What is the integral of 4e^{-x}cos(x) ?

The integral of 4e^{-x}cos(x) is 4((e^{-x}sin(x))/2-(e^{-x}cos(x))/2)+C

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integral of 4e^{-x}cos(x)

\int 4 e^{- x} cos (x) d x

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

Solution steps

\int 4 e^{- x} cos (x) d x

Take the constant out:

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

= 4 \cdot \int e^{- x} cos (x) d x

Apply Integration By Parts

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

= 4 (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

= 4 (e^{- x} sin (x) - (- \int e^{- x} sin (x) d x))

Apply Integration By Parts

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

= 4 (e^{- x} sin (x) - (- (- e^{- x} cos (x) - \int e^{- x} cos (x) d x)))

Therefore

4 \cdot \int e^{- x} cos (x) d x = 4 (e^{- x} sin (x) - (- (- e^{- x} cos (x) - \int e^{- x} cos (x) d x)))

Isolate

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

= 4 (\frac{e ^{- x} sin ( x )}{2} - \frac{e ^{- x} cos ( x )}{2})

Add a constant to the solution

= 4 (\frac{e ^{- x} sin ( x )}{2} - \frac{e ^{- x} cos ( x )}{2}) + C

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What is the integral of 4e^{-x}cos(x) ?
The integral of 4e^{-x}cos(x) is 4((e^{-x}sin(x))/2-(e^{-x}cos(x))/2)+C