Find $\mathcal{L}\left\{\cos^3\left(t\right)\right\}$

Question

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?

Answer 1

Hint :

Use triple-angle formula: $$\cos^3t=\frac{\cos3t+3\cos t}{4}$$

Answer 2

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 $\cos t$.)