考研数学 第三章 一元函数积分学（不定积分）

1第三章一元函数积分学不定积分一求下列不定积分1DXX1LN2解L2CXXD21LN4L1LN22CXDXX22ART2ARCTARCT1ARCTN3OSIOSI2解CXXDDXX22OS1INCOS1INSIC1INC1I48D解方法一令,TX1CTTDTXD1LN811887828CX8LN方法二DXXDXD11187878CXLN|L88CX81LN5DXDXCOSI12I2SIN2COSIN1DXIN1COSI12XXX2COS2SIN22TAN1T2|COSIN1|L2XDXXC|L|I|L二求下列不定积分1212XXD解11222XDXXTXAN令TDSECO2CCTTDSINSICO22241X解令XTANT,CTTDTDTTDXDSIN1I3SINISINCOETANC124434224CXX23213221XD解令TANTDTTDTTX222222SIN1COSINSEC1ANXCT2RTSINAR4A02XD解令TASINCTATDTATADX2SIN412COS1COSI2223CXAXA222RCSIN5DX321解令TSINDTTDTTD42COS2142COS1COS432TTTTTTIN3SI838IN1CX2OS41SARC83TTINCIN2ICXX518ARCSI326DX421解令TDTTDTTDX22424211USIN令UD2COSICXCU323OS7DXX12解令TDTANSEC,SECCTDTTTDXXSINCOS11122XARCOS2三求下列不定积分41DXEX1243解CEEDXEXXXXARCTN1122224321XD解令,T2LNTCTTDTTTDDX2LNARL112LN14122CXXARTLN四求下列不定积分1DX1052解DXXXXD9495951052229DX9839849542962973984953452525XXXC9495646783241XD解24424111/TDTDTTTX令CXCUDUUT2422LN|SETAN|LSECAN令5五求下列不定积分1XD2COS解XDXD2SIN412COS12XSIIN42CX8S2XD3SEC解XDXXTANSECTTANSECTASECXDDX32SEC|T|L1TNCXXD|TASEC|LTASEC1SEC33X2LN解DXXDD23323LNL1LNLX223L6LLDX236LNLLCXLNLLN234DXCOSL解DXXXDXCOSLNSINLCOLSINLCOLNX2CSL5DXXXDXD2SIN81SI2SIN81COSIN81SINO23434CXOT4I2I4S2六求下列不定积分61DXX21LN解2221LN1LXDXXX222LN1TA令TDT222SEC1AN1LDTX22SICO1LNT22IN1LCTXSIL41LN2X21LNL22DX21ARCTN解DXXXDX22221ARCTN1ARCTNRTCLRT1RT1223DXE2ARCTN解DXEEDEXXXXX22221ARCTN1ARCTN1RTEXXX2RT1XXXXRT22CEEDEEXXXXXXARTNARCTN11ARCTN2227七设,求XEXF321LN0DXF解DXDFL21222243LN1LNCEXCXX0考虑连续性,所以C1C1,C11CDXFEXCXX431LN2LN220八设,A,B为不同时为零的常数,求FXAEFXCOSSIN解令,所以TTL，COSLNINLTTAFDXXXFCSSICBBASLL2九求下列不定积分1DXX32解CXXX3LN232222D153232解523523XDXXXC22513DXX21LN8解CXXDXDXX1LN21LN1LN1LN222241L22X解CXXDXD|1LN|1LNLN12222十求下列不定积分1DXARCT2解12222ARCTN11ARCTN1RTNXDXDDXX2222TARCTRTTXTDXTCOS1ARCTN1OS1RTNA222令CECTTXIN4TI841RC2XE221ARNT2DX1ARCSIN解令TT2AN,RI则CTTTDTTDTXANA1ARCSIN2222XXCX1RCSI1ARCSINRI3DXX221ACSIN解DTTDTTTX1CSCOSIN1SINRI22222令9CTDTDTT21COCO21|SIN|LCXXX2ARSIN|LARCI4DX1TN2解DTTDTTTX1CSSECANARCT2222令22OTT1OSTTDTDTCXXXCTT22ARTN1|LNAR1|IN|LCOCXX22ART1LART十一求下列不定积分1D234解DTTDTTTXX233COSINCOS2SIN8I令DT532COCS132XX2523442A2解DTADTATTX22COS1NSECSEC令XATRTN23DXEX21解UDUTDTTDTXXCOSIN1SI11222令令10CECUXX21ARSINOS4A0DX2解XU令DUA24TASIN2令TD42SI8TTDTACOS124COS1822CTATTADTATT4SINSI34COSIN222CTTTTAIN1SINCI43222ACOOSN3CAXXAXA222RCSI32CA3RIN2十二求下列不定积分1XDCOS1SIN解XDXDXD222COS1COS1SINCO1INSICOS122UUX令CDU|LN12XX|COS12|LCOS12D2IN解XDXXCOS2COS2COS11TX2TAN令|COS2|LN32|COS|LN12XTDXTDTCT|S|L2TA1ART4|S|L3RC43DXOSIN解DXXCOSIN121CDXDXCOSIN12OSINS2124SINCSIXCX|82TA|LOSIN21十三求下列不定积分1DX1解CTTDTTX33321411令C2342DXE1解DTDTTTEDXEXX1SECAN1SEC12令CTTXXXROLN|ASC|LN23DXXAR1解令TDXTXTTANSEC2,SEC,1AN,RC212DTTDTDTTTDXX222COS1ANANSECA1ARCTN22TTTOSTTTCT2|S|LNACXXX21ARTN|L1RCT第三章一元函数积分学定积分一若FX在A，B上连续,证明对于任意选定的连续函数X,均有,0BADXF则FX0证明假设F0,A0因为FX在A，B上连续,所以存在0,使得在,上FX0令M按以下方法定义A，B上X在INXFX,上X,其它地方X0所以22X02DXFDXFBA和矛盾所以FX00BAF二设为任意实数,证明20TAN1DXI4COT120DX证明先证4COSSIN20XFF20SSIFF令T,所以X20COSSINDXFF02SINCOTDFTF20INTFTF0IXFXF于是20COSSINDXFXF20COSSINDXFF20SINCODXFXF132COSSIN020DXFXF所以4I20FF0COSSINDXFXF所以20TAN1DXI4INCOSI12020DX同理4COT120DXI三已知FX在0，1上连续,对任意X,Y都有|FXFY|0,00,证明对于满足00因为F0F10X00,1使FX0FX1MAX所以（1）DF10DFF0在（0，X0）上用拉格朗日定理0FF）,0X16在（X0,1）上用拉格朗日定理0XFF1,0X所以410010XFXFFFDXFDD（因为）AB2所以104DXFXF由（1）得0XF十设FX在0,1上有一阶连续导数,且F1F01,试证1102DXF证明102XF101221002FDXF十一设函数FX在0,2上连续,且0,A0证明0,2,使20XF20F|F|A解因为FX在0,2上连续,所以|FX|在0,2上连续,所以0,2,取使|F|MAX|FX|0X2使|F|FX|所以|1|1|1|202020FDXFDXFXDFA第三章一元函数积分学广义积分一计算下列广义积分1232031DXEX0241DXX231XD45610