Problem 03 | Division by t

$f(t) = \dfrac{\sin^2 t}{t}$

$f(t) = \dfrac{\frac{1}{2}(1 - \cos 2t)}{t}$

$f(t) = \dfrac{1}{2} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left( \dfrac{1}{t} \right) - \dfrac{1}{2} \left( \dfrac{\cos 2t}{t} \right)$
 

Since
$\mathcal{L} (1) = \dfrac{1}{s}$   and

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
 

Then,
$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \int_s^\infty \dfrac{du}{u} - \dfrac{1}{2} \int_s^\infty \dfrac{u \, du}{u^2 + 4}$

$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \int_s^\infty \dfrac{du}{u} - \dfrac{1}{4} \int_s^\infty \dfrac{2u \, du}{u^2 + 4}$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \Big[ \ln u \Big]_s^\infty - \dfrac{1}{4} \Big[ \ln (u^2 + 4) \Big]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \left[ \dfrac{2\ln u - \ln (u^2 + 4)}{4} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \left[ \dfrac{\ln u^2 - \ln (u^2 + 4)}{4} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln \dfrac{u^2}{u^2 + 4} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln \dfrac{\dfrac{u^2}{u^2}}{\dfrac{u^2}{u^2} + \dfrac{4}{u^2}} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln \dfrac{1}{1 + \dfrac{4}{u^2}} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln \dfrac{1}{1 + \dfrac{4}{\infty^2}} - \ln \dfrac{1}{1 + \dfrac{4}{s^2}} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln \dfrac{1}{1 + 0} - \ln \dfrac{1}{\dfrac{s^2 + 4}{s^2}} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ \ln 1 - \ln \dfrac{s^2}{s^2 + 4} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ 0 - \ln \dfrac{s^2}{s^2 + 4} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \left[ -\ln \dfrac{s^2}{s^2 + 4} \right]$

$\mathcal{L} \left\{ f(t) \right\} = -\frac{1}{4} [ \, \ln s^2 - \ln (s^2 + 4) \, ]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} [ \, \ln (s^2 + 4) - \ln s^2 \, ]$

$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{4} \ln \left( \dfrac{s^2 + 4}{s^2} \right)$           answer