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:

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

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:

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

To solve \(\int_{0}^{\frac{1}{2}} xe^{-2x} \, dx\), we use integration by parts. Let \(u = x\) and \(dv = e^{-2x} \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2} e^{-2x}\).

Applying integration by parts:

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

\(\int xe^{-2x} \, dx = -\frac{1}{2} xe^{-2x} - \int \left(-\frac{1}{2} e^{-2x}\right) \, dx\)

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

Integrate \(\int e^{-2x} \, dx\):

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

Substitute back:

\(-\frac{1}{2} xe^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) = -\frac{1}{2} xe^{-2x} - \frac{1}{4} e^{-2x}\)

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

\(\left[-\frac{1}{2} \left(\frac{1}{2}\right)e^{-1} - \frac{1}{4} e^{-1}\right] - \left[-\frac{1}{2}(0)e^{0} - \frac{1}{4} e^{0}\right]\)

\(= \left(-\frac{1}{4} e^{-1} - \frac{1}{4} e^{-1}\right) - \left(0 - \frac{1}{4}\right)\)

\(= -\frac{1}{2} e^{-1} + \frac{1}{4}\)

Thus, the exact value is \(\frac{1}{4} - \frac{1}{2} e^{-1}\).

To solve \(\int_{1}^{4} \frac{\ln x}{\sqrt{x}} \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{\sqrt{x}} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = 2\sqrt{x}\).

Applying integration by parts:

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

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x \bigg|_{1}^{4} - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \int x^{-1/2} \, dx\)

Integrate \(\int x^{-1/2} \, dx\):

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 2 \cdot 2x^{1/2} \bigg|_{1}^{4}\)

\(= 2\sqrt{x} \ln x \bigg|_{1}^{4} - 4\sqrt{x} \bigg|_{1}^{4}\)

Evaluate at the limits:

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

\(= (4 \ln 4 - 8) - (0 - 4)\)

\(= 4 \ln 4 - 8 + 4\)

\(= 4(\ln 4 - 1)\)

To solve \(\int_{2}^{4} 4x \ln x \, dx\), we use integration by parts.

Let \(u = \ln x\) and \(dv = 4x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = 2x^2\).

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

\(\int 4x \ln x \, dx = 2x^2 \ln x - \int 2x^2 \cdot \frac{1}{x} \, dx\).

Simplifying the integral, \(\int 2x^2 \cdot \frac{1}{x} \, dx = \int 2x \, dx = x^2\).

Thus, \(\int 4x \ln x \, dx = 2x^2 \ln x - x^2\).

Evaluate from 2 to 4:

\(\left[ 2x^2 \ln x - x^2 \right]_{2}^{4} = \left( 2(4)^2 \ln 4 - (4)^2 \right) - \left( 2(2)^2 \ln 2 - (2)^2 \right)\).

Calculate each part:

For \(x = 4\): \(2(16) \ln 4 - 16 = 32 \ln 4 - 16\).

For \(x = 2\): \(2(4) \ln 2 - 4 = 8 \ln 2 - 4\).

Combine results:

\((32 \ln 4 - 16) - (8 \ln 2 - 4) = 32 \ln 4 - 16 - 8 \ln 2 + 4\).

\(= 32 \ln 4 - 8 \ln 2 - 12\).

Since \(\ln 4 = 2 \ln 2\), substitute:

\(= 32(2 \ln 2) - 8 \ln 2 - 12\).

\(= 64 \ln 2 - 8 \ln 2 - 12\).

\(= 56 \ln 2 - 12\).

Thus, the given result is confirmed: \(56 \ln 2 - 12\).