※ 引述《icola (Br)》之銘言：
: ∫1/(1+sinx)dx=?
: 請版上大大告訴我詳解好嗎?
: 謝謝
: ans : tanx-secx+C
1 + sin(x) = 2 sin(π/4 + x/2 ) cos(π/4 - x/2 )
= 2 sin ( (π/2) - (π/4 - x/2) ) cos(π/4 - x/2 )
= 2 ( cos(π/4 - x/2 ) )^2
=> ∫1/(1+sinx)dx = - tan(π/4 - x/2) + C
- cos(x)
= ---------------- + C
1 + sin(x)
= tan(x) - sec(x) + C
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