《高等數(shù)學(xué)》第04章不定積分習(xí)題詳解

1、第四章不定積分習(xí)題詳解181.求下列不定積分:解:(上5xVx)dxx(2)解:(2x3x)2dx4x(3)略.(4)解:,2cot(6)(8)(10)第四章(x21n2x)dx不定積分題4-135x2)dx52x2C=arcsinxcotx解:解:6x9x1n612'x21n3-dx1/2(cscx1)dx10x23xdx10x8xdx80xdx80xCln80sin2xdx=21、,-(1cosx)dx1sinx2cos2x,dxcosxsinx2cos.2xsinx,dxcosxsinx(cosxsinx)dxsinxcosxC解:解:cos2x22dxcosxsinxcotxt

2、anxC2一2cosxsinx2cos.-2xsinx1.2sinx1、，)dxcosxsecx(secxtanx)dx2secxdxsecxtanxdxtanxsecxC解：設(shè)f(x)max1,x,則f(x)x,1,“*)在()上連續(xù)Ci,則必存在原函數(shù)F(x),F(x)1又F(x)須處處連續(xù)，有-1xlim1(xC2)xlim1(2xCi)1-x2C3,1C2Ci,lim(1x2x12一一1C3)lim(xC2),即萬(wàn)C3CiC,可得C22+C,C3C.C,x故max1,xdxxC,1.1C,2.解：設(shè)所求曲線方程為yf(x),其上任一點(diǎn)(x,y)處切線的斜率為dydx從而dx由y(0)0

3、,得C0,因此所求曲線方程為3.解:因?yàn)?ci2sinx2sinxcosx，cos2x2cosxsinx1cos2x41.八sin2x2sinxcosx所以1sin2x212-cosx、21一cos2x者B是sinxcosx的原函數(shù).4題4-21.填空.4dx=xd(1+C)x(2)1-dx=d(lnx+C)x+C)(4)2secxdx=d(tanx+C)sinxdx=d(cosx+C)(6)cosxdx=d(sinx+C)d(arcsinx+C)(8),xdx.1x2=d(1x2+C)tanxsecxdx=d(secx+C)(10)六dx=d(arctanx+C)1.(11)dx=d(2ar

4、ctan.x+C)(12)(x1),xxdx=d(+C)2.求下列不定積分:解:xdx,x24.1d(4)2,x42(x214)2d(x24)(2)解:(x2ln4解:(4)解:解:(6)解:解:(8)解:(9)解:(10)解:(11)解:14)2.x24Cxdxx1ex.-dxxln4xd(lnx)11exd(-)x2x3x(e2edx.49x21lnx2dx(xlnx)xlnxlnlnx-1-xdxeexdx2x23Jdxx23x94x22)exdx1ex2x(e2e3x2)d(ex)3x-e4x2exC23x3x22J(/3311吟3x2(2)-arcsin-2d(xlnx)(xlnx)

5、xlnxdxe122xInxlnInxd(lnx)InInxd(lnInx)In|InInx|C12x113x21dx2d(ex)arctanexd(12x2xdxx2)12x2d(12x2)2.12x2Cdx94x2.2xarcsin-3x233x25d(x23dx123)-x2xdx4x23-ln(32x2)d(94x2).2,sin(t-dx2(x2)(x1)1ln3)dt(13廠2dx(arccosx)3.1x1,、2-(arccosx)Clncotxdxsin2xarctan.x,-=dx.x(1x)4cosxdxIncotcos2(t1sincos2(arccosx)dx2sinx

6、cosxIncotxdlncotx(arctanx)2C1cos2x2()dx)dtdtcos2(t)d2(t)lncotx2cscxdxcotx13darccosx(arccosx)12-(lncotx)C4arctan.x一(TTVxlncotxdcotxcotxarctan.xdarctan.x2-12cos2xcos2xdx(43xcos2x2、cos2x)dxsin2x1cos4x,dx2sin2xsin4xsinxcosx,3dx.sinxcosxcos3xdxarccosx10.dx21xarcsinx,dx,1x2cosx,dx.sinx3sinxcosxd(sinx2cosx

7、)2(sinxcosx)3C2cos2xcosxdx1sinxd(sinx)sinxarccosx_10d(arccosx),-arccosx10ln102arcsinx-arcsinxd(arcsinx)C1d(sinx)2sinxC.-sinx(12)解:(13)解:(14)解:(15)解:(16)解:(17)解：(18)解：(19)解：(20)解：(21)解：(22)解：(23)解:sin3xcos5xdxsin2xcos5xdcosx(1cos2x)cos5xdcosx(24)解:(25)解:(26)解:(27)(28)(29)1816-cosx-cosx86tan3xse(5xdxs

