Calculus: Early Transcendentals 8th Edition

$\frac{\sin(\frac{2}{t})-\frac{2}{t}}{4}+C$
$\int\frac{\sin^2(1/t)}{t^2}dt$ 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 $u=\frac{1}{t}$. $=\frac{\sin 2u-2u}{4}+C$ $=\boxed{\frac{\sin(\frac{2}{t})-\frac{2}{t}}{4}+C}$