Answer
$$\frac{2x+\sin \left(2x\right)-4\cos ^3\left(x\right)\sin \left(x\right)}{16}+C$$
Work Step by Step
Given
$$\int \sin ^{2} x \cos ^{2} x d x$$
Since
\begin{align*}
\int \sin ^{2} x \cos ^{2} x d x&=\int (1-\cos ^{2} x ) \cos ^{2} x d x\\
&=\int (\cos ^{2} x -\cos ^{4} x ) d x\\
\end{align*}
Use
$$ \int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$$
Then
\begin{align*}
\int \sin ^{2} x \cos ^{2} x d x&=\int (1-\cos ^{2} x ) \cos ^{2} x d x\\
&=\int (\cos ^{2} x -\cos ^{4} x ) d x\\
&=\frac{\cos x \sin x}{2}+\frac{1}{2} \int d x-\frac{\cos ^{3} x \sin x}{4}-\frac{3}{4} \int \cos ^{2} x d x\\
&=\frac{\cos x \sin x}{2}+\frac{1}{2} x-\frac{\cos ^{3} x \sin x}{4}-\frac{3}{4}\left(\frac{\cos x \sin x}{2}+\frac{1}{2} x\right)+C \\
&=\frac{2x+\sin \left(2x\right)-4\cos ^3\left(x\right)\sin \left(x\right)}{16}+C
\end{align*}
