求,不定积分.∫上限兀/2 下限0（cosx/2-sinx/2）dx+∫上限 兀下限 兀/2（sinx/2-cosx/2

∫(0,兀/2)（cosx/2-sinx/2）dx+∫(兀/2.兀)（sinx/2-cosx/2）dx
=2[∫(0,兀/2)（cosx/2-sinx/2）dx/2+∫(兀/2.兀)（sinx/2-cosx/2）dx/2 ]
=2(sinx/2+cosx/2)|(0,兀/2)-2(sinx/2+cosx/2)|(兀/2.兀)
=2(√2-1)-2(1-√2)
=4(√2-1)