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

The integral of e^{-3x}sin(3x) is -(e^{-3x}cos(3x))/6-(e^{-3x}sin(3x))/6+C

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

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

- \frac{e ^{- 3 x} cos ( 3 x )}{6} - \frac{e ^{- 3 x} sin ( 3 x )}{6} + C

Solution steps

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

Apply Integration By Parts

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{e ^{- 3 x} cos ( 3 x )}{6} - \frac{e ^{- 3 x} sin ( 3 x )}{6}

Add a constant to the solution

= - \frac{e ^{- 3 x} cos ( 3 x )}{6} - \frac{e ^{- 3 x} sin ( 3 x )}{6} + C

\frac{\partial}{\partial y} (sin (x - y))

\frac{d}{d x} (2 x^{2} - 4 x + 6)

y = - 3 - 2 e^{x}

\int ln ((2 x)^{2}) d x

\int \frac{1}{7 x ^{2} - 14 x + 28} d x

What is the integral of e^{-3x}sin(3x) ?
The integral of e^{-3x}sin(3x) is -(e^{-3x}cos(3x))/6-(e^{-3x}sin(3x))/6+C