D:y ≤ x ≤ √(πy/2) ,0 ≤ y ≤ π/2
视为X型区域,D：2x²/π ≤ y ≤ x,0 ≤ x ≤ π/2
I = ∫[0,π/2] sinx /x dx ∫[2x²/π,x] dy
= ∫[0,π/2] (sinx /x) ( 2x²/π ﹣x ) dx
= ∫[0,π/2] [ (2/π) x sinx - sinx ] dx
= [ (2/π) (﹣x cosx + sinx) + cosx ) ] | [0,π/2]
= 2/π ﹣1