The integral $$\int_{\pi/6}^{\pi/4} \frac{dx}{\sin 2x(\tan^5 x + \cot^5 x)}$$ equals:

We have to evaluate

$$I=\displaystyle\int_{\pi/6}^{\pi/4}\frac{dx}{\sin 2x\bigl(\tan^{5}x+\cot^{5}x\bigr)}.$$

First we express every trigonometric quantity in terms of $$\tan x$$. The double-angle identity gives $$\sin 2x=2\sin x\cos x$$ and, by definition, $$\tan x=\dfrac{\sin x}{\cos x},\quad\cot x=\dfrac{\cos x}{\sin x}.$$ Putting these together,

$$\sin 2x=2\sin x\cos x=\frac{2\tan x}{1+\tan^{2}x}.$$

Now set $$t=\tan x.$$ Then $$dt=(1+\tan^{2}x)\,dx=(1+t^{2})\,dx,$$ so $$dx=\dfrac{dt}{1+t^{2}}.$$ The limits change as follows:

$$$ x=\frac{\pi}{6}\;\Longrightarrow\;t=\tan\! \frac{\pi}{6}=\frac1{\sqrt3},\qquad x=\frac{\pi}{4}\;\Longrightarrow\;t=\tan\! \frac{\pi}{4}=1. $$$

Substituting $$t,\;dt$$ and $$\sin 2x$$ into the integral, we obtain

$$$ \begin{aligned} I&=\int_{1/\sqrt3}^{1} \frac{\displaystyle\frac{dt}{1+t^{2}}} {\displaystyle\frac{2t}{1+t^{2}}\;\bigl(t^{5}+t^{-5}\bigr)} =\int_{1/\sqrt3}^{1}\frac{dt}{2t\bigl(t^{5}+t^{-5}\bigr)}.\\[4pt] \end{aligned} $$$

Multiplying numerator and denominator by $$t^{4}$$ (to clear the negative power) gives

$$ I=\frac12\int_{1/\sqrt3}^{1}\frac{t^{4}\,dt}{t^{10}+1}. $$

At this stage the integrand is a simple rational function in $$t$$ whose denominator is a quadratic in $$t^{5}$$. We make the direct substitution

$$u=t^{5}\quad\Longrightarrow\quad du=5t^{4}\,dt\;\;\Longrightarrow\;\;t^{4}dt=\frac{du}{5}.$$

Under this substitution the limits transform to

$$$ t=1\;\Longrightarrow\;u=1^{5}=1,\qquad t=\frac1{\sqrt3}\;\Longrightarrow\;u=\Bigl(\frac1{\sqrt3}\Bigr)^{5} =\frac1{(\sqrt3)^{5}} =\frac1{9\sqrt3}. $$$

Replacing $$t^{4}dt$$ and the limits we get

$$$ \begin{aligned} I&=\frac12\int_{u=1/(9\sqrt3)}^{1}\frac{\dfrac{du}{5}}{u^{2}+1} =\frac1{10}\int_{1/(9\sqrt3)}^{1}\frac{du}{1+u^{2}}. \end{aligned} $$$

We recall and state the standard formula

$$\displaystyle\int\frac{du}{1+u^{2}}=\tan^{-1}u+\text{constant}.$$

Applying this formula,

$$$ I=\frac1{10}\bigl[\tan^{-1}u\bigr]_{\,u=\;\frac1{9\sqrt3}}^{\,u=1} =\frac1{10}\left(\tan^{-1}1-\tan^{-1}\!\frac1{9\sqrt3}\right). $$$

Because $$\tan^{-1}1=\dfrac{\pi}{4},$$ the value finally becomes

$$ I=\frac1{10}\left(\frac{\pi}{4}-\tan^{-1}\!\frac1{9\sqrt3}\right). $$

Hence, the correct answer is Option 2.