Integrating over the unit circle

Parametrization. Let $z=e^{i\theta}$, $\theta\in[0,2\pi]$.

Trigonometric. $\cos^2\theta=\frac{1+\cos2\theta}{2}$, so
$$\int_0^{2\pi}\cos^2\theta\,d\theta=\int_0^{2\pi}\frac{1+\cos2\theta}{2}\,d\theta=\frac{1}{2}\cdot2\pi+\frac12\int_0^{2\pi}\cos2\theta\,d\theta=\pi+0=\pi.$$

Complex approach. $\cos\theta=(z+z^{-1})/2$ with $z=e^{i\theta}$, $d\theta=dz/(iz)$. Then
$$\int_0^{2\pi}\cos^2\theta\,d\theta=\oint_{|z|=1}\frac{(z+z^{-1})^2}{4}\cdot\frac{dz}{iz}.$$
The integral evaluates to $\pi$ by the residue theorem.

Conclusion. $\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$.