Solve the IVP:
$x''(t) + x(t) + sin^2t = 0$ subject to $x(0) = 0$ and $x'(0) = 0$ using Laplace.
I have found out that $L[sin^2t](s)$ is $\frac{1} {2s}- \frac1 2(\frac{s}{s^2+4})$ but I need a guide through in solving the actual equation.
Any help will be appreciated.
Based on the comments, we take the Laplace transform of both sides of the Differential Equation. Doing this gives
$$s^2 \mathcal{L}(x(t)) + \mathcal{L}(x(t))+\mathcal{L}(\sin^2(t))=0,$$
which means
$$s^2 \mathcal{L}(x(t)) + \mathcal{L}(x(t))=-\mathcal{L}(\sin^2(t)).$$
We have $$-\mathcal{L}(\sin^2(t))= -\frac{1}{2s} + \frac{s}{2(s^2+4)}.$$
so
$$\mathcal{L} (x(t))= \frac{1}{s^2+1} \left ( -\frac{1}{2s} + \frac{s}{2(s^2+4)} \right )=\frac{2s}{3(s^2+1)} -\frac{s}{6(s^2+4)} -\frac{1}{2s}$$ by partial fractions. Use tables to find out what the inverse Laplace Transform of the right hand side is. Does this guide you in the right direction ?