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

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

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

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

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

Solution steps

\int 2 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

= 2 \cdot \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

= 2 (- \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

= 2 (- \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

= 2 (- \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

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

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

Add a constant to the solution

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

\frac{d y}{d x} - 2 y tan (x) = y^{2} tan^{2} (x)

\int \frac{x ^{2}}{x - 6} d x

f (x) = \frac{x ^{2}}{e ^{x}}

y = \frac{2}{ln ( x )}

f (x) = \frac{1}{x ^{7}}

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