✔ 最佳答案
考慮 [e^-xcosx]'=e^-x[-cosx-sinx]
[e^-xsinx]'=e^-x[-sinx+cosx]
∫e^-xcosx dx=-e^-xcosx-∫e^-xsinx dx
=-e^-xcosx-[-e^-xsinx+∫e^-xcosx dx]
=-e^-xcosx+e^-xsinx-∫e^-xcosx dx
-->
∫e^-xcosx dx=(1/2)e^-x(sinx-cosx)
y'=dy/dx=4e^-xcosx
y=2e^-x(sinx-cosx)+C