What is the integral of e^{-st}cos(5t) ?

The integral of e^{-st}cos(5t) is (5e^{-st}sin(5t))/(25+s^2)-(se^{-st}cos(5t))/(25+s^2)+C

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integral of e^{-st}cos(5t)

\int e^{- s t} cos (5 t) d t

\frac{5 e ^{- s t} sin ( 5 t )}{25 + s ^{2}} - \frac{s e ^{- s t} cos ( 5 t )}{25 + s ^{2}} + C

Solution steps

\int e^{- s t} cos (5 t) d t

Apply Integration By Parts

: \frac{1}{5} e^{- s t} sin (5 t) - \int - \frac{1}{5} s e^{- s t} sin (5 t) d t

= \frac{1}{5} e^{- s t} sin (5 t) - \int - \frac{1}{5} s e^{- s t} sin (5 t) d t

Take the constant out:

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

= \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{1}{5} s \cdot \int e^{- s t} sin (5 t) d t)

Simplify

- \frac{1}{5} s : - \frac{s}{5}

= \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{s}{5} \cdot \int e^{- s t} sin (5 t) d t)

Apply Integration By Parts

: - \frac{1}{5} e^{- s t} cos (5 t) - \int \frac{1}{5} s e^{- s t} cos (5 t) d t

= \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{s}{5} (- \frac{1}{5} e^{- s t} cos (5 t) - \int \frac{1}{5} s e^{- s t} cos (5 t) d t))

Take the constant out:

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

= \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{s}{5} (- \frac{1}{5} e^{- s t} cos (5 t) - \frac{1}{5} s \cdot \int e^{- s t} cos (5 t) d t))

Simplify

\frac{1}{5} s : \frac{s}{5}

= \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{s}{5} (- \frac{1}{5} e^{- s t} cos (5 t) - \frac{s}{5} \cdot \int e^{- s t} cos (5 t) d t))

Therefore

\int e^{- s t} cos (5 t) d t = \frac{1}{5} e^{- s t} sin (5 t) - (- \frac{s}{5} (- \frac{1}{5} e^{- s t} cos (5 t) - \frac{s}{5} \cdot \int e^{- s t} cos (5 t) d t))

Isolate

\int e^{- s t} cos (5 t) d t

= \frac{5 e ^{- s t} sin ( 5 t )}{25 + s ^{2}} - \frac{s e ^{- s t} cos ( 5 t )}{25 + s ^{2}}

Add a constant to the solution

= \frac{5 e ^{- s t} sin ( 5 t )}{25 + s ^{2}} - \frac{s e ^{- s t} cos ( 5 t )}{25 + s ^{2}} + C

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What is the integral of e^{-st}cos(5t) ?
The integral of e^{-st}cos(5t) is (5e^{-st}sin(5t))/(25+s^2)-(se^{-st}cos(5t))/(25+s^2)+C