Find the inverse Laplace transform f(t)=L^(-1){F(s)} of the function F(s)=frac(8e^(-4s))(s^2+64)

Find the inverse Laplace transform $$f(t)=L^{-1}\left\{F(s)\right\}$$ of the function $$\displaystyle{F}{\left({s}\right)}={\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}$$
Use h(t-c) for the Heaviside function h_c(t) if necessary.
$$f(t)=L^{-1}\left\{\frac{8e^{-4s}}{s^2+64}\right\}=$$

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StrycharzT

Step 1
Consider the provided function,
$$\displaystyle{F}{\left({s}\right)}={\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}$$
The inverse laplace transform of F(s) is find as,
$$\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={L}^{{-{1}}}{\left({\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\right)}$$
Apply inverse tranform rule,
if $$\displaystyle{L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={f{{\left({t}\right)}}}\ {t}{h}{e}{n}\ {L}^{{-{1}}}{\left({e}^{{-{c}{s}}}{F}{\left({s}\right)}\right)}={h}{\left({t}-{c}\right)}{f{{\left({t}-{c}\right)}}}$$
where $$\displaystyle{h}_{{c}}{\left({t}\right)}$$ is Heaviside funtion.
For $$\displaystyle{\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}.{F}{\left({s}\right)}={\frac{{{8}}}{{{s}^{{2}}+{64}}}},{c}={4}$$
Step 2
Use the inverse Laplace transform table,
$$\displaystyle{L}^{{-{1}}}{\left({\frac{{{a}}}{{{s}^{{2}}+{a}^{{2}}}}}\right)}={\sin{{\left({a}{t}\right)}}}$$
So, $$\displaystyle{L}^{{-{1}}}{\left({\frac{{{8}}}{{{s}^{{2}}+{64}}}}\right)}={L}^{{-{1}}}{\left({\frac{{{8}}}{{{s}^{{2}}+{8}^{{2}}}}}\right)}={\sin{{\left({8}{t}\right)}}}$$
Thus, $$\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left({\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\right)}={h}{\left({t}-{4}\right)}{\sin{{\left({8}{\left({t}-{4}\right)}\right)}}}$$