Answer
$\sqrt {5}[\dfrac{(2 \pi)^3}{3}+2\pi]$ or, $2 \pi\sqrt {5} [1+\dfrac{(2 \pi)^2}{3}]$
Work Step by Step
$I=\int_{0}^{2 \pi} (t^2+\cos^2 2t+\sin^2 2t) \times \sqrt{{{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}+(\dfrac{dz}{dt})^{2}}dt$
or, $=\int_{0}^{2 \pi} (t^2+\cos^2 2t+\sin^2 2t) \times \sqrt{{{(1)^{2}}+{(-2 \sin 2t)^{2}}}+(2 \cos 2t)^{2}}dt$
or, $=(\sqrt {5}) \times \int_{0}^{2 \pi} (t^2+\cos^2 2t+\sin^2 2t) dt$
Hence, we have $I=(\sqrt {5}) \times [\dfrac{(2 \pi)^3}{3}+2\pi]=2 \pi\sqrt {5} [1+\dfrac{(2 \pi)^2}{3}]$
