The 'traditional' way I always see this integral calculated is with the identity
$$\cos^2t=\frac{1+\cos(2t)}{2}$$
My alternative method uses $\cos^2t+\sin^2t=1$. It's obvious that
$$\int_0^{2\pi}\cos^2t+\sin^2t\,dt=\int_0^{2\pi}1\,dt=2\pi$$
The interval of integration is an integer multiple of the periods of each function ($\cos^2t$ has a period of $\pi$), and so it seems reasonable to me that given the Pythagorean identity used above, $\cos^2t$ and $\sin^2t$ contribute, for lack of a better word, equally to this final answer of $2\pi$ above, and so the integral in the title should be half of $2\pi$, or just $\pi$.
Is this method valid? What additional statements, if any, are necessary to make it rigorous enough to be valid?