Question

求不定积分 求大神帮忙 急!!!

Answer 1

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''