题目

积分计算 ∫√（ax-bx²）dx

提问时间：2021-01-25

答案

∫ √(ax-bx²) dx
= ∫ √{-[√b*x - a/(2√b)]² + a²/(4b)} dx
Let u = √b*x - a/(2√b) and du = √b dx
==> (1/√b)∫ √[-u² + a²/(4b)] du
Let u = a*siny/(2√b) and du = a*cosy/(2√b) dy,√[-u² + a²/(4b)] = a*cosy/(2√b)
==> a²/[4b^(3*2)] * ∫ cos²y dy
= a²/[4b^(3/2)] * (1/2)∫ (1 + cos2y) dy
= a²/[4b^(3/2)] * (1/2)(y + 1/2 * sin2y) + C
= a²/[8b^(3/2)] * y + a²/[8b^(3/2)] * sinycosy + C
= a²/[8b^(3/2)] * arcsin(2√b*u / a) + a²/[8b^(3/2)] * (2√b*u / a) * √(a²-4bu²) / a
= -a/[8b^(3/2)] * [(2/a)(a-2bx)√(abx-b²x²) + a*arcsin(1-2bx/a)] + C