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

please help if you can..

Merlyn.

Printable View

- Dec 10th 2008, 11:00 PMmercouPlease help with calculus fourier problem
I have included a pdf attachment I made with mathematica to help try and explain my problem.

please help if you can..

Merlyn. - Dec 11th 2008, 02:22 AMmr fantastic
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. - Dec 11th 2008, 03:47 AMCaptainBlack
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