Show that $J_0(x)=\sum\limits_0^\infty \frac{(-1)^n x^{2n}}{4^n(n!)^2}$ is $J_0(x)=\frac{1}{\pi}\int_0^\pi \cos(x\sin\theta)d\theta$

Question

A question from Introduction to Analysis by Arthur Mattuck:

Let $n!!=n(n-2)(n-4)\cdot…\cdot k$, where $k=1$ or $2$,depending on whether n is odd or even. (define $0!!=1$.)

Assume $\int_0^\pi \sin^{2n}\theta d\theta=\frac{(2n-1)!!}{2n!!}\pi.(n=0,1,2,…)$

Using this, prove by using term-by-term integration that the Bessel function $$J_0(x)=\sum_0^\infty \frac{(-1)^n x^{2n}}{4^n(n!)^2}$$ can also be represented as the following definite integral:$$J_0(x)=\frac{1}{\pi}\int_0^\pi \cos(x\sin\theta)d\theta.$$

By multiplying the series with $\frac{(2n-1)!!\pi}{(2n-1)!!\pi}$ term by term, I get $J_0(x)={1 \over \pi}\sum_0^\infty \frac{(-1)^nx^{2n}}{2n!}\int_0^\pi \sin^{2n}\theta d\theta.$ Then I get stuck.