\(\sqrt{\quad}\)(2ax-\(x^{2}\)) = \(\sqrt{\quad}\)( \(a^{2}\) - (x-a\()^{2}\) ) と考え、三角関数での置換を行う

x-a = asinθ ( -π/2≦θ≦π/2 ) と置くと、
dx = acosθdθ
\(\sqrt{\quad}\)(2ax-\(x^{2}\)) = acosθ
θ=arcsin( (x-a)/a )

∫x\(\sqrt{\quad}\)(2ax-\(x^{2}\)) dx
= ∫\(a^{3}\)(1+sinθ)(cosθ\()^{2}\) dθ
= \(a^{3}\)∫( (cosθ\()^{2}\) + sinθ(cosθ\()^{2}\) ) dθ
= \(a^{3}\)∫( (1+cos2θ)/2 + sinθ(cosθ\()^{2}\) ) dθ
= \(a^{3}\)( \(\frac{1}{2}\)・θ + \(\frac{1}{4}\)・sin2θ - \(\frac{1}{3}\)・(cosθ\()^{3}\) ) +C
= \(a^{3}\)( \(\frac{1}{2}\)・θ + \(\frac{1}{2}\)・sinθcosθ - \(\frac{1}{3}\)・(cosθ\()^{3}\) ) +C
= \(\frac{1}{2}\)・\(a^{3}\)θ + \(\frac{1}{2}\)・a( asinθ・acosθ ) - \(\frac{1}{3}\)・(acosθ\()^{3}\) + C
= \(\frac{1}{2}\)・\(a^{3}\) arcsin( (x-a)/a )
　　+ \(\frac{1}{2}\)・a(x-a)\(\sqrt{\quad}\)(2ax-\(x^{2}\))
　　- \(\frac{1}{3}\)・(2ax-\(x^{2}\))\(\sqrt{\quad}\)(2ax-\(x^{2}\)) +C
= \(\frac{1}{2}\)・\(a^{3}\) arcsin(\(\frac{x}{a}\) - 1) + \(\frac{1}{6}\)・(2\(x^{2}\)-ax-3\(a^{2}\))\(\sqrt{\quad}\)(2ax-\(x^{2}\)) + C
　　…(答え)