Answer
$$\frac{1}{2}\left(\frac{1}{2}\sin \frac{2}{t}-\frac{1}{t} \right)+c$$
Work Step by Step
Given $$ \int \frac{\sin^2(1/t)}{t^2}dt $$
Let $x=\dfrac{1}{t}\ \to\ dx=\dfrac{-dt}{t^2}$, then
\begin{align*}
\int \frac{\sin^2(1/t)}{t^2}dt &=-\int \sin^2 xdx\\
&=\frac{-1}{2}\int (1-\cos 2x)dx\\
&=\frac{-1}{2}\left(x-\frac{1}{2}\sin 2x\right)+c\\
&=\frac{1}{2}\left(\frac{1}{2}\sin \frac{2}{t}-\frac{1}{t} \right)+c
\end{align*}
