凑微分法-课件（PPT·精·选）.ppt

(A) : dx 1x2 1)4( dx )1e()3( dx 5)2( dx x)1( x x 7??????解: ?? dx x 7)1(?? dx x5)2(??? dx e x)1()3( dx 1x2 1)4(??: 设y=f(u),u =φ(x). 则y=f[ φ(x)] 故dy=f'(u)φ'(x)dx dx 1x2 12)1x2(d1x2 1)1x2 ln( d????????如 Cx? 88 1C x?5 ln 5Cxe x??定义: dx???)( )]([xdxf????du uf)(CxF?)]([?)(' )]([xxf???φ(x)=u 注:上述积分方法中关键是将被积表达式 g(x ) 写成 f[φ( x)]d φ(x) ,所以第一换元积分法也叫凑微分法。这样的积分方法称为第一换元积分法. ??dxxg)(CuF?)(回代例题与练****A) 例1:求下列不定积分. dx 1x2 1)1(??解: du 2 1dx ,u1x2???令 C1x2 ln2 1?? C1x2 ln2 1)1x2(d1x2 12 1???????或原式回代 dx )1x sin( )2(???解: du 1dx ,u1x?????令Cu cos 1udu sin 1???????原式回代 C)1x cos 1?????( C)1x cos( 1)1x(d)1x( sin 1?????????????或原式)b ax(d)b ax(fa 1dx )b ax(f)(???????注: (A) 练****1)∫e 2x+1 dx 2) ∫(2x+1) 5dx Cu ln2 1du u 12 1????原式(B) 例 dx )x31(x)1( 10 2??解: )x31(d)x31(6 1 210 2?????原式 C)x31(66 1 11 2???? dx ex)2( 3x2? Ce3 1dx e3 1 3 3x3x????解:原式)b ax(d)b ax(f)1n(a 1dx x)b ax(f)( 1n1n n1n?????????????(B) 练****2. 求下列不定积分?? dx )1x cos( x)1( 2 dx ex 1)2( x 12?注: ( B)例3:求下列不定积分 dx x x ln)1(?解: ?原式 dx x lnx 1)2( 2??原式注: ???x lnd)x (ln f dx )x (ln fx 1)(Ⅲ练****3:求下列不定积分? dx x x ln1 2) (??dx x )1x sin(ln 2) (??x ln1x dx 3 2) ( (B) (C) (C) Cx? 2 ln2 1Cx ?? ln 1?? xdx ln ln 1 2??x xd ln ln解: 例 ? dx xe sin )1( x cos解: (1) ?原式解: (2) 解: (3) ? xdx tan )2(? xdx cot )3(?原式?原式注:(Ⅳ)???????)bx cos a(d)bx cos a(fa 1dx )bx cos a(xf sin??????)bx sin a(d)bx sin a(fa 1dx )bx sin a(xf cos 或??? dxx xC xdx xB sin 2 cos )2 ).( cos sin )1 ).(练****4:求下列不定积分: ? xdx sin )4( 3 ?原式解: (4) Cx cos 3 1x cos 3???? Ce x?? cos??? xde x cos cosCx?? cos ln ??? xdx cos cos 1?? dxx x cos sinCx? sin ln ?? xdx sin sin 1?? dxx x sin cos???xdx cos ) cos 1( 2??? dxxx) cos 1( sin 2 (C) ( B)( C) 例 dx ax 1)1( 22??解(1) : ?原式)ax(dax 1a2 1)ax(dax 1a2 1?????????? Cax lnax lna2 1????? Cax ax lna2 1???? dx )1x(x 1)2(??解(2) : ?原式Cxx????1 ln ln 另解: ?原式 dxaxaxa ) 11(2 1???????? dxaxax) )( 1()1(1 1 ln????? xdx xdx xx )1 11???( Cx x???1 lnCx x??1 ln ????? 22)2 1()2 1( )2 1(x xd???