By convolution theorem

$\phi_{1}(s) = \frac{s}{s^2 + a^2}$

$f_{1}(t) = cosat$

$\phi_{2}(s) = \frac{s}{s^2 + b^2}$

$f_{2}(t) = cosbt$

$\therefore$ According to definition convolution

$f(t) = \int\limits_{0}^{t}cosau*cosb(t - u)du$

$\therefore f(t) = \int\limits_{0}^{t}\frac{1}{2}(cos(au - bt = bu) + cos(au + bt - bu))du$

$\therefore f(t) = \frac{1}{2} \bigg[\frac{sin[u(a + b) - bt]}{a + b} + \frac{sin[u(a - b) + bt]}{a - b}\bigg]_{0}^{t}$

$\therefore f(t) = \frac{1}{2}\bigg[\frac{sinat + bt - bt}{a + b} - \frac{sin(-bt)}{a + b} + \frac{sinta -tb + bt}{a - b} - \frac{sin(bt)}{a - b}\bigg]_{0}^{t}$

$\therefore f(t) = \frac{1}{2}\bigg[\frac{sinat + sinbt}{a + b} + \frac{sinat - sinbt}{a - b}\bigg]$

$\therefore f(t) = \frac{1}{2}\bigg[\frac{asinat - bsinbt + asinbt - bsinbt + asinat - asinbt - asinat + bsinat - bsinbt}{a^2 - b^2}\bigg]$

$\therefore f(t) = \frac{1}{2}\bigg[\frac{2asinat - 2bsinbt}{a^2 - b^2}\bigg]$

$\therefore f(t) = \frac{asinat - bsinbt}{a^2 - b^2}$