f" f(x)dx =- f' f(-t)dt =f" f(-t)dt,①f(x)为偶函数，则 f(-t)=f(t)" f(x)dx = J" f(x)dx + J" f(x)dx= 2f" f(t)dt;②f(x)为奇函数，则f(-t)=-f(t),", f(x)dx = J" f(x)dx + f" f(x)dx= 0
− = 0 ( ) a f x dx − − = 0 ( ) a f t dt ( ) , 0 − a f t dt ① f (x)为偶函数，则 f (−t) = f (t), − − = + 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
2x2+xcosxCI例6计算dx.1+ /1-x222xxcosxT原式=ldxdx +解1+/1-x1+V1-x奇函数偶函数2xx(1-V1-x4dx=dx2/1-x1-(1-x)1+~= 4f (1- ~1-x2)dx =4-4/ /1-x"dx单位圆的面积=4-元.合
奇函数 例6 计算 解 . 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 偶函数 + − = 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 = 4 − . 单位圆的面积
例7未若f(x)在[0,1]上连续，证明(1) f, f(sin x)dx =f, f(cos x)dx;" f(sin x)dx.(2) (" xf(sin x)dx=2J0xsinxC元由此计算dx.2101+cosx元证（(1)= dx =-dt,x=T2设元元=t=0,x=x=0=t=2'2台合
例 7 若 f ( x)在[0,1]上连续，证明 （1） = 2 2 0 0 f (sin x)dx f (cos 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 （ 1 ） 设 x − t = 2 dx = −dt, x = 0 , 2 t = 2 x = t = 0