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

Let \(u = x^2\) and \(dv = \sin 2x \, dx\). Then \(du = 2x \, dx\) and \(v = -\frac{1}{2} \cos 2x\).

Applying integration by parts:

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

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

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

Now, integrate \(\int x \cos 2x \, dx\) by parts again:

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

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

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

Substitute back:

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

Evaluate from 0 to \(\frac{1}{2}\pi\):

At \(x = \frac{1}{2}\pi\):

\(-\frac{1}{2} \left( \frac{1}{2}\pi \right)^2 \cdot (-1) + \frac{1}{2} \cdot \frac{1}{2}\pi \cdot 0 + \frac{1}{4} \cdot (-1)\)

\(= \frac{1}{8} \pi^2 - \frac{1}{4}\)

At \(x = 0\):

\(0 + 0 + \frac{1}{4}\)

Subtract the two results:

\(\left( \frac{1}{8} \pi^2 - \frac{1}{4} \right) - \frac{1}{4} = \frac{1}{8} (\pi^2 - 4)\)