Question

问：数学 secx不定积分 怎么做啊 从这步怎么推到下一步？求大神解答

Answer 1

∫d(sinx)/(1-sin²x)
=∫d(sinx)/[(1-sinx)(1+sinx)]
=∫d(sinx) * [1/(1-sinx) + 1/(1+sinx)] * (1/2)
=(1/2)*[∫d(sinx)/(1-sinx) + ∫d(sinx)/(1+sinx)]
=1/2*[-∫d(1-sinx)/(1-sinx) + ∫d(1+sinx)/(1+sinx)]
=1/2*[-ln|1-sinx| + ln|1+sinx|] + C
=1/2*ln|(1+sinx)/(1-sinx)| + C
=1/2*ln|(1+sinx)²/(1-sin²x)| + C 注：分子、分母同乘以 (1+sinx)
=1/2*ln|(1+sinx)²/cos²x| + C
=1/2*ln|(1+sinx)/cosx|² + C
=1/2 * 2 * ln|(1+sinx)/cosx| + C
=ln|(1/cosx + sinx/cosx)| + C
=ln|secx + tanx| + C

Answer 2