Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_{\pi /16}^{\pi /8} {8{{\csc }^2}4x} dx \cr
& = \int_{\pi /16}^{\pi /8} {2{{\csc }^2}4x} \left( 4 \right)dx \cr
& {\text{take out the constant 2}} \cr
& = 2\int_{\pi /16}^{\pi /8} {{{\csc }^2}4x} \left( 4 \right)dx \cr
& {\text{use }}\int {{{\csc }^2}udu} = - \cot u + C \cr
& = 2\left( { - \cot 4x} \right)_{\pi /16}^{\pi /8} \cr
& {\text{using The Fundamental Theorem}} \cr
& = - 2\left( {\cot 4\left( {\frac{\pi }{8}} \right) - \cot 4\left( {\frac{\pi }{{16}}} \right)} \right) \cr
& {\text{simplify}} \cr
& = - 2\left( {\cot \left( {\frac{\pi }{2}} \right) - \cot \left( {\frac{\pi }{4}} \right)} \right) \cr
& = - 2\left( {0 - 1} \right) \cr
& = 2 \cr} $$
