求不定积分∫[x√(4-x²)]dx

问题描述:

求不定积分∫[x√(4-x²)]dx

原式=1/2∫根号下(4-x^2)dx^2=1/2∫根号下(4-t)dt=-1/2*2/3*(4-t)^(3/2)+C=-1/3*(4-x^2)^(3/2)+C