Answer
$$\int^{-1/2}_{-1}t^{-2}\sin^2\Big(1+\frac{1}{t}\Big)dt=\frac{1}{2}-\frac{\sin2}{4}$$
Work Step by Step
$$A=\int^{-1/2}_{-1}t^{-2}\sin^2\Big(1+\frac{1}{t}\Big)dt$$
We set $u=1+\frac{1}{t}$, which means $$du=-\frac{1}{t^2}dt=-t^{-2}dt$$ $$t^{-2}dt=-du$$
For $t=-1/2$, we have $$u=1+\frac{1}{-1/2}=1-2=-1$$
For $t=-1$, we have $$u=1+\frac{1}{-1}=1-1=0$$
Therefore, $$A=-\int^{-1}_0\sin^2udu$$
Recall the identity: $$\sin^2x=\frac{1-\cos2x}{2}$$
So, $$A=-\int^{-1}_0\frac{1-\cos2u}{2}du$$ $$A=-\Big(\int^{-1}_0\frac{1}{2}du-\frac{1}{2}\int^{-1}_0\cos2udu\Big)$$ $$A=-\Big(\frac{u}{2}\Big]^{-1}_0-\frac{1}{2}\times\frac{1}{2}\sin2u\Big]^{-1}_0\Big)$$ $$A=-\Big(-\frac{1}{2}-\frac{1}{4}(\sin(-2)-\sin0)\Big)$$ $$A=-\Big(-\frac{1}{2}-\frac{1}{4}(\sin(-2)-0)\Big)$$ $$A=\frac{1}{2}+\frac{\sin(-2)}{4}=\frac{1}{2}-\frac{\sin2}{4}$$
