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

= \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} cos (4 x) - \int - \frac{1}{2} 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) - \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

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

\int 2 x^{\frac{1}{3}} d x

x \to \infty lim (cos (x))

y^{^{''}} + 12 y^{^{'}} + 36 y = e^{- 5 x} (x^{2} - 2 x + 3)

\frac{e ^{x - 2}}{x - 1}

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