8、ecxx-sec75cos5xsin4xdxtan3xsec4xdx1tanx6x1tan5、x(13x)=66x解：設(shè)解:tan2xsec4xdsecxt6dx6arctanVx(sec21)sec4xdsecxsin9xsinx,dx1ccos9x18-cosxC2.32tanxsecxdtanx,3tanx(tanx1)dtanx_5.一.一,dx6tdt,代入原式得t3(115升6t5dt6t2)t21Jdt6tt216arctantC2,xtant,dxsectdt,1貝U23dx(x21)31(tan211)32.sectdt1dtsectsintC-x-C1x2x21dx2xx(

9、1)21xd(-)xx(1)2xd(-)xd(1)2x1)2（x）2(30)解:設(shè)x3sect,dx3secttantdt,則2x2x9sec2193secttantdt6sect2.tantdt，2一(sect1)dt3,-(tant1)C：(3、x29arccos-)x(31)解:設(shè)x2sint,dx2costdt,則2x4sin2144sin2t一.2.2costdt=4sintdtx2(1cos2t)dt=2t2sintcostC2arcsin-2(32)解:12x3-dx1112(x3)8dx9(33)解:(34)解:(22丁)2(xd(x3)213)2.2(3x1)一一arctan

10、C12x2x二dx322x2二dx3222/1、2/(x4)(234d(x-)24-2,xdx2ln(x21I)c求下列不定積分解:xsin2xdxx八cos2x2(2)解:-xxedx(3)解:lnxdx略.(5)解:(6)解:lnx32.xcosxdxx2sinxarcsin-5(2x1)512)2二dx1(xx)4-3xd(cos2x)x八cos2x2cos2xdxsin2x4xdex3xlnxd()3xxedxxxex2dsinx2xdcosx3xIlnx3x2sinxx2sinx2xsinx2xcosx2sinxC因?yàn)閑xsin2xdxsin2xdex3xd(lnx)3sinxdx2

11、2xcosx23xIlnx32x7dx3x2sinxcosxdx2xsinxdxxxesin2xed(sin2x)(8)(9)解:xsin2x2cos2xd(ex)xsin2x2excos2x4eexsin2xdxx2arctanxdx3x.arctanx33xXarctanx3解：xcos2exsin2xexsin2xxsin2xdx2excos2x3xarctanxd32ex3xarctanxcos2x2exd(cos2x)3xdarctanx3xdxln(11x-xdsin2x1一xsin2x41解：1x3x.arctanx3x2)Ccos2xdx1一cos2x8(xxdxxcos2x)

12、dx1xcos2xdx21一xsin2x4sin2xdxarcsin.xdx2arcsin.xdx2xarcsin.x、xdarcsinx2、xarcsin.x1.dx2，xarcsin,1x.1(10)斛：xedx一323xxe3xxedx23xxe3xxde3x2xe93x2一e273x(11)解：因?yàn)閏oslnxdxxcoslnxdcoslnxxcoslnxsinlnxdxxcoslnxxsinlnxxdsinlnxxcoslnxxsinlnxcoslnxdx于是coslnxdxxcoslnxxsinlnx-C2(12)解：xf(x)dxxdf(x)xf(x)f(x)dxxf(x)f(x

13、)C習(xí)題4-4求下列不定積分(1)解:33xx112dxdx(xx1)dxx1x1xInx1C54c(2)解：土。一8dx3xx2(xx1)dx2(xx1)dx2xx83xx/843、人(x71”)dxx8lnx4lnx13lnx1C(3)解:2x22x(x2)(x2131)2dx1,x2,3x4,dx；dx-；dxx2x21(x21)2ln x4r2.t-2(x 1)dx21d(xx2 11)2-21dxx2123d(x21)2(x21)2lnx211n(x221)2arctanx3-22(x1)2xx212arctanxC（上式最后一個(gè)積分用積分表公式28）(4)解:6x211x44212

14、dx-2dxx(x1)xx1(x1)4lnx2lnx11八2I1-C2lnx2(x1)Cx1x1(5)解：-2dxx3x2x12dx(x1)(x21)1dx1x1,一dx2x12x2111nx241n(x21)arctanxC2(6)解:dxsin2x72dx,utanxcos2xdu4u2(8)1arctan2.32tanx1解:1解:dtt17dtlnt1CIn2arctant習(xí)題4-5t2)dtd(x 2).1 (x 2)2利用積分表計(jì)算下列不定積分:/、dx(1):54xx,5 4x-dx解：因?yàn)閐x在積分表中查得公式(73)一ln(xx2a2)Ca22,于dx54xx2ln(x54x

