Answer: (a) The coefficient of \(x^2\) is \(\frac{1}{2}\).

(b) The arc length is \(\ln(\sqrt{2}+1)\).

(a) Let \(y=-\ln(\cos x)\).

For a Maclaurin series, the coefficient of \(x^2\) is \(\frac{y''(0)}{2!}\).

Differentiate:

\(y' = -\frac{1}{\cos x}(-\sin x)=\tan x\).

Differentiate again:

\(y''=\sec^2 x\).

So \(y''(0)=\sec^2 0=1\).

Hence the coefficient of \(x^2\) is \(\frac{y''(0)}{2!}=\frac{1}{2}\).

(b) The length of arc from \(x=0\) to \(x=\frac{\pi}{4}\) is

\(s=\int_0^{\pi/4} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\).

Since \(\frac{dy}{dx}=\tan x\),

\(s=\int_0^{\pi/4} \sqrt{1+\tan^2 x}\,dx=\int_0^{\pi/4} \sec x\,dx\).

Using \(\int \sec x\,dx=\ln(\sec x+\tan x)+C\),

\(s=\left[\ln(\sec x+\tan x)\right]_0^{\pi/4}\).

Now \(\sec \frac{\pi}{4}=\sqrt{2}\), \(\tan \frac{\pi}{4}=1\), \(\sec 0=1\), and \(\tan 0=0\).

So

\(s=\ln(\sqrt{2}+1)-\ln 1=\ln(\sqrt{2}+1)\).