令x=sinu,则√(1-x²)=cosu,dx=cosudu∫ [√(1-x²)]³ dx=∫ (cosu)^4 du=(1/4)∫ (1+cos2u)² du=(1/4)∫ (1+2cos2u+cos²2u) du=(1/4)∫ [1+2cos2u+(1/2)(1+cos4u)] du=(1/4)[(3/2)u + sin2u...