高等数学(北大版)答案习题3.4

习题3.41221231323311ln2ln2ln2ln2ln200000:111..54,13,11.54,(5),,42541(5)111114(5)5.28836ln212.(1ln22xxxxxxdxIxuxuxudxuduxuIuduudxuuuxedxxdexeedxe求下列各定积分1/2222200/22224000000442220022).3.1sincos(sin)131sin(1sin).2422164.sincoscoscossin.995.99ln(9)10ln3.2226.xxdxttdtxtttdtIIxxdxxdxxxxdxxxxdxxxxx12266602000112200943393322233000032211113sin(1cos2)(sin2).222264137.442arcsin.2223113458.111(1).22889.coscos2cosdxtdttdtttxxxxdxxxxdxxdxuduuxxdx32200322200/222000/2/2/2/20cos2cossin442coscoscos.3311110.cos2cos22coscos()22220,21;(1)sin()(1)!!2sin().!!2nnnnnnnxxdxxxdxxdxxxdxxdxudutdtnktdtntdtn2212200!!(1)!!11.()(sin)cos!!(1)!!2nannnnaxdxxattdtnnn是偶数;是奇数./2110/2660022320000320033010!!15612.sin.11!!69353513.sin2sin2.2642216111114.(sin)(1cos2)sin2223411sin2sin264211cos2cos26464xdxxdxuduxxdxxxdxxxdxxxxxdxxdxxx00330/4/442200/4/422200/4/42200/43/400101cos241sin2.6486415.tantan(sec1)tansectantantan(sec1)112tantan|1.34343416.arcsinaxdxxxdxxxdxxxdxxdxxdxxdxxxxdxx110011220022222200022222222002222rcsin|arcsin11.222117.ln()ln()ln()ln()ln()ln()||.18.()[,].()baxxdxxdxxxxxadxxxxaxdxxaxadxaxaxaaaafxabfxd设在连续证明21011003200223222200()(()).(),0,1,(),()()(())()(()).119.()().2,0,0,,1()()2baaaaaxbafabaxdxxabatabdxbadtfxdxbafabatdtbafabaxdxxfxdxxfxdxxtxtxataxfxdxxfxdx令则故证明令则时时故证证220011()().22aatftdtxfxdt110020.(1)(1).1,0,1,0.,mnnmxxdxxxdxxtxxtdxdt证明令则时时故证10110100000000000000(1)(1)(1)(1).21.,(),()()().()()()()()()()mnmnmnnmxtxxxttxtaxxxxxxdxttdtttdtxxdxfxfxdxdtftxtdxfxdxdttfxdxtfxdxdtxfxdxtftdtxftdttftdt利用分部积分公式证明若连续则证0/2000000000()().22.(sin)(sin).,0,,,,(sin)()(sin())()(sin)(sin)(sin)(sin)(sin).2xftxtdtxfxdxfxdxxtxtxdxdtxfxdxtftdttftdtftdttftdtftdtxfxdxx利用换元积分法证明时时故证0000/20/20/2/2/200(sin)(sin),1(sin)(sin)211(sin)(sin)22,/2,/2,,0,,(sin)(sin())(sin),(sin)(sinfxdxftdtxfxdxftdtftdtftdtuttutududtftdtfudufuduxfxdxfx令则时时/20).dx20/2/2222000/20000sin23..1cossinsincos1cos1cos1cosarctancos|.424.()(,),,:(1)()()()11(2)lim()(TxxxxxdxxxxxdxdxdxxxxxfxTxFxfxdxftdtTTftdtfxT利用上题结果求设函数在上连续以为周期证明函数也以为周期;解000000000000000).(1)()()()()()()()()()()()()()().11(2)()()1()TTxTTTxxTxTTxTTxxTTxdxxTFxTfxdxftdtTxfxdxfxdxftdtftdtTxfxdxfxdxftdtftdtTxfxdxftdtFxTftdtfxdxxTxfxdxxT证00000()().()(,),11()lim()()lim0.11lim()().xxTxxxTxFxftdtxFxTFxftdtfxdxxTxftdtfxdxxT在上连续以为周期,故有界,于是000000000000000025.(),()0,()0,()(,).()0.(,),,,),.0()0.()()TxTxTxTTxxxfxfxfxdxfxxxTfxfxxTfxxTmmfxdxmdxmTfxdxfxd设是以T为周期的连续函数且证明:在区间内至少有两个根为明确起见设如果在没有根则由连续函数的中间值定理在(恒正设其最小值为则,由周期性和假设证,010001001000001100110000,.,).,),()()0,(,)(,),()()()0,.(,)(,)()(,).xTxxTxxxxfxxTxfxxTfxTfxfxxxxTfxdxfxdxfxdxfxxxxTfxxxT矛盾故在(至少有一个根若在(再无其它根由于在和恒正矛盾故在或至少还有一个根根,即在区间内至少有两个根26.求定积分244022444400222440022222/2200222sincos..sincossincossincossincos2sincos4(sin2)1121sin21sin22282sin2mmdxdxxxmdxdxdxmdxxxxxdxdxdxdxxxxxxxdxdxdxdxxxxdxx其中为正整数被积函数以2为周期故周期为解,/2/2200/22202440cot242csc21cot2444arctan2|22.2cot2112222.sincosmdxdxxdxduuxudxdxmxx