例4求 3 5 sInx-sin xax 解:原式- cos x (sin x)2x 兀 cos x(sin x) dx- cos x(sin x)2dx 2 J3 (sin x)'id sin x-J (sin x)id sinx (sin x)2 sInx 5 2x xdx  −  0 3 5 例4 求 sin sin 解: 原式= x x dx   0 2 3 | cos |(sin ) x x dx x x dx   = −    2 2 3 2 0 2 3 cos (sin ) cos (sin ) (sin x) d sin x (sin x) d sin x 2 2 3 2 0 2 3   = −       2 2 2 5 0 2 5 (sin ) 5 2 (sin ) 5 2 = x − x 5 4 =