What is the integral of e^{-x}cos(9x) ?

The integral of e^{-x}cos(9x) is (9e^{-x}sin(9x))/(82)-(e^{-x}cos(9x))/(82)+C

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integral of e^{-x}cos(9x)

\int e^{- x} cos (9 x) d x

\frac{9 e ^{- x} sin ( 9 x )}{82} - \frac{e ^{- x} cos ( 9 x )}{82} + C

Solution steps

\int e^{- x} cos (9 x) d x

Apply Integration By Parts

: \frac{1}{9} e^{- x} sin (9 x) - \int - \frac{1}{9} e^{- x} sin (9 x) d x

= \frac{1}{9} e^{- x} sin (9 x) - \int - \frac{1}{9} e^{- x} sin (9 x) d x

Take the constant out:

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

= \frac{1}{9} e^{- x} sin (9 x) - (- \frac{1}{9} \cdot \int e^{- x} sin (9 x) d x)

Apply Integration By Parts

: - \frac{1}{9} e^{- x} cos (9 x) - \int \frac{1}{9} e^{- x} cos (9 x) d x

= \frac{1}{9} e^{- x} sin (9 x) - (- \frac{1}{9} (- \frac{1}{9} e^{- x} cos (9 x) - \int \frac{1}{9} e^{- x} cos (9 x) d x))

Take the constant out:

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

= \frac{1}{9} e^{- x} sin (9 x) - (- \frac{1}{9} (- \frac{1}{9} e^{- x} cos (9 x) - \frac{1}{9} \cdot \int e^{- x} cos (9 x) d x))

Therefore

\int e^{- x} cos (9 x) d x = \frac{1}{9} e^{- x} sin (9 x) - (- \frac{1}{9} (- \frac{1}{9} e^{- x} cos (9 x) - \frac{1}{9} \cdot \int e^{- x} cos (9 x) d x))

Isolate

\int e^{- x} cos (9 x) d x

= \frac{9 e ^{- x} sin ( 9 x )}{82} - \frac{e ^{- x} cos ( 9 x )}{82}

Add a constant to the solution

= \frac{9 e ^{- x} sin ( 9 x )}{82} - \frac{e ^{- x} cos ( 9 x )}{82} + C

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5 e^{- t}

What is the integral of e^{-x}cos(9x) ?
The integral of e^{-x}cos(9x) is (9e^{-x}sin(9x))/(82)-(e^{-x}cos(9x))/(82)+C