Integral of \( \sin^2(x) \cos^2(x) \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \sin^2(x) \cos^2(x) \, dx \]

Use the trigonometric identities \( \sin^2 x = \dfrac{1}{2}(1 - \cos(2x)) \) and \( \cos^2 x = \dfrac{1}{2}(1 + \cos(2x)) \) to rewrite the integral:

\[ \int \sin^2(x) \cos^2(x) \, dx = \dfrac{1}{4} \int (1 - \cos(2x))(1 + \cos(2x)) \, dx \]

Expand the integrand of the integral on the right side using the difference of squares:

\[ = \dfrac{1}{4} \int (1 - \cos^2(2x)) \, dx \]

Use the trigonometric identity \( \cos^2 \theta = \dfrac{1}{2}(1 + \cos(2\theta)) \), which for \( \theta = 2x \) gives \( \cos^2(2x) = \dfrac{1}{2}(1 + \cos(4x)) \):

\[ = \dfrac{1}{4} \int \left( 1 - \dfrac{1}{2}(1 + \cos(4x)) \right) dx \]

Simplify the integrand inside the parentheses:

\[ = \dfrac{1}{4} \int \left( \dfrac{1}{2} - \dfrac{1}{2}\cos(4x) \right) dx = \dfrac{1}{8} \int (1 - \cos(4x)) \, dx \]

Use standard integrals \( \int 1 \, dx = x \) and \( \int \cos(kx) \, dx = \dfrac{1}{k}\sin(kx) \) to evaluate the integral:

\[ = \dfrac{1}{8} \left( x - \dfrac{1}{4} \sin(4x) \right) + c \]

where \( c \) is the constant of integration.

The final result is given by: