What is the Laplace transform of $\sin^2(t)$? 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? 
 A: 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 $\sin^2 (t) =\frac{1 - \cos (2t)}{2}$.
A: 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 $\mathcal{L}\left(\sin(2t)\right)=\frac{2}{4+s^2}$.
