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What is the Laplace transform of sin^5(t)?
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Unnikrishnan Menon, Studied Laplace Transform in college
Answered December 23, 2017 &middot; Upvoted by Sharangdhar Bodas, M.Sc.,B.Ed. Mathematics &amp; Physical Sciences, University of
Just break it down!
$\begin{equation}\begin{split}L\left[\sin^5t\right]&amp;=L\left[\sin^2t\times\sin^3t\right]\\&amp;=L\left[\dfrac {(1-\cos 2t)}{2}\times\dfrac{(3\sin t-sin 3t)}{4}\right]\\&amp;=\dfrac{1}{8}L\left[3\sin t-\sin 3t-3\sin t\cos 2t+\sin 3t\cos 2t\right]\\&amp;=\dfrac{1}{8}L\left[3\sin t-\sin 3t-\dfrac{3}{2}(\sin 3t-\sin t)+\dfrac{1}{2}(\sin 5t+\sin t)\right]\\&amp;=\dfrac{1}{8}L\left[5\sin t-\dfrac{5}{2}\sin 3t+\dfrac{1}{2}\sin 5t\right]\\&amp;=\dfrac{5} {8}L\left[\sin t\right]-\dfrac{5}{16}L\left[\sin 3t\right]+\dfrac{1}{16}L\left[\sin 5t\right]\\&amp;=\boxed{\dfrac{5} Continue Reading {8(s^2+1)}-\dfrac{15}{16(s^2+9)}+\dfrac{5$
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