Evaluate the definite integral: integral0^pi/2 8 4 y 2 y\,dy

To solve the integral $\int_0^{\pi/2} 8 \sin^4 y \cos^2 y\,dy$, we will use trigonometric identities to reduce the powers of sine and cosine.

Step 1: Rewrite the integrand using half-angle and product identities.
The integrand is $8 \sin^4 y \cos^2 y$.
We can express $\sin^4 y \cos^2 y$ as $\sin^2 y (\sin y \cos y)^2$.
Using the identities:
$\sin^2 y = \frac{1 - \cos(2y)}{2}$
$\sin y \cos y = \frac{\sin(2y)}{2}$
Substitute these into the expression:

$$8 \sin^4 y \cos^2 y = 8 \left(\frac{1 - \cos(2y)}{2}\right) \left(\frac{\sin(2y)}{2}\right)^2$$ $$= 8 \left(\frac{1 - \cos(2y)}{2}\right) \left(\frac{\sin^2(2y)}{4}\right)$$

Now, apply the half-angle identity for $\sin^2(2y)$:
$\sin^2(2y) = \frac{1 - \cos(2 \cdot 2y)}{2} = \frac{1 - \cos(4y)}{2}$
Substitute this back:

$$= 8 \left(\frac{1 - \cos(2y)}{2}\right) \left(\frac{1 - \cos(4y)}{2 \cdot 4}\right)$$ $$$$