Solution
= Solution
Put $q(x)=\sqrt{\pi^2/16+x}$. The original PDF confirms that the $+x$ is inside the radical. For the composite <cosine>,
$$
f'(x)=-\frac{\sin q}{2q},\qquad
f''(x)=\frac{\sin q}{4q^3}-\frac{\cos q}{4q^2}.
$$
At $x=0$, $q=\pi/4$ and $\sin q=\cos q=\sqrt2/2$. Substituting into $f(0)+f'(0)x+f''(0)x^2/2$ gives
$$
\boxed{f(x)=\frac{\sqrt2}{2}-\frac{\sqrt2}{\pi}x
+\frac{\sqrt2(4-\pi)}{\pi^3}x^2+O(x^3)}.
$$
All three displayed coefficients are nonzero.