清華大學(xué)微積分高等數(shù)學(xué)第4講不定積分二ppt課件

1、 作作 業(yè)業(yè) P P 習(xí)題習(xí)題5.45.4 1(2)(6)(10). 1(2)(6)(10). 2(4)(13). 3.2(4)(13). 3.P142 P142 習(xí)題習(xí)題5.55.5 1(3)(12). 2(3). 1(3)(12). 2(3). 3(2). 7(4). (10).3(2). 7(4). (10). 復(fù)習(xí)復(fù)習(xí): P141: P141 預(yù)習(xí)預(yù)習(xí): P143: P143155155第十四講第十四講 不定積分二不定積分二一、變量代換法一、變量代換法二、分部積分法二、分部積分法 dxxf)( dxxxf)()( )(tx 令令經(jīng)常遇到相反的情況經(jīng)常遇到相反的情況 )() )(xdxf

2、 dtttf)()( 一、變量代換法一、變量代換法湊微分法湊微分法難求難求 ！容易求容易求 ！難求難求 ！容易求容易求 ！dxx 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(2則則有有有有反反函函數(shù)數(shù)且且若若),()(,)()()(1xttxCtFdtttf CxFdxxf )()(1 定理定理2：變量代換法：變量代換法證證dxdtdtdFtFdxdcxFdxd )()(1 dtdxdtdF1 )()(1)()(xftttf dxeIx211求求例例解解22tex 令令),2ln(2 tx即即

3、dtttdx222 dttttI2212 dtt2122Ct 2arctan212Cex 22arctan2 dxxI242求求例例解解txsin2 令令)22( tttttxcos2cos2cos2sin124222 tdtI2cos4 dtt22cos14Ctt )2sin21(2tdtdxcos2 改改寫寫為為將將為為了了作作變變量量回回代代I,CtttI )cossin(2直直角角三三角角形形作作一一個(gè)個(gè)根根據(jù)據(jù)代代換換函函數(shù)數(shù),sin2tx x224x t dxxI24Cxxx 2422arcsin2 932xdxI求求例例解解txtan3 令令tttxsec3sec31tan392

4、22 Ctt tansecln dttdtttxdxsecsec3sec3922tdtdx2sec3 xt392 x39sec2 xt122339ln9cxxxdx 1239lncxx cxx 9ln2CttI tansecln的的方方法法嗎嗎？還還有有其其他他問(wèn)問(wèn)：二二次次根根式式去去掉掉根根號(hào)號(hào) )0(,22aaxdxI求求例例如如)t0( chtax令令 dtdtshtashtaI11“雙曲代換雙曲代換 和和 “倒數(shù)代換倒數(shù)代換 )0(222axaxdxI例例如如：求求)1(1)(11212222xdaxxaxdxx txx1,0 令令時(shí)時(shí)當(dāng)當(dāng) dttatxdaxIx1)1(1)(112

5、2212cxxaactaa 222222111udvvduuvd )( udvvduuvd)( vduuvudv二、分部積分法二、分部積分法難求難求 ！容易求容易求 ！容易求容易求 ！難求難求 ！分部積分公式分部積分公式 dxxex計(jì)計(jì)算算例例 1?dvu 和和關(guān)關(guān)鍵鍵：如如何何正正確確選選擇擇dvxdxuex ,若若選選擇擇 dxexxedxxexxx2222則則解解更難求更難求 ！dvdxeuxx ,故故選選擇擇 dxexedxxexxxCexexx Cxex )1( dxexx22容易求容易求 ！ dxex xdxx sin22計(jì)計(jì)算算例例)(cossin22xdxxdxx )(cosc

6、os22xxdxx)(sin2cos2 xxdxx xdxxxxcos2cos2 xdxxxxxsin2sin2cos2Cxxxxx cos2sin2cos2解解 xdxx ln3計(jì)計(jì)算算例例dvxdxux ,ln選選擇擇dxxuxv1,22 則則 dxxxxxxdxx12ln2ln22于于是是 dxxxx21ln22Cxx )ln21(42解解 xdxxarctan4計(jì)計(jì)算算例例)2(arctanarctan2 xxdxdxxdxxxxx 222121arctan2dxxxxx 22211121arctan2Cxxxx arctan2121arctan22解解 dxx3sec5計(jì)計(jì)算算例例

7、dxxxdxx23secsecsec dxxxxxsec)1(sectansec2 dxxdxxxxsecsectansec3 )(tansecxdx dxxxxxxtansectantansecCxxxx tansecln21tansec21解解出現(xiàn)方程式出現(xiàn)方程式 dxx3sec回歸回歸 xdxexcos6計(jì)計(jì)算算例例 xdexdxexxsincos xdxexexxsinsin xdexexxcossinCxxex )cos(sin21 xdxexexexxxcoscossin回歸回歸解法一解法一 xdxexcos)(coscosxxedxxdxe xdxexexxsincos xxxd

