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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?

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    $\begingroup$ Wolfram Alpha says $\frac{2s}{(s^2+1)(s^2+9)}$ for the transform of the second half there. I don't know how it got it though. $\endgroup$ – Dan Uznanski Nov 20 '14 at 13:58
  • $\begingroup$ Gosh :P I should always check Wolfram before I post lol, my apologies. So I must have messed up somewhere. I will try it again to see where I'm going wrong $\endgroup$ – jm324354 Nov 20 '14 at 13:58
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Hint :

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

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  • $\begingroup$ I see, so that will just let me split it up from the get-go... $\endgroup$ – jm324354 Nov 20 '14 at 15:22
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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 $\cos t$.)

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  • $\begingroup$ Oh this is cool, I never thought to split a trig function up like that, because that definitely makes things easier to integrate. $\endgroup$ – jm324354 Nov 20 '14 at 15:01
  • $\begingroup$ Variations of this trick turn out to be repeatedly useful, including for proving identities like this one for arbitrary powers of $\sin$ and $\cos$. $\endgroup$ – Travis Willse Nov 20 '14 at 15:13
  • $\begingroup$ Out of curiosity...you say, the integral above can also be evaluated using these identities? I've looked further into the complex realm and a few identities, and I'm curious. Perhaps you mean the exponential "outside" of the cosine could be combined with cosine's exponential identity and the integal further divided? $\endgroup$ – jm324354 Nov 20 '14 at 22:15
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    $\begingroup$ More precisely I mean that $\int e^u \cos au \,du = \int e^u (e^{iau} + e^{-iau}) du = \int e^{(1 + 2ia)u} du + \int e^{(1 - 2ia)u} du$, which is a sum of integrals of exponential functions. In some sense the harder part is recombining the antiderivatives into an expression that is manifestly real, although there's a standard trick that helps here too. $\endgroup$ – Travis Willse Nov 21 '14 at 1:47

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