What is the integral of e^{-x}cos(6x) ?

The integral of e^{-x}cos(6x) is (6e^{-x}sin(6x))/(37)-(e^{-x}cos(6x))/(37)+C

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integral of e^{-x}cos(6x)

\int e^{- x} cos (6 x) d x

\frac{6 e ^{- x} sin ( 6 x )}{37} - \frac{e ^{- x} cos ( 6 x )}{37} + C

Solution steps

\int e^{- x} cos (6 x) d x

Apply Integration By Parts

: \frac{1}{6} e^{- x} sin (6 x) - \int - \frac{1}{6} e^{- x} sin (6 x) d x

= \frac{1}{6} e^{- x} sin (6 x) - \int - \frac{1}{6} e^{- x} sin (6 x) d x

Take the constant out:

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

= \frac{1}{6} e^{- x} sin (6 x) - (- \frac{1}{6} \cdot \int e^{- x} sin (6 x) d x)

Apply Integration By Parts

: - \frac{1}{6} e^{- x} cos (6 x) - \int \frac{1}{6} e^{- x} cos (6 x) d x

= \frac{1}{6} e^{- x} sin (6 x) - (- \frac{1}{6} (- \frac{1}{6} e^{- x} cos (6 x) - \int \frac{1}{6} e^{- x} cos (6 x) d x))

Take the constant out:

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

= \frac{1}{6} e^{- x} sin (6 x) - (- \frac{1}{6} (- \frac{1}{6} e^{- x} cos (6 x) - \frac{1}{6} \cdot \int e^{- x} cos (6 x) d x))

Therefore

\int e^{- x} cos (6 x) d x = \frac{1}{6} e^{- x} sin (6 x) - (- \frac{1}{6} (- \frac{1}{6} e^{- x} cos (6 x) - \frac{1}{6} \cdot \int e^{- x} cos (6 x) d x))

Isolate

\int e^{- x} cos (6 x) d x

= \frac{6 e ^{- x} sin ( 6 x )}{37} - \frac{e ^{- x} cos ( 6 x )}{37}

Add a constant to the solution

= \frac{6 e ^{- x} sin ( 6 x )}{37} - \frac{e ^{- x} cos ( 6 x )}{37} + C

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

(2010, 73.1), (2000, 71.3)

\int cos^{2} (y) d y

\frac{\partial}{\partial x} (x^{2} + x y - 10)

\frac{d}{d x} (2 x^{3} cos (x) sin (x))

What is the integral of e^{-x}cos(6x) ?
The integral of e^{-x}cos(6x) is (6e^{-x}sin(6x))/(37)-(e^{-x}cos(6x))/(37)+C