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

The integral of-e^{-x}cos(x) is -1/2 e^{-x}sin(x)+1/2 e^{-x}cos(x)+C

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

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

- \frac{1}{2} e^{- x} sin (x) + \frac{1}{2} e^{- x} cos (x) + C

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Therefore

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

Isolate

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

= - (\frac{e ^{- x} sin ( x )}{2} - \frac{e ^{- x} cos ( x )}{2})

Simplify

- (\frac{e ^{- x} sin ( x )}{2} - \frac{e ^{- x} cos ( x )}{2}) : - \frac{1}{2} e^{- x} sin (x) + \frac{1}{2} e^{- x} cos (x)

= - \frac{1}{2} e^{- x} sin (x) + \frac{1}{2} e^{- x} cos (x)

Add a constant to the solution

= - \frac{1}{2} e^{- x} sin (x) + \frac{1}{2} e^{- x} cos (x) + C

What is the integral of-e^{-x}cos(x) ?
The integral of-e^{-x}cos(x) is -1/2 e^{-x}sin(x)+1/2 e^{-x}cos(x)+C