第五章_第3节定积分的换元法和分部积分法

1和分部积分法定积分的换元法第三节一、换元公式二、分部积分公式三、小结2定理假设（1）)(xf在],[ba上连续；（2）函数)(tx在],[上是单值的且有连续导数；（3）当t在区间],[上变化时，)(tx的值在],[ba上变化，且a)(、b)(，则有dtttfdxxfba)()]([)(.一、换元公式3证设)(xF是)(xf的一个原函数，),()()(aFbFdxxfba)],([)(tFt令dtdxdxdFt)()()(txf),()]([ttf)(t是)()]([ttf的一个原函数.),())(()(aFF则)())(()(bFF4)]([)]([FF),()(aFbF)()()(aFbFdxxfba即.)()]([dtttf),()()()]([dtttf:注意;,换元公式仍成立时、当1２、换元要换限。5例1.计算).(0022axdxaa解:令,sintax则,costdtaxd且当时时0xax2t∴原式=2atdta)cos(202212)sin(tta221220242a,0t20tdt2cos6例2.计算.xdxx40122解:令,12xt则tdtxdtx,212且当时时0x4x3t∴原式=tdttt312122tdt)(312321)(tt33121313322,1t7例３计算.sincos205xdxx解令,cosxt2x,0t0x,1t205sincosxdxx015dtt1066t.61,sinxdxdt、不换元就不换限。3205xdxxsincos如205xxdcoscos20661xcos6184例dxx40122401212xxd)(40121121211)(x4132)(5例2221xxdx222211)(xxdx222111)()(xxd221xarcsin12469例６计算解.sinsin053dxxxxxxf53sinsin)(23sincosxx053sinsindxxx023sincosdxxx2023sincosdxxx223sincosdxxx2023sinsinxdx223sinsinxdx2025sin52x225sin52x.5410例７计算解.)ln1(ln43eexxxdx原式43)ln1(ln)(lneexxxd43)ln1(ln)(lneexxxd432)ln(1ln2eexxd43)lnarcsin(2eex.611例８计算解aadxxax022)0(.1令,sintaxax,2t0x,0t,costdtadx原式2022)sin1(sincosdttatata20cossincosdtttt20cossinsincos121dttttt20cossinln21221tt.412例9当)(xf在],[aa上连续，且有①)(xf为偶函数，则aaadxxfdxxf0)(2)(；②)(xf为奇函数，则aadxxf0)(.证,)()()(00aaaadxxfdxxfdxxf在0)(adxxf中令tx,130)(adxxf0)(adttf,)(0adttf①)(xf为偶函数，则),()(tftfaaaadxxfdxxfdxxf00)()()(;)(20adttf②)(xf为奇函数，则),()(tftfaaaadxxfdxxfdxxf00)()()(.014dxxxxx55224232)(sin如010例dxxxx)sin(cos1111422dxx20112cosdxx2022212cos)(cos2212202xdx2022xtan2042)tan(tan15奇函数例11计算解.11cos21122dxxxxx原式1122112dxxx11211cosdxxxx偶函数1022114dxxx10222)1(1)11(4dxxxx102)11(4dxx102144dxx.4单位圆的面积16例12若)(xf在]1,0[上连续，证明（1）2200)(cos)(sindxxfdxxf;证（1）设tx2,dtdx0x,2t2x,0t20)(sindxxf022sindttf20)(cosdttf;)(cos20dxxf17设tx,dtdx0x,tx,0t0)(sindxxxf0)][sin()(dttft,)(sin)(0dttft0022dxxfdxxxf)(sin)(sin)(021dxxxxcossin由此计算:证明180)(sindttf0)(sindtttf0)(sindxxf,)(sin0dxxxf.)(sin2)(sin00dxxfdxxxf02cos1sindxxxx02cos1sin2dxxx02)(coscos112xdx0)arctan(cos2x.42)44(21913例证明为周期的周期函数是以设,)(lxf.)(无关的值与adxxflaa:证明llalalaadxxfdxxfdxxfdxxf00)()()()(ltxlaldxxf令而)(dtltfa0)(dttfa0)(dxxfa0)(dxxfa0)(llaadxxfdxxf0)()(.无关与a2014例:,)](,[)(证明上连续在设020aaxfdxxafxfdxxfaa2002)]()([)(:证明dxxfdxxfdxxfaaaa2002)()()(dxxafdxxfaaa022)()(只要证明xat2令dttfdxxafaaa20)()2(aadxxf2)(dxxafxfdxxfaa2002)]()([)(2115例,),(,],[)(内可导在上连续在设babaxf试证及且,)(,)(0afMxfbadxxfbaM)()(22:证明,)(0af而,)()()(dttfafxfxadttfxfxa)()(baxabadxdttfdxxf])([)(dxdttfbaxa])([dxaxMba)(22)(abMbadxxfbaM)()(22:问题?能否用中值定理证明22设函数)(xu、)(xv在区间ba,上具有连续导数，则有bababavduuvudv.定积分的分部积分公式推导,vuvuuv,)(bababadxvudxvudxuv.bababavduuvudv二、分部积分公式,][bababaudvduvuv即231例dxxex1010xxdedxeexxx101010xee12例.arcsin210xdx210arcsinxx21021xxdx621)1(112120221xdx1221021x.1231224例3解.cos4021xxdx,cos22cos12xx402cos1xxdx402cos2xxdxxdxtan24040tan21xxxdxtan214040218xcosln.42ln825例4计算解.)2()1ln(102dxxx102)2()1ln(dxxx1021)1ln(xdx102)1ln(xx10)1ln(21xdx32lndxxx101121xx211110)2ln()1ln(32lnxx.3ln2ln3526例5设求解21,sin)(xdtttxf.)(10dxxxf因为ttsin没有初等形式的原函数，无法直接求出)(xf，所以采用分部积分法10)(dxxxf102)()(21xdxf102)(21xfx102)(21xdfx)1(21f102)(21dxxfx2721,sin)(xdtttxf,sin22sin)(222xxxxxxf10)(dxxxf)1(21f102)(21dxxfx102sin221dxxx1022sin21dxx102cos21x).11(cos21,0sin)1(11dtttf28例6证明定积分公式2200cossinxdxxdxInnnnnnnnnnnnn,3254231,22143231为正偶数为大于1的正奇数证,tx2令dtdx02022dttxdxnn)(sinsin20tdtncos20xdxncos29201xxncossinx2sin10dxxndxxnnn2200211sin)(sin)(nnInIn)1()1(221nnInnI积分关于下标的递推公式nI4223nnInnI,直到下标减到0或1为止20xdxInnsin201xxdncossindxxxnn20221cossin)(30,2200dxI,1sin201xdxI于是nnnnnnnnnn,3254231,22143231为正偶数为大于1的正奇数21nnInnI208xsin如221436587dxx205cos1325431dxxx010sindtttx2210222)(sindtt22102cosdtt2010cos221436587109dxx0102sin327例有证明对且上连续在设Nnxfxf,)(,],[)(020)]()([sin)(02220ffnnxdxxf:证明20nxdxxfsin)(201nxdxfncos)(20201nxdxxfxnxfncos)(cos)(})()]()({[dxxfffn20021)]()([022ffn8例dxxfdtttxfx00)(,sin)(计算设解dxxf0)(dxxfxxxf00)()(dxxxxf0sin)(xdxxxfsin)()(0xdxxfsin)()(01dxxxf02sin)()()(ff22还有更好的解法吗?348例dxxfdtttxfx00)(,sin)(计算设另解dxxf0)(