Question

cosθ+isinθ=1

Answer 1

问：cosθ+isinθ=1

答：亲，你好 cosθ+isinθ=1x = asinθ、dx = acosθ dθ∫[0→a] dx/[x + √(a² - x²)]= ∫[0→π/2] acosθ/[asinθ + acosθ] dθ= (1/2)∫[0→π/2] 2cosθ/[sinθ + cosθ] dθ= (1/2)∫[0→π/2] [(sinθ + cosθ) - (sinθ - cosθ)]/(sinθ + cosθ) dθ= (1/2)∫[0→π/2] dθ - (1/2)∫[0→π/2] d(- cosθ - sinθ)/(sinθ + cosθ)= θ/2 |[0→π/2] + (1/2)∫ d(sinθ + cosθ)/(sinθ + cosθ)= π/4 + (1/2)ln[sinθ + cosθ] |[0→π/2]= π/4 + (1/2){ln(1 + 0) - ln(0 + 1)}= π/4 |cosθ+isinθ| =√(cos²θ+sin²θ) =√1 =1 所以原式=1²=1希望我的回答能给您帮助。