Answer
\begin{aligned}
\int 16 \sin ^{2} x \cos ^{2} x d x=2 x-4 \sin x \cos ^{3} x+2 \sin x \cos x+C
\end{aligned}
Work Step by Step
Given $$ \int 16 \sin ^{2} x \cos ^{2} x d x $$
So, we have
\begin{aligned}
I&= \int 16 \sin ^{2} x \cos ^{2} x d x\\
&\text{since} \ \ \sin ^{2} x=\frac{1-\cos 2 x}{2} \ \ \cos ^{2}x=\frac{1+\cos 2 x}{2} ,\text{ we get}\\
&=16 \int\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1+\cos 2 x}{2}\right) d x\\
&=4 \int\left(1-\cos ^{2} 2 x\right) d x\\
&=4 \int\left( \sin ^{2} 2 x\right) d x\\
&\text{since} \ \ \sin ^{2} 2 x=\frac{1-\cos 4 x}{2} \ \ ,\text{ we get}\\
&= 4 \int\left(\frac{1-\cos 4 x}{2}\right) d x\\
&=2 \int d x-\frac{1}{2} \int 4\cos 4 x d x\\
&=2 x-\frac{1}{2} \sin 4 x+C\\
&=2 x-\frac{1}{2} \sin 4 x+C\\
&=2 x-\sin 2 x \cos 2 x+C\\
&=2 x-2 \sin x \cos x\left(2 \cos ^{2} x-1\right)+C\\
&=2 x-4 \sin x \cos ^{3} x+2 \sin x \cos x+C
\end{aligned}