Let $I=\int\limits_{0}^{\pi/2}\cos^2 x dx$ ... (i) and $I=\int\limits_{0}^{\pi/2}\cos^2(\pi/2 - x) dx$ $=\int\limits_{0}^{\pi/2}\sin^2 x dx$ ... (ii) $\left[\because \int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx \right]$ Adding (i) and (ii), we get $2I = \int\limits_{0}^{\pi/2}\cos^2 x dx + \int\limits_{0}^{\pi/2}\sin^2 x dx$ $= \int\limits_{0}^{\pi/2} (\cos^2 x + \sin^2 x) dx$ $= \int\limits_{0}^{\pi/2} dx = [x]^{\pi/2}_0 = \pi/2$$\Rightarrow I = \pi/4$