What is the integral of e^{-3x}sin(2x) ?

The integral of e^{-3x}sin(2x) is -(2e^{-3x}cos(2x))/(13)-(3e^{-3x}sin(2x))/(13)+C

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integral of e^{-3x}sin(2x)

\int e^{- 3 x} sin (2 x) d x

- \frac{2 e ^{- 3 x} cos ( 2 x )}{13} - \frac{3 e ^{- 3 x} sin ( 2 x )}{13} + C

Solution steps

\int e^{- 3 x} sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} e^{- 3 x} cos (2 x) - \int \frac{3}{2} e^{- 3 x} cos (2 x) d x

= - \frac{1}{2} e^{- 3 x} cos (2 x) - \int \frac{3}{2} e^{- 3 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^{- 3 x} cos (2 x) - \frac{3}{2} \cdot \int e^{- 3 x} cos (2 x) d x

Apply Integration By Parts

: \frac{1}{2} e^{- 3 x} sin (2 x) - \int - \frac{3}{2} e^{- 3 x} sin (2 x) d x

= - \frac{1}{2} e^{- 3 x} cos (2 x) - \frac{3}{2} (\frac{1}{2} e^{- 3 x} sin (2 x) - \int - \frac{3}{2} e^{- 3 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^{- 3 x} cos (2 x) - \frac{3}{2} (\frac{1}{2} e^{- 3 x} sin (2 x) - (- \frac{3}{2} \cdot \int e^{- 3 x} sin (2 x) d x))

Therefore

\int e^{- 3 x} sin (2 x) d x = - \frac{1}{2} e^{- 3 x} cos (2 x) - \frac{3}{2} (\frac{1}{2} e^{- 3 x} sin (2 x) - (- \frac{3}{2} \cdot \int e^{- 3 x} sin (2 x) d x))

Isolate

\int e^{- 3 x} sin (2 x) d x

= - \frac{2 e ^{- 3 x} cos ( 2 x )}{13} - \frac{3 e ^{- 3 x} sin ( 2 x )}{13}

Add a constant to the solution

= - \frac{2 e ^{- 3 x} cos ( 2 x )}{13} - \frac{3 e ^{- 3 x} sin ( 2 x )}{13} + C

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What is the integral of e^{-3x}sin(2x) ?
The integral of e^{-3x}sin(2x) is -(2e^{-3x}cos(2x))/(13)-(3e^{-3x}sin(2x))/(13)+C