Evaluate the integral
\[ \int \sin^2(x) \cos^2(x) \; dx \]
Use trigonometric identities \( \sin^2 x = \dfrac{1}{2}(1 - \cos (2 x)) \) and \( \cos^2 x = \dfrac{1}{2}(1 + \cos (2 x)) \) on the right and write
\[ \int \sin^2(x) \cos^2(x) \; dx = \dfrac{1}{4} \int (1 - \cos (2 x)) (1 + \cos (2 x)) \; dx \]
Expand the integrand of the integral on the right side
\[ = \dfrac{1}{4} \int (1 - \cos^2(2 x)) \; dx \]
Use trigonometric identity \( \cos^2 x = \dfrac{1}{2}(1 + \cos (2 x)) \) which also gives \( \cos^2 (2x) = \dfrac{1}{2}(1 + \cos (4x)) \)
\[ = \dfrac{1}{8} \int (1 - \dfrac{1}{2}(1 + \cos (4x)) ) \; dx \]
Expand and simplify the integrand
\[ = \dfrac{1}{8} \int (1 - \dfrac{1}{2}(1 - \cos (4x)) ) \; dx \]
Use the common integrals \( \int 1 dx = x + c\) and \( \int \cos (kx) dx = \dfrac{1}{k} \sin x + c\) to evaluate the above integral and obtain the final answer
\[ \boxed { \int \sin^2(x) \cos^2(x) \; dx = \dfrac{1}{8} \left(x - \dfrac{1}{4} \sin (4x) \right) + c } \]
