∫(x)=/()+(xk=2/( ②f(x)为奇函数,则f(-1)=-f(),f(x)d=f(x)atx+f(x)d=0 例6计算 1+ 2 原式= dx x coSx dx(被积函数奇偶性) 41==4xy=如=x =4-4√1-x2dx (减去单位圆面积) 例7若f(x)在[O,1]上连续,证明 (ISr(sin x)dr='/(cosx)dx (2)5 xf(sin x)dx=f/(sin x) 由此计 xsIn x 证明:(1)设x t→ax=-dt.x=0→t →t=0, 2 f∫(sinx)dhx f(cosn)dt =f(cos x)dxr; 2 (2)设x=x-1→ax=-dt,x=0→t=丌.X=丌 xf(sin x)dr =-(-DfTsin(I-D]dt =I(r-D)f(sin D)dt, xf(sn x)a f∫(snt)dt tf(sin t )dt =t. f(sin x)dx -oxf(sin r(sin x)dr= sin x 丌 d(cos x) 1+cos x 1+cos x ctan( cos x4 − −  = + a a a a f x dx f x dx f x dx 0 0 ( ) ( ) ( ) 2 ( ) ; 0 = a f t dt ② f (x) 为奇函数，则 f (−t) = − f (t), − −  = + a a a a f x dx f x dx f x dx 0 0 ( ) ( ) ( ) = 0. 例 6 计算 . 1 1 1 2 cos 1 2 2 − + − + dx x x x x 原式 − + − = 1 1 2 2 1 1 2 dx x x − + − + 1 1 2 1 1 cos dx x x x （被积函数奇偶性）  + − = 1 0 2 2 1 1 4 dx x x  − − − − = 1 0 2 2 2 1 (1 ) (1 1 ) 4 dx x x x  = − − 1 0 2 4 (1 1 x )dx  = − − 1 0 2 4 4 1 x dx （减去单位圆面积） = 4−. 例 7 若 f (x) 在 [0,1] 上连续，证明 （1）   = 2 2 0 0 (sin ) (cos )   f x dx f x dx ; （2）   =    0 0 (sin ) 2 xf (sin x)dx f x dx . 由此计算  +  0 2 1 cos sin dx x x x . 证明: （1）设 x = −t 2   dx = −dt, x = 0 , 2   t = 2  x =  t = 0,  2 0 (sin )  f x dx              = − − 0 2 2 sin   f t dt  = 2 0 (cos )  f t dt (cos ) ; 2 0 =  f x dx （2）设 x =  −t  dx = −dt, x = 0 t =, x =  t = 0,   0 xf(sin x)dx  = − − − 0 ( ) [sin( )]   t f  t dt ( ) (sin ) , 0 = −   t f t dt   0 xf(sin x)dx  =   0 f (sin t)dt  −  0 tf (sin t)dt  =   0 f (sin x)dx (sin ) , 0 −  xf x dx (sin ) . 2 (sin ) 0 0  =    xf x dx f x dx  +  0 2 1 cos sin dx x x x  + =   0 2 1 cos sin 2 dx x x  + = −   0 2 (cos ) 1 cos 1 2 d x x     0 arctan(cos ) 2 = − x ) 4 4 ( 2    = − − − . 4 2  =