TITLE: Find $\mathcal{L}\left\{\cos^3\left(t\right)\right\}$
QUESTION [9 upvotes]: I began by breaking the problem up as follows:
\begin{align}
\mathcal{L}\left\{\cos^3\left(t\right)\right\}=\int_0^\infty e^{-st}\cos^3\left(t\right)\:dt & = \int_0^\infty e^{-st}\cos\left(t\right)\:dt+\int_0^\infty e^{-st}\cos\left(t\right)\sin^2\left(t\right)\:dt,
\end{align}
which then simplifies to 
\begin{align}
\frac{s}{s^2+1}+\int_0^\infty e^{-st}\cos\left(t\right)\sin^2\left(t\right)\:dt,
\end{align}
because I've memorized the table (hopefully) and I know that the $\mathcal{L}\left\{\cos\left(t\right)\right\}$ eventually simplifies to that for $s>0$. But now the second half. After very tedious calculations I've arrived at $\mathcal{L}\left\{\cos\left(t\right)\sin^2\left(t\right)\right\}=0$? Does that make sense?

REPLY [2 votes]: One option is to use the complex identity
$$\cos t = \frac{1}{2}(e^{it} + e^{-it}),$$
which gives
\begin{align}\cos^3 t
&= \frac{1}{8}(e^{3it} + 3 e^{it} + e^{-it} + e^{-3it}) \\
&= \frac{1}{4}\left[\frac{1}{2}(e^{3it} + e^{-3it}) + 3 \cdot \frac{1}{2}(e^{it} + e^{-it})  \right] \\
&= \frac{1}{4}(\cos 3t + 3 \cos 3t),
\end{align}
which reduces the computation to the standard integral
$$\int e^u \cos au \,du .$$
(In fact, this integral too can be readily handled using the above complex formula for $\cos t$.)