I'm trying to show that $J_n(x)=\frac{1}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta) d\theta$ is solution for $x^2y''+xy'+(x^2-n^2)y=0$.
I try the following: $$J'_n(x)=\frac{1}{\pi}\int_0^{\pi} \sin (n\theta - x\sin \theta)\sin (\theta)\, d\theta$$
and,
\begin{align*} J_n''(x)&=-\frac{1}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta)\sin^2 (\theta)\, d\theta \\ &=-\frac{1}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta)(1-\cos ^2\theta) \, d\theta \\ &=-J_n(x)+\frac{1}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta)\cos^2 (\theta)\, d\theta \\ \Rightarrow J_n''(x)+J_n(x)&=\frac{1}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta)\cos^2 (\theta)\, d\theta\\ \Rightarrow x^2J_n''(x)+x^2J_n(x)&=\frac{x^2}{\pi}\int_0^{\pi} \cos(n\theta - x\sin \theta)\cos^2 (\theta)\, d\theta \end{align*}
but I don't know how to continue from here. How can I continue?.