What is the integral of e^{2x}cos(4x) ?

The integral of e^{2x}cos(4x) is (e^{2x}sin(4x))/5+(e^{2x}cos(4x))/(10)+C

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integral of e^{2x}cos(4x)

\int e^{2 x} cos (4 x) d x

\frac{e ^{2 x} sin ( 4 x )}{5} + \frac{e ^{2 x} cos ( 4 x )}{10} + C

Solution steps

\int e^{2 x} cos (4 x) d x

Apply Integration By Parts

= \frac{1}{4} e^{2 x} sin (4 x) - \int \frac{1}{2} e^{2 x} sin (4 x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} e^{2 x} sin (4 x) - \frac{1}{2} \cdot \int e^{2 x} sin (4 x) d x

Apply Integration By Parts

= \frac{1}{4} e^{2 x} sin (4 x) - \frac{1}{2} (- \frac{1}{4} e^{2 x} cos (4 x) - \int - \frac{1}{2} e^{2 x} cos (4 x) d x)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} e^{2 x} sin (4 x) - \frac{1}{2} (- \frac{1}{4} e^{2 x} cos (4 x) - (- \frac{1}{2} \cdot \int e^{2 x} cos (4 x) d x))

Therefore

\int e^{2 x} cos (4 x) d x = \frac{1}{4} e^{2 x} sin (4 x) - \frac{1}{2} (- \frac{1}{4} e^{2 x} cos (4 x) - (- \frac{1}{2} \cdot \int e^{2 x} cos (4 x) d x))

Isolate

\int e^{2 x} cos (4 x) d x

= \frac{e ^{2 x} sin ( 4 x )}{5} + \frac{e ^{2 x} cos ( 4 x )}{10}

Add a constant to the solution

= \frac{e ^{2 x} sin ( 4 x )}{5} + \frac{e ^{2 x} cos ( 4 x )}{10} + C

What is the integral of e^{2x}cos(4x) ?
The integral of e^{2x}cos(4x) is (e^{2x}sin(4x))/5+(e^{2x}cos(4x))/(10)+C