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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

y = (x^{2} - 3 x + 3), (2, 1)

\int 45 + 4 x - x^{2} d x

y = (ln (x))^{x}

\int \frac{6 x ^{2} - 3 x}{x ^{2}} d x

y^{^{'}} = 2 t

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