Integrals — Class 12 Maths Solution

: $\int\limits_0^{\pi /2} {{{\cos }^2}x} dx = \int\limits_0^{\pi /2} {\cfrac{{1 + \cos 2x}}{2}} dx$

$= \left[ {\cfrac{1}{2}\left( {x + \cfrac{{\sin 2x}}{2}} \right)} \right]_0^{\pi /2} = \cfrac{1}{2}\left[ {\left( {\cfrac{\pi }{2} - 0} \right) + \left( {\cfrac{{\sin x}}{2} - \cfrac{{\sin 0}}{2}} \right)} \right] = \cfrac{\pi }{4}$