∫2,0√(4-x^2)dx怎么算

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∫2,0√(4-x^2)dx怎么算

∫2,0√(4-x^2)dx怎么算
∫2,0√(4-x^2)dx怎么算

∫2,0√(4-x^2)dx怎么算
令x=2sint,dx=2costdt,x=2时,t= π/2,x=0时,t=0
∫(2,0)√(4-x^2)dx
=∫(π/2,0)2cost*2costdt
=2∫(π/2,0)2cos²tdt
=2∫(π/2,0)(1-cos2t)dt
=2t-sin2t│(π/2,0)

令x = 2sinθ、dx = 2cosθ dθ
∫(0→2) √(4 - x²) dx
= ∫(0→π/2) 4cos²θ dθ
= 2∫(0→π/2) (1 + cos2θ) dθ
= 2[ θ + (1/2)sin(2θ) ]:(0→π/2)
= 2[ π/2 + 0 ]
= π