Question

求定积分∫ (0~a) 1/[x+√(a^2-x^2)] dx

Answer 1

令x = asinθ，dx = acosθdθ
原式= ∫(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) (sinθ - cosθ)/(sinθ + cosθ) dθ
= (1/2)(π/2) - (1/2)∫(0→π/2) - d(cosθ + sinθ)/(sinθ + cosθ) dθ
= π/4 + (1/2)ln(sinθ + cosθ) |(0→π/2)
= π/4 + (1/2)[ln(1 + 0) - ln(0 + 1)]
= π/4