Answer
$$\frac{1}{4}\cot 8x + C$$
Work Step by Step
$$\eqalign{
& - \int {2{{\csc }^2}8x} dx \cr
& {\text{set }}u = 8x{\text{ then }}\frac{{du}}{{dx}} = 8,\,\,\,\,\,\,\,\,\frac{{du}}{8} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& - \int {2{{\csc }^2}8x} dx = - \int {2{{\csc }^2}u} \left( {\frac{{du}}{8}} \right) \cr
& = - \int {{{\csc }^2}u} \left( {\frac{{du}}{4}} \right) \cr
& = - \frac{1}{4}\int {{{\csc }^2}u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {{{\csc }^2}x} dx = - \cot x + C \cr
& = - \frac{1}{4}\left( { - \cot u} \right) + C \cr
& = \frac{1}{4}\cot u + C \cr
& {\text{write in terms of }}x\cr
& = \frac{1}{4}\cot 8x + C \cr} $$
