• We know,$L[\cos{at}]=\dfrac{s}{s^2+a^2}$

$\therefore L[\cos{t}]=\dfrac{s}{s^2+1^2}$

• By effect of multiplication by 't',

$L[t\cos{t}]=(-1)^1.\dfrac{\mathrm{d}\bigg[\dfrac{s}{s^2+1}\bigg]}{\mathrm{d}s}$

$\therefore L[t\cos{t}]=\dfrac{(s^2+1)(1)-(s)(2s)}{(s^2+1)^2}$

$\therefore L[t\cos{t}]=\dfrac{s^2+1-2s^2}{(s^2+1)^2}$

$\therefore L[t\cos{t}]=\dfrac{1-s^2}{(1+s^2)^2}$

• Now, substituting s=3, we get,

$L[t\cos{t}]=\dfrac{1-(3)^2}{[1+(3)^2]^2}$

$\therefore L[t\cos{t}]=\dfrac{-2}{25}$

$\therefore \int_{0}^{\infty}e^{-3t}\ t\cos{t}\ \mathrm dt=\dfrac{-2}{25}$