Answer
$2 \pi^2$
Work Step by Step
Here, we have $W=\int_C F\cdot dr$
$=\int_0^{2 \pi} [(t-\sin t) i+(3-\cos t)j) \cdot ((1-\cos t) i+\sin t j) dt$
$=\int_0^{2 \pi} t-\sin t-t \cos t+\cos t \sin t+3\sin t-\sin t \cos t dt$
$=\int_0^{2 \pi} t-t \cos t+2 \sin t dt$
$=\int_0^{2 \pi} -t \cos t+[\dfrac{t^2}{2}-2 \cos t ]_0^{2 \pi}$
$=[\int t \cos t dt]_0^{2 \pi}+2 \pi^2$
$W=\int_C F\cdot dr=[\int t \cos t dt]_0^{2 \pi}+2 \pi^2$
Solve $I=[\int t \cos t dt]_0^{2 \pi}=[ t \int\cos t dt-(\int \dfrac{dt}{dt} \int \cos t dt) dt]_0^{2 \pi}$
$=-0+2 \pi^2 $
$=2 \pi^2$
