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

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

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

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

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

Solution steps

\int cos (x) e^{- 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

y = 3 x, y = 5 x^{2}

\frac{d ^{2}}{d x ^{2}} (x e^{7 x})

\int \frac{5 x ^{2}}{x ^{2} - 4} d x

\int \frac{2 s + 2}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

n = 1 \sum \infty 0.4 5^{n}

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