Evaluate the integral: $\int\limits_{0}^{\frac{\pi}{4}} \sin^3 2t \cos 2t \, dt$

$\frac{1}{8}$

The correct answer is Option (2) → $\frac{1}{8}$

Let $I = \int\limits_{0}^{\frac{\pi}{4}} \sin^3 2t \cos 2t \, dt$. Consider $\int \sin^3 2t \cos 2t \, dt$

Put $\sin 2t = u$ so that $2 \cos 2t \, dt = du$ or $\cos 2t \, dt = \frac{1}{2} du$

So $\int \sin^3 2t \cos 2t \, dt = \frac{1}{2} \int u^3 \, du = \frac{1}{8} [u^4] = \frac{1}{8} \sin^4 2t = F(t) \text{ say}$

Therefore, by the second fundamental theorem of integral calculus

$I = F\left(\frac{\pi}{4}\right) - F(0) = \frac{1}{8} \left[ \sin^4 \frac{\pi}{2} - \sin^4 0 \right] = \frac{1}{8} \text{}$