Integrals — Class 12 Maths Solution

$\int\limits_0^{\pi /4} {\sin 2x} \,dx = \left[ { - \cfrac{1}{2}\cos 2x} \right]_0^{\pi /4}$

$= \cfrac{{ - 1}}{2}\left( {\cos \cfrac{\pi }{2} - \cos 0} \right) = - \cfrac{1}{2}\left( { - 1} \right) = \cfrac{1}{2}$