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What is the Laplace transform of the function $\sin^2(\omega t)$

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2 Answers 2

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Write it as $$\frac{1-\cos 2\omega t}2$$ and apply the linearity property individually for each term.

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$$ \begin{align} \mathcal{L}\left[\sin^2\omega t\right]&=\mathcal{L}\left[\frac{1-\cos2\omega t}{2}\right]\\ &=\frac{1}{2}\mathcal{L}\left[1\right]-\frac{1}{2}\mathcal{L}\left[\cos2\omega t\right]\\ &=\frac{1}{2}\cdot\frac{1}{s}-\frac{1}{2}\cdot\frac{s}{s^2+(2\omega)^2}\\ &=\frac{1}{2s}\left(\frac{4\omega^2}{s^2+4\omega^2}\right). \end{align} $$

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