The value of the integral $\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}cos^2x\, dx $ is:

$\frac{\pi}{2}$

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

$I=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos^2x\, dx$

$\cos^2x$ → even function

$I=2\int\limits^{\frac{\pi}{2}}_{0}\cos^2x\, dx=2\int\limits^{\frac{\pi}{2}}_{0}\frac{\cos 2x+1}{2}dx$

$I=\left[\frac{\sin 2x}{2}+x\right]^{\frac{\pi}{2}}_{0}$

$=\frac{\pi}{2}$