Yes, one can use a periodic extension of a function in order to express it as a Fourier series. A typical exercise on the subject is: consider the function $f(x)=x^2$ on the interval $[0,L]$, extended periodically... This means you'll have an expansion in terms of $\cos\frac{2\pi}{L}n x$ and $\sin\frac{2\pi}{L} nx$ (as these are periodic with period $L$), and the coefficients are
$$
A_n = \frac{2}{L}\int_0^L f(x)\cos \frac{2\pi n}{L} x\,dx, \quad
B_n = \frac{2}{L}\int_0^L f(x)\sin \frac{2\pi n}{L} x\,dx
$$
for $n=0,1,2,\dots$ (by convention, the series contains $A_0/2$ as the constant term, which allows the above formula to work also when $n=0$.) The interval need not be symmetric ($[-a,a]$) in order for the Fourier series to exist; however, it's often more convenient to integrate over symmetric intervals.
The quality of approximation near the endpoints may be poor if the behavior of the function at both ends does not match. To mitigate this issue, and also to avoid the problem of non-symmetric interval, I suggest using even periodic extension: that is, first extend the function to $[-L,L]$ as an even function, $f(-x)=f(x)$, and then extend periodically as a function with period $2L$, meaning $f(x+2L)=f(x)$. An advantage of doing so is that $f(-L)=f(L)$ automatically, so the values at the end of the interval $[-L,L]$ match.
In practice, the above means using the cosine Fourier series
$$
f(x) = \frac{A_0}{2} + \sum_{n=1}^\infty A_n\cos \frac{\pi n}{L} x
$$
where
$$
A_n = \frac{2}{L}\int_0^L f(x)\cos \frac{\pi n}{L} x\,dx, \quad n=0,1,2,\dots
$$
My blog post illustrates the advantage of using cosines in such situations on the example of non-periodic function $e^x$.
Best Answer
We need to extend the function $f(x)$ so that $f(x)=f(x+2)$ and $f(x)=x^2$ for $-1\leq x \leq 1$. Note that the following function works: $$f(x)=\left(x-2\left\lfloor{\frac{x+1}{2}}\right\rfloor\right)^2$$ Let's try to compute $f(x+2)$: \begin{align*} f(x+2)&=\left(x+2-2\left\lfloor{\frac{x+3}{2}}\right\rfloor\right)^2\\ &=\left(x+2-2\left\lfloor{1+\frac{x+1}{2}}\right\rfloor\right)^2\\ &=\left(x+2-2\left(1+\left\lfloor{\frac{x+1}{2}}\right\rfloor\right)\right)^2\\ &=\left(x+\left\lfloor{\frac{x+1}{2}}\right\rfloor\right)^2\\ &=f(x) \end{align*} For $-1\leq x< 1$, we note that $\left\lfloor{\frac{x+1}{2}}\right\rfloor=0$. Ergo, $f(x)=x^2$ on the domain $[-1,1)$. We can also confirm that $f(1)=1$, implying that $f(x)=x^2$ on the domain $[-1,1]$. So we have confirmed that $f(x)$ meets both of our requirements. If we graph this function, we get exactly what we expect — parabolas that repeat every $2$ units:
To compute the Fourier series, we note first that $f(x)$ is even, so its Fourier series will only contain cosines. We have to compute the following integral: $$\int_{-1}^1 x^2\cos\left(n\pi x\right)\, dx$$ When $n=0$, the integral is simple: \begin{align*} \int_{-1}^1 x^2\cos(0)\,dx&=\int_{-1}^1 x^2\,dx\\ &=\frac{x^3}{3}\Bigg|_{-1}^{1}\\ &=\frac{2}{3} \end{align*} In cases in which $n\neq 0$, we can integrate by parts: \begin{align*} \int_{-1}^1 x^2\cos\left(n\pi x\right)\,dx&=\frac{2x\sin\left(n\pi x\right)}{n\pi}\Bigg|_{-1}^1-\frac{2}{n\pi}\int_{-1}^1x\sin\left(n\pi x\right)\,dx\\ &=0-\frac{2}{n\pi}\int_{-1}^1x\sin\left(n\pi x\right)\,dx\\ &=\frac{2x\cos\left(n\pi x\right)}{n^2 \pi^2}\Bigg|_{-1}^{1}-\frac{2}{n^2\pi^2}\int_{-1}^{1}\cos\left(n\pi x\right)\,dx\\ &=\frac{2(-1)^n}{n^2\pi^2}+\frac{2(-1)^n}{n^2\pi^2}-\frac{2\sin\left(n\pi x\right)}{n^3 \pi^3}\Bigg|_{-1}^{1}\\ &=\frac{4(-1)^n}{n^2\pi^2} \end{align*} Ergo, our Fourier series is the following: $$\frac{1}{3}+\frac{4}{\pi^2}\sum_{n=1}^\infty (-1)^n\frac{\cos\left(n\pi x\right)}{n^2}$$