Answer
$$\frac{1}{16} \left( 2x- \sin 2 x\cos 2x \right) +c $$
Work Step by Step
\begin{aligned}
\int \sin ^{2} x \cos ^{2} x d x &=\int \frac{1}{2}(1-\cos 2 x) \cdot \frac{1}{2}(1+\cos 2 x) d x \\
&=\frac{1}{4} \int\left(1-\cos ^{2} 2 x\right) d x \\
&=\frac{1}{4} \int\left(1-\frac{1}{2}(1+\cos 4 x)\right) d x \\
&=\frac{1}{4} \int\left( \frac{1}{2}- \frac{1}{2}\cos 4 x \right) d x\\
&=\frac{1}{4} \left( \frac{1}{2}x- \frac{1}{8}\sin 4 x \right) +c \\
&=\frac{1}{16} \left( 2x- \sin 2 x\cos 2x \right) +c
\end{aligned}
