第1个回答 2018-03-19
1、先求∫e^x*cos2x dx
∫e^x*cos2x dx = (1/2)∫e^x d(sin2x)
= (1/2)(e^x)(sin2x) - (1/2)∫e^x*sin2x dx
= (1/2)(e^x)(sin2x) - (1/2)(-1/2)∫e^x d(cos2x)
= (1/2)(e^x)(sin2x) + (1/4)(e^x)(cos2x) - (1/4)∫e^x*cos2x dx，将最后那个积分移到左边得
(1+1/4)∫e^x*cos2x dx = (1/4)(e^x)(2sin2x+cos2x)
∫e^x*cos2x dx = (1/5)(e^x)(2sin2x+cos2x) + C
∫e^x*sin²x dx
= ∫e^x*(1/2)(1-cos2x) dx
= (1/2)∫e^x dx - (1/2)∫e^x*cos2x dx，代入上面的结果
= (1/2)(e^x) - (1/2)(1/5)(e^x)(2sin2x+cos2x) + C''
= (1/10)(e^x)(5-2sin2x-cos2x) + C''追问
∫e^x*cos2x dx = (1/2)∫e^x d(sin2x)
= (1/2)(e^x)(sin2x) - (1/2)∫e^x*sin2x dx
= (1/2)(e^x)(sin2x) - (1/2)(-1/2)∫e^x d(cos2x)
= (1/2)(e^x)(sin2x) + (1/4)(e^x)(cos2x) - (1/4)∫e^x*cos2x dx，将最后那个积分移到左边得
(1+1/4)∫e^x*cos2x dx = (1/4)(e^x)(2sin2x+cos2x)
∫e^x*cos2x dx = (1/5)(e^x)(2sin2x+cos2x) + C
∫e^x*sin²x dx
= ∫e^x*(1/2)(1-cos2x) dx
= (1/2)∫e^x dx - (1/2)∫e^x*cos2x dx，代入上面的结果
= (1/2)(e^x) - (1/2)(1/5)(e^x)(2sin2x+cos2x) + C''
= (1/10)(e^x)(5-2sin2x-cos2x) + C''追问
大佬你是不是看错题目了？哪来的e？
第2个回答 2018-03-19
如图所示
我自己算也是这个数，但答案是1/2tcost^2
追答可以全部拍上来看看吗？
追问应该是答案错了，我倒着验算一下是对的，谢谢
本回答被提问者采纳
相似回答