• December 10th 2008, 11:00 PM
mercou
I have included a pdf attachment I made with mathematica to help try and explain my problem.

Merlyn.
• December 11th 2008, 02:22 AM
mr fantastic
Quote:

Originally Posted by mercou
I have included a pdf attachment I made with mathematica to help try and explain my problem.

Merlyn.

I assume $f(x) = \cos x$ ....?

Then $\int_0^{2 \pi} \cos^2 x \, dx = \int_0^{2 \pi} \frac{1}{2} [ \cos (2x) + 1] \, dx$

substituting from the usual double angle formula

$= \frac{1}{2} \int_0^{2 \pi} \cos (2x) + 1 \, dx = \pi$.

If you're studying Fourier Series it's expected you're familiar with basic techniques of integration.
• December 11th 2008, 03:47 AM
CaptainBlack
Quote:

Originally Posted by mercou
I have included a pdf attachment I made with mathematica to help try and explain my problem.

Because of the symmetries of sinusoids the integral of $\sin^2$ or $\cos^2$ over a complete period is half of the area of the rectangle of height $1$ 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 $\cos^2(x)$ from $0$ to $2 \pi$).