求解∫√(x-4x∧2)dx
问题描述：
求解∫√(x-4x∧2)dx
答
答：
∫ √(x - 4x²) dx
= ∫ √[1/16 - (2x - 1/4)²] dx
令2x - 1/4 = (1/4)siny、siny = 8x - 1、8 dx = cosy dy
= ∫ √(1/16 - 1/16·sin²y)·(1/8)cosy dy
= ∫ (1/4)(1/8)cos²y dy
= (1/64)∫ (1 + cos2y) dy
= (1/64)(y + 1/2·sin2y) + C
= (1/64)y + (1/64)sinycosy + C
= (1/64)arcsin(8x - 1) + (1/64)(8x - 1)√[1 - (8x - 1)²] + C
= (1/64)arcsin(8x - 1) + (1/16)(8x - 1)√(x - 4x²) + C