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

Let \(u = \theta\) and \(dv = \sin \frac{1}{2} \theta \, d\theta\).

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

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

\(\int \theta \sin \frac{1}{2} \theta \, d\theta = -2\theta \cos \frac{1}{2} \theta + 2 \int \cos \frac{1}{2} \theta \, d\theta\).

The integral \(\int \cos \frac{1}{2} \theta \, d\theta\) is:

\(2 \sin \frac{1}{2} \theta\).

Thus, the indefinite integral is:

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

Now, evaluate from 0 to \(\frac{1}{2}\pi\):

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

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

At \(\theta = 0\):

\(0\).

Thus, the definite integral is:

\(\left(-\pi \frac{1}{\sqrt{2}} + 4 \frac{1}{\sqrt{2}}\right) - 0 = \frac{4 - \pi}{\sqrt{2}}\).