Consider the function $f$ defined on $\mathbb{R}$ by $$f(x)=\int_0^\pi\cos\left(x\sin\theta \right)d\theta.$$ I showed that this function satisfies the following differential equation: $$xf''(x)+f'(x)+xf(x)=0$$ this implies that $$f''(x)+\frac{f'(x)}{x} +f(x)=0$$ since $f'$ is bounded then $$\lim_{x\to\infty}f''(x)+f(x)=0$$ how to continue to prove that $$\lim_{x\to\infty}f(x)=0.$$
I am also interested if there's another method to prove it without the differential equation.
$$ f(x) = 2 \int_0^{\pi/2} \cos(x\sin\theta)\;d\theta $$ change variables, $t = \sin\theta$ $$ f(x) = 2 \int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\;dt $$ But $\frac{1}{\sqrt{1-t^2}}$ is integrable on $[0,1]$, so we may conclude from the Riemann-Lebesgue lemma that: $$ \lim_{x \to \infty} \int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\;dt = 0 . $$