Evaluate the following integral: $\int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$

(1/8) Let $I = \int_{0}^{\frac{\pi}{4}} \sin ^{3} 2 t \cos 2 t \,d t$.
Put $\sin 2 t = u$. Then,differentiating both sides with respect to $t$,we get $2 \cos 2 t \,d t = d u$,which implies $\cos 2 t \,d t = \frac{1}{2} d u$.
Now,change the limits of integration:
When $t = 0$,$u = \sin(2 \times 0) = \sin 0 = 0$.
When $t = \frac{\pi}{4}$,$u = \sin(2 \times \frac{\pi}{4}) = \sin \frac{\pi}{2} = 1$.
Substituting these into the integral:
$I = \int_{0}^{1} u^{3} \cdot \frac{1}{2} d u$
$I = \frac{1}{2} \left[ \frac{u^{4}}{4} \right]_{0}^{1}$
$I = \frac{1}{2} \left( \frac{1^{4}}{4} - \frac{0^{4}}{4} \right)$
$I = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.