Answer
$$\int_{1/2}^1\frac{\cos(x^{-2})}{x^3}dx=\frac{1}{2}(\sin4-\sin1)$$
Work Step by Step
To evaluate the integral
$$\int_{1/2}^1\frac{\cos(x^{-2})}{x^3}dx$$
we will use substitution $x^{-2}=t$ which gives us $-2x^{-3}dx=dt\Rightarrow\frac{dx}{x^3}=-\frac{dt}{2}$ and the integration bounds would be: for $x=1/2$ we have $t=4$ and for $x=1$ we have $t=1$. Putting this into the integral we get:
$$\int_{1/2}^1\frac{\cos(x^{-2})}{x^3}dx=\int_4^1\cos t\left(-\frac{dt}{2}\right)=
-\frac{1}{2}\int_4^1\cos t dt=\frac{1}{2}\int_1^4\cos t dt=
\left.\frac{1}{2}\sin t\right|_1^4=\frac{1}{2}(\sin4-\sin1)$$