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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: e^{- y} \frac{1}{x} sin (x y) - \int - e^{- y} \frac{1}{x} sin (x y) d y

= e^{- y} \frac{1}{x} sin (x y) - \int - e^{- y} \frac{1}{x} sin (x y) d y

Take the constant out:

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

= e^{- y} \frac{1}{x} sin (x y) - (- \frac{1}{x} \cdot \int e^{- y} sin (x y) d y)

Apply Integration By Parts

: - e^{- y} \frac{1}{x} cos (x y) - \int e^{- y} \frac{1}{x} cos (x y) d y

= e^{- y} \frac{1}{x} sin (x y) - (- \frac{1}{x} (- e^{- y} \frac{1}{x} cos (x y) - \int e^{- y} \frac{1}{x} cos (x y) d y))

Take the constant out:

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

= e^{- y} \frac{1}{x} sin (x y) - (- \frac{1}{x} (- e^{- y} \frac{1}{x} cos (x y) - \frac{1}{x} \cdot \int e^{- y} cos (x y) d y))

Therefore

\int e^{- y} cos (x y) d y = e^{- y} \frac{1}{x} sin (x y) - (- \frac{1}{x} (- e^{- y} \frac{1}{x} cos (x y) - \frac{1}{x} \cdot \int e^{- y} cos (x y) d y))

Isolate

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

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

Add a constant to the solution

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

4

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