How do you integrate #e^(-ax)cos(bx)#?

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Answer 1 · 2016-07-12T10:13:45

#=- e^(-ax)/(a^2+ b^2) qquad (a cos bx - b sin bx) #

This is not IBP but gives an alternative more compact way of doing it

#int dx qquad e^{-ax} cos bx#

#= mathcal(Re) int dx qquad e^{-ax} e^{ i bx}#

#= mathcal(Re) int dx qquad e^{(-a+ i b)x} #

#= mathcal(Re) qquad 1/(-a+ i b)e^{(-a+ i b)x} #

#= mathcal(Re) qquad -(a+ i b)/(a^2+ b^2)e^(-ax) e^{ i bx} #

#=- e^(-ax)/(a^2+ b^2) quad mathcal(Re) qquad (a+ i b)(cos bx + i sin bx) #

#=- e^(-ax)/(a^2+ b^2) qquad (a cos bx - b sin bx) #

How do you integrate #e^(-a*x)cos(b*x)#?