Answer
$$-\frac{1}{10} \cot ^{5} 2 x+\frac{1}{6}\cot^3{2x}-\frac{1}{2} \cot 2 x-x+C$$
Work Step by Step
We integrate as follows:
\begin{align*}
\int \cot ^{6} 2 x d x&=\int \cot ^{4} 2 x \cot ^{2} 2 x d x\\
&=\int \cot ^{4} 2 x\left(\csc ^{2} 2 x-1\right) d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{4} 2 x d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{2} 2 x \cot ^{2} 2 x d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{2} 2 x\left(\csc ^{2} 2 x-1\right) d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{2} 2 x \csc ^{2} 2 x d x+\int \cot ^{2} 2 x d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{2} 2 x \csc ^{2} 2 x d x+\int\left(\cot ^{2} 2 x-1\right) d x\\
&=\int \cot ^{4} 2 x \csc ^{2} 2 x d x-\int \cot ^{2} 2 x \csc ^{2} 2 x d x+\int \csc ^{2} 2 x d x-\int d x\\
&=-\frac{1}{10} \cot ^{5} 2 x+\frac{1}{6}\cot^3{2x}-\frac{1}{2} \cot 2 x-x+C
\end{align*}
