Apply the formula: $\int\sin\left(\theta \right)^n\cos\left(\theta \right)^mdx$$=\frac{-\sin\left(\theta \right)^{\left(n-1\right)}\cos\left(\theta \right)^{\left(m+1\right)}}{n+m}+\frac{n-1}{n+m}\int\sin\left(\theta \right)^{\left(n-2\right)}\cos\left(\theta \right)^mdx$, where $m=2$ and $n=4$
The integral $\frac{1}{2}\int\sin\left(x\right)^{2}\cos\left(x\right)^2dx$ results in: $\frac{1}{4}x+\frac{1}{8}\sin\left(2x\right)+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{8}-\frac{3}{8}\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)$