I am attempting to find the fourier series for $f(x) = x\cos x$ for $-\pi < x < \pi$. The method is fine however I find that for $n = 1$ I am unable to evaluate the coefficient $b_1$.
$f(x) = \sum_{n=1}^{\infty}{b_n\sin(nx)}$ since $f(x)$ is an odd function.
Calculating my coefficients yields $$b_n = -\left(\frac{\cos((n+1)\pi)}{n+1}+\frac{\cos((n-1)\pi)}{n-1}\right)$$
This is clearly not defined at $n=1$, how should I proceed?
Note: this is an intermediary step in order to find the fourier series of $x(1+\cos x)$ if you wanted context.
Your assertion comes from produt-to-sum formula $$\sin s\cos t=\frac12\sin(s+t)+\frac12\sin(s-t)$$
So the integrand of interest in the various passages is still $\frac12\sin((n+1)x)+\frac12\sin((n-1)x)$. The only difference is that for $n=1$ its antiderivative is no longer $-\frac{\cos((n-1)x)}{2(n-1)}-\frac{\cos((n+1)x)}{2(n+1)}$.