∫(0→2)(4-x^2)^1.5dx答案是3∏,怎么会有∏

Question

∫(0→2)(4-x^2)^1.5dx
答案是3∏,怎么会有∏

Answer 1

令x = 2sinθ,dx = 2cosθ dθ
当x = 0,θ = 0,当x = 2,θ = π/2
∫(0→2) (4 - x²)^(1.5) dx
= ∫(0→π/2) (4 - 4sin²θ)^(1.5)(2cosθ) dθ
= ∫(0→π/2) (4cos²θ)^(1.5)(2cosθ) dθ
= 16∫(0→π/2) cos⁴θ dθ
= 16∫(0→π/2) (cos²θ)² dθ
= 16∫(0→π/2) [(1 + cos2θ)/2]² dθ
= 4∫(0→π/2) (1 + 2cos2θ + cos²2θ) dθ
= 4∫(0→π/2) (1 + 2cos2θ) + 4∫(0→π/2) (1 + cos4θ)/2 dθ
= 4[θ + sin2θ]：[0→π/2] + 2[θ + (1/4)sin4θ]：[0→π/2]
= 4(π/2) + 2(π/2)
= 2π + π
= 3π