u212du1 u1 32152du2 u331arctan3 u51C1arctan3tanx125551311dx令3x1u11u3 u2du3u111u du3 ln1x3u23 u3ln1|u|C33x123 3x120. 22x311dxx2x1dx1x22x3xC. 2x2321. x11dx令x1uu12 udu2u22 1dux11u1u22. 21u22u2ln| u1 |C2x14x14l n x11C. xdxx令xu4u21u4 u3du423. 4 u111du2 u24 u4ln1|u|Cu2x4x4ln14xC. 1xdx; 1xx令1 1xu, 就x

23、1u2, dx 14 uu 2 2du, 1u2x1xdxu1u2 14 u2du2u11112du1xx1u2u22uln|u1|2arctanuC. x3x4dx2dxu1ln|1x1x|2arctan1xC1x1x1x24 2x 22x13dx. x12dxxx2dxx2 x2122121 2lnx21dx21 2x11dx3dx21422 x122x21 2x21lnx21lnx21 2arctanx2 x31 x2x2arctanxC2221（上式最終一個(gè)積分用積分表公式28）測(cè)試題 A 一、挑選題 . 1、D; 2、D；3、B; 4 、B；5、A; 6 、C；7、B； 8、D；9、

24、B.；10、 A 二、填空 . 1 、1sinbxC； 2 、-2x3C;3 、 函 數(shù) 族FxC,這 里Fxfx4 、2b3sinxcosxC；5、arctanx,dx；6、yx21；7、xcosx 12 secx dx8、sinx e1 ex1C,9、asec t,t0 ,2；10、2x.三、求以下不定積分. （1）解：1cos2x d xx11cos2x1dx11cos2x dx1cos22cos2x2cos2xxtanxC（2）解：1dxxexdxdex1 lnx e1C1ex1ex52xCe（3）解：23x4x52xdx23xdx51xdx2 3x4442ln3lnln2（4）解：a

25、rcsinx2dxx2 arcsinx2arcsinx1xx2dxxarcsin2x2arcsinxd1x2xarcsin2x21x2arcsinx21x2darcsinxCx22dxxarcsin 2x21x2arcsinx2 xC（5）解：令tx1,就x2t1,于是xdx12 tdtt2dtt11t1 1dtlnt1xt21t21t1（6）解：1x322dx1x21x22dx1x2dxxxxx1x1ln1x22 11x2CxC2（7）解：arcsindx12 xarcsinx2darcsinx1x2arcsin（8）解：1x2dx91x2dx9xx2dx94x443112x2d2x1914

26、x2d94x232 3381arcsin2x194x2C234,1x0四、設(shè)fx x,10 x1,求fxdx. 2x ,x1解：fx 在, 上連續(xù),就必存在原函數(shù)F x ,使得Fx xC 1,xC2,0 xx01,1x2x2x12C 3又Fx 須到處連續(xù),有l(wèi)im x 0 xC1lim x 01x2xC2,即C 1C2,故,故物體運(yùn)動(dòng)規(guī)律為2lim x 1x2C3lim x 11x2xC2,即1C33C222聯(lián)立并令C 1C,可得C21C,C31C.2故fxdxxC,.,C ,0 xx01 . 1x2xxx12 2t,1221C2五、應(yīng)用題解 設(shè)運(yùn)動(dòng)規(guī)律為st,就s tv t4s t2 t21

27、 dt2t3tC,由s 1 ,3求得C33st2t3t4、33n11tann1xIn2. 六、如Intannxdx,n2 ,3,證明：In證明：由于故Inntannxdxtann2xtan2xdxn1tann22xsec2x1 dxn2xdxtann2xsec 2xdxItann2xdxtannnxdtanxtan11tann1xIn2nn11tanxI2. 測(cè)試題 B 一、單項(xiàng)挑選題1、選（） . 函數(shù)fx的任意兩個(gè)原函數(shù)之間相差一個(gè)常數(shù). 2、選（ B） 兩邊對(duì)x 3dxx3C微分得ffx33x2,f2 t3 t3fxfxdx3x2dx9x5Cx2CxfC.x eC2,3353、選（ B）

28、 原式xdFxxdxlnxx2lnxxlnxdxx2x2lnxxdxx21lnx1C22244、選（ C）xf1x2dx1f 1x2d 1x21 1225、選（ D）x e1dxexx112dx1ex21dxx e1ex11dx6、選（ C）提示：分子的1 看成是分母兩個(gè)因子之和7、選 A原式1x11 2x11dx1dxx112dxxxfx dx,ln|x|x11ln|x1|C .8、選A 2 x2ex 2ex2C.fxx dxx d fx xfx 2,1依據(jù)題意fxdxex2C,再留意到fx dxfx,兩邊同時(shí)對(duì)x求導(dǎo),得fx2xex2,xfx dxxfxfx dx2x2ex2ex 2C.9

29、、選（ B）2ln1ex1xC .令t1exextex1xlnt21 ,dx2 tdt,t2111x edxt21dtt11t11dtlnt1C2ln12t110、 選（ B）設(shè)tln x ,就當(dāng)0 x1時(shí), t0,ft1 .t eC2,即x于是f tf tdttC 1,即fxxC1當(dāng)1x時(shí) , 0t,ftte,于是ftftdt得fx xC 1,0 xx0exC2,得C21.又f0 0 ,C 10,再由fx在x0處連續(xù) , f0 lim x 0fx所以fx1x ,10 x0 .xfxC; ;3、22 sin x dxx；4、1lnx；5、x ex二、填空 .1 、tanxC； 2 、xf24x54x3c 1xc2.6 、f x