Find $\int\cos^2x\,dx$

$\frac{x}{2} + \frac{\sin 2x}{4} + C$

The correct answer is Option (3) → $\frac{x}{2} + \frac{\sin 2x}{4} + C$

From the identity $\cos 2x = 2 \cos^2 x - 1$, which gives

$\cos^2 x = \frac{1 + \cos 2x}{2}$

Therefore, $\displaystyle \int \cos^2 x \, dx = \frac{1}{2} \int (1 + \cos 2x) \, dx = \frac{1}{2} \int dx + \frac{1}{2} \int \cos 2x \, dx$

$= \frac{x}{2} + \frac{1}{4} \sin 2x + C$