How to find the Laplace transform of $\sin^2 (2t)$?

Question

How to find the Laplace transform of $\sin^2 (2t)$?

So far I have: $\sin^2(x) = \frac12 - \frac12 \cos(2x)$.

Answer 1

Chat defeated me! You are pretty much finished

$$sin^2(2t) = \frac{1}{2} - \frac{1}{2}cos(4t)$$

Transforming $$L(sin^2(2t)) = \frac{1}{2}L(1) - \frac{1}{2}L(cos(4t))=\frac{1}{2}(\frac{1}{s}) - \frac{1}{2}(\frac{s}{s^2+16})$$

Your x seems to have become a 2t since yesterday?

Answer 2

You can also do this using this theorem,

$ If,L(x(t))=X(s) \\ L(\frac{d(x(t))}{dt})=sX(s)-x(0^{-})\\\\ Let, x(t)=sin^{2}(wt)\\ So, \frac{dx(t)}{dt}=wsin(2wt)\\\\ If, L(sin(2wt))=\frac{2w}{4w^{2}+s^{2}}\\ Hence, L(sin^{2}(wt))=\frac{2w^{2}}{s(s^{2}+4w^{2})} $