## Thomas' Calculus 13th Edition

$\displaystyle \frac{1}{2}-\frac{1}{4}\sin 2$
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: $=\displaystyle \frac{1}{2}-\frac{1}{4}\sin 2$