What is the integral of e^{-x}cos(8x) ?

The integral of e^{-x}cos(8x) is (8e^{-x}sin(8x))/(65)-(e^{-x}cos(8x))/(65)+C

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integral of e^{-x}cos(8x)

\int e^{- x} cos (8 x) d x

\frac{8 e ^{- x} sin ( 8 x )}{65} - \frac{e ^{- x} cos ( 8 x )}{65} + C

Solution steps

\int e^{- x} cos (8 x) d x

Apply Integration By Parts

: \frac{1}{8} e^{- x} sin (8 x) - \int - \frac{1}{8} e^{- x} sin (8 x) d x

= \frac{1}{8} e^{- x} sin (8 x) - \int - \frac{1}{8} e^{- x} sin (8 x) d x

Take the constant out:

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

= \frac{1}{8} e^{- x} sin (8 x) - (- \frac{1}{8} \cdot \int e^{- x} sin (8 x) d x)

Apply Integration By Parts

: - \frac{1}{8} e^{- x} cos (8 x) - \int \frac{1}{8} e^{- x} cos (8 x) d x

= \frac{1}{8} e^{- x} sin (8 x) - (- \frac{1}{8} (- \frac{1}{8} e^{- x} cos (8 x) - \int \frac{1}{8} e^{- x} cos (8 x) d x))

Take the constant out:

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

= \frac{1}{8} e^{- x} sin (8 x) - (- \frac{1}{8} (- \frac{1}{8} e^{- x} cos (8 x) - \frac{1}{8} \cdot \int e^{- x} cos (8 x) d x))

Therefore

\int e^{- x} cos (8 x) d x = \frac{1}{8} e^{- x} sin (8 x) - (- \frac{1}{8} (- \frac{1}{8} e^{- x} cos (8 x) - \frac{1}{8} \cdot \int e^{- x} cos (8 x) d x))

Isolate

\int e^{- x} cos (8 x) d x

= \frac{8 e ^{- x} sin ( 8 x )}{65} - \frac{e ^{- x} cos ( 8 x )}{65}

Add a constant to the solution

= \frac{8 e ^{- x} sin ( 8 x )}{65} - \frac{e ^{- x} cos ( 8 x )}{65} + C

\frac{\partial}{\partial x} (12 x^{4} - 6 x^{2} y + 4 y^{3})

\frac{d}{d x} (x^{\frac{3}{5}} (4 - x))

arccos (e^{2 x})

\int_{0}^{2} \frac{1}{x ^{1.9}} d x

f (x) = \frac{x + 3}{x + 5}

What is the integral of e^{-x}cos(8x) ?
The integral of e^{-x}cos(8x) is (8e^{-x}sin(8x))/(65)-(e^{-x}cos(8x))/(65)+C