Solution:

By method of induction,

$L[f(t)] = f(s) = \int_0^{\infty{}}e^{-st}\ f(t)\ dt$

Differentiating both sides with respect to s and applying the rule of differentiation under integral sign,

$\displaystyle f'(s) = \int_0^{\infty{}}\frac{\partial{}}{\partial{}s}\ \left[e^{-st}\ f(t)\ \right]dt = -\int_0^{\infty{}}e^{-st}\ tf(t)\ dt = L[t f(t)]$

$\therefore L[t f(t)] = (-1)\ \dfrac{d}{\ ds} f(s)$

Thus the rule is true for n=1

Now assume that the rule is true for n = m and prove that it is true for n = m + 1,

$L[ t^mf(t)] = {(-1)}^m\ \dfrac{d^m}{{ds}^m} f(s) = \int_0^{\infty{}}e^{-st}\ t^m\ f(t)\ dt$

Differentiating both sides with respect to s and applying the rule of differentiation under integral sign,

$\displaystyle {(-1)}^m\ \frac{d^{m+1}}{{ds}^m+1}f(s) = \int_0^{\infty{}}\ \frac{\partial{}}{\partial{}s}\left[e^{-st}\ t^m\ f(t)\ dt\right]$

$=-\int_0^{\infty{}}e^{-st}\ t^{m+1}f(t)\ dt$

$-L[ t^{m+1}f(t)]$

$L[t^{m+1}f(t)] = {(-1)}^{m+1}\ \dfrac{d^{m+1}}{{ds}^m+1} f(s)$

Thus the property is true for n = m+ 1

Hence it is true for any value of n.

$L[ t^nf(t)] = {(-1)}^n\ \dfrac{d^n}{{ds}^n}f(s)$

$f(t) = t \cos^2t$

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

$\dfrac{1}{2}L[t] + \dfrac{1}{2}L[t\ cos{2t}]$

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

$L[t \cos{2t}] = (-1)\dfrac{d}{ds} L[\cos{2t}]$

$=(-1)\dfrac{d}{ds}\left(\dfrac{s}{s^2+\ 2^2}\right)$

$=-\left[\dfrac{s^2+\ 2^2\ –\ s.2s}{{\left(s^2+\ 2^2\right)}^2}\right]$

$= +\dfrac{s^2-\ 2^2\ }{{\left(s^2+\ 2^2\right)}^2}$

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