∫[0→4] x²√(4x-x²) dx
=∫[0→4] x²√[4-(x-2)²] dx
令x-2=2sinu,則√[4-(x-2)²]=2cosu,dx=2cosudu,u：-π/2→π/2
=∫[-π/2→π/2] (2sinu+2)²(2cosu)² du
=16∫[-π/2→π/2] (sinu+1)²cos²u du
=16∫[-π/2→π/2] (sin²ucos²u+2sinucos²u+cos²u) du
其中：2sinucos²u為奇函數(shù),其余部分為偶函數(shù)
=32∫[0→π/2] (sin²ucos²u+cos²u) du
=8∫[0→π/2] sin²2u du + 16∫[0→π/2] (1+cos2u) du
=4∫[0→π/2] (1-cos4u) du + 16∫[0→π/2] (1+cos2u) du
=4u - sin4u + 16u + 4sin2u |[0→π/2]
=10π
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