\( \int \frac { \cos ^ { 3 } 3 x } { ( 1 - \sin 3 x ) } d x \)

To tackle the integral \( \int \frac {\cos^{3}(3x)}{1 - \sin(3x)} \, dx \), we can start with a substitution to simplify our integrand. Let \( u = \sin(3x) \). Then, the derivative \( du = 3\cos(3x) \, dx \) or \( dx = \frac{du}{3\cos(3x)} \). Now we need to express \(\cos^3(3x)\) in terms of \(u\). Notice that \(\cos^2(3x) = 1 - \sin^2(3x) = 1 - u^2\), therefore: \[ \cos^3(3x) = \cos(3x)(1 - u^2) = \frac{1 - u^2}{\sqrt{1 - u^2}} \cdot \sqrt{1 - u^2} = (1 - u^2) \cos(3x). \] So we need to adapt our integral: \[ \int \frac{\cos^{3}(3x)}{1 - u} \cdot \frac{du}{3\cos(3x)} = \frac{1}{3} \int \frac{(1 - u^2)}{1 - u} \, du. \] Now we simplify the integrand: \[ \frac{(1 - u^2)}{(1 - u)} = \frac{(1 - u)(1 + u)}{(1 - u)} = 1 + u. \] Thus, the integral becomes: \[ \frac{1}{3} \int (1 + u) \, du = \frac{1}{3} \left( u + \frac{u^2}{2} \right) + C = \frac{1}{3} \left( \sin(3x) + \frac{\sin^2(3x)}{2} \right) + C. \] Finally, we can rewrite it as: \[ \frac{1}{3} \sin(3x) + \frac{1}{6} \sin^2(3x) + C. \] So the result of the integral is: \[ \int \frac {\cos^{3}(3x)}{1 - \sin(3x)} \, dx = \frac{1}{3} \sin(3x) + \frac{1}{6} \sin^2(3x) + C. \]