What is the integral of e^{-st}sin(4t) ?

The integral of e^{-st}sin(4t) is -(4e^{-st}cos(4t))/(16+s^2)-(se^{-st}sin(4t))/(16+s^2)+C

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integral of e^{-st}sin(4t)

\int e^{- s t} sin (4 t) d t

- \frac{4 e ^{- s t} cos ( 4 t )}{16 + s ^{2}} - \frac{s e ^{- s t} sin ( 4 t )}{16 + s ^{2}} + C

Solution steps

\int e^{- s t} sin (4 t) d t

Apply Integration By Parts

: - \frac{1}{4} e^{- s t} cos (4 t) - \int \frac{1}{4} s e^{- s t} cos (4 t) d t

= - \frac{1}{4} e^{- s t} cos (4 t) - \int \frac{1}{4} s e^{- s t} cos (4 t) d t

Take the constant out:

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

= - \frac{1}{4} e^{- s t} cos (4 t) - \frac{1}{4} s \cdot \int e^{- s t} cos (4 t) d t

Simplify

\frac{1}{4} s : \frac{s}{4}

= - \frac{1}{4} e^{- s t} cos (4 t) - \frac{s}{4} \cdot \int e^{- s t} cos (4 t) d t

Apply Integration By Parts

: \frac{1}{4} e^{- s t} sin (4 t) - \int - \frac{1}{4} s e^{- s t} sin (4 t) d t

= - \frac{1}{4} e^{- s t} cos (4 t) - \frac{s}{4} (\frac{1}{4} e^{- s t} sin (4 t) - \int - \frac{1}{4} s e^{- s t} sin (4 t) d t)

Take the constant out:

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

= - \frac{1}{4} e^{- s t} cos (4 t) - \frac{s}{4} (\frac{1}{4} e^{- s t} sin (4 t) - (- \frac{1}{4} s \cdot \int e^{- s t} sin (4 t) d t))

Simplify

- \frac{1}{4} s : - \frac{s}{4}

= - \frac{1}{4} e^{- s t} cos (4 t) - \frac{s}{4} (\frac{1}{4} e^{- s t} sin (4 t) - (- \frac{s}{4} \cdot \int e^{- s t} sin (4 t) d t))

Therefore

\int e^{- s t} sin (4 t) d t = - \frac{1}{4} e^{- s t} cos (4 t) - \frac{s}{4} (\frac{1}{4} e^{- s t} sin (4 t) - (- \frac{s}{4} \cdot \int e^{- s t} sin (4 t) d t))

Isolate

\int e^{- s t} sin (4 t) d t

= - \frac{4 e ^{- s t} cos ( 4 t )}{16 + s ^{2}} - \frac{s e ^{- s t} sin ( 4 t )}{16 + s ^{2}}

Add a constant to the solution

= - \frac{4 e ^{- s t} cos ( 4 t )}{16 + s ^{2}} - \frac{s e ^{- s t} sin ( 4 t )}{16 + s ^{2}} + C

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What is the integral of e^{-st}sin(4t) ?
The integral of e^{-st}sin(4t) is -(4e^{-st}cos(4t))/(16+s^2)-(se^{-st}sin(4t))/(16+s^2)+C