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

Jul 12, 2016

$= - {e}^{- a x} / \left({a}^{2} + {b}^{2}\right) q \quad \left(a \cos b x - b \sin b x\right)$

Explanation:

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

$\int \mathrm{dx} q \quad {e}^{- a x} \cos b x$

$= m a t h c a l \left(R e\right) \int \mathrm{dx} q \quad {e}^{- a x} {e}^{i b x}$

$= m a t h c a l \left(R e\right) \int \mathrm{dx} q \quad {e}^{\left(- a + i b\right) x}$

$= m a t h c a l \left(R e\right) q \quad \frac{1}{- a + i b} {e}^{\left(- a + i b\right) x}$

$= m a t h c a l \left(R e\right) q \quad - \frac{a + i b}{{a}^{2} + {b}^{2}} {e}^{- a x} {e}^{i b x}$

$= - {e}^{- a x} / \left({a}^{2} + {b}^{2}\right) \quad m a t h c a l \left(R e\right) q \quad \left(a + i b\right) \left(\cos b x + i \sin b x\right)$

$= - {e}^{- a x} / \left({a}^{2} + {b}^{2}\right) q \quad \left(a \cos b x - b \sin b x\right)$