Expanding a function in a Fourier Series

Question

I am having an issue integrating the sin function with the variable of n, any help would be appreciated. I have deduced it to an odd sine series with the following for B_n and I am unsure how to integrate the functions with the variable of n

Answer 1

You can take care of the remaining integration by a simple substitution.

Consider the first term of the integrand. You can rewrite it as $$\cos\left(\frac{n\pi x}{5}-x\right)=\cos\bigg(\left(\frac{n\pi}{5}-1\bigg)x\right)$$ and then substitute $t=\left(\frac{n\pi}{5}-1\right)x$, so that $\mathrm{d}t=\left(\frac{n\pi}{5}-1\right)\,\mathrm{d}x$. This means you have $$\frac{1}{5}\int_0^5 \cos\bigg(\left(\frac{n\pi}{5}-1\right)x\bigg)\,\mathrm{d}x=\frac{1}{5\left(\frac{n\pi}{5}-1\right)}\int_0^{n\pi-5} \cos t\,\mathrm{d}t$$ The other integrand is handled nearly identically.