Answer
$2 \pi^2$
Work Step by Step
Here, we have the work done:
$W=\int_C F\cdot dr=\int_0^{2 \pi} [(t-\sin t) i+(3-\cos t)j) \cdot ((1-\cos t) i+\sin t j) dt$
This implies that:
$\int_0^{2 \pi} t-\sin t-t \cos t+\cos t \sin t+3\sin t-\sin t \cos t dt=\int_0^{2 \pi} (t-t \cos t+2 \sin t) dt$
and $[\int t \cos t dt]_0^{2 \pi}+2 \pi^2=[ t \int\cos t dt-(\int \dfrac{dt}{dt} \int \cos t dt) dt]_0^{2 \pi}+2 \pi^2$
Hence, we have: $W=0+2 \pi^2 =2 \pi^2$
