What is the integral of-e^{-x}cos(3x) ?

The integral of-e^{-x}cos(3x) is -3/10 e^{-x}sin(3x)+1/10 e^{-x}cos(3x)+C

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integral of-e^{-x}cos(3x)

\int - e^{- x} cos (3 x) d x

- \frac{3}{10} e^{- x} sin (3 x) + \frac{1}{10} e^{- x} cos (3 x) + C

Solution steps

\int - e^{- x} cos (3 x) d x

Take the constant out:

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

= - \int e^{- x} cos (3 x) d x

Apply Integration By Parts

= - (\frac{1}{3} e^{- x} sin (3 x) - \int - \frac{1}{3} e^{- x} sin (3 x) d x)

Take the constant out:

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

= - (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} \cdot \int e^{- x} sin (3 x) d x))

Apply Integration By Parts

= - (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} (- \frac{1}{3} e^{- x} cos (3 x) - \int \frac{1}{3} e^{- x} cos (3 x) d x)))

Take the constant out:

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

= - (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} \cdot \int e^{- x} cos (3 x) d x)))

Therefore

- \int e^{- x} cos (3 x) d x = - (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} \cdot \int e^{- x} cos (3 x) d x)))

Isolate

\int e^{- x} cos (3 x) d x

= - (\frac{3 e ^{- x} sin ( 3 x )}{10} - \frac{e ^{- x} cos ( 3 x )}{10})

Simplify

- (\frac{3 e ^{- x} sin ( 3 x )}{10} - \frac{e ^{- x} cos ( 3 x )}{10}) : - \frac{3}{10} e^{- x} sin (3 x) + \frac{1}{10} e^{- x} cos (3 x)

= - \frac{3}{10} e^{- x} sin (3 x) + \frac{1}{10} e^{- x} cos (3 x)

Add a constant to the solution

= - \frac{3}{10} e^{- x} sin (3 x) + \frac{1}{10} e^{- x} cos (3 x) + C

x \to \infty lim (x^{6 x})

d er i v a t i v e ln (x + 9)

\frac{d}{d x} (\frac{x ^{9}}{9} - \frac{x ^{7}}{7})

x \to 1 lim (\frac{e ^{x^{2} - 1}}{x ^{2} - 3 x + 2})

\frac{d y}{d x} = \frac{( y - 1 )}{x ^{2}}

What is the integral of-e^{-x}cos(3x) ?
The integral of-e^{-x}cos(3x) is -3/10 e^{-x}sin(3x)+1/10 e^{-x}cos(3x)+C