Evaluate $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx$

$\frac{\pi}{4}-\frac{1}{2}$

The correct answer is Option (2) → $\frac{\pi}{4}-\frac{1}{2}$

We know that $\sin^2 x$ is an even function. Therefore, we get

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx$

$= 2 \int_{0}^{\frac{\pi}{4}} \frac{(1 - \cos 2x)}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) \, dx$

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