∫(0→π/2)sinx/(sinx+cosx)dx
= ∫(0→π/2)cosx/(sinx+cosx)dx
=1/2·[∫(0→π/2)sinx/(sinx+cosx)dx+∫(0→π/2)cosx/(sinx+cosx)dx]
=1/2·∫(0→π/2)1·dx
=1/2·π/2
=π/4
sinx/（sinx+cosx）在0~丌/2上的定积分?用求定积分的公式:f（sinx）在0~丌／2上的定积分等于f（c
∫(0→π/2)sinx/(sinx+cosx)dx
= ∫(0→π/2)cosx/(sinx+cosx)dx
=1/2·[∫(0→π/2)sinx/(sinx+cosx)dx+∫(0→π/2)cosx/(sinx+cosx)dx]
=1/2·∫(0→π/2)1·dx
=1/2·π/2
=π/4