Question

高数，用分部积分法求不定积分，红色部分接下来该怎么写啊

Answer 1

∫ (e^x)sin²x dx
= (1/2)∫ (e^x)(1 - cos2x) dx
= (1/2)∫ e^x dx - (1/2)∫ (e^x)cos2x dx
= (1/2)e^x - (1/2) • I
I = ∫ (e^x)cos2x = (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
(1 + 1/4) • I = (1/2)(e^x)sin2x + (1/4)(e^x)cos2x
I = (2/5)(e^x)sin2x + (1/5)(e^x)cos2x = (1/5)(e^x)(2sin2x +cos2x)
∴原式= (1/2)e^x - (1/2)(1/5)(e^x)(2sin2x +cos2x) + C
= (1/10)(5 - 2sin2x - cos2x)(e^x) + C

Answer 2

A=∫cos2xde^x
=cos2xe^x+2∫sin2xde^x
=(cos2x+2sin2x)e^x-4∫cos2xde^x
所以∫cos2xde^x=1/4（cos2x+2sin2x)e^x+C

追问

第三个等式看不懂

追答

A=∫cos2xde^x
=cos2xe^x+2∫sin2xde^x
=(cos2x+2sin2x)e^x-4A
所以A=∫cos2xde^x=1/4（cos2x+2sin2x)e^x+C

追问

哦，我会了，刚才绕来绕去把我绕糊涂了