Find Laplace transform of $f(t)=\sin^2 t$ using $\mathcal{L}(\frac{d}{dt}f(t))=\mathcal{L}(f'(t))$. (Check my answer)

Question

I want to find Laplace transform of $f(t)=\sin^2 t$ using $\mathcal{L}(\dfrac{d}{dt}f(t))=\mathcal{L}(f'(t))$.

This is my answer.

\begin{alignat*}{2} && \mathcal{L} (f'(t)) &= \mathcal{L} (\sin(2t))\\ \iff& & sF(s)-f(0)&= \frac{2}{s^2+4}\\ \iff& & sF(s)-0&= \frac{2}{s^2+4}\\ \iff& & F(s)&= \frac{2}{s(s^2+4)}. \end{alignat*}

and we have $$F(s)=\mathcal{L} \left(\sin^2 t\right)=\dfrac{2}{s(s^2+4)}.$$

Is this right or wrong answer?

Answer 1

Yes, this is correct. (For what it's worth, WolframAlpha also happens to agree.) Good job!