Answer
$\displaystyle \frac{1}{2}-\frac{1}{4}\sin 2$
Work Step by Step
Applying th.7, let $u=g(t)=1+\displaystyle \frac{1}{t}=1+t^{-1} \quad$ (a continuous function).
Then, $g'(t)=-t^{-2}$
Then, $\quad \left[\begin{array}{ll}
u=1+t^{-1} & du=g'(t)dt\\
& du=-t^{-2}dt \\
& t^{-2}dt=-du
\end{array}\right]$,
and, the borders change to
$g(-1)=1-1=0$ and $\displaystyle \quad g(-\frac{1}{2})=1-2=-1$
$\displaystyle \int_{-1}^{-1/2}t^{-2}\sin^{2}(1+\frac{1}{t})dt=-\int_{0}^{-1}\sin^{2}udu$
Using the double angle formula for $\cos 2u$
$\cos 2u=\cos^{2}u-\sin^{2}u=-(1-\sin^{2}u)-\sin^{2}u=1-2\sin^{2}u$,
We transform $\displaystyle \sin^{2}u= \frac{1}{2}(1-\cos 2u) $
$=- \displaystyle \frac{1}{2}\int_{0}^{-1}(1-\cos 2u)du$
$=- \displaystyle \frac{1}{2} \int_{0}^{-2}dt +\frac{1}{2}\int_{0}^{-1}\cos 2udu$
For the second integral, substitute: $\left[\begin{array}{ll}
t=2u, & dt=2du\\
& du=dt/2 \\
u=0\rightarrow t=0, & u=-1\rightarrow t=-2
\end{array}\right]$
$=- \displaystyle \frac{1}{2} [ \int_{0}^{-1}dt +\frac{1}{2}\int_{0}^{-2}\frac{\cos tdt}{2}]$
$=-\displaystyle \frac{1}{2}\left[ t \right]_{0}^{-1}+\frac{1}{4}\int_{0}^{-2}\cos tdt$
$=-\displaystyle \frac{1}{2}(-1)+\frac{1}{4}\left[ \sin t \right]_{0}^{-2}$
$=\displaystyle \frac{1}{2}+\frac{1}{4}[\sin(-2)-\sin 0]$
$=\displaystyle \frac{1}{2}+\frac{1}{4}\sin(-2)$
sine is an odd function, so:
$=\displaystyle \frac{1}{2}-\frac{1}{4}\sin 2$
