(1+cos 2u)d2u =-(2u+sin 2u) +1) 0 例14.设f(x)={1 求f(x-2) -1<x<0 + cOSx 解设x=t+2,则t=x-2,dx=dt 且x=1,t=-1,x=4,t=2 ∫/(x=2)k=/(0=O+. 1+c01×八 d(t)=tan cOS11 例14.设 2 4 1 , 0 ( ) , ( 2) . 1 , 1 0 1 cos x xe x f x f x dx x x −    = −   −    + 求 解 设x = t +2, 则 t = x–2, d x = d t 且x t x t = = − = = 1, 1, 4, 2 4 0 1 (1 cos 2 ) 2 2 u d u  = +  1 1 4 (2 sin 2 ) ( 1) 2 2 2 0 u u   = + = + 4 2 1 1 f x dx f t dt ( 2) ( ) − − =   0 2 1 0 f t dt f t dt ( ) ( ) − = +   1 1 1 4 tan 2 2 2 e − = − + 0 2 2 1 0 1 1 cos t dt te dt t − − = + +   0 2 2 2 1 0 2 1 1 ( ) 2 2cos 2 t dt e d t t − − = − −  