I have included a pdf attachment I made with mathematica to help try and explain my problem.
please help if you can..
Merlyn.
I assume $\displaystyle f(x) = \cos x$ ....?
Then $\displaystyle \int_0^{2 \pi} \cos^2 x \, dx = \int_0^{2 \pi} \frac{1}{2} [ \cos (2x) + 1] \, dx$
substituting from the usual double angle formula
$\displaystyle = \frac{1}{2} \int_0^{2 \pi} \cos (2x) + 1 \, dx = \pi$.
If you're studying Fourier Series it's expected you're familiar with basic techniques of integration.
Because of the symmetries of sinusoids the integral of $\displaystyle \sin^2$ or $\displaystyle \cos^2$ over a complete period is half of the area of the rectangle of height $\displaystyle 1$ and length equal to the period.
Look at your diagram and you will see that this is obvious (draw a rectangle around the plot of $\displaystyle \cos^2(x)$ from $\displaystyle 0$ to $\displaystyle 2 \pi$).
Alternatively use the double angle formula to reduce the integral to one that can be done easily and by this tedious process you will arrive at the same result.
CB