What is the integral of cos(t)e^{-st} ?

The integral of cos(t)e^{-st} is (e^{-st}sin(t))/(1+s^2)-(se^{-st}cos(t))/(1+s^2)+C

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integral of cos(t)e^{-st}

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

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

Solution steps

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

Apply Integration By Parts

= e^{- s t} sin (t) - \int - s e^{- s t} sin (t) d t

Take the constant out:

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

= e^{- s t} sin (t) - (- s \cdot \int e^{- s t} sin (t) d t)

Apply Integration By Parts

= e^{- s t} sin (t) - (- s (- e^{- s t} cos (t) - \int s e^{- s t} cos (t) d t))

Take the constant out:

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

= e^{- s t} sin (t) - (- s (- e^{- s t} cos (t) - s \cdot \int e^{- s t} cos (t) d t))

Therefore

\int e^{- s t} cos (t) d t = e^{- s t} sin (t) - (- s (- e^{- s t} cos (t) - s \cdot \int e^{- s t} cos (t) d t))

Isolate

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

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

Add a constant to the solution

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

What is the integral of cos(t)e^{-st} ?
The integral of cos(t)e^{-st} is (e^{-st}sin(t))/(1+s^2)-(se^{-st}cos(t))/(1+s^2)+C