Answer
$\frac{1}{4}\sin \left( {\frac{2}{t}} \right) - \frac{1}{{2t}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sin }^2}\left( {1/t} \right)}}{{{t^2}}}} dt \cr
& {\text{Use the identity si}}{{\text{n}}^2}\theta = \frac{{1 - \cos 2\theta }}{2},{\text{ Let }}\theta = \frac{1}{t} \cr
& \int {\frac{{{{\sin }^2}\left( {1/t} \right)}}{{{t^2}}}} dt = \int {\frac{{1 - \cos 2\left( {1/t} \right)}}{2}\left( {\frac{1}{{{t^2}}}} \right)} dt \cr
& {\text{Rewrite the integrand}} \cr
& = \frac{1}{2}\int {\left( {1 - \cos \left( {\frac{2}{t}} \right)} \right)\left( {\frac{1}{{{t^2}}}} \right)} dt \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \frac{2}{t},{\text{ }}du = - \frac{2}{{{t^2}}}dt,{\text{ }} - \frac{1}{2}du = \frac{1}{{{t^2}}}dt \cr
& {\text{Substituting}} \cr
& = \frac{1}{2}\int {\left( {1 - \cos u} \right)\left( { - \frac{1}{2}} \right)} du \cr
& = \frac{1}{4}\int {\left( {\cos u - 1} \right)} du \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\left( {\sin u - u} \right) + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}\frac{2}{t}{\text{ for }}u \cr
& = \frac{1}{4}\left( {\sin \left( {\frac{2}{t}} \right) - \frac{2}{t}} \right) + C \cr
& = \frac{1}{4}\sin \left( {\frac{2}{t}} \right) - \frac{1}{{2t}} + C \cr} $$
