Taylor expansion of sin(x)=
x-x³/3!+x⁵/5!-...(-1)ⁿx²ⁿ⁺¹/(2n+1)!
Therefore Taylor expansion of sin(t²)=
t²-t⁶/3!+t¹⁰/5!-...(-1)ⁿt⁴ⁿ⁺²/(2n+1)! for integer n≥0.
Now apply Laplace Transform to the series, ℒ{sin(t²)}.
ℒ(tʳ)=r!/sʳ⁺¹; let r=4n+2, then:
ℒ{sin(t²)}=∑(-1)ⁿ/(2n+1)!ℒ{t⁴ⁿ⁺²}=∑((-1)ⁿ/(2n+1)!)(4n+2)!/s⁴ⁿ⁺³.
So let’s see what the Laplace series looks like:
2!/s³-(6!/3!)/s⁷+(10!/5!)/s¹¹-(14!/7!)/s¹⁵+...
2/s³-120/s⁷+30240/s¹¹-17297280/s¹⁵+...
You didn’t mess up!