(1)先求S(sin2x)^2dx=x/2-(1/4)sina2xcos2xS(sin2x)^2dx=-(1/2)Ssin2xdcos2x=-(1/2){sin2xcos2x-2Scos2xcos2xdx}=-(1/2){sin2xcos2x-2S(1-(sinx)^2)dx}=-(1/2)sin2xcos2xS(1-(sinx)^2)dx=-(1/2)sin2xcos2xx-S((sinx)^2)dxS(sin2x)^2dx=-(1/4)sin2xcos2x(1/2)x(2)S(sinx)^4dx=-S(sinx)^3dcosx=-(sinx)^3cosx3Scos^2xsin^2xdx=-(sinx)^3cosx(3/4)S4cos^2xsin^2xdx=-(sinx)^3cosx(3/4)Ssin^22xdx=-(sinx)^3cosx(3/4)(x/2-(1/4)sina2xcos2x)下面自己整理吧。