What is the integral of e^{-3x}sin(1/3 x) ?

The integral of e^{-3x}sin(1/3 x) is -(3e^{-3x}cos(1/3 x))/(82)-(27e^{-3x}sin(1/3 x))/(82)+C

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integral of e^{-3x}sin(1/3 x)

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

- \frac{3 e ^{- 3 x} cos ( \frac{1}{3} x )}{82} - \frac{27 e ^{- 3 x} sin ( \frac{1}{3} x )}{82} + C

Solution steps

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

Apply Integration By Parts

: - 3 e^{- 3 x} cos (\frac{1}{3} x) - \int 9 e^{- 3 x} cos (\frac{1}{3} x) d x

= - 3 e^{- 3 x} cos (\frac{1}{3} x) - \int 9 e^{- 3 x} cos (\frac{1}{3} x) d x

Take the constant out:

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

= - 3 e^{- 3 x} cos (\frac{1}{3} x) - 9 \cdot \int e^{- 3 x} cos (\frac{1}{3} x) d x

Apply Integration By Parts

: 3 e^{- 3 x} sin (\frac{1}{3} x) - \int - 9 e^{- 3 x} sin (\frac{1}{3} x) d x

= - 3 e^{- 3 x} cos (\frac{1}{3} x) - 9 (3 e^{- 3 x} sin (\frac{1}{3} x) - \int - 9 e^{- 3 x} sin (\frac{1}{3} x) d x)

Take the constant out:

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

= - 3 e^{- 3 x} cos (\frac{1}{3} x) - 9 (3 e^{- 3 x} sin (\frac{1}{3} x) - (- 9 \cdot \int e^{- 3 x} sin (\frac{1}{3} x) d x))

Therefore

\int e^{- 3 x} sin (\frac{1}{3} x) d x = - 3 e^{- 3 x} cos (\frac{1}{3} x) - 9 (3 e^{- 3 x} sin (\frac{1}{3} x) - (- 9 \cdot \int e^{- 3 x} sin (\frac{1}{3} x) d x))

Isolate

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

= - \frac{3 e ^{- 3 x} cos ( \frac{1}{3} x )}{82} - \frac{27 e ^{- 3 x} sin ( \frac{1}{3} x )}{82}

Add a constant to the solution

= - \frac{3 e ^{- 3 x} cos ( \frac{1}{3} x )}{82} - \frac{27 e ^{- 3 x} sin ( \frac{1}{3} x )}{82} + C

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What is the integral of e^{-3x}sin(1/3 x) ?
The integral of e^{-3x}sin(1/3 x) is -(3e^{-3x}cos(1/3 x))/(82)-(27e^{-3x}sin(1/3 x))/(82)+C