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

\int \frac{1}{48 x + 6 x ^{2}} d x

x \to 6 lim (8 x^{2} - 7 x + 1)

\frac{d}{d x} (3 cos (- x^{3} + 3 x + 1))

\int x - \frac{5 - 25 - 4 x}{2} d x

\int \frac{1}{n + 3}

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