4.3 分部积分法分部积分公式 设函数uu(x)及vv(x)具有连续导数.那么,(uv)uvuv,移项得 uv(uv)uv.对这个等式两边求不定积分,得 分部积分过程 这两个公式称为分部积分公式.vdxuuvdxvu,或vduuvudv,vdxuuvvduuvudvdxvu vdxuuvvduuvudvdxvu vdxuuvvduuvudvdxvu vdxuuvvduuvudvdxvu.1)v 容易求得;xvuxvudd)2比容易计算.:)d(的原则或及选取vvu 例1 x sin xcos xC.例2 例3 x2ex2xex2exCex(x22x2)C.vdxuuvvduuvudvdxvu.分部积分过程：例 1 xdxxxxxdxdxxsinsinsincosxdxxxxxdxdxxsinsinsincosxdxxxxxdxdxxsinsinsincosxdxxxxxdxdxxsinsinsincos 例 2 CexedxexexdedxxexxxxxxCexedxexexdedxxexxxxxxCexedxexexdedxxexxxxxxCexedxexexdedxxexxxxxxCexedxexexdedxxexxxxxx.例 3 2222dxeexdexdxexxxxxxxxxxdeexdxxeex2222dxexeexxxx2222222dxeexdexdxexxxxx2222dxeexdexdxexxxxx2222dxeexdexdxexxxxx xxxxxdeexdxxeex2222dxexeexxxx222xxxxxdeexdxxeex2222dxexeexxxx222 例4 例5 vdxuuvvduuvudvdxvu.分部积分过程：例 4 dxxxxxxdxxdxx121ln21ln21ln222Cxxxxdxxx22241ln2121ln21dxxxxxxdxxdxx121ln21ln21ln222dxxxxxxdxxdxx121ln21ln21ln222dxxxxxxdxxdxx121ln21ln21ln222Cxxxxdxxx22241ln2121ln21.例 5 xxdxxxdxarccosarccosarccosdxxxxx211arccos)1()1(21arccos2212xdxxx Cxxx21arccos.xxdxxxdxarccosarccosarccosxxdxxxdxarccosarccosarccos 例6 vdxuuvvduuvudvdxvu.分部积分过程：例 6 2arctan21arctanxdxxdxxdxxxxx2221121arctan21dxxxx)111(21arctan2122Cxxxxarctan2121arctan212.2arctan21arctanxdxxdxx dxxxxx2221121arctan21 dxxxx)111(21arctan2122 解 因为 例7 例 7 求xdxexsin.xdexexdexdxexxxxsinsinsinsinxxxxxdexexdxexecossincossinxdexexexxxcoscossin xdexexexxxcoscossin xdxexexexxxsincossin,所以 Cxxexdxexx)cos(sin21sin.xdexexdexdxexxxxsinsinsinsinxdexexdexdxexxxxsinsinsinsin xxxxxdexexdxexecossincossin vdxuuvvduuvudvdxvu.分部积分过程：vdxuuvvduuvudvdxvu.分部积分过程：解 因为 例8 例 8 求xdx3sec.xxdxdxxxdxtansecsecsecsec23xdxxxx2tansectansec dxxxxx)1(secsectansec2 xdxxdxxxsecsectansec3 xdxxxxx3sec|tansec|lntansec,所以 xdx3secCxxxx|)tansec|lntan(sec21.xxdxdxxxdxtansecsecsecsec23xxdxdxxxdxtansecsecsecsec23 于是 )32()()1(2111222nnnInaxxnaI.解 当n1时,用分部积分法,有 例9 例 9 求nnaxdxI)(22,其中 n 为正整数.dxaxxnaxxaxdxnnn)()1(2)()(222122122dxaxaaxnaxxnnn)()(1)1(2)(222122122解 CaxaaxdxIarctan1221 )(1(2)(211221nnnnIaInaxxI,即dxaxxnaxxaxdxnnn)()1(2)()(222122122dxaxxnaxxaxdxnnn)()1(2)()(222122122 dxaxaaxnaxxnnn)()(1)1(2)(222122122,解法一 于是 解法二 例10 例 10 求dxex.令xt2,则dx2tdt.dxexCxeCtedttextt)1(2)1(22xdexxdedxexxx2)(2xdeexdexxxx222CxeCeexxxx)1(222dxexCxeCtedttextt)1(2)1(22dxexCxeCtedttextt)1(2)1(22dxexCxeCtedttextt)1(2)1(22.xdexxdedxexxx2)(2xdexxdedxexxx2)(2 xdeexdexxxx222 CxeCeexxxx)1(222.注：在后者中u(x)不是以v(x)为中间变量的复合函数,故用分部积分法.在前者中f(x)是以(x)为中间变量的复合函数,故用换元积分法.第一步都是凑微分第一换积分元法与分部积分法的比较 )()()()()()(duufuxxdxfdxxxf令 )()()()()()()()(xduxvxvxuxdvxudxxvxu )()()()()()(duufuxxdxfdxxxf令,)()()()()()()()(xduxvxvxuxdvxudxxvxu.第一步都是凑微分第一换积分元法与分部积分法的比较 )()()()()()(duufuxxdxfdxxxf令 )()()()()()()()(xduxvxvxuxdvxudxxvxu )()()()()()(duufuxxdxfdxxxf令,)()()()()()()()(xduxvxvxuxdvxudxxvxu.2222 duedxedxxeuxx 2222 dxeexdexdxexxxxx提问：下列积分已经过凑微分,下一步该用什么方法？2222 duedxedxxeuxx 2222 dxeexdexdxexxxxx 2222 duedxedxxeuxx,2222 dxeexdexdxexxxxx.提示：可用分部积分法的积分小结 (1)被积函数为幂函数与三角函数或指数函数的积:(2)被积函数为幂函数与对数函数或反三角函数的积:(3)被积函数为指数函数与三角函数的积:xdxxcos,dxxex,dxexx2 xdxxln,xdxarccos,xdxxarctan xdxexsin,xdx3sec.解题技巧:的一般方法及选取vu把被积函数视为两个函数之积,按“反对幂指三反对幂指三”的顺序,前者为 后者为u.v内容小结内容小结 分部积分公式xvuvuxvudd1.使用原则:xvuvd易求出,易积分2.使用经验:“反对幂指三反对幂指三”,前 u 后v3.题目类型:分部化简;循环解出;递推公式思考与练习思考与练习1.下述运算错在哪里?应如何改正?xxxdsincosxxxxxdsin)sin1(sinsinxxxxdsinsincos12xxxdsincos1,1dsincosdsincosxxxxxx得 0=1答答:不定积分是原函数族,相减不应为 0.求此积分的正确作法是用换元法.xxsinsindCx sinln2.已知 xfxxcosCxxxCxxxxdxdxcos2sincoscos dxxfxxfxfxxxxfdd xxxfd的一个原函数是，求解解:。