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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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

\int_{1}^{3} (x - 1) ln (x) d x

\frac{d}{d x} (\frac{4}{3} p (x) i r (x) (x)^{3})

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 7)

(\frac{4}{3}) p i x^{3}

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