Fourier tricks for $f(x)=\sin(x)+x\cos(x)$ Let $f(x)=\sin(x)+x\cos(x)$, $x\in(-\pi,\pi)$.

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*Find the Fourier series of $f$

I know all the formulas for the coefficients $a_0,b_0,c_0$ and $a_n,b_n,c_n$ and the property for odd functions that then $c_k(f)=c_{-k}(f)$ i.e $a_k(f)=0$ and that we can write it as a sine series
$$\sum_{k=1}^{\infty}b_k(f)\sin(kx)$$
Is there a trick I am oveseeing as brute forcing it seems really tough and pages long?

I will just show the FS for $x\cos(x)$. I will use the FS with complex coefficients $(c_n)_{n \in \mathbb{Z}}$.
$$2\pi c_n=\int_{(-\pi,\pi]}x\cos(x)e^{-ixn}dx=\int_{(-\pi,\pi]}\frac{xe^{ix}+xe^{-ix}}{2}e^{-ixn}dx$$
$$\int_{(-\pi,\pi]}xe^{ix(1-n)}dx=i\int_{(-\pi,\pi]}x\sin(x(1-n))dx=2\pi i\frac{\cos(\pi n)}{(1-n)}$$
$$\int_{(-\pi,\pi]}xe^{-ix(n+1)}dx=-i\int_{(-\pi,\pi]}x\sin(x(n+1))dx=-2\pi i\frac{\cos(\pi n)}{(n+1)}$$
$\forall n \neq -1,1$, therefore
$$2\pi c_n=\pi i\bigg(\frac{\cos(\pi n)}{(1-n)}-\frac{\cos(\pi n)}{(n+1)}\bigg)$$
$$2\pi c_1=\int_{(-\pi,\pi]}\frac{xe^{ix}+xe^{-ix}}{2}e^{-ix}dx=\frac{i}{2}\pi$$
$$2\pi c_{-1}=\int_{(-\pi,\pi]}\frac{xe^{ix}+xe^{-ix}}{2}e^{ix}dx=-\frac{i}{2}\pi$$
Here is a truncated one:

