Evaluate the integral
\[ \int \sin^3 x \; dx \]
Write the integrand \( \sin^3 x \) as a product of as \( \sin^2 x \) and \( \sin x \)
\[ \int \sin^3 x \; dx = \int \sin^2 x \sin x \; dx\]
Use of the trigonometric identity \( \; \sin^2 x = 1 - \cos^2 x \) to write the integral as
\[ \int \sin^3 x \; dx = \int (1 - \cos^2 x) \sin x \; dx\]
Expand \( (1 - \cos^2 x) \sin x \) and write
\[ \int \sin^3 x \; dx = \int \sin x \; dx - \int \cos^2 x \sin x \; dx \]
Integration by Substitution: Let \( u = \cos x \) and hence \( \dfrac{du}{dx} = - \sin x \) or \( du = - \sin x \; dx \) and substitute to obtain
\[ \int \sin^3 x \; dx = \int \sin x \; dx + \int u^2 \; du \]
Use integral formulas to evaluate the above integral and write
\[ \int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} u^3 + c \]
Substitute back \( u = \cos x \) to find the final answer
\[ \boxed { \int \sin^3 x \; dx = - \cos x + \dfrac{1}{3} \cos^3 x + c } \]