How do you integrate $e^{- a \cdot x} cos (b \cdot x)$ $e^(-ax)cos(bx)$ ?

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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

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

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

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

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

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

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

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

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