Because the integrand is not periodic with period 2L you have to integrate over -L,L here, also you divide by 2L in the second sin term and you should not it should just be L. You could also do this the way shown in the MathWorld page.
CB
The question ask to expand the following function of period on the interval as a Fourier sine series.
This function is periodic on a period of and thus the coefficients are given by
Upon evaluating the integral, I split the sine terms into two seperate cosine terms using a trig identity. After integration I am unable to get a single cosine as given in the correct answer
Have I made an error or am I on the correct track?
Because the integrand is not periodic with period 2L you have to integrate over -L,L here, also you divide by 2L in the second sin term and you should not it should just be L. You could also do this the way shown in the MathWorld page.
CB