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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

= 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}

Add a constant to the solution

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

\frac{d}{d x} (a x^{2} + b)

\frac{d}{d x} (4 x ln (7 x) - 4 x)

\int - \frac{sec ^{2} ( x ) - 1}{tan ( x )} d x

\frac{s + 3}{s ^{2} + s + 1}

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{5}{x ^{4}}) (x + 9 x^{3}))

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