What is the integral of e^{-sx}sin(2x) ?

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

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integral of e^{-sx}sin(2x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Simplify

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Simplify

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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