Let \(u = \sin 4x\), then \(\frac{du}{dx} = 4\cos 4x\) or \(du = 4\cos 4x \, dx\).
Thus, \(dx = \frac{du}{4\cos 4x}\).
Substitute into the integral:
\(\int \cos^3 4x \, dx = \int \cos^3 4x \cdot \frac{du}{4\cos 4x} = \frac{1}{4} \int \cos^2 4x \, du\).
Since \(\cos^2 4x = 1 - \sin^2 4x = 1 - u^2\), the integral becomes:
\(\frac{1}{4} \int (1 - u^2) \, du\).
Integrate to obtain:
\(\frac{1}{4} \left( u - \frac{u^3}{3} \right) + C\).
Evaluate from \(u = \sin 4(0) = 0\) to \(u = \sin 4\left(\frac{1}{24}\pi\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\).
\(\frac{1}{4} \left[ \left( \frac{1}{2} - \frac{(\frac{1}{2})^3}{3} \right) - (0 - 0) \right]\).
\(= \frac{1}{4} \left( \frac{1}{2} - \frac{1}{24} \right)\).
\(= \frac{1}{4} \cdot \frac{11}{24}\).
\(= \frac{11}{96}\).
