Answer
$\frac{\sin(\frac{2}{t})-\frac{2}{t}}{4}+C$
Work Step by Step
$\int\frac{\sin^2(1/t)}{t^2}dt$
Let $u=\frac{1}{t}=t^{-1}$. Then $du=-t^{-2}dt=-\frac{1}{t^2}dt$, and $\frac{1}{t^2}dt=-du$.
$=\int\sin^2 u*(-1)du$
$=-\int\sin^2 u\ du$
Use the half-angle identity $\sin^2 u=\frac{1}{2}(1-\cos 2u)$.
$=-\int\frac{1}{2}(1-\cos 2u)du$
$=-\frac{1}{2}\int(1-\cos 2u)du$
Let $v=2u$. Then $dv=2du$, and $du=\frac{1}{2}dv$.
$=-\frac{1}{2}\int(1-\cos v)*\frac{1}{2}dv$
$=-\frac{1}{4}(v-\sin v+C)$
$=\frac{\sin v-v}{4}+C$
Substitute back $v=2u$ and then $u=\frac{1}{t}$.
$=\frac{\sin 2u-2u}{4}+C$
$=\boxed{\frac{\sin(\frac{2}{t})-\frac{2}{t}}{4}+C}$
