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

My question is use $L(f^{'}(t))=sF(s)-f(0)$ to show that $L(\sin^{2}(t))=2/(s^2+4)$.

My approach was to integrate $\sin^2(t)$ to find $f(t)$, then use this to find $F(s)$ and $f(0)$.

However, using that method, I find that the Laplace transform of $\sin^2(t)$ is $(s/2-1/2(s/(s^2+4))$, not $2/(s^2+4)$.

This is also the answer I get using an online Laplace calculator.

So where am I going wrong, or is the question wrong?

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    $\begingroup$ try writing $\sin^2 (t) =\frac 1 2 (1 - \cos 2t)$ $\endgroup$ Aug 19 '16 at 6:40
  • $\begingroup$ With Wolfram Alpha online calculator you can see the comparison between Linkha's hint and the corresponding Laplace transform with the codes LaplaceTransform[(sin t)^2,t,s] and on the other hand this LaplaceTransform[1/2(1-cos(2t)),t,s] . Thus the right way is use the hint from the user using (know the calculations of) a table of Laplace transforms and use the rules of calculate, combined with trigonometric idenitites, in this case. $\endgroup$
    – user243301
    Aug 19 '16 at 6:55
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We have that $$\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin^2(x) \quad\Rightarrow \quad\sin^2(x)=\frac{1 - \cos (2x)}{2}.$$ Hence $$f(t):=\int_0^t\sin^2(x) dx=\int_0^t \frac{1 - \cos (2x)}{2} \ dx =\frac{t}{2}-\frac{\sin(2t)}{4},$$ and $$F(s)=\frac{1}{2s^2}−\frac{2}{4(s^2+4)}.$$ Therefore $$L(\sin^2(t))=L(f^{'}(t))=sF(s)-f(0)=\frac{1}{2s}−\frac{s}{2(s^2+4)}-0=\frac{2}{s(s^2+4)}.$$ Check the final result here.

P.S. There is an easier way to obtain the Laplace Transform of $\sin^{2}(t)$: transform directly the identity $\sin^2 (t) =\frac{1 - \cos (2t)}{2}$.

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  • $\begingroup$ Hi guys. Thanks, both these responses are excellent. However, my lecturer insists that I have to use the L(f'(t)=sF(s)-f(0) identity, but that I CANT START by integrating sin^2(t). That's what has me so confused. $\endgroup$ Aug 20 '16 at 1:10
  • $\begingroup$ @Rockwallaby Is it better now? $\endgroup$
    – Robert Z
    Aug 20 '16 at 7:55
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Just using the definition of Laplace transform and De Moivre's identity,

$$\begin{eqnarray*} \int_{0}^{+\infty}\sin^2(t)e^{-st}\,dt &=& -\frac{1}{4}\int_{0}^{+\infty}\left(e^{it}-e^{-it}\right)^2 e^{-st}\,dt\\[0.2cm]&=&-\frac{1}{4}\left(\int_{0}^{+\infty}e^{-(s-2i)t}+e^{-(s+2i)t}\,dt-2\int_{0}^{+\infty}e^{-st}\,dt\right)\\[0.2cm]&=&-\frac{1}{4}\left(\frac{1}{s-2i}+\frac{1}{s+2i}-\frac{2}{s}\right)\\[0.2cm]&=&\frac{1}{4}\left(\frac{2}{s}-\frac{2s}{s^2+4}\right)=\color{red}{\frac{2}{s(s^2+4)}}. \end{eqnarray*}$$ That also follows from $\frac{d}{dt}\sin^2(t)=\sin(2t)$ and $\mathcal{L}\left(\sin(2t)\right)=\frac{2}{4+s^2}$.

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  • $\begingroup$ Hi guys. Thanks, both these responses are excellent. However, my lecturer insists that I have to use the L(f'(t)=sF(s)-f(0) identity, but that I CANT START by integrating sin^2(t). That's what has me so confused. $\endgroup$ Aug 20 '16 at 1:10
  • $\begingroup$ @Rockwallaby: you may compute a Laplace transform in any way you like, disregarding what your instructor tell you to do or not to do. To practice math is not to be restrained, and here you have plenty of (not so) different approaches. $\endgroup$ Aug 20 '16 at 1:14
  • $\begingroup$ I like the cut of your jibe Jack - but my lecturer seemed pretty sure. Thanks for your help mate. $\endgroup$ Aug 20 '16 at 1:47

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