Evaluate the definite integral integral0^pi 16 2 x 2 x \,dx.

We know that $\sin(2x) = 2 \sin x \cos x$.
Squaring both sides gives $\sin^2(2x) = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x$.
Therefore, $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x)$.

Substitute this into the integral:

$$\int_0^\pi 16 \left( \frac{1}{4} \sin^2(2x) \right) \,dx$$ $$\int_0^\pi 4 \sin^2(2x) \,dx$$

The identity is $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$.
In our case, $\theta = 2x$, so $2\theta = 4x$.

$$\sin^2(2x) = \frac{1 - \cos(4x)}{2}$$

Substitute this into the integral:

$$\int_0^\pi 4 \left( \frac{1 - \cos(4x)}{2} \right) \,dx$$ $$\int_0^\pi 2 (1 - \cos(4x)) \,dx$$ $$\int_0^\pi (2 - 2\cos(4x)) \,dx$$

$$\int (2 - 2\cos(4x)) \,dx = 2x - 2 \left( \frac{\sin(4x)}{4} \right) + C$$ $$= 2x - \frac{1}{2} \sin(4x) + C$$

$$\left[ 2x - \frac{1}{2} \sin(4x) \right]_0^\pi$$ $$= \left( 2(\pi) - \frac{1}{2} \sin(4\pi) \right) - \left( 2(0) - \frac{1}{2} \sin(4(0)) \right)$$

We know that $\sin(4\pi) = 0$ and $\sin(0) = 0$.

$$= \left( 2\pi - \frac{1}{2}(0) \right) - \left( 0 - \frac{1}{2}(0) \right)$$ $$= 2\pi - 0 - 0 + 0$$ $$= 2\pi$$

The final answer is $2\pi$.