To solve \(\int_{0}^{\frac{1}{4}\pi} x \sec^2 x \, dx\), we use integration by parts.

Let \(u = x\) and \(dv = \sec^2 x \, dx\).

Then \(du = dx\) and \(v = \tan x\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx\).

Now, integrate \(\tan x\):

\(\int \tan x \, dx = \ln |\sec x|\).

Thus, \(\int x \sec^2 x \, dx = x \tan x - \ln |\sec x|\).

Evaluate from 0 to \(\frac{1}{4}\pi\):

\(\left[ x \tan x - \ln |\sec x| \right]_{0}^{\frac{1}{4}\pi}\).

At \(x = \frac{1}{4}\pi\):

\(\frac{1}{4}\pi \tan \left( \frac{1}{4}\pi \right) - \ln \left( \sec \left( \frac{1}{4}\pi \right) \right) = \frac{1}{4}\pi - \ln \left( \sqrt{2} \right)\).

At \(x = 0\):

\(0 \cdot \tan(0) - \ln(\sec(0)) = 0\).

Thus, the integral evaluates to:

\(\frac{1}{4}\pi - \ln \left( \sqrt{2} \right) = \frac{1}{4}\pi - \frac{1}{2}\ln 2\).