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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} ((sin (m x + n y))^{2})

\int 2 e^{- x^{2}} d x

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

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{4}}})

x \to - 2 lim ((5 x + 7)^{4})

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