What is the integral of e^{-2x}sin(4x) ?

The integral of e^{-2x}sin(4x) is -(e^{-2x}cos(4x))/5-(e^{-2x}sin(4x))/(10)+C

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integral of e^{-2x}sin(4x)

\int e^{- 2 x} sin (4 x) d x

- \frac{e ^{- 2 x} cos ( 4 x )}{5} - \frac{e ^{- 2 x} sin ( 4 x )}{10} + C

Solution steps

\int e^{- 2 x} sin (4 x) d x

Apply Integration By Parts

: - \frac{1}{4} e^{- 2 x} cos (4 x) - \int \frac{1}{2} e^{- 2 x} cos (4 x) d x

= - \frac{1}{4} e^{- 2 x} cos (4 x) - \int \frac{1}{2} e^{- 2 x} cos (4 x) d x

Take the constant out:

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

= - \frac{1}{4} e^{- 2 x} cos (4 x) - \frac{1}{2} \cdot \int e^{- 2 x} cos (4 x) d x

Apply Integration By Parts

: \frac{1}{4} e^{- 2 x} sin (4 x) - \int - \frac{1}{2} e^{- 2 x} sin (4 x) d x

= - \frac{1}{4} e^{- 2 x} cos (4 x) - \frac{1}{2} (\frac{1}{4} e^{- 2 x} sin (4 x) - \int - \frac{1}{2} e^{- 2 x} sin (4 x) d x)

Take the constant out:

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

= - \frac{1}{4} e^{- 2 x} cos (4 x) - \frac{1}{2} (\frac{1}{4} e^{- 2 x} sin (4 x) - (- \frac{1}{2} \cdot \int e^{- 2 x} sin (4 x) d x))

Therefore

\int e^{- 2 x} sin (4 x) d x = - \frac{1}{4} e^{- 2 x} cos (4 x) - \frac{1}{2} (\frac{1}{4} e^{- 2 x} sin (4 x) - (- \frac{1}{2} \cdot \int e^{- 2 x} sin (4 x) d x))

Isolate

\int e^{- 2 x} sin (4 x) d x

= - \frac{e ^{- 2 x} cos ( 4 x )}{5} - \frac{e ^{- 2 x} sin ( 4 x )}{10}

Add a constant to the solution

= - \frac{e ^{- 2 x} cos ( 4 x )}{5} - \frac{e ^{- 2 x} sin ( 4 x )}{10} + C

6 (t - 4)

\frac{\partial}{\partial x} (e^{\frac{1}{x}})

\int \frac{x + 1}{x ^{2} + x + 1} d x

\frac{d y}{d t} + 3 y = 11, y (0) = 1

\int \frac{- 4 x}{2 x + 1} d x

What is the integral of e^{-2x}sin(4x) ?
The integral of e^{-2x}sin(4x) is -(e^{-2x}cos(4x))/5-(e^{-2x}sin(4x))/(10)+C