1、不定积分 )?()(xF)()?(xf)()(xfxF)(xF)(xfIx)(xF)(xf)()()(),()()(xfxFxFxfxFxF而)(xf)(xF)()()(),()()(xfxFxFxfxFxF而)(xF)(xf)()(xFTxF)()(xfTxf)()(xfxFx()d f xx()d()f xxF xC,()()xf x()()()()0F xxf xf x()()F xxC()()F xxC)()(xfxFCxF)()(xfsin dcosx xxC cos dsinx xxC2secdtansec tan dsecx xxCxx xxC2cscdcotcsc cot dc
2、scx xxCxx xxC 21darcsinarccos1xxCxCx 21darctanarccot1xxCxCx e dexxxCdlnxxaa xCa11d1xxxC11221dd22xxxxCxCx74333d4xxxC 211dxCxx 1dlnxxCxxxx111)ln（d0 daxa xCxCxxx1)ln0时，（()d()()f x xF xCf x dfxxfxC时，0 x1)()()d()d()df xg xxf xxg xx2)()d()dkf x x k f x x2(e5)d.xxx解解 原式=(2e)5 2)d(2e)d5 2 dxxxxxxx(2e)ln(2e)
3、x2ln25xe52ln2 1ln2xxCC dsincos dxx xdcossin dxx x 1dlndxxxxxxxxx22222sin22cos1cot1csccossin1xxxxxxxcossin22sincos22cos1tan1sec2222cos2sin2sinsin2cos2sin2sinsin2sin2sin2coscos2cos2cos2coscos2d2 dxx x()()d()d()fxxxfxxtatataxaxxasec,tan,sin:,.1)2222三角代换将有 txx1,.2)令的幂分母中有ueeuxx设有有根式设根式,.3)22d.sincosxxx2
4、22222sincos1dddsincoscossinxxxxxxxxxCxxcottan42d.1xxx421 1d1xxx Cxxxarctan313221(1)dd1xxxx 22)(1d1axxa.d22xax解解 22dxax,axu 令则xaud1d21uuda1Cuaarctan1Caxa)arctan(1例例 求求).0(d22axax解解 2)(1daxax2)(1)(daxaxCax arcsin22dxax Caxaxaln21.d22axx解解 221ax)(axax)()(axaxa21)11(21axaxa 原式原式=a21axxaxxdda21axax)(da21
5、ax lnax lnCaxax)(d .dtanxx解解 xxxdcossinxxcoscosdCx cosln?dcotxxxxxsindcosCx sinlnxxsinsindxxdtan类似 22ln(1)d.1xxxx)1221(11)1ln(222xxxxxx211x22ln1d ln1xxxxCxx2231ln32322d3x xxc1d.1 sinxx2221 sinsindsecddcoscosxxxx xxxx2dcos1tantancoscosxxxCxx 21ln1d.1xxxx)1(ln()1(ln(11lnxxxx2111dd ln121xxxx111lnd ln21
6、1xxxx21d2x xxCCxx11ln4122tand.x x2(sec1)dtanxxxxC2121111xxx.dsecxx xxtansecxxdsecxxdsecxxtansec)tan(secxxxxxxxxdtansectansecsec2)tan(secdxx Cxxtansecln同样可证xxdcscCxxcotcscln或xxdcscCx2tanln.d3cossin22xxx解解 xx3cossin22221)2sin4(sinxx xxxx2sin2sin4sin24sin24141241)8cos1(81xxx2cos2sin2)4cos1(81x原式原式=xd41
7、)8d(8cos641xx)2(sind2sin221xx)4d(4cos321xxx41x8sin641x2sin361x4sin321C .)0(d22axxa解解 令22sin,(,),xatt 则taaxa22222sintacosttaxdcosd 原式tacosttadcosttadcos22Ca242sin2ttax22xa taxarcsinCxax222122atttcossin22sin2axaxa22 原式.)0(d22aaxx解解 令22tan,(,),xat t 则22222tanataaxtasecttaxdsecd,2 原式 ta2sectasectdttdsec
8、1tanseclnCttax22ax tln22ax a)ln(1aCCCaxx22lnxa1C .)0(d22aaxx解解,时当ax 令2sec,(0,),xat t则22222secataaxtatanxdtttadtansec 原式td ttatansectatanttdsec1tanseclnCttax22ax t1 lnCCaxx22ln)ln(1aCC22ax axa,时当ax令,ux,au 则于是22daxx22dauu122lnCauu122lnCaxx1222lnCaxxa)ln2(1aCCCaxx22ln.d422xxxa解解 令令tx1 xxdtanxxdcotxxdse
9、cxxdcscCx coslnCx sinlnCxx tanseclnCxxcotcsclnxxad122xxad122xaxd122xaxd122Caxaarctan1Caxaxaln21CaxarcsinCaxx)ln(22xaxd122Caxx22lnddddduvv uu vv uuvxdddd()duvxuvv uuvvuxdddduvxuvv uuvvux.dcosxxx解解 令令,xu,cosxv 则原式xxsinxxdsinCxxxcossin.dlnxxx解解 令,ln xu xv 则原式xx ln212xxd21Cxxx2241ln21xx sind21ln d2xx.da
