I need help solving an Integral Equation that my professor showed without teaching to us. I've tried emulating the textbooks example but I still can't crack it. $$ f(t) = \cos\, t + 4 \, e^{-2t} - \int \sin\, (t-\tau)f(\tau)\, dt $$
My work so far on the problem: \begin{align} f(t) &= \cos\, t + 4 \, e^{-2t} - \int \sin\, (t-\tau)f(\tau)\, dt \\ \mathcal{L}\{f(t)\} &= \mathcal{L}\{\cos\, t\} + \mathcal{L}\{4 \, e^{-2t}\} - \mathcal{L}\left\{\int \sin\, (t-\tau)f(\tau)\, dt \right\} \\ F(s) &= \frac{s}{s^2 + 1} + \frac{4}{s+2} - F(s)\frac{1}{s^2 + 1} \\ F(s) &= \frac{s}{(s^2+2)} + \frac{4 s^2+1}{(s+2)(s^2+2)} \end{align}
I know that from here I'm supposed to take the inverse Laplace Transform to find $f(t)$ but everything I've tried on the second term (Partial fractions, expanding it out) either fails me or gives me a super ugly equation that I'm sure is false
Any and all help is appreciated.
First, the equation should read: $$ F(s) = \frac{s}{s^2+2} + 4 \, \frac{s^2+1}{(s+2)(s^2+2)} $$ which leads to $$ F(s) = \frac{s}{s^2+2} + \frac{4}{s+2} - \frac{4}{(s+2)(s^2+2)}. $$ Using partial fractions, $$ \frac{1}{(s+2)(s^2 + 2)} = \frac{A}{s+2} + \frac{B s + C}{s^2+2}$$ and leads to $1 = A \, (s^2 + 2) + (B \, s + C)(s+2) = (A + B) \, s^2 + (C + 2 B) \, s + 2 (A + C)$. From this then \begin{align} F(s) &= \frac{s}{s^2+2} + \frac{4}{s+2} - \frac{4}{(s+2)(s^2+2)} \\ &= \frac{4}{s+2} + \frac{s}{s^2 + 2} - \frac{2}{3} \, \left( \frac{1}{s+2} - \frac{s -2}{s^2 + 2} \right) \\ &= \frac{10}{3} \, \frac{1}{s+2} + \frac{5}{3} \, \frac{s}{s^2 + 2} - \frac{\sqrt{2}}{3} \, \frac{\sqrt{2}}{s^2 + 2}. \end{align} Inverting this gives $$ f(t) = \frac{10}{3} \, e^{-2 t} + \frac{5}{3} \, \cos(\sqrt{2} \, t) - \frac{\sqrt{2}}{3} \, \sin(\sqrt{2} \, t). $$