1、2022-12-31 作作 业业 P137 习题习题5.4 1(2)(6)(10).2(4)(13).3.P142 习题习题5.5 1(3)(12).2(3).3(2).7(4).(10).复习复习:P135141 预习预习:P1431552022-12-32第十四讲第十四讲 不定积分不定积分（二）（二）一、变量代换法一、变量代换法二、分部积分法二、分部积分法2022-12-33 dxxf)(dxxxf)()()(tx 令令常常遇到相反的情况常常遇到相反的情况)()(xdxf dtttf)()(一、变量代换法一、变量代换法凑微分法凑微分法难求难求！容易求容易求！难求难求！容易求容易求！2022
2、-12-34dxx 11求求例例于于是是令令,2txtx dtttdxx 1211dttt 11)1(211 2dttdt Ctt )1ln(2解解Cxx )1ln(22022-12-35则则有有有有反反函函数数且且若若),()(,)()()(1xttxCtFdtttf CxFdxxf )()(1 定理定理2：（变量代换法）：（变量代换法）证证dxdtdtdFtFdxdcxFdxd )()(1 dtdxdtdF1 )()(1)()(xftttf 2022-12-36 dxeIx211求求例例解解22tex 令令),2ln(2 tx即即dtttdx222 dttttI2212 dtt2122Ct
3、 2arctan212Cex 22arctan22022-12-37 dxxI242求求例例解解txsin2 令令)22(tttttxcos2cos2cos2sin124222 tdtI2cos4 dtt22cos14Ctt )2sin21(2tdtdxcos2 2022-12-38改改写写为为将将为为了了作作变变量量回回代代I,CtttI )cossin(2直角三角形直角三角形作一个作一个根据代换函数根据代换函数,sin2tx x224x t dxxI24Cxxx 2422arcsin22022-12-39 932xdxI求求例例解解txtan3 令令tttxsec3sec31tan3922
4、2 Ctt tansecln dttdtttxdxsecsec3sec3922tdtdx2sec3 2022-12-310 xt392 x39sec2 xt122339ln9cxxxdx 1239lncxx cxx 9ln2CttI tansecln2022-12-311的的方方法法吗吗？还还有有其其他他问问：二二次次根根式式去去掉掉根根号号 )0(,22aaxdxI求求例例如如)t0(chtax令令 dtdtshtashtaI11“双曲代换双曲代换”和和 “倒数代换倒数代换”2022-12-312 )0(222axaxdxI例例如如：求求)1(1)(11212222xdaxxaxdxx tx
5、x1,0 令令时时当当 dttatxdaxIx1)1(1)(1122212cxxaactaa 2222221112022-12-313udvvduuvd )(udvvduuvd)(vduuvudv二、分部积分法二、分部积分法难求难求！容易求容易求！容易求容易求！难求难求！分部积分公式分部积分公式2022-12-314 dxxex计计算算例例 1?dvu 和和关关键键：如如何何正正确确选选择择dvxdxuex ,若若选选择择 dxexxedxxexxx2222则则解解更难求更难求！dvdxeuxx ,故故选选择择 dxexedxxexxxCexexx Cxex )1(dxexx22容易求容易求！
6、dxex2022-12-315 xdxx sin22计计算算例例)(cossin22xdxxdxx )(coscos22xxdxx)(sin2cos2 xxdxx xdxxxxcos2cos2 xdxxxxxsin2sin2cos2Cxxxxx cos2sin2cos2解解2022-12-316 xdxx ln3计计算算例例dvxdxux ,ln选选择择dxxuxv1,22 则则 dxxxxxxdxx12ln2ln22于于是是 dxxxx21ln22Cxx )ln21(42解解2022-12-317 xdxxarctan4计计算算例例)2(arctanarctan2 xxdxdxxdxxxxx
7、 222121arctan2dxxxxx 22211121arctan2Cxxxx arctan2121arctan22解解2022-12-318 dxx3sec5计计算算例例 dxxxdxx23secsecsec dxxxxxsec)1(sectansec2 dxxdxxxxsecsectansec3 )(tansecxdx dxxxxxxtansectantansecCxxxx tansecln21tansec21解解出现方程式出现方程式 dxx3sec回归回归2022-12-319 xdxexcos6计计算算例例 xdexdxexxsincos xdxexexxsinsin xdexex
8、xcossinCxxex )cos(sin21 xdxexexexxxcoscossin回归回归解法一解法一 xdxexcos2022-12-320)(coscosxxedxxdxe xdxexexxsincos xxxdexesincos xdxexexexxxcossincosCxxex )cos(sin21解法二解法二2022-12-321)(coscosxxedxxdxe xdxexexxsincos xdexexxcoscos xdxexexexxxcoscoscos xdxexcos出现恒等式出现恒等式问题出在此问题出在此解法三不可取！解法三不可取！解法三解法三2022-12-32
