(a) Let \(u = \\sin x\), then \(du = \\cos x \, dx\). The integral becomes \(\int \\sin 2x \\cos^2 x \, dx = \int 2u(1-u^2) \, du\).

Integrate to get \(\int 2(u - u^3) \, du = 2 \left( \frac{u^2}{2} - \frac{u^4}{4} \right) = u^2 - \frac{u^4}{2}\).

Evaluate from \(u = 0\) to \(u = 1\):

\(\left[ u^2 - \frac{u^4}{2} \right]_0^1 = \left( 1 - \frac{1}{2} \right) - (0 - 0) = \frac{1}{2}\).

(b) Differentiate \(y = \\sin 2x \\cos^2 x\) using the product rule:

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

Set \(\frac{dy}{dx} = 0\) and solve:

\(2 \\cos 2x \\cos^2 x = 2 \\sin 2x \\cos x \\sin x\).

Using double angle formulas, simplify to \(4 \\sin^2 x = 1\), giving \(\\sin^2 x = \frac{1}{4}\).

Thus, \(\\sin x = \frac{1}{2}\), so \(x = \frac{1}{6}\pi\).