We can choose to let \(u\) be either \(e^t\) or \(\cos(t)\text{;}\) we pick \(u = \cos(t)\text{,}\) and thus \(dv = e^t \, dt\text{.}\) With \(du = -\sin(t) \, dt\) and \(v = e^t\text{,}\) integration by parts tells us that
\begin{equation*}
\int e^t \cos(t) \, dt = e^t \cos(t) - \int e^t (-\sin(t))\, dt\text{,}
\end{equation*}
or equivalently that
\begin{equation}
\int e^t \cos(t) \, dt = e^t \cos(t) + \int e^t \sin(t) \, dt\text{.}\tag{5.4.4}
\end{equation}
The new integral has the same algebraic structure as the original one. While the overall situation isn’t necessarily better than what we started with, it hasn’t gotten worse. Thus, we proceed to integrate by parts again. This time we let \(u = \sin(t)\) and \(dv = e^t \, dt\text{,}\) so that \(du = \cos(t) \, dt\) and \(v = e^t\text{,}\) which implies
\begin{equation}
\int e^t \cos(t) \, dt = e^t \cos(t) + \left( e^t \sin(t) - \int e^t \cos(t) \, dt \right)\text{.}\tag{5.4.5}
\end{equation}
We seem to be back where we started, as two applications of integration by parts has led us back to the original problem,
\(\int e^t \cos(t) \, dt\text{.}\) But if we look closely at
Equation (5.4.5), we see that we can use algebra to solve for the value of the desired integral. Adding
\(\int e^t \cos(t) \, dt\) to both sides of the equation, we have
\begin{equation*}
2 \int e^t \cos(t) \, dt = e^t \cos(t) + e^t \sin(t)\text{,}
\end{equation*}
and therefore
\begin{equation*}
\int e^t \cos(t) \, dt = \frac{1}{2} \left( e^t \cos(t) + e^t \sin(t) \right) + C\text{.}
\end{equation*}
Note that since we never actually encountered an integral we could evaluate directly, we didn’t have the opportunity to add the integration constant
\(C\) until the final step.
