Evaluate the indefinite integral: \[ \int \frac{\sin^3(x)}{\cos(x) - \cos^3(x)} \, dx. \]
Solution
1. On the domain \(\{x \in \mathbb{R} : \sin(x) \neq 0, \cos(x) \neq 0\}\) of the integrand, \(\frac{\sin^3(x)}{\cos(x) - \cos^3(x)} = \tan(x)\).
Factoring and the Pythagorean identity \(1 - \cos^2(x) = \sin^2(x)\) give \[\cos(x) - \cos^3(x) = \cos(x)(1 - \cos^2(x)) = \cos(x) \sin^2(x),\] so \[\frac{\sin^3(x)}{\cos(x) - \cos^3(x)} = \frac{\sin^3(x)}{\cos(x) \sin^2(x)} = \frac{\sin(x)}{\cos(x)} = \tan(x).\]
2. \(\int \tan(x) \, dx = -\ln|\cos(x)| + C\).
Substitute \(u = \cos(x)\), so \(du = -\sin(x) \, dx\): \[\int \frac{\sin(x)}{\cos(x)} \, dx = -\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C.\]
3. Q.E.D.
By steps 1 and 2, \[\int \frac{\sin^3(x)}{\cos(x) - \cos^3(x)} \, dx = \boxed{-\ln|\cos(x)| + C} = \ln|\sec(x)| + C.\]