$\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{(\tan x+\cot x)}^3}dx=$

$\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{(\tan x+\cot x)}^3}dx$

$=\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{\left(2\cos ec2x\right)}^3}dx$

$$\begin{gathered} =\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{8{\left(\cos ec2x\right)}^3}dx \\ =\underset{\pi /12}{\overset{\pi /4}{\int }}\cos 2x{\left(\sin 2x\right)}^{3}dx \end{gathered}$$

$\begin{array}{l}Put\sin 2x=t\\ 2\cos 2x dx=dt\\ x=\frac{\pi }{12}\Rightarrow t=\frac{1}{2}\\ x=\frac{\pi }{4}\Rightarrow t=1\end{array}$

$\begin{array}{l}=\underset{1/2}{\overset{1}{\int }}\cos 2x.t^3.\frac{dt}{2\cos 2x}\\ =\underset{1/2}{\overset{1}{\int }}\frac{t^3}{2}dt=\frac{1}{2}{\left[\frac{t^4}{4}\right]}_{1/2}^1=\frac{1}{2}\left[\frac{1}{4}-\frac{1}{64}\right]\\ =\frac{1}{8}-\frac{1}{128}=\frac{15}{128}\end{array}$