The integral ∫ π 12 π 4 8 cos 2 x tan x + cot x 3 d x equals

Given

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

$=\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}\frac{\cos 2x}{{\left(\frac{1}{\sin 2x}\right)}^3}=\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}\cos 2x·\sin 2x·{\sin }^2\left(2x\right)dx$

$=\frac{1}{4}\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}\sin 4x·(1-\cos 4x)dx$

$=\frac{1}{4}\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}\sin 4x-\frac{1}{8}\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}\sin 8x$

$=-\frac{1}{16}{\left[\cos 4x\right]}_{\frac{\pi }{12}}^{\frac{\pi }{4}}+\frac{1}{8\times 8}{\left[\cos 8x\right]}_{\frac{\pi }{12}}^{\frac{\pi }{4}}$

$=-\frac{1}{16}\left[-1-\frac{1}{2}\right]+\frac{1}{64}\left[1+\frac{1}{2}\right]$

$=\frac{15}{128}$.