Question

大学定积分的计算题，求高人指点，最好写出详细的计算过程

Answer 1

先求出不定积分，需用万能换元法
令z = tan(x/2)，dx = 2dz/(1 + z²)，sinx = 2z/(1 + z²)
∫ 1/(2 + sinx) dx
= ∫ [2/(1 + z²)]/[2 + (2z)/(1 + z²)] dz
= ∫ 1/[(1 + z²) + z] dz
= ∫ 1/[(z + 1/2)² + 3/4] dz
= (2/√3)arctan[(z + 1/2) · 2/√3] + C
= (2/√3)arctan[(2tan(x/2) + 1)/√3] + C

分区间，注意若ƒ(x) = 1/(2 + sinx)，ƒ(0) = ƒ(π) = ƒ(2π)
F(x) = (2/√3)arctan[(2tan(x/2) + 1)/√3]

∫(0→2π) 1/(2 + sinx) dx
= ∫(0→π) 1/(2 + sinx) dx + ∫(π→2π) 1/(2 + sinx) dx
= [F(π) - F(0)] - [F(2π) - F(π)]，x = π是间断点，分左右极限做
= [lim(x→0) F(x) - lim(x→π⁺) F(x)] - [lim(x→2π) F(x) - lim(x→π⁻) F(x)]
= [π/(3√3) - (- π/√3)] - [π/(3√3) - π/√3]
= 2π/√3