8、exesincos xdxexexexxxcossincosCxxex )cos(sin21解法二解法二)(coscosxxedxxdxe xdxexexxsincos xdexexxcoscos xdxexexexxxcoscoscos xdxexcos出現(xiàn)恒等式出現(xiàn)恒等式問(wèn)題出在此問(wèn)題出在此解法三不可?。〗夥ㄈ豢扇?！解法三解法三 利用分部積分推導(dǎo)遞推公式利用分部積分推導(dǎo)遞推公式), 2, 1(sin7 nxdxInn求求積積分分例例Cxxdx cossin的的情情形形下下面面討討論論3 n解解Cxxxxdx cossin212sin2 )cos(sinsinsin11xxdxdxxInn

9、n xdxxxnxxnncossincos) 1(cossin21 dxxxnxxnn)sin1 (sin) 1(cossin221 xdxnxdxnxxnnnsin) 1(sin) 1(cossin21 xdxnnxxnInnn21sin1cossin1)2( n5,4 nn例例如如： xdxxxxdx234sin43cossin41sinCxxxxx )cossin(83cossin413sin32cossin3154cossin5124 xdxxxxx xdxxxxdx345sin54cossin51sinCxxxxx cos158cossin154cossin5124 nnaxdxI)

10、(822求求積積分分例例),0(Nna )(1()()(1221221221nnnnaxxdaxxaxdxICaxaaxdxI arctan1221 dxaxxnaxxnn)() 1( 2)(222122 1221)(nnaxdxI dxaxaaxnaxxnn)()() 1( 2)(22222122解解nnnIanInaxx21122)1(2)1(2)( 121222)22(32)()22(1 nnnIannaxxanI2 n例例如如： 22222222121axdxaaxxaI得遞推公式得遞推公式Caxaaxxa )arctan1(21222 axdxxPaxdxxPdxexPnnaxnco

11、s)(sin)()( 小結(jié)小結(jié)：以下積分可以用分部積分法：以下積分可以用分部積分法 xdxxPxdxxPxdxxPnnnarctan)(arcsin)(ln)( xdxxdxxdxarctanarcsinln dxaxxdxax)ln(2222 dxxdxbxedxbxeaxax3seccossin nnnaxdxxdxxdx)(cossin22dxeeIxx arctan9求求例例dttdxtxtex1,ln 則則令令 )1(arctan1arctanttddttttI)(arctan1arctan1tdttt dttttt )1(1arctan12解解dtttttttI )1()1(arc

12、tan1222dttdtttt 211arctan1Ctttt )1ln(21lnarctan12Cexeexxx 21lnarctan1dxxexx 22)2(10求求例例解解 )21()2(222xdexdxxexxx dxexxxxexxx)2(21)21(22 dxxexexxx22cexxexxx ) 1(22)()()(xQxPxRmn mmmmmnnnnnbxbxbxbxQaxaxaxaxP 11101110)()(其其中中真真分分式式多多項(xiàng)項(xiàng)式式代代數(shù)數(shù)有有理理函函數(shù)數(shù) 12111223 xxxxxx例例如如：三、有理函數(shù)的積分三、有理函數(shù)的積分一代數(shù)有理函數(shù)的積分一代數(shù)有理函

13、數(shù)的積分假假分分式式時(shí)時(shí)當(dāng)當(dāng)真真分分式式時(shí)時(shí)當(dāng)當(dāng),;,mnmn 簡(jiǎn)簡(jiǎn)分分式式的的和和真真分分式式可可分分解解為為四四類類最最 axA )1(naxA)()2( qpxxCBx 2)3(nqpxxCBx)()4(2 caxAdxaxAln)1(caxnAdxaxAnn 1)(1()()2( 四類最簡(jiǎn)分式的積分四類最簡(jiǎn)分式的積分 dxqpxxCBppxBdxqpxxCBx221212)2()3( qpxxdxCBpqpxxB2222ln21 )()(22ln2142222qpqxdxCBpqpxxB dxqpxxCBppxBdxqpxxCBxnn)()2()()4(22121212)(1)1 ( 2 nqpxxnB nqpqxdxCBp)()(224222等等函函數(shù)數(shù)下下列列積積分分不不能能表表示示為為初初 xkdxdxxkdxxdxxdxxxdxxxdxxdxxdxxdxex2222223sin1,sin1cos,sincos,sin,sin1,ln1,2 dxxfCxFdxxfxfxfdxxx)()()(,)(,)(2)1ln(111求求且且是是它它的的反反函函數(shù)數(shù)單單調(diào)調(diào)連連續(xù)續(xù)設(shè)設(shè)練習(xí)練習(xí)以下標(biāo)