对一个复合函数积分~~对这样一个函数积分：（R^2-y^2)^(3/2)谢啦!请各路高手帮忙做一下啦~谢谢谢谢。实在是够

∫（R^2-y^2)^(3/2) dy
=y(R^2-y^2)^(3/2)-∫y*(3/2)*[(R^2-y^2)^(1/2)]*(-2y)dy
=y(R^2-y^2)^(3/2)+3∫y^2*[(R^2-y^2)^(1/2)]dy
...(1)
令y=RsinA,A属于[-pi/2,pi/2]
则∫y^2*[(R^2-y^2)^(1/2)]dy
=∫R^2[(sinA)^2]*RcosA*d(RsinA)
=∫R^4[(sinAcosA)^2]dA
=∫R^4[(sin2A)^2/4]dA
=∫(R^4)*[(1-cos4A)/8]dA
=(A-0.25*sin4A)*R^4/8+C
=[arcsin(y/R)]-{y*[(R^2-y^2)^(3/2)]/R^4}+C...(2)
将（2）代入（1）得：∫（R^2-y^2)^(3/2) dy
=y(R^2-y^2)^(3/2)+[arcsin(y/R)]-{y*[(R^2-y^2)^(3/2)]/R^4}+C

对了