不定积分 x^2(cosx)^2

Question

如题所述

Answer 1

∫x²cos²x dx
= (1/2)∫x²(cos2x+1) dx
= (1/2)∫x²cos2x + (1/2)∫x² dx
= (1/2)(x³/3) + (1/4)∫x² d(sin2x)，分部积分法
= (x³/6) + (1/4)x²sin2x - (1/2)∫xsin2x dx，分部积分法
= ... + (1/4)∫x d(cos2x)，分部积分法
= ... + (1/4)xcos2x - (1/4)∫cos2x dx，分部积分法
= ... + (1/4)xcos2x - (1/8)sin2x + C
= x³/6 + (1/4)x²sin2x + (1/4)xcos2x - (1/8)sin2x + C

Answer 2

=(x*cos(2*x))/4 - sin(2*x)/8 + (x^2*sin(2*x))/4 + x^3/6+C本回答被提问者采纳