The Jacobi-Anger expansion is represented as
$$ e^{ix \cos\theta} = \sum_{n=-\infty}^{\infty} i^{n}J_{n}e^{in \theta} $$
With this known value, I'm attempting to show that
$$ \int_{- \pi}^{\pi} |J_{n}(x)|^{2}dx = 1 $$
I know that with the relation of $ J_{-n}(x) = (-1)^n J_{n}(x) $, we have
$$ e^{ix \cos\theta} = J_{0}(x) + 2 \sum_{n=1}^{\infty} i^{n}J_{n} \cos(n \theta) $$
I do not know how to go from here.
I know that having this expression of $e^{ix \cos\theta}$, I can acquire the integral form of the Bessel function of the first kind.
$$ J_n(x) = \frac{1}{\pi} \int_{0}^{\pi} \cos(x \sin\theta - n\theta)\ d\theta $$