Question

e的2x次方与sin²x乘积的不定积分求解，详细过程

分部积分求不下去了，后面一直循环，求解🙆题目写错了，是e的x次方与sin²x的乘积

Answer 1

希望有所帮助

追答

拍个清楚的

追问

谢谢，已经弄懂啦😜

追答

望采纳

Answer 2

∫e^x*sin^2xdx
=(1/2)*∫e^x*(1-cos2x)dx
=(1/2)*∫e^xdx-(1/2)*∫e^x*cos2xdx
令∫e^x*cos2xdx=A
则(1/2)*∫e^xdx-(1/2)*∫e^x*cos2xdx
=(1/2)*e^x-(1/2)*A
=(1/2)*e^x-(1/2)*∫cos2xd(e^x)
=(1/2)*e^x-(1/2)*cos2x*e^x+(1/2)*∫e^xd(cos2x)
=e^x*sin^2x-∫e^x*sin2xdx
=e^x*sin^2x-∫sin2xd(e^x)
=e^x*sin^2x-e^x*sin2x+∫e^xd(sin2x)
=e^x*(sin^2x-sin2x)+2∫e^x*cos2xdx
=e^x*(sin^2x-sin2x)+2A
即(1/2)*e^x-(1/2)*A=e^x*(sin^2x-sin2x)+2A
(5/2)*A=(1/2-sin^2x+sin2x)*e^x
A=e^x*(1-2sin^2x+2sin2x)/5+C，其中C是任意常数
所以∫e^x*sin^2xdx=(1/2)*e^x-(1/2)*A+C
=(1/2)*e^x*[1-(1-2sin^2x+2sin2x)/5]+C
=e^x*(2+sin^2x-sin2x)/5+C

Answer 3

Answer 4

1/2∫e^xcos2xdx=1/2e^xcos2x+∫e^xsin2xdx
=1/2e^xcos2x+e^xsin2x-2∫e^xcos2xdx
把-2∫e^xcos2xdx移到左边
5/2∫e^xcos2xdx=1/2e^xcos2x+e^xsin2x
∫e^xcos2xdx=1/5e^x(cos2x+2sin2x)

Answer 5

I = ∫e^x(sinx)^2 dx = (1/2)∫e^x(1-cos2x)dx
= (1/2)∫e^xdx - (1/2)∫e^xcos2xdx
= (1/2)e^x - (1/2)∫e^xcos2xdx
I1= ∫e^xcos2xdx = ∫cos2xde^x = e^xcos2x + 2∫e^xsin2xdx
= e^xcos2x + 2∫sin2xde^x = e^xcos2x + 2e^xsin2x - 4I1
解得 I1 = (1/5)e^x(cos2x+2sin2x)
则 I = (1/2)e^x - (1/10)e^x(cos2x+2sin2x) + C