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

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) - (- \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))

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

\frac{d}{d x} y + y x^{7} = 0

y^{^{'}} + x y = x y^{4}

x \to 2 lim (- f (x))

\int_{0}^{2} x^{5} d x

\int cos^{4} (2 t) d t

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