Answer
\[
\begin{aligned}
\int e^{-y}\cos(y)\,dy = \frac{e^{-y}(\sin(y) - \cos(y))}{2} + C
\end{aligned}
\]
Work Step by Step
\[
\begin{aligned}
I &= \int e^{-y}\cos(y)\,dy.
\end{aligned}
\]
\[
\textbf{First integration by parts:} \quad u = e^{-y}, \quad dv = \cos(y)\,dy.
\]
Thus:
\[
du = -e^{-y}\,dy, \quad v = \sin(y).
\]
Applying the formula \(\int u\,dv = uv - \int v\,du\):
\[
\int e^{-y}\cos(y)\,dy = e^{-y}\sin(y) - \int \sin(y)(-e^{-y})\,dy.
\]
This simplifies to:
\[
I = e^{-y}\sin(y) + \int e^{-y}\sin(y)\,dy.
\]
\[
\textbf{Second integration by parts for } \int e^{-y}\sin(y)\,dy:
\]
Let:
\[
u = e^{-y}, \quad dv = \sin(y)\,dy.
\]
Then:
\[
du = -e^{-y}\,dy, \quad v = -\cos(y).
\]
Again:
\[
\int e^{-y}\sin(y)\,dy = e^{-y}(-\cos(y)) - \int(-\cos(y))(-e^{-y})\,dy.
\]
Simplify:
\[
\int e^{-y}\sin(y)\,dy = -e^{-y}\cos(y) - \int e^{-y}\cos(y)\,dy.
\]
Notice that \(\int e^{-y}\cos(y)\,dy = I\). Thus:
\[
\int e^{-y}\sin(y)\,dy = -e^{-y}\cos(y) - I.
\]
Substitute back into our expression for \(I\):
\[
I = e^{-y}\sin(y) + [-e^{-y}\cos(y) - I].
\]
Combine like terms:
\[
I = e^{-y}\sin(y) - e^{-y}\cos(y) - I.
\]
Add \(I\) to both sides:
\[
2I = e^{-y}\sin(y) - e^{-y}\cos(y).
\]
Divide both sides by 2:
\[
I = \frac{e^{-y}\sin(y) - e^{-y}\cos(y)}{2}.
\]
Factor out \(e^{-y}\):
\[
I = \frac{e^{-y}(\sin(y) - \cos(y))}{2} + C.
\]
\[
\boxed{\int e^{-y}\cos(y)\,dy = \frac{e^{-y}(\sin(y) - \cos(y))}{2} + C}.
\]
