不定积分计算 ∫sinx cosx /sinx+cosx dx

Question

如题所述

Answer 1

∫sin2xdx/(sinx+cosx)
=∫cos(π/2-2x)dx/[√2cos(π/4-x)]
=√2∫cos(π/4-x)dx -(1/√2)∫dx/cos(π/4-x)
=√2sin(x-π/4)-(1/√2)ln|sec(x-π/4)+tan(x-π/4)|+C

Answer 2

简单分析一下，答案如图所示