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

 

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 $\boxed { \int \sin^2(x) \cos^2(x) \; dx = \dfrac{1}{8} \left(x - \dfrac{1}{4} \sin (4x) \right) + c }$