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

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

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

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

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

Solution steps

\int sin (2 x) e^{- 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

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