All trigo together

$\begin{aligned} I & = \int_2^{2018} \left(\color{#3D99F6}\csc(\sec^{-1} x^2) \color{#D61F06} \cos (\tan^{-1}(\sin (\cot^{-1}x)))\right)^2 dx \\ & = \int_2^{2018} \left({\color{#3D99F6}\frac {x^2}{\sqrt{x^4-1}}} \times \color{#D61F06} \cos \left(\tan^{-1}\left(\frac 1{\sqrt{1+x^2}}\right)\right)\right)^2 dx \\ & = \int_2^{2018} \left({\color{#3D99F6}\frac {x^2}{\sqrt{x^4-1}}} \times \color{#D61F06} \sqrt{\frac {1+x^2}{2+x^2}} \right)^2 dx \\ & = \int_2^{2018} \left(\frac {x^2}{\sqrt{(x^2-1)(x^2+2)}} \right)^2 dx \\ & = \int_2^{2018} \frac {x^4}{(x^2-1)(x^2+2)} dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \int_2^{2018} \left(1 - \frac 4{3(x^2+2)} + \frac 1{6(x-1)} - \frac 1{6(x+1)} \right) dx \\ & = x - \frac {2\sqrt 2}3 \tan^{-1} \frac x{\sqrt 2} + \frac 16 \ln \left(\frac {x-1}{x+1}\right) \bigg|_2^{2018} \\ & \approx 2015.603 \end{aligned}$

Therefore, $\lfloor I \rfloor = \boxed{2015}$ .

You can consider a triangle where cot€= x, thus arccotx = €, then find sin of that angle and work your way down. Do the same for x^2 and the entire expression should get reduced to x^4/(x^2-1)(x^2+2), whose integration from 2 to 2018 would give the answer as 2015.6