Answer
$8 \pi \sin (1)$
Work Step by Step
Here, we have
$x=2 \cos t \implies dx=-2 \sin t dt, y= 2 \sin t \implies dy=2\cos t dt$ and $z=-1 \implies dz=(0) dt$
Plug the above values in the given integral.
$\int_{2 \pi}^{0}2 \sin (1) \sin t-2 \sin 1 \cos t +4 \cos 1 \sin t \cos t = -4 \sin (1)\int_{2 \pi}^{0} (sin^2 t+\cos^2 t) $
Thus, we have
$-4 \sin (1)\int_{2 \pi}^{0} dt=8 \pi \sin (1)$
