Determine a parameter from two exact integrals

$\int_0^1dx/(4-x^2)=[\tfrac14\ln|(2+x)/(2-x)|]_0^1=\tfrac14\ln3>0$. The right integral is undefined for $p=0$ and negative for $p<0$ because its upper limit is then negative while its integrand is positive. Therefore any solution has $p>0$.

For $p>0$, the right integral is $[\sin^{-1}(px)/p]_0^{1/(2p)}=\pi/(6p)$. Equating gives $\tfrac14\ln3=\pi/(6p)$, hence $p=2\pi/(3\ln3)$.