Fourier Series integral proof

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Consider a Fourier series representation of a function $f(x).$ Let $S_{N}(x)$ be the $N^{\text{th}}$ partial sum defined by $$S_{N}(x) = \frac{a_{0}}{2} + \sum_{n=1}^{N}a_{n}\cos\frac{n\pi x}{L} + b_{n}\sin\frac{n\pi x}{L}$$ where $L$ is the period of the $f(x).$ I want to show that

$$J_{N} = \int_{-L}^{L} \left [ S_{N}(x) \right ]^2 dx = \frac{La_{0}^2}{2} + L\sum_{n=1}^{N}a_{n}^2+b_{n}^2.$$

I already know that $$\int_{-L}^{L}S_{N}(x)dx = La_{0}.$$ I proved this by writing out the sum and integrating each term, which gives $$\int_{-L}^{L}S_{N}(x)dx = La_{0} = \int_{-L}^{L}\frac{a_{0}}{2}dx + \int_{-L}^{L}a_{1}\cos\frac{\pi x}{L}+ b_{1}\sin\frac{\pi x}{L}dx \ +...+$$ $$\int_{-L}^{L}a_{n}\cos\frac{n\pi x}{L}+ b_{n}\sin\frac{n\pi x}{L}dx.$$ From here, the integral is easily deduced. How can I use this to solve the problem? I'm pretty sure this has something to do with Bessel's Inequality and Paresval's Identity, but I don't really know how to continue.

Thanks in advance!

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It actually is Parseval's identity. A sketch of the argument- First, convince yourself that by switching to complex coefficients $c_n\in \mathbb C$ that satisfy certain things so that the function is real valued, its enough to consider the finite sum of complex exponentials

$$ S_N(x) = \sum_{n=-N}^N c_n e^{in\pi x/L}$$

next, show that the collection of functions $e_n(x) := e^{in\pi x/L}$ for $n=-N,\dots,N$ are orthogonal with respect to the inner product $$\langle f,g\rangle = \int_{-L}^L f(x) \overline{g(x)} dx $$ and each vector has length $\|e_n\| := \langle e_n , e_n \rangle^{1/2} = L$. This allows you to use essentially a version of Pythagoras's theorem $$ \|S_N\|^2 := \langle S_N , S_N\rangle = \sum_{n=-N}^N \sum_{m=-N}^Nc_n \overline c_m \langle e_n,e_m\rangle = \sum_{n=-N}^N \sum_{m=-N}^Nc_n \overline c_m L \delta_{mn} = L\sum_{n=-N}^N c_n \overline c_n = L\sum_{n=-N}^N |c_n|^2$$ the conditions you derived that relates $c_n$ and $a_n,b_n$ will give your expression.

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The statement you want to show is exactly the Parseval's Identity (up to rescaling).