To solve \(\int_0^{\frac{1}{4}\pi} x^2 \cos 2x \, dx\), we use integration by parts. Let \(u = x^2\) and \(dv = \cos 2x \, dx\).

Then \(du = 2x \, dx\) and \(v = \frac{1}{2} \sin 2x\).

Applying integration by parts:

\(\int u \, dv = uv - \int v \, du\)

\(= \frac{1}{2} x^2 \sin 2x - \int \frac{1}{2} \sin 2x \cdot 2x \, dx\)

\(= \frac{1}{2} x^2 \sin 2x - \int x \sin 2x \, dx\).

Now, apply integration by parts again to \(\int x \sin 2x \, dx\).

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

Then \(du = dx\) and \(v = -\frac{1}{2} \cos 2x\).

\(\int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \int \frac{1}{2} \cos 2x \, dx\)

\(= -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x\).

Substitute back:

\(\frac{1}{2} x^2 \sin 2x - \left( -\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x \right)\)

\(= \frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x\).

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

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

\(\frac{1}{2} \left( \frac{1}{4}\pi \right)^2 \sin \left( \frac{1}{2}\pi \right) + \frac{1}{2} \left( \frac{1}{4}\pi \right) \cos \left( \frac{1}{2}\pi \right) - \frac{1}{4} \sin \left( \frac{1}{2}\pi \right)\)

\(= \frac{1}{2} \cdot \frac{1}{16} \pi^2 \cdot 1 + 0 - \frac{1}{4} \cdot 1\)

\(= \frac{1}{32} \pi^2 - \frac{1}{4}\).

At \(x = 0\):

\(0 + 0 - 0 = 0\).

Thus, the integral evaluates to:

\(\frac{1}{32} \pi^2 - \frac{1}{4}\).

Rewriting \(\frac{1}{4}\) as \(\frac{8}{32}\), we have:

\(\frac{1}{32} \pi^2 - \frac{8}{32} = \frac{1}{32}(\pi^2 - 8)\).