Question

1/(1+sin x)求不定积分

Answer 1

∫ dx/(1+sinx) = ∫ (1-sinx)/cos²x dx
= ∫ (sec²x - secx tanx) dx
= tanx - secx + C

追问

∫ dx/(1+sinx) = ∫ (1-sinx)/cos²x dx
这是怎么化简的？

追答

分子分母同时乘以 1 - sinx

Answer 2

∫dx/(1+sinx)
=∫dx/(1+cos(x-π/2))
=∫d(x-π/2)/[2cos(x/2-π/4)^2]
=tan(x/2-π/4)+C