答：
原式
=∫ (1+sinx-1)/(1+sinx) dx
=∫ 1-1/(1+sinx)dx
=x- ∫ 1/(1+cos(x-π/2))dx
=x- ∫ 1/(1+2[cos(x/2-π/4)]^2-1) dx
=x-1/2*∫ 1/[(cos(x/2-π/4)]^2 dx
=x-1/2*2tan(x/2-π/4) + C
=x-tan(x/2-π/4) + C
注：答案里tan(x/2-π/4)可以做不同的化简,即答案的形式不唯一.
答：
原式
=∫ (1+sinx-1)/(1+sinx) dx
=∫ 1-1/(1+sinx)dx
=x- ∫ 1/(1+cos(x-π/2))dx
=x- ∫ 1/(1+2[cos(x/2-π/4)]^2-1) dx
=x-1/2*∫ 1/[(cos(x/2-π/4)]^2 dx
=x-1/2*2tan(x/2-π/4) + C
=x-tan(x/2-π/4) + C
注：答案里tan(x/2-π/4)可以做不同的化简,即答案的形式不唯一.