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Find Laplace transform of $(\sin^2 2t)$

How do I go about this ?

do I spilt them up like

$ L ( \sin 2t) \cdot L (\sin 2t) $ ?

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    $\begingroup$ No, you can't do that. Use that $\sin^2(2t) = \frac{1-\cos(4t)}{2}$ $\endgroup$
    – Jakobian
    Aug 16, 2018 at 12:37
  • $\begingroup$ Also note that multiplication on time domain becomes convolution on Laplace domain. $\endgroup$
    – Mefitico
    Aug 16, 2018 at 12:42

3 Answers 3

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No! The Laplace transform of fg is NOT L(f)L(g). Instead use the basic definition: $\int_0^\infty sin^2(2t)e^{-st}dt$. You might find it simplest to use the fact that $sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$ so that $sin(2t)= \frac{e^{2it}- e^{-2it}}{2i}$ and $sin^2(2t)= \left(\frac{e^{2it}- e^{-2it}}{2i}\right)^2= -\frac{e^{4it}- 2+ e^{-4it}}{4}$.

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  • $\begingroup$ We can use the convolution property too, right? $\endgroup$
    – Zacky
    Aug 16, 2018 at 13:17
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Given $\sin^22t$

$$\sin^22t=\dfrac{1-\cos4t}{2}$$ $$L(\sin^22t)=\dfrac{L(1)-L(\cos4t)}{2}=\dfrac{1}{2s}-\dfrac12\left(\dfrac{s}{s^2+16}\right)$$

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Hint: $$\sin^2 2t= \dfrac{1-\cos 4t}{2}$$

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  • $\begingroup$ How did you get this ? $\endgroup$
    – user185692
    Aug 16, 2018 at 12:53
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    $\begingroup$ @user185692 you can use $ \cos^2 x + \sin^2 x = 1$ and $\cos (a+b)=\cos a \cos b - \sin a \sin b$ $\endgroup$
    – RBH
    Aug 16, 2018 at 12:56

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