第二節(jié)換元積分法-在線瀏覽
2024-08-30 21:13本頁面
【正文】 ) [ ( ) ] ( )xtf x d x f t t d t? ??? ?????? 令1 ()1( ) ( ( ) )txF t C F x C? ????? ?? ? ?令定理21( ) , , ( ) 0 , [ ( ) ] ( )( ) ,( ) [ ( ) ] ( ) ( ( ) )x t t f t tFtf x dx f t t dt F x C? ? ? ?? ? ? ??????? ? ???設(shè) 單調(diào) 可導(dǎo) 并且 又具有原函數(shù) 則第二換元公式 1( ) ( ) ,()dtF t f td x t?? ?? ?證: ? ?1[ ( ) ]F x C? ? ??? 1[ ( ) ] dtFx dx? ????() dtFt dx???1[ ( ) ] ( )()f t t t?? ???? ?[ ( )]ft?? ()fx?證畢n a x b t??根式代換 令法3131x dxx???例1 33 213 1 , ,3tx t x d x t d t?? ? ? ?解: 令41 ( 2 )3 t t dt???原式 521115 3 t C? ? ?523311( 3 1 ) ( 3 1 )1 5 3x x C? ? ? ? ?( 1 ) 1dxxx???例2 21 , 1 , 2x t x t d x td t? ? ? ? ? ?解: 令22( 2 )t dttt????原式 222 dtt? ??22ln2 2 2t Ct????2 1 2ln2 12x Cx??????注: 通常根式內(nèi)為一次多項式時, 可考慮用根式代換三角代換 s i n , t a n , s e cx a t x a t x a t? ? ?令22 ( 0)a x dx a???例3 sin , ( ) , c os22x a t t dx a tdt??? ? ? ? ?解:令a x22ax?t22c osa tdt? ?原式2 (1 c o s 2 )2a t d t???2 1( s in 2 )22a t t C? ? ?22 s i n c o s22aat t t C? ? ? ?2 221a r c s i n22at x a x Ca? ? ? ?22 ( 0 )dx axa ???例4 2ta n ( ) , se c22x a t t dx a tdt??? ? ? ? ?解:令 ax22xa?tse c tdt? ?原式 1ln | s e c t a n |t t C? ? ?221lnx a x Caa?? ? ?22 1l n ( ) ( l n )x x a C C C a? ? ? ? ? ?22 ( 0 )dx axa ???例5 , se c ( 0 ) ,2x a x a t
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