What is the integral of sin(2t)e^{-st} ?

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

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integral of sin(2t)e^{-st}

\int sin (2 t) e^{- s t} d t

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

Solution steps

\int sin (2 t) e^{- s t} d t

Apply Integration By Parts

: - \frac{1}{2} e^{- s t} cos (2 t) - \int \frac{1}{2} s e^{- s t} cos (2 t) d t

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

Take the constant out:

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

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

Simplify

\frac{1}{2} s : \frac{s}{2}

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

Apply Integration By Parts

: \frac{1}{2} e^{- s t} sin (2 t) - \int - \frac{1}{2} s e^{- s t} sin (2 t) d t

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

Take the constant out:

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

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

Simplify

- \frac{1}{2} s : - \frac{s}{2}

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

Therefore

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

Isolate

\int e^{- s t} sin (2 t) d t

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

Add a constant to the solution

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

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