∫cos²2xdx，高等数学，不定积分

Question

如题所述

Answer 1

倍角加分步
cos^2x=(cos2x+1)/2
原因为化为
∫1/2*x^2dx+1/4∫x^2dsin2x
=1/6x^3+1/4sin2x*x^2-1/2∫xsin2xdx
=1/6x^3+1/4sin2x*x^2+1/4xcos2x-1/4∫cos2xdx
=1/6x^3+1/4sin2x*x^2+1/4xcos2x+1/8sin2x
思路是这样,错没错不晓得