∫1/(x+根号下的(a^2-x^2))dx 0到a的积分

1个回答

  • 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