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)

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) - (- \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)))

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}{d x} ((2 x^{2} + 3)^{2})

\frac{2}{s + 1} + 8 a \cdot \frac{s}{s ^{2} + 9}

\frac{d}{d x} ((1 - 2 x)^{5} (3 x - 2)^{6})

2 e^{2 t} + 8 t e^{2 t} + 4 t^{2} e^{2 t}

\frac{d}{d x} (\frac{ln ( x )}{2 x})

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