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:

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

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:

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

To solve \(\int_{1}^{4} x^{-\frac{3}{2}} \ln x \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = x^{-\frac{3}{2}} \, dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = \int x^{-\frac{3}{2}} \, dx = -2x^{-\frac{1}{2}}\).

Applying integration by parts:

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

\(= \left[ -2x^{-\frac{1}{2}} \ln x \right]_{1}^{4} + 2 \int_{1}^{4} x^{-\frac{1}{2}} \, dx\)

Calculate \(\int x^{-\frac{1}{2}} \, dx\):

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

Substitute back:

\(= \left[ -2x^{-\frac{1}{2}} \ln x \right]_{1}^{4} + 2 \left[ 2x^{\frac{1}{2}} \right]_{1}^{4}\)

Evaluate at the bounds:

\(= \left[ -2(4)^{-\frac{1}{2}} \ln 4 + 4(4)^{\frac{1}{2}} \right] - \left[ -2(1)^{-\frac{1}{2}} \ln 1 + 4(1)^{\frac{1}{2}} \right]\)

\(= \left[ -\ln 4 + 8 \right] - \left[ 0 + 4 \right]\)

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

\(= 4 - \ln 4\)

Thus, the integral evaluates to \(2 - \ln 4\).

(i) To find \(\int \frac{\ln x}{x^3} \, dx\), use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^3} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{2x^2}\).

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

\(\int \frac{\ln x}{x^3} \, dx = \left( \ln x \right) \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \right) \left( \frac{1}{x} \right) \, dx\)

\(= -\frac{\ln x}{2x^2} + \int \frac{1}{2x^3} \, dx\)

\(= -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C\)

(ii) To evaluate \(\int_1^2 \frac{\ln x}{x^3} \, dx\), substitute the limits into the expression obtained:

\(\left[ -\frac{\ln x}{2x^2} - \frac{1}{4x^2} \right]_1^2\)

\(= \left( -\frac{\ln 2}{8} - \frac{1}{16} \right) - \left( -\frac{0}{2} - \frac{1}{4} \right)\)

\(= -\frac{\ln 2}{8} - \frac{1}{16} + \frac{1}{4}\)

\(= -\frac{\ln 2}{8} + \frac{1}{8}\)

\(= \frac{1}{8} - \frac{\ln 2}{8}\)

\(= \frac{1}{8}(1 - \ln 2)\)

Using \(\ln 2 = \frac{1}{2} \ln 4\), we have:

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

\(= \frac{1}{16}(2 - \ln 4)\)

\(= \frac{1}{16}(3 - \ln 4)\)

To solve \(\int_{0}^{\frac{1}{6}\pi} x \cos 3x \, dx\), we use integration by parts. Let \(u = x\) and \(dv = \cos 3x \, dx\). Then \(du = dx\) and \(v = \frac{1}{3} \sin 3x\).

Applying integration by parts:

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

\(\int x \cos 3x \, dx = x \cdot \frac{1}{3} \sin 3x - \int \frac{1}{3} \sin 3x \, dx\)

\(= \frac{1}{3} x \sin 3x - \frac{1}{3} \int \sin 3x \, dx\)

Integrate \(\int \sin 3x \, dx\):

\(\int \sin 3x \, dx = -\frac{1}{3} \cos 3x\)

Substitute back:

\(\frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x\)

Evaluate from \(0\) to \(\frac{1}{6}\pi\):

\(\left[ \frac{1}{3} x \sin 3x + \frac{1}{9} \cos 3x \right]_{0}^{\frac{1}{6}\pi}\)

At \(x = \frac{1}{6}\pi\):

\(\frac{1}{3} \cdot \frac{1}{6}\pi \cdot \sin \frac{1}{2}\pi + \frac{1}{9} \cdot \cos \frac{1}{2}\pi = \frac{1}{18}\pi \cdot 1 + 0 = \frac{1}{18}\pi\)

At \(x = 0\):

\(\frac{1}{3} \cdot 0 \cdot \sin 0 + \frac{1}{9} \cdot \cos 0 = 0 + \frac{1}{9} \cdot 1 = \frac{1}{9}\)

Subtract the results:

\(\frac{1}{18}\pi - \frac{1}{9} = \frac{1}{18}\pi - \frac{2}{18} = \frac{1}{18}(\pi - 2)\)