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

To solve the integral \( \int \frac{\sin^3 x}{1 - \cos x} \, dx \), we can use a substitution. Notably, since \( \sin^2 x = 1 - \cos^2 x \), we can express \( \sin^3 x \) as \( \sin x (1 - \cos^2 x) \). Now, simplifying the integral, we have: \[ \sin^3 x = \sin x (1 - \cos^2 x) = \sin x - \sin x \cos^2 x. \] So, our integral can be split into two parts: \[ \int \left( \frac{\sin x}{1 - \cos x} - \frac{\sin x \cos^2 x}{1 - \cos x} \right) \, dx. \] Next, we can use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \). Therefore, integrating both parts becomes much simpler in terms of \( u \). By carefully employing partial fractions and integral identities, we can resolve the integral into manageable parts. But don't forget to revert back to the \( x \) variable after integrating. This approach showcases the beauty of trigonometric identities and substitutions in calculus! And remember, while transformations often simplify integrals, the key is to be thorough with checks after changes of variables to avoid common pitfalls.