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