（xsinx)²在0到π上的定积分，求答案，急急急！！！！！！1高分！！！！！！！！！！

原式=0.5∫x^2 (1-cos2x)dx
=0.5∫x^2dx-0.5∫x^2cos2xdx
=0.5x^3/3-0.5[ x^2 *0.5sin2x-∫xsin2xdx]
=0.5x^3/3-0.25x^2sin2x+0.5[-0.5xcos2x+∫0.5cos2xdx]
=[0.5x^3/3-0.25x^2sin2x-0.25xcos2x+0.125sin2x]
=[0.5π^3/3-0.25π]-[0]
=π^3/6-π/4本回答被提问者采纳