10、rctanxxx解解 令令,arctan xu xv 则原式xx arctan212xxxd12122xx arctan212xxd)111(212xx arctan212Cxx)arctan(2121arctan d2xxesind.xxx解解 令令,sin xu exv,则原式e sinxxe cosdxxxe sincos dexxxxe cose sindxxxxx故 原式=1e(sincos)2xxxC注注:分部积分法可循环使用分部积分法可循环使用,每次使用都按每次使用都按“反对幂三指反对幂三指”sin dexxe sinxx例例 求求.darccosxx解解 令,arccosxu
11、1 v,则xv 原式=xxarccosxxxd21xxarccos)1d()1(222121xxxxarccosCx 21例例 1ln dlndx xxxxxxCxxxlned.xx解解 令,tx则,2tx ttxd2d 原式2e dttt2(ett2e(1)xxCe)tC2dett2 e2 e dtttt练习练习:ede1xxxx(先分部,再换元)d(e1)e1xxxx2d(e1)x2e1xx2e1dxx.)(d22nnaxxI解解 xaxxnnd)(21222naxx)(22xaxnnd)(2122naxx)(22nIn2122nIan222)(aaxnaxx)(22nInaxx)(1d2
12、2naxx)(22已知CaxaIarctan11利用递推公式可求得.nI如：如：3I2222)(41axxa2243Ia2222)(41axxa243a22221axxa1221Ia2222)(41axxa22483axxaCaxaarctan835得递推公式得递推公式nnnIannaxxanI22221212)(21 00()mmmnnnPxa xaf xQxb xb有理函数假分式有理函数假分式相除多项式多项式+真分真分 式式分解若干部分分式之和若干部分分式之和部分分式的形式为部分分式的形式为kkqxpxNxMaxA)(;)(2)04,N(2qpk;)1(1)1(2xx;653)2(2xxx
13、.)1)(21(1)3(2xx解解(1)用拼凑法用拼凑法22)1()1(1xxxx2)1(1x)1(1xx2)1(1x)1(xx2)1(1x11xx1)1(xx)1(xx6532xxx)3)(2(3xxx2xA3xB原式)2(xA33xx5原式)3(xB2x23xx6故25x原式36x3x)1)(21(12xx xA2121xCBx原式)21(xA21x54，得，得代入等式两端代入等式两端分别令分别令1,0 xC541215461CB52B51C原式=x214512112xxCaxAln)1(nCaxnAn1)(1xaxAd.1xaxAnd)(.2xqxpxNxMd.32xqxpxNxMnd)
14、(.42)1,04(2nqp变分子为)2(2pxM2pMN 再分项积分 dxpxqpxxd)2()(2.d3222xxxx解解 原式原式xxxd3223)22(21x32)32d(2122xxxx32ln212xx22)2()1()1d(3xxCx21arctan23思考思考:如何求)1?(d)32(22nxxxxn提示提示:变形方法同前例,并利用.)(d22nnauuI解解 原式xxd14)1(2x)1(2 x214d.1xx 2arctan2211xx21221 ln21xx21xxCxxxxd12122121xxxxd121221212)(2121xx)d(1xx 2)(2121xx)d
15、(1xx 注意寻找解题技巧注意寻找解题技巧按常规方法较繁按常规方法较繁.xx21arctan2212Cxxxx1212ln24122)0(x设设)cos,(sinxxR表示三角函数有理式表示三角函数有理式,xxxRd)cos,(sin令2tanxt t 的有理函数的积分则则22122tan12tan2sinttxxx2211cos,ttx,arctan2tx 22dd1xtt.d)cos1(sinsin1xxxx解解 令,2tanxt 则2211cos,ttx212sinttx22,dd1xtt 2121tt212tt)1(2211ttttd212tttd122121221tt 2tlnC2t
16、an412x2tanxCx2tanln21 原式原式41dsinxx22csc(1cot)dxxx222cscdcotcscdx xxx xd(cot)x.cot31cot3Cxx 万能置换不一定是最佳方法万能置换不一定是最佳方法,故三角函故三角函数有理式的计算中先考虑其它手段数有理式的计算中先考虑其它手段,不不得已才用万能置换得已才用万能置换.41dsinxx解解 sin,cosdRxxxxxxx2222tan1sec1cossinxxxxx22cot1csccossin22sinxxx22sin211cos22cosxxRxxRcos,sincos,sinxsintx cos)cos,(s
17、in)cos,(sinxxRxxRxcostx sin)cos,(sin)cos,sin(xxRxxRxx cos,sin,2222212dtan,sin,cos,d2111xttttxxxttttx tan,d),(xbaxxRn令nbxat,d),(xxRndxcbxa令ndxcbxat,d),(xbaxbaxxRmn,pbxat令.,的最小公倍数为nmp.21d3xx解解 令,23xu则,23 uxuuxd3d2原式u123uuduuud11)1(32uuud)111(33221uuu1lnC3223)2(x323x321ln3xC.d3xxx,6tx 则有原式23tttt d65ttt
18、td)111(626331t221ttt1lnCCxxxx)1(ln6632663解解 令6tx 5,d6dxtt.d11xxxx解解,1xxt则,122ttx22)1(d2dtttx原式原式ttt221tttd)1(222ttd112211lnttCCxxx1)1(22ln原式原式).0(d)1(1xxxx令先化为先化为 的形式的形式 .ndcxbax)()(,)(1tghtftxg xhxgf.f x()()()()f tF tCf xF xC ,xf x d.xfxx dddxfxxx fxxfxfxx xxxC,xf xxC 12()()xf xx00 xxxx d.f xx 12dddxxf xxxx00 xxxx 1122F xCFxC00 xxxx 1212d,max,d,min,d,sgndf xxxxf xx 022011)()()(xFcxFxFcxF00 xxxxsincosdsincosAxBxxxx xF0 x 2222d,df xxax g xxxaB 22,dxacxf xtaxtaxtaxtan,sin,sec设 21dd,d2df x g xf x g xf x g xf x g xxxxxxxxx 344443334dd 1ln 1114xxxxCxx 例例
