∫[1/2→1] 1/[x²√(1-x²)] dx令x=sinu,则dx=cosudu,√(1-x²)=cosu=∫[π/6→π/2] cosu/[sin²ucosu] du=∫[π/6→π/2] 1/sin²u du=∫[π/6→π/2] csc²u du=-cotu |[π/6→π/2]=√...