$y=\sqrt{\cot 2x} \;$ revolved around x-axis between $x = \frac{\pi }{12} \;$ and $x=\frac{\pi }{4} \;$ find volume The region between the curve $y=\sqrt{\cot 2x} \;\;$ and the x-axis, from $x=\frac{\pi }{12} \;\;$ to $x=\frac{\pi }{4} \;\;$ is revolved around the x-axis to generate a solid. Find the volume of the solid.
I tried finding critical points, by taking derivative, but it seems y' ≠ 0
so I just plugged in $\frac{\pi }{12}$  and $\frac{\pi }{4}$ into $\sqrt{\cot 2x}$ and it seems that at $\frac{\pi }{12}$ function has the maximum y-value of $3^\frac{1}{4}$.
My radius will vary, r = $(3^\frac{1}{4} - \sqrt{\cot2x})$ 
Area = $\pi \cdot r^2$
Volume = $\pi \cdot r^2 \cdot dx$
Integration: 
$$ \int^{3^\frac{1}{4}}_{0} \pi (\sqrt{3} - 2\cdot3^\frac{1}{4} \cdot \sqrt{\cot2x} + \cot2x) $$
but wolfram can't integrate it...
is it even correct?

HINT: The volume of the solid generated by rotating the curve $y=f(x)$ about the x-axis between $x=a$ & $x=b$ is $$=\color{red}{\int_{a}^{b}\pi y^2dx}$$
