(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum of the function \(y = 5 \sin^3 x \cos^2 x\). First, find the derivative using the product and chain rule:

\(\frac{dy}{dx} = 15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x\).

Set the derivative to zero to find critical points:

\(15 \sin^2 x \cos^3 x - 10 \sin^4 x \cos x = 0\).

Factor out common terms:

\(5 \sin^2 x \cos x (3 \cos^2 x - 2 \sin^2 x) = 0\).

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

Using \(\sin^2 x + \cos^2 x = 1\), solve for \(x\):

\(2 \tan^2 x = 3\) leads to \(x = 0.886\) radians.

(ii) To find the area of the shaded region, use the substitution \(u = \cos x\), so \(du = -\sin x \, dx\).

The integral becomes:

\(\int 5 \sin^3 x \cos^2 x \, dx = \int 5 (1-u^2) u^2 (-du)\).

\(= \int 5 (u^2 - u^4) \, du\).

Integrate and use limits \(u = 1\) and \(u = 0\) (corresponding to \(x = 0\) and \(x = \frac{1}{2} \pi\)):

\(\int 5 (u^2 - u^4) \, du = 5 \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1\).

\(= 5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5 - 3}{15} \right) = \frac{2}{3}\).