(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of \(y = e^{\cos x} \sin^3 x\). Differentiate using the product rule and chain rule:

\(\frac{dy}{dx} = e^{\cos x} \left( -\sin x \sin^3 x + 3\cos x \sin^2 x \right)\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(-\sin x \sin^3 x + 3\cos x \sin^2 x = 0\).

Factor out \(\sin^2 x\):

\(\sin^2 x (-\sin x + 3\cos x) = 0\).

Thus, \(\sin x = 0\) or \(-\sin x + 3\cos x = 0\).

For \(-\sin x + 3\cos x = 0\), solve for \(\cos^2 x + 3\cos x - 1 = 0\).

Using the quadratic formula, find \(\cos x\) and then \(x\):

\(x = 1.26\) (to 2 decimal places).

(ii) Use the substitution \(u = \cos x\), then \(du = -\sin x \, dx\).

The integral becomes \(\int e^u (u^2 - 1) \, du\).

Integrate by parts:

\(\int e^u (u^2 - 1) \, du = e^u (u^2 - 1) - 2\int ue^u \, du\).

Complete the integration to obtain:

\(e^u (u^2 - 2u + 1)\).

Substitute back the limits \(u = 1\) and \(u = -1\) (or \(x = 0\) and \(x = \pi\)):

The area \(R\) is \(\frac{4}{e}\).

(i) Let \(x = t^2 + 1\). Then \(dx = 2t \, dt\).

When \(t = 0\), \(x = 0^2 + 1 = 1\).

When \(t = 2\), \(x = 2^2 + 1 = 5\).

Substitute into the integral:

\(I = \int_0^2 4t^3 \ln(t^2 + 1) \, dt = \int_1^5 (2x - 2) \ln x \, dx\).

(ii) Use integration by parts:

Let \(u = \ln x\) and \(dv = (2x - 2) \, dx\).

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

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

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

\(= (x^2 - 2x) \ln x - \left( \frac{1}{2}x^2 - 2x \right)\).

Evaluate from 1 to 5:

\(= [(5^2 - 2 \cdot 5) \ln 5 - \left( \frac{1}{2} \cdot 5^2 - 2 \cdot 5 \right)] - [(1^2 - 2 \cdot 1) \ln 1 - \left( \frac{1}{2} \cdot 1^2 - 2 \cdot 1 \right)]\).

\(= [15 \ln 5 - (\frac{25}{2} - 10)] - [0 - (\frac{1}{2} - 2)]\).

\(= 15 \ln 5 - \frac{5}{2} + \frac{3}{2}\).

\(= 15 \ln 5 - 4\).

(i) The minimum point \(A\) occurs where \(x = 1\) because \((\ln x)^2\) is minimized when \(\ln x = 0\), i.e., \(x = 1\).

(ii) First, find \(f'(x) = 2 \frac{\ln x}{x}\). Then, differentiate again to find \(f''(x) = 2 \left( \frac{1}{x} \cdot \frac{-1}{x^2} + \frac{\ln x}{x^2} \right) = 2 \left( \frac{-1}{x^2} + \frac{\ln x}{x^2} \right)\). At \(x = e\), \(f''(e) = 2 \left( \frac{-1}{e^2} + \frac{1}{e^2} \right) = 0\).

(iii) The area is given by \(\int_1^e (\ln x)^2 \, dx\). Using the substitution \(x = e^u\), \(dx = e^u \, du\), and the limits change from \(x = 1\) to \(u = 0\) and \(x = e\) to \(u = 1\). The integral becomes \(\int_0^1 u^2 e^u \, du\).

(iv) Evaluate \(\int_0^1 u^2 e^u \, du\) using integration by parts. Let \(v = u^2\) and \(dw = e^u \, du\), then \(dv = 2u \, du\) and \(w = e^u\). The integral becomes \(\left[ u^2 e^u \right]_0^1 - \int_0^1 2u e^u \, du\). Apply integration by parts again to \(\int_0^1 2u e^u \, du\) with \(v = u\) and \(dw = 2e^u \, du\). The result is \(e - 2\).