Question

eˣsin²x的积分？

Answer 1

简单计算一下，答案如图所示

Answer 2

∫ e^x. (sinx)^2 dx
=(1/2)∫ e^x. [1-cos(2x)] dx
=(1/2)e^x -(1/2)∫ e^x.cos(2x) dx
=(1/2)e^x -(1/2)[(2/5)e^x.sin(2x) + (1/5)e^x.cos(2x) ] +C
=(1/2)e^x -(1/5)e^x.sin(2x) - (1/10)e^x.cos(2x) +C
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∫ e^x.cos(2x) dx
=(1/2)∫ e^x.dsin(2x)
=(1/2)e^x.sin(2x) - (1/2)∫ e^x.sin(2x) dx
=(1/2)e^x.sin(2x) + (1/4)∫ e^x dcos(2x)
=(1/2)e^x.sin(2x) + (1/4)e^x.cos(2x) -(1/4)∫e^x.cos(2x) dx
(5/4) ∫ e^x.cos(2x) dx=(1/2)e^x.sin(2x) + (1/4)e^x.cos(2x)
∫ e^x.cos(2x) dx=(2/5)e^x.sin(2x) + (1/5)e^x.cos(2x) +C