9、2 利用分部积分推导递推公式利用分部积分推导递推公式),2,1(sin7 nxdxInn求求积积分分例例Cxxdx cossin的的情情形形下下面面讨讨论论3 n解解Cxxxxdx cossin212sin22022-12-323 )cos(sinsinsin11xxdxdxxInnn xdxxxnxxnncossincos)1(cossin21 dxxxnxxnn)sin1(sin)1(cossin221 xdxnxdxnxxnnnsin)1(sin)1(cossin21 xdxnnxxnInnn21sin1cossin1)2(n2022-12-3245,4 nn例例如如：xdxxxxdx2
10、34sin43cossin41sinCxxxxx )cossin(83cossin413sin32cossin3154cossin5124 xdxxxxx xdxxxxdx345sin54cossin51sinCxxxxx cos158cossin154cossin51242022-12-325 nnaxdxI)(822求求积积分分例例),0(Nna )(1()()(1221221221nnnnaxxdaxxaxdxICaxaaxdxI arctan1221 dxaxxnaxxnn)()1(2)(222122 1221)(nnaxdxI dxaxaaxnaxxnn)()()1(2)(22222
11、122解解nnnIanInaxx21122)1(2)1(2)(2022-12-326121222)22(32)()22(1 nnnIannaxxanI2 n例例如如：22222222121axdxaaxxaI得递推公式得递推公式Caxaaxxa )arctan1(212222022-12-327 axdxxPaxdxxPdxexPnnaxncos)(sin)()(小结小结：下列积分可以用分部积分法：下列积分可以用分部积分法 xdxxPxdxxPxdxxPnnnarctan)(arcsin)(ln)(xdxxdxxdxarctanarcsinln dxaxxdxax)ln(2222 dxxdxb
12、xedxbxeaxax3seccossin nnnaxdxxdxxdx)(cossin222022-12-328dxeeIxx arctan9求求例例dttdxtxtex1,ln 则则令令 )1(arctan1arctanttddttttI)(arctan1arctan1tdttt dttttt )1(1arctan12解解2022-12-329dtttttttI )1()1(arctan1222dttdtttt 211arctan1Ctttt )1ln(21lnarctan12Cexeexxx 21lnarctan12022-12-330dxxexx 22)2(10求求例例解解 )21()2
13、(222xdexdxxexxx dxexxxxexxx)2(21)21(22 dxxexexxx22cexxexxx )1(22352022-12-331)()()(xQxPxRmn mmmmmnnnnnbxbxbxbxQaxaxaxaxP 11101110)()(其其中中真真分分式式多多项项式式代代数数有有理理函函数数 12111223 xxxxxx例例如如：三、有理函数的积分三、有理函数的积分（一）代数有理函数的积分（一）代数有理函数的积分假假分分式式时时当当真真分分式式时时当当,;,mnmn 2022-12-332简简分分式式的的和和真真分分式式可可分分解解为为四四类类最最 axA)1(
14、naxA)()2(qpxxCBx 2)3(nqpxxCBx)()4(2 caxAdxaxAln)1(caxnAdxaxAnn 1)(1()()2(四类最简分式的积分四类最简分式的积分2022-12-333 dxqpxxCBppxBdxqpxxCBx221212)2()3(qpxxdxCBpqpxxB2222ln21 )()(22ln2142222qpqxdxCBpqpxxB2022-12-334 dxqpxxCBppxBdxqpxxCBxnn)()2()()4(22121212)(1)1(2 nqpxxnB nqpqxdxCBp)()(2242222022-12-335等等函函数数下下列列积积
15、分分不不能能表表示示为为初初 xkdxdxxkdxxdxxdxxxdxxxdxxdxxdxxdxex2222223sin1,sin1cos,sincos,sin,sin1,ln1,22022-12-336 dxxfCxFdxxfxfxfdxxx)()()(,)(,)(2)1ln(111求求且且是是它它的的反反函函数数单单调调连连续续设设练习练习2022-12-337以下题目不用计算立即写出结果以下题目不用计算立即写出结果dxexx121.1 dueudxxxar 231)sin(.2 duu3 dxxx2.32 udu dxxxsin1.4 udusin2022-12-338 xxdx2costan.7duu 21dxxxx 1arcsin.8 udu dxxx2ln.6 duu 294.5xxdx udu2022-12-339dxxx lnln.9xxxlnlnlnln xdxxexcossin.102sin dueu
