Question

用分部积分法求积分

Answer 1

原式=∫(0,1)√(1-x²)arcsinxdx²
=-∫(0,1)(1-x²)^(1/2)arcsinxd(1-x²)
=-2/3*∫(0,1)arcsinxd(1-x²)^(3/2)
=-2/3*arcsinx*(1-x²)^(3/2)(0,1)+2/3*∫(0,1)(1-x²)^(3/2)darcsinx
=-2/3*arcsinx*(1-x²)^(3/2)(0,1)+2/3*∫(0,1)(1-x²)*√(1-x²)*1/√(1-x²)dx
=-2/3*arcsinx*(1-x²)^(3/2)(0,1)+2/3*∫(0,1)(1-x²)dx
=[-2/3*arcsinx*(1-x²)^(3/2)+2/3*(x-x³/3)](0,1)
=[-2/3*arcsin1*0^(3/2)+2/3*(1-1³/3)]-[-2/3*arcsin0*(1-0²)^(3/2)+2/3*(0-0³/3)]
=4/9

Answer 2

Answer 3

令arcsinx=u，则x=sinu
x：0→1，则u：0→π/2
∫[0：1]2x√(1-x²)arcsinxdx
=∫[0：π/2]2sinu·√(1-sin²u)·ud(sinu)
=∫[0：π/2]2sinu·cosu·u·cosudu
=2∫[0：π/2]u·sinu·cos²udu
=-⅔∫[0：π/2]ud(cos³u)
=-⅔ucos³u|[0：π/2] +⅔∫[0：π/2]cos³udu
=-⅔[(π/2)cos³(π/2)-0·cos³0]+⅔∫[0：π/2](1-sin²u)cosudu
=⅔∫[0：π/2](1-sin²u)d(sinu)
=⅔(sinu-⅓sin³u)|[0：π/2]
=⅔[sin(π/2)-⅓sin³(π/2)]-⅔(sin0-⅓sin³0)
=⅔(1-⅓)-0
=4/9