Example 6.
\(\ \ \)\(\lap{\cos(3t)}\text{.}\)Solution. Solution
We start by applying the definition of the Laplace transform:
\begin{equation*}
\lap{ \cos(3t)} = \int_0^{\infty} e^{-st} \cdot \cos(3t)\ dt \text{.}
\end{equation*}
Rather than directly integrating, we will use a modified Euler’s Formula to express cosine in terms of \(e\)
\begin{equation*}
\cos(3t) = \frac12\left(e^{3it} + e^{-3it}\right).
\end{equation*}
Substituting this into the integral gives:
\begin{align*}
\lap{ \cos(3t)}
=\amp\ \frac12 \int_0^{\infty} e^{-st} \frac12\left(e^{3it} + e^{-3it}\right)\ dt \\
=\amp\ \frac12\left[\int_0^{\infty} e^{-st}\cdot e^{3it}\ dt + \int_0^{\infty} e^{-st}\cdot e^{-3it}\ dt\right]\\
=\amp\ \frac{1}{2} \left[\lap{e^{3it}} + \lap{e^{-3it}}\right]\\
=\amp\ \frac{1}{2} \left[\frac{1}{s - 3i} + \frac{1}{s + 3i}\right] \qquad (\text{by } \knowl{./knowl/xref/L2.html}{\text{\(L_2\)}})\\
=\amp\ \frac{1}{2} \left[\frac{s + 3i + s - 3i}{(s - 3i)(s + 3i)}\right]\\
=\amp\ \frac{s}{s^2 + 9}.
\end{align*}
Therefore, the Laplace transform of \(\cos(3t)\) is:
\begin{equation*}
\lap{\cos(3t)} = \frac{s}{s^2 + 9}.
\end{equation*}
