Question

求1/(sinx+2)的不定积分

Answer 1

u = tanx/2，dx = 2/(1 + u²)，sinx = 2u/(1 + u²)
∫ 1/(sinx + 2) dx
= ∫ [2/(1 + u²)]/[2u/(1 + u²) + 2] du
= ∫ 1/(u² + u + 1) du
= ∫ 1/[(u + 1/2)² + 3/4]
= (2/√3)arctan[(u + 1/2)(2/√3)] + C
= (2/√3)arctan[(2tanx/2 + 1)/√3] + C

Answer 2

或者这样
∫dx/(sinx+2)
=∫dx/(2+cos(π/2-x))
=∫dx/(1+2cos(π/4-x/2)^2)
= -∫d(π/4-x/2)/[cos(π/4-x/2)^2 *(2+sec(π/4-x/2)^2)]
=-∫dtan(π/4-x/2)/(3+tan(π/4-x/2)^2)
=(-1/√3)∫d[tan(π/4-x/2)/√3]/[1+tan(π/4-x/2)^2/3]
=(-1/√3)arctan [ tan(π/4-x/2) /√3] +C