Answer
$$ - \frac{1}{3}\csc 3t + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cot }^2}3t\sec 3t} dt \cr
& {\text{Use the identities }}\cot x = \frac{{\cos x}}{{\sin x}},\,\,\,\,\sec x = \frac{1}{{\cos x}} \cr
& = \int {{{\left( {\frac{{\cos 3t}}{{\sin 3t}}} \right)}^2}\left( {\frac{1}{{\cos 3t}}} \right)} dt \cr
& {\text{Simplifying}} \cr
& = \int {\frac{{{{\cos }^2}3t}}{{{{\sin }^2}3t}}\left( {\frac{1}{{\cos 3t}}} \right)} dt \cr
& = \int {\frac{{\cos 3t}}{{{{\sin }^2}3t}}} dt \cr
& \cr
& {\text{Use the substitution method}}{\text{, }}u = \sin 3t,\,\,\,du = 3\cos 3tdt \cr
& = \int {\frac{{\cos 3t}}{{{u^2}}}\left( {\frac{{du}}{{3\cos 3t}}} \right)} \cr
& = \frac{1}{3}\int {\frac{{du}}{{{u^2}}}} \cr
& {\text{Integrate}} \cr
& = - \frac{1}{{3u}} + C \cr
& {\text{Write in terms of }}t \cr
& = - \frac{1}{{3\sin 3t}} + C \cr
& = - \frac{1}{3}\csc 3t + C \cr} $$
