Right answer is (a) \(\frac{3}{8s}\)
To explain: In the given question,
sin^4 t=\(\frac{1+cos^2(2t)-2cos(2t)}{4}\)
L(sin^4 t)=\(L\left (\frac{1+cos^2 (2t)-2 cos(2t)}{4}\right )\)
=\(\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)}\)
By First Shifting Property
L(e^2t sin^4 t)=\(\frac{3}{8(s-2)}-\frac{s}{2((s-2)^2+4)}+\frac{s}{8((s-2)^2+16)}\)