#### Answer

$$ - \frac{1}{6}cot6x + C$$

#### Work Step by Step

$$\eqalign{
& \int {{{\csc }^2}6x} dx \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& \int {{{\csc }^2}ax} dx = - \frac{1}{a}cotax + C \cr
& {\text{letting }}a = 6,{\text{ we have}} \cr
& = - \frac{1}{6}cot6x + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( { - \frac{1}{6}cot6x + C} \right) \cr
& {\text{ = }}\frac{d}{{dx}}\left( { - \frac{1}{6}cot6x} \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{ = }} - \frac{1}{6}\left( { - {{\csc }^2}6x} \right)\left( 6 \right) + 0 \cr
& = {\csc ^2}6x \cr} $$