The integral $$\displaystyle\int_{\pi/12}^{\pi/4} \frac{8\cos 2x}{(\tan x + \cot x)^3}\,dx$$ equals:

We have to evaluate the definite integral $$I=\displaystyle\int_{\pi/12}^{\pi/4}\frac{8\cos2x}{(\tan x+\cot x)^3}\,dx.$$

First we simplify the denominator. The well-known trigonometric identity is $$\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin^2x+\cos^2x}{\sin x\cos x}=\frac{1}{\sin x\cos x}.$$ Using $$\sin2x=2\sin x\cos x,$$ the last fraction becomes $$\frac{1}{\sin x\cos x}=\frac{2}{\sin2x}.$$ So $$\tan x+\cot x=\frac{2}{\sin2x}.$$

Now we raise this to the third power: $$(\tan x+\cot x)^3=\left(\frac{2}{\sin2x}\right)^3=\frac{8}{(\sin2x)^3}.$$

Substituting this into the original integrand gives $$\frac{8\cos2x}{(\tan x+\cot x)^3}=\frac{8\cos2x}{\dfrac{8}{(\sin2x)^3}}=\cos2x\;(\sin2x)^3.$$

Thus the integral becomes $$I=\int_{\pi/12}^{\pi/4}(\sin2x)^3\cos2x\,dx.$$

Now we perform a simple substitution. Let $$u=\sin2x.$$ Then $$\frac{du}{dx}=2\cos2x\quad\Longrightarrow\quad\cos2x\,dx=\frac{du}{2}.$$

Expressing the integral in terms of $$u$$, we get $$I=\int_{x=\pi/12}^{x=\pi/4}u^3\left(\frac{du}{2}\right)=\frac{1}{2}\int u^3\,du.$$

The limits also change. For the lower limit $$x=\pi/12,$$ we have $$u=\sin\left(2\cdot\frac{\pi}{12}\right)=\sin\frac{\pi}{6}=\frac{1}{2}.$$ For the upper limit $$x=\pi/4,$$ we have $$u=\sin\left(2\cdot\frac{\pi}{4}\right)=\sin\frac{\pi}{2}=1.$$ Hence

$$I=\frac{1}{2}\int_{1/2}^{1}u^3\,du.$$

We integrate using the power formula $$\displaystyle\int u^n\,du=\frac{u^{\,n+1}}{n+1}+C.$$ So

$$I=\frac{1}{2}\left[\frac{u^4}{4}\right]_{1/2}^{1}=\frac{1}{8}\left(u^4\right)\Bigl|_{1/2}^{1}.$$

Substituting the limits, $$I=\frac{1}{8}\left(1^4-\left(\frac{1}{2}\right)^4\right)=\frac{1}{8}\left(1-\frac{1}{16}\right)=\frac{1}{8}\cdot\frac{15}{16}=\frac{15}{128}.$$

Hence, the correct answer is Option D.