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

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

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

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

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

Solution steps

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{d y}{d x} + 2 x y = 4

\frac{d}{d x} (ln (13 - x^{2}))

\int (csc^{2} (\frac{x}{2})) d x

y = (t - 5) (t^{2} - 3), (2, - 3)

x \to 0 lim (x^{18} ln (x))

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