The question is:
Show that the function
$y={2\over\pi}\int_0^{\pi\over2}\cos(x\sin(\theta))d\theta$
satisfy the ODE $y''+{y'\over x}+y=0$.
I tried use the Leibniz rule for integration and end up with the following equation:
$y''+{y'\over x}+y={2\over \pi}\int_0^{\pi\over2}\left[\cos(x\sin(\theta))\cos^2(\theta)-{\sin(x\sin(\theta))\sin(\theta)\over x}\right]d\theta$
But I don't know what to do to proceed.
Consider $$ \frac{d}{dθ}\Bigl[\sin(x\sin(θ))\cos(θ)\Bigr] =x\cos(x\sin(θ))\cos^2(θ)-\sin(x\sin(θ))\sin(θ) $$ and apply this and the fundamental theorem to your last integral.