(i) To find the \(x\)-coordinate of \(M\), we first find the derivative of \(y = 5 \sin^2 x \cos^3 x\) using the product rule. Let \(u = \sin^2 x\) and \(v = \cos^3 x\). Then \(\frac{du}{dx} = 2 \sin x \cos x\) and \(\frac{dv}{dx} = -3 \cos^2 x \sin x\).
The derivative \(\frac{dy}{dx} = 5 \left( u \frac{dv}{dx} + v \frac{du}{dx} \right) = 5 \left( \sin^2 x (-3 \cos^2 x \sin x) + \cos^3 x (2 \sin x \cos x) \right)\).
Simplifying, \(\frac{dy}{dx} = 5 \left( -3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x \right)\).
Setting \(\frac{dy}{dx} = 0\), we get \(-3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x = 0\).
Factoring out \(\sin x \cos^2 x\), we have \(\sin x \cos^2 x (-3 \sin^2 x + 2 \cos^2 x) = 0\).
Thus, \(-3 \sin^2 x + 2 \cos^2 x = 0\) leads to \(3 \tan^2 x = 2\).
Solving for \(x\), \(\tan^2 x = \frac{2}{3}\), so \(x = \arctan \left( \sqrt{\frac{2}{3}} \right) \approx 0.685\).
(ii) Using the substitution \(u = \sin x\), \(du = \cos x \, dx\). The limits change from \(x = 0\) to \(x = \frac{1}{2} \pi\) to \(u = 0\) to \(u = 1\).
The integral becomes \(\int 5 \sin^2 x \cos^3 x \, dx = 5 \int u^2 (1-u^2) \, du\).
This simplifies to \(5 \int (u^2 - u^4) \, du\).
Integrating, \(5 \left( \frac{1}{3} u^3 - \frac{1}{5} u^5 \right)\) from 0 to 1.
Evaluating, \(5 \left( \frac{1}{3} - \frac{1}{5} \right) = 5 \left( \frac{5}{15} - \frac{3}{15} \right) = 5 \times \frac{2}{15} = \frac{10}{15} = \frac{2}{3}\).