15、x22)ln3xdx解：在積分表中查得公式（135）lnn xdxx(ln x)nn lnn 1 xdx現(xiàn)在n3,重復(fù)利用此公式三次,ln3 xdx xln3x 3x ln2 x6xln x 6x1(3)1TTdx(1x2)2第四章不定積分習(xí)題詳解解：在積分表中查得公式（28）1 2-2- d x (b ax2)2x-22b(ax2b)1dx22bax于是現(xiàn)在a1,b1,于1-dx(1x)x-22(x1)dxx21x-22(x1)arctanxC/、dx4),2x、x1解：在積分表中查得公式51)1-arccosadxx<x211arccos-x(5)x2Jx22xdx解：令tx1,因?yàn)?/p>

16、x2x22xdxx2,(x1)21dx(t22t1)t21dt由積分表中公式（56）、(55)、(54)2221x.xadx京2xxx22In8x2,x22xdxxa1,2(x85a28Inx解：在積分表中查得公式（1)2a.In2a2),(x1)2(x1)2a216)、(15)1023a)34(x1)2a23C第四章不定積分習(xí)題詳解dxaxbadxdxx、ax bx2、axbbx2bx、axb,axbarctan,b于是現(xiàn)在a2,b1,于是dx . 2x 12 arctan . ； 2x 1 Cx“2x 1 xdx2x1x2,2x1xcos6xdx解：在積分表中查得公式(135)31cosn

17、xdx1n-cosn1xsinxn2cosxdx現(xiàn)在n6,重復(fù)利用此公式三次，得615cos xdx cos xsinx653?cos xsinx2415/1x.(sin2x)24 42(8)e2xsin3xdx解：在積分表中查得公式(128)eax sin bxd x現(xiàn)在a 2 , b 3 ,于是e 2x sin3xd x1 ax-2e (asinbx bcosbx) Ca b1 ce ( 2sin3x 3cos3x) C131 °eax(2sin3x 3cos3x) C .本章復(fù)習(xí)題A一、填空.cii2y x2 1.(1)已知F(x)是snB的一個(gè)原函數(shù)，則d(F(x2)=2sn

18、工dx.xx(2)已知函數(shù)yf(x)的導(dǎo)數(shù)為y2x,且x1時(shí)y2,則此函數(shù)為如果f(x)dxxlnxC，則f(x)=lnx1.(4)已知f(x)dxsinxxC,則exf(ex1)dx=sin(ex1)ex1C.(5)如果f(sinx)cosxdxsin2xC,則f(x)=2x.二、求下列不定積分解:21cosxdx1cos2x21cosx(2)解:(3)解:解:(5)解:(6)(8)2cos2x-dx1tanxxdxxed(ex1)ln(e3x2xdx3x(4)dx21cosx2cos1)2(1secx)dx/、22(arcsinx)dxxarcsinx2-2xarcsin-2xarcsin

19、2xarcsin解:解:解:(11ln(122arcsinxd1x22.1x2arcsinx22tdt2(t1)t2、2dxx)x2)dx11(2)dxCxln3ln425Cln2arcsinx12xdarcsinxarcsinx2xCt212dtt2112(1x2)日Z(yǔ)E)dtIn(arcsinx)2-1x22(1x)(arcsinx)dx(12d(arcsinx)arcsinx1x22dx.94x294x2dxxdx94x2I222d(3x)1(3幻2x一arcsin-4x2(9)解：tan54xsecxdx(10)(11)(12)/7(sec解：令x解:解:5x2secxsint,1xt

20、tan23x2,xedxlnlnx,dxx1,三、設(shè)f(x)x1,0解:“*)在(limx01d(994x2,43,tanxsecxdsecx3secx)dsecx7t2costdt1costarcsinx1cost1cost4x2)8secx1dt(sec2x6secxdt231)sec4secx1costxdsecxdg)2tcos一t.t2sin-sin-22dt.t2cos-sin-2x2xde2212x2xe2arcsinx-InInxdInxInInxf(x)dx.2x,x)上連續(xù)F(x)F(x)須處處連續(xù)，有一12(xC1)lim(xx02,則必存在原函數(shù)xCi,1-x22xC2

21、)2xC2,C3,即C1C2x22edx12x2xe21x2e2F(x),使得2 lim(x2 C3) x 1聯(lián)立并令Ci故 f(x)dxx C, 1 2 -x 221x 2x C,C,x 00x1.x 112八八3八lim(-xxC2)，即1C3-C2xi2322，一1C,可得C2-+C,C31C.2四、若Intannxdx,n2,3,證明：In-tann1xIn2n1證明：因?yàn)镮n tann xdxtann 2 xtan2 xd xtann 2 x(sec2 x 1)dxtann 2 xsec2 xd x tann 2 xdxtann 2 xdtanxtann 2 xd xn1tanxIn2n1n1tanxIn2n1本