解将∫(x)作周期延拓,延拓后为偶函数,则 bn=0(n=1,2,3… 2—π2 (x) 2 Soda 2丌 丌(3 F(x)cos ndx x cos nxdx=-(xsin nx xsin ndr) 4 (x cos nx) cos ndx nn Jo 4—n (-1)(n=1,2,3,… 高等应用数学CAⅠ电子教案 上页下页迅回解 将f (x)作周期延拓，延拓后为偶函数，则 b = 0 (n =1,2,3, ) n 0 0 2 a f x x ( )d   =  a F x nx x n ( ) cos d 2 0 =   ( sin 2 sin d ) 2 cos d 2 0 0 2 0 2 x nx x nx x n x nx x   = = −      2 2 0 0 4 4 ( cos ) cos d x nx nx x n n     = −  (n =1,2,3, ) 2 4 ( 1)n n = − 2 2 3  = 2 0 2 x xd   =  3 0 2 3 x     =    