Answer
$$\int16\sin^2x\cos^2xdx=\frac{4x-\sin4x}{2}+C$$
Work Step by Step
$$A=\int16\sin^2x\cos^2xdx$$ $$A=\int(4\sin x\cos x)^2dx$$
Recall the identity: $2\sin x\cos x=\sin2x$
$$A=\int(2\sin2x)^2dx=4\int(\sin2x)^2dx$$
This is an example of Case 3, so we need to rewrite $$(\sin2x)^2=\frac{1-\cos4x}{2}$$
$$A=4\int\frac{1-\cos4x}{2}dx$$ $$A=4\Big(\frac{x-\frac{1}{4}\sin4x}{2}\Big)+C$$ $$A=\frac{4x-\sin4x}{2}+C$$
