求不定积分 |1/(sin2x+4cosx)dx-|||-A 1

Question

如题所述

Answer 1

方法如下，请作参考：

若有帮助，请采纳。

Answer 2

∫dx/(sin2x+4cosx) = (1/2)∫dx/[cosx(2+sinx)]
= (1/2)∫cosxd/[(cosx)^2(2+sinx)]
= (1/2)∫dsinx/[(1-sinx)(1+sinx)(2+sinx)]
= (1/12)∫[1/(1-sinx)+3/(1+sinx)-2/(2+sinx)]dsinx
= (1/12)[-ln(1-sinx)+3ln(1+sinx)-2ln(2+sinx)] + C
= (1/12)ln{[(1+sinx)^3]/[(1-sinx)(2+sinx)^2]} + C