Answer
$\frac{1}{8}(1-\cos 4t)$
Work Step by Step
Given: $F(s)=\frac{2}{s(s^2+16)}$
Rewrite as:
$F(s)=\frac{1}{2s}.\frac{4}{s^2+16}$
Using the Convolution Theorem
$L^{-1}[F(s)]=L^{-1}[\frac{1}{2s}.\frac{4}{s^2+16}]\\
=L^{-1}[\frac{1}{2s}].L^{-1}[\frac{4}{s^2+16}]\\
=\frac{1}{2} * \sin 4t\\
=\int^t_0 \frac{1}{2}\sin 4\omega d\omega\\
=[-\frac{1}{8}(\cos 4\omega)]^t_0\\
=\frac{1}{8}(1-\cos 4t)$
