Answer
$\pi \sqrt 5$
Work Step by Step
$I=\int_{0}^{\pi} (2 \sin t) (t) (-2 \cos t) \times \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}dt=\int_{0}^{\pi} (2 \sin t) (t) (-2 \cos t) \times \sqrt {{(2 \cos t)^{2}}+1^2+(2 \sin t)^2}dt=\int_{0}^{\pi} (2 \sin t) (t) (-2 \cos t) \times \sqrt 5 dt$
$=-2 \sqrt 5 \int_{0}^{\pi} t \sin (2t) dt$
$=-2 \sqrt 5[t \int \sin (2t) dt-\int [\dfrac{dt}{dt} \int \sin 2t dt]dt]_0^{\pi}$
$=-2 \sqrt 5[\dfrac{-t \cos 2t}{2}+\dfrac{\sin 2t}{4}]_0^{\pi}$
$=-2 \sqrt 5[-\dfrac{\pi \cos 2 \pi}{2}+\dfrac{\sin 2\pi}{4}]+2 \sqrt 5[-\dfrac{(0) \cos (0)}{2}+\dfrac{\sin (0)}{4}]$
$= -2 \sqrt 5(\dfrac{-\pi}{2}$
$=\pi \sqrt 5$
