$\displaystyle \cos^2(2 \pi f_0 t) = (\cos(4 \pi f_0 t)+1)/2$
then as $\displaystyle \cos$ is symetrical in all sorts of ways the right hand side will integrate up to half the area eonclosed in the rectangle enclosing the curve, which has height $\displaystyle 1$ and length $\displaystyle T_0$.
(Draw a picture of the curve, it will then be obvious)
CB