![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
求不定积分∫xcos2xdx 求过程答案!谢谢!!!
3个回答
展开全部
∫ xcos2x dx
= (1/2)∫ xcos2x d2x
= (1/2)∫ x dsin2x
= (1/2)xsin2x - (1/2)∫ sin2x dx
= (1/2)xsin2x - (1/2)(1/2)(-cos2x) + C
= (1/2)xsin2x + (1/4)cos2x + C
= (1/2)∫ xcos2x d2x
= (1/2)∫ x dsin2x
= (1/2)xsin2x - (1/2)∫ sin2x dx
= (1/2)xsin2x - (1/2)(1/2)(-cos2x) + C
= (1/2)xsin2x + (1/4)cos2x + C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询