Question

求个不定积分1/(1+sinx)，谢谢！

Answer 1

∫dx/(1+sinx)使用代换,令F=tan(x/2),x=2arctanF
dx=2/(F²+1)dF
原式=2∫[{1/(F²+1)]/[1+sin(2arctanF)]}dF
=2∫{[1/(F²+1)]/[1+2F/(1+F²)]}dF
=2∫{[1/(F²+1)]/[(1+F²)+2F/(1+F²)]}dF
=2∫[dF/(1+2F+F²)]
=2∫[dF/(1+F)²]
=-2/(1+F)+C
代回=-2/[1+tan(x/2)]+C
做到这里或许看起来和你的答案不同，我们不妨化简下你给的答案

tan（x/2-π/4）+c= (tan x/2 -1)/(1+tanx/2) +c
=(tanx/2+1-2)/(tanx/2+1) +c
=1-2/[tan(x/2)+1]+c
=-2/[tan(x/2)+1]+c+1
由于c是常数，这和我求的其实一样的，对吧~

Answer 2

详细解题过程见下图：

Answer 3

如图，为了100分，拼了~~~