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

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

Applying integration by parts:

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

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

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

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

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

Substitute back:

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

Evaluate from 0 to 1:

At \(x = 1\):

\(-\frac{1}{2}(2-1)e^{-2} - \frac{1}{4}e^{-2} = -\frac{1}{2}e^{-2} - \frac{1}{4}e^{-2} = -\frac{3}{4}e^{-2}\)

At \(x = 0\):

\(-\frac{1}{2}(2-0)e^{0} - \frac{1}{4}e^{0} = -1 - \frac{1}{4} = -\frac{5}{4}\)

Subtract the results:

\(\left(-\frac{3}{4}e^{-2}\right) - \left(-\frac{5}{4}\right) = \frac{5}{4} - \frac{3}{4}e^{-2}\)

Thus, the exact value is \(\frac{1}{4}(3 - e^{-2})\).

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 \(v = \frac{2}{5} x^{\frac{5}{2}}\).

Applying integration by parts:

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

\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{5}{2}} \cdot \frac{1}{x} \, dx\)

\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \int x^{\frac{3}{2}} \, dx\)

Integrate \(\int x^{\frac{3}{2}} \, dx\):

\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{2}{5} \cdot \frac{2}{5} x^{\frac{5}{2}}\)

\(= \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{4}{25} x^{\frac{5}{2}}\)

Evaluate from 1 to 4:

\(\left[ \frac{2}{5} x^{\frac{5}{2}} \ln x - \frac{4}{25} x^{\frac{5}{2}} \right]_{1}^{4}\)

\(= \left( \frac{2}{5} \cdot 32 \ln 4 - \frac{4}{25} \cdot 32 \right) - \left( \frac{2}{5} \cdot 1 \cdot 0 - \frac{4}{25} \cdot 1 \right)\)

\(= \frac{64}{5} \ln 4 - \frac{128}{25} + \frac{4}{25}\)

\(= \frac{64}{5} \ln 4 - \frac{124}{25}\)

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

\(= \frac{128}{5} \ln 2 - \frac{124}{25}\)

To solve \(\int x \sec^2 x \, dx\), we use integration by parts. Let \(u = x\) and \(dv = \sec^2 x \, dx\). Then \(du = dx\) and \(v = \tan x\).

Applying integration by parts:

\(\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx\)

The integral \(\int \tan x \, dx\) is \(-\ln |\cos x|\).

Thus,

\(\int x \sec^2 x \, dx = x \tan x + \ln |\cos x|\)

Now, evaluate from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{3}\):

\(\left[ x \tan x + \ln |\cos x| \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)

Calculate at \(x = \frac{\pi}{3}\):

\(\frac{\pi}{3} \tan \frac{\pi}{3} + \ln |\cos \frac{\pi}{3}| = \frac{\pi}{3} \sqrt{3} + \ln \frac{1}{2}\)

Calculate at \(x = \frac{\pi}{6}\):

\(\frac{\pi}{6} \tan \frac{\pi}{6} + \ln |\cos \frac{\pi}{6}| = \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{\sqrt{3}}{2}\)

Subtract the two results:

\(\left( \frac{\pi}{3} \sqrt{3} + \ln \frac{1}{2} \right) - \left( \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{\sqrt{3}}{2} \right)\)

Simplify using logarithm properties:

\(\frac{\pi}{3} \sqrt{3} - \frac{\pi}{6} \frac{1}{\sqrt{3}} + \ln \frac{1}{2} - \ln \frac{\sqrt{3}}{2}\)

\(= \frac{5}{18} \sqrt{3} \pi - \frac{1}{2} \ln 3\)

(i) Let \(y = \frac{\cos x}{\sin x} = \cot x\). Differentiate using the quotient rule:

\(\frac{dy}{dx} = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{\sin^2 x}\)

\(= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\)

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

\(= \frac{-1}{\sin^2 x} = -\csc^2 x\)

(ii) Integrate by parts: Let \(u = x\) and \(dv = \csc^2 x \, dx\).

Then \(du = dx\) and \(v = -\cot x\).

\(\int x \csc^2 x \, dx = -x \cot x + \int \cot x \, dx\)

\(= -x \cot x + \ln |\sin x|\)

Evaluate from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\):

\(\left[ -x \cot x + \ln |\sin x| \right]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\)

At \(x = \frac{1}{2}\pi\), \(-x \cot x = 0\) and \(\ln |\sin x| = 0\).

At \(x = \frac{1}{4}\pi\), \(-x \cot x = -\frac{1}{4}\pi\) and \(\ln |\sin x| = \ln \frac{1}{\sqrt{2}} = -\frac{1}{2} \ln 2\).

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

\(= \frac{1}{4}\pi + \frac{1}{2} \ln 2\)

\(= \frac{1}{4}(\pi + \ln 4)\)

(a) To find the equation of the tangent, we first need the derivative of \(y = x \cos 2x\). Using the product rule:

\(\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos 2x + x \cdot \frac{d}{dx}(\cos 2x)\)

\(= \cos 2x - 2x \sin 2x\)

At \(x = \frac{1}{2} \pi\), \(y = x \cos 2x = \frac{1}{2} \pi \cdot \cos \pi = -\frac{\pi}{2}\).

\(\frac{dy}{dx} = \cos \pi - 2 \cdot \frac{1}{2} \pi \cdot \sin \pi = -1\).

The equation of the tangent is \(y - (-\frac{\pi}{2}) = -1(x - \frac{1}{2} \pi)\).

Simplifying gives \(x + y = 0\).

(b) To find the area of the shaded region, integrate \(y = x \cos 2x\) from \(x = 0\) to \(x = \frac{\pi}{4}\).

Using integration by parts, let \(u = x\) and \(dv = \cos 2x \, dx\), then \(du = dx\) and \(v = \frac{1}{2} \sin 2x\).

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

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

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

\(\left[ \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right]_0^{\frac{\pi}{4}}\)

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

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