What is the integral of e^{-st}cosh(at) ?

The integral of e^{-st}cosh(at) is 1/2 (1/(-s-a)e^{-st-at}+1/(-s+a)e^{-st+at})+C

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integral of e^{-st}cosh(at)

\int e^{- s t} cosh (a t) d t

\frac{1}{2} (\frac{1}{- s - a} e^{- s t - a t} + \frac{1}{- s + a} e^{- s t + a t}) + C

Solution steps

\int e^{- s t} cosh (a t) d t

Rewrite using trig identities

= \int \frac{1}{2} e^{- s t} (e^{- a t} + e^{a t}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{- s t} (e^{- a t} + e^{a t}) d t

Expand

e^{- s t} (e^{- a t} + e^{a t}) : e^{- s t - a t} + e^{- s t + a t}

= \frac{1}{2} \cdot \int e^{- s t - a t} + e^{- s t + a t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int e^{- s t - a t} d t + \int e^{- s t + a t} d t)

\int e^{- s t - a t} d t = \frac{1}{- s - a} e^{- s t - a t}

\int e^{- s t + a t} d t = \frac{1}{- s + a} e^{- s t + a t}

= \frac{1}{2} (\frac{1}{- s - a} e^{- s t - a t} + \frac{1}{- s + a} e^{- s t + a t})

Add a constant to the solution

= \frac{1}{2} (\frac{1}{- s - a} e^{- s t - a t} + \frac{1}{- s + a} e^{- s t + a t}) + C

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What is the integral of e^{-st}cosh(at) ?
The integral of e^{-st}cosh(at) is 1/2 (1/(-s-a)e^{-st-at}+1/(-s+a)e^{-st+at})+C