Answer
$$\pi - \frac{{\sin 4\pi k}}{{4k}}$$
Work Step by Step
$$\eqalign{
& \int_0^{2\pi } {{{\sin }^2}kx} dx \cr
& {\text{identity si}}{{\text{n}}^2}\theta = \frac{{1 - \cos 2\theta }}{2} \cr
& = \int_0^{2\pi } {\frac{{1 - \cos 2kx}}{2}} dx \cr
& = \int_0^{2\pi } {\left( {\frac{1}{2} - \frac{{\cos 2kx}}{2}} \right)} dx \cr
& {\text{find antiderivatives}} \cr
& = \left[ {\frac{x}{2} - \frac{{\sin 2kx}}{{2\left( {2k} \right)}}} \right]_0^{2\pi } \cr
& = \left[ {\frac{x}{2} - \frac{{\sin 2kx}}{{4k}}} \right]_0^{2\pi } \cr
& {\text{fundamental theorem of calculus}} \cr
& = \left[ {\frac{{2\pi }}{2} - \frac{{\sin 4\pi k}}{{4k}}} \right] - \left[ {\frac{0}{2} - \frac{{\sin 2k\left( 0 \right)}}{{4k}}} \right] \cr
& {\text{simplifying}} \cr
& = \left[ {\frac{{2\pi }}{2} - \frac{{\sin 4\pi k}}{{4k}}} \right] - \left[ 0 \right] \cr
& = \pi - \frac{{\sin 4\pi k}}{{4k}} \cr} $$
