What is the integral of e^{-x}cos(5x) ?

The integral of e^{-x}cos(5x) is (5e^{-x}sin(5x))/(26)-(e^{-x}cos(5x))/(26)+C

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integral of e^{-x}cos(5x)

\int e^{- x} cos (5 x) d x

\frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26} + C

Solution steps

\int e^{- x} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} e^{- x} sin (5 x) - \int - \frac{1}{5} e^{- x} sin (5 x) d x

= \frac{1}{5} e^{- x} sin (5 x) - \int - \frac{1}{5} e^{- x} sin (5 x) d x

Take the constant out:

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

= \frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} \cdot \int e^{- x} sin (5 x) d x)

Apply Integration By Parts

: - \frac{1}{5} e^{- x} cos (5 x) - \int \frac{1}{5} e^{- x} cos (5 x) d x

= \frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \int \frac{1}{5} e^{- x} cos (5 x) d x))

Take the constant out:

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

= \frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \frac{1}{5} \cdot \int e^{- x} cos (5 x) d x))

Therefore

\int e^{- x} cos (5 x) d x = \frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \frac{1}{5} \cdot \int e^{- x} cos (5 x) d x))

Isolate

\int e^{- x} cos (5 x) d x

= \frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26}

Add a constant to the solution

= \frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26} + C

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

\frac{\partial}{\partial x} (x e^{x - y})

\frac{d}{d x} (\frac{4 x ^{3} - 7 x + 8}{x})

\frac{d y}{d x} = \frac{1 + 3 x ^{2}}{( 1 + y ) ^{2}}, y (1) = 2

\int e^{- s \cdot t} d t

What is the integral of e^{-x}cos(5x) ?
The integral of e^{-x}cos(5x) is (5e^{-x}sin(5x))/(26)-(e^{-x}cos(5x))/(26)+C