$(1$ point$)$

$L\{cos\}(s)={\displaystyle \underset{0}{\overset{}{}}}{e}^{-st}cos(at)dt$

$t=dv=$

$du=v=$

$={|}_0^{}-{\displaystyle \underset{0}{\overset{}{}}}||$

u $dv=$

du $v=$

$=-\frac{a^2}{s^2}{\displaystyle \underset{0}{\overset{}{}}}$

Notice that this last integral is the same as the integral we started with. Treat this like an equation and move the integral to the left hand side:

$$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{-st}cos(at)dt=\frac{1}{s}$

Therefore

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{e}^{-t}cos(at)dt=$

$$