The value of $\int\limits_{0}^{\frac{\pi}{4}} (\sin 2x) dx$ is:

$\frac{1}{2}$

The correct answer is Option (3) → $\frac{1}{2}$

$\int\limits_{0}^{\frac{\pi}{4}} (\sin 2x) dx = \int\limits_{0}^{\frac{\pi}{4}} 2 \sin x \cos x \, dx$

Let $\sin x = t$

$\cos x \, dx = dt$

When $x = 0$ then $t = 0$

When $x = \frac{\pi}{4}$ then $t = \frac{1}{\sqrt{2}}$

$= 2 \int\limits_{0}^{\frac{1}{\sqrt{2}}} t \, dt$

$= 2 \left[ \frac{t^2}{2} \right]_{0}^{\frac{1}{\sqrt{2}}}$

$= \frac{1}{2} - 0 = \frac{1}{2}$