Question

求证： sinθ+ cosθ=1

Answer 1

x = 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