Find the inverse Laplace transform of F(s)=frac{(s+4)}{(s^2+9)} a)cos(t)+frac{4}{3}sin(t) b)non of the above c) cos(3t)+sin(3t) d) cos(3t)+frac{4}{3} sin(3t) e)cos(3t)+frac{2}{3} sin(3t) f) cos(t)+4sin(t)

Step 1
The given function is $F(s)=\frac{(s+4)}{(s^2+9)}$
The following formulae are used in finding the inverse laplace transform of the given function.
$L^{-1}\{f(s)+g(s)\}=L^{-1}\{f(t)\}+L^{-1}\{g(t)\}$
$L^{-1}\{a\cdot f(s)\}=a\cdot L^{-1}\{f(t)\}$
$L^{-1}\left\{\frac{s}{(s^2+a^2)}\right\}=\cos at$
$L^{-1}\left\{\frac{1}{(s^2+a^2)}\right\}=\sin at$
Step 2
Evaluate the inverse laplace transform of the given function as follows.
$L^{-1}\{s+4s2+9\}=L^{-1}\left\{\frac{s}{(s^2+9)}\right\}+L^{-1}\left\{\frac{4}{(s^2+9)}\right\}$
$=L^{-1}\left\{\frac{s}{(s^2+3^2)}\right\}+4L^{-1}\left\{\frac{1}{(s^2+3^2)}\right\}$
$=\cos 3t+4(\frac{1}{3}\sin 3t)$
$=\cos 3t+\frac{4}{3}\sin 3t$
Therefore, the inverse Laplace transform of the given function is $\cos 3t+\frac{4}{3}\sin 3t$
Thus, the correct option is (d).