1积分计算不定积分1.22211xxxxxdxdxdxxexexe；22111xxxxxdxdeedxxexxxe，所以211xxxdxCxexe2.221111xxxxxedxedxedxeCxxxx3.lnln1lnxaxbxaxbxaxbdxdxxaxbxbxalnlnln()ln()xaxbdxaxbxdxxbxa所以1lnln()ln()xaxbxaxbdxaxbxCxaxb4.因为cossin2sincosxxxxxx2222sincoscos2sincossinsincoscossincossinxxxxxxxxxxxdxdxxxxxxx22cos2sincossincossincossincossinxxxxxxxxdxdxxxxxxx1sincoscossincossinxxddxxxxxxxcoscossinxCxxx5.因为1xxxee所以222111xxxxxxxxxeexexxxedxdxdxdxxexexexe11xxxxdxxdCxexexe6.因为arcsinarccos2xx所以2arcsinarccosarcsinarcsin2xxdxxdxxdx22222arcsinarcsinarcsin211xxxxxdxxxdxxx2222211arcsinarcsin1arcsin12211xxxdxxxdxxx222arcsin1arcsin2arcsin122xxxxxxdx2222arcsin1arcsin21arcsin21arcsin22xxxxxxxxdx222arcsin1arcsin21arcsin222xxxxxxxxC7.arcsinarccosxxdx22arcsinarccos1xxxxdxx2arcsinarccos21xxxxC8.被积函数定义域为1x或1x，当1x时，设sec02xtt2111sectanarccossectan1dxttdttCCttxxx当1x时，设xt，1221111arccosarccos11tdtdtCCtxtttt211arccos1dxCxxx9.225551455444555111211111155111xxxxdxdxdxxxx2355511135111Cxxx10.542222884111111112121212xxxttdxdxdtdtxxttt322111212dtttt或54222884422111111121212212xxxttdxdxdtdtxxtttt222211212tdttt2221122121tdttttt221212142121ttdttttt定积分计算定积分定义：1.122222201111limlim1411nnnknnndxnnnnnxkn2.122222201111111limlim41641414nnnkdxnnnnnxkn3.3xfxa，3334411limln12lim1ln2lnlnnnfffnaanann313011lnlnlimln4nnkkaaaxdxnn4.101111111limlim1211nnnkdxknnnnnxn利用定积分是数值，与积分变量选取无关：设连续函数()fx满足122300()()()fxxxfxdxxfxdx，求()fx解：令1200(),()afxdxbfxdx，则23()fxxaxbx4则112300222300()()afxdxxaxbxdxbfxdxxaxbxdx，即1234816234abaabb，解得381ab所以233()8fxxxx变限积分1.222332000001111lim1lim1lim33xxxttxxxeedtedtxxx2.42225650000012111lim1lim1lim63xtxuxxtxuxxxexedteduxxx3.求242256500000112sin1limsin()limsinlim63txuttttttttxdxuduttt+++====蝌4.设fx有连续导数，00,00ff，求20020limxxxftdtxftdt22200002200022limlimlim22xxxxxxxftdtxfxfxxftdtxftdtxfxftdtxfx204lim3xxfxfxxfx204lim103xfxfxffxx5.当0x时，220xFxxtftdx的导数与2x为等价无穷小，求0f。2222000333000limlimlimxxxxxxxtftdxxftdxtftdxFxxxx220202lim3xxxftdxxfxxfxx0022lim033xxftdxfx又，32001limlim33xxFxFxxx，所以2110,0332ff56.设fxx，sin0202xxgxx，求0xFxftgxtdt。解：000xtuxxxftgxtdttgxtdtxugudu00xxxguduugudu002200sinsin02sinsin2xxxuduuuduxxuduuuduxsin0212xxxxx定积分1.2420sincosxxdxp=ò()()222200111sin2cossin3sin442xxdxxxdxpp轾犏=+犏臌蝌()22201sin32sin3sinsin16xxxxdxp=++ò()()()()2011cos62cos2cos41cos232xxxxdxp=-+-+-ò32p=21202arctan1xdxxtan12244240002arctanseccossec1xtxtdxtdtttdttx22440012cos21112cos22464168xtdxtdt3.120arctan1xdxx4.22211222104400arctan1arctan1arctan121464xxxdxdxxxx5.设101101xxxfxxe，求201fxdx。121011xtfxdxftdt6222001sin1tan1cos2xxxxedxedxx22012tantan22xxxedx220sec2tan22xxxedx22220002tantan2tan2222xxxxxxeedxedxe7.1ln()(0)1xtfxdtxt，求1()fxfx.解：，令1ut，121111ln1ln1ln11111xxxtuufdtduduxtuuuu1()fxfx1111lnln1lnln11111xxxxtuttdtdudtdttuuttt221111lnlnln1ln1lnln122xxxxtxttdtdttdtttt8.设40tannnIxdx，求证1122121nInnn解：221444200011tantansec1tan11nnnnnnIxdxxxdxxIInn所以211nnIIn，又0,4x时，0tan1x，所以nI单调减少，所以2222nnnnIIII即21221nnIIn，所以1122121nInnn9.设0()cosxsxtdt1)证明:当n为正整数,且1nxn时,221nsxn2)求0()limxsxx证明：1）对任意给定的0x，必存在正整数n，使得1nxn时：110111cos()coscosnnixiiiiitdtsxtdttdt7令1tiu，则2100coscos2cos2iitdttdttdt所以21()2nsxn，即有当n为正整数,且1nxn时,221nsxn2）000000coscos()coslimlimlimlim11xxxxxxt...
