工程数学题（复变函数）求积分∫0到2π1/（4+cosθ）dθ

Question

工程数学题（复变函数）
求积分∫0到2π1/（4+cosθ）dθ

Answer 1

令θ=2x,则dθ = 2dx,积分区域变为0到π,cos2x = [1-(tanx)^2]/[1+(tanx)^2]
积分=2∫ 1/(4+cos2x) dx = 2∫ d(tanx) / {4 + 4(tanx)^2 + 1 - (tanx)^2}
=2∫d(tanx) / [5+3(tanx)^2] = (2/5)·(√5/3)∫d[√(3/5)·(tanx)] / {1 + [√(3/5)·tanx]^2}
=(2/√15)arctan[√(3/5)·tanx]
积分区域分为 0到π/2 ,和π/2到π
积分值=2部分的和 = (2/√15)·(π/2) ·2 = 2π/(√15)