令x=2a(sin x)^2,则dx=4a(sin t)(cos t)dt,原式：=8a^2∫(sin t)^4 dt=8a^2∫[(1-cos 2t)/2]^2 dt=a^2∫(3-4cos 2t+cos 4t)dt=3a^2 t-2a^2(sin 2t)+a^2(sin 4t)/4+C=3a^2(arcsin√(x/(2a)))-2a√x(2a-x)+((a-x)√x(2a-x))/2+C