Is 「1 d sin ot11(0 sin ot= cost」=L 0 s+0 o dt ∵8(0-)=0 ∵sin0=0 s+0 (2):8(t) de(t) e(t)]=∴L[(t)] d e(t)=s-0=1 dt S dt 3积分性质若]=F则[ ir(E)E]- F(S) 证令u=∫f(dtdy=e"dt则du=f()dtv if(t.e lt= F(s t是变量,S是常量, 积分上限t用0-代替    ( )         ( )  ( ) s F(s) dt s e f(t) - s e f( )d e dt f( )d - s e u f(t)dt dv e dt du f(t)dt v - S F(S) 3 L f(t) F(t) L f( )d - 0 1 S 1 (t) s dt d L (t) L S 1 L (t) dt d (t) 2 (t) (0-) 0 sin0 0 s s 0 s s 1 dt 1 dsin t L cos t L s L sin t 0 -s t 0 -s t t 0 0 -s t t 0 -s t -s t t 0 2 2 2 2 2 2  =       −        =   = = = = =   = = =        =   =    =  = = +  =       − +    =           = +    =        −  − −  − − − 证 令 则 积分性质 若 则     t是变量，S是常量, 积分上限t用0-代替