$\int\limits_{0.}^{\pi /2} {\cfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx}$

Let$I = \int\limits_{0.}^{\pi /2} {\cfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx}$

…..(i)
Then $I = \int\limits_{0.}^{\pi /2} {\cfrac{{\sin \left( {\cfrac{\pi }{2} - x} \right) - \cos \left( {\cfrac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\cfrac{\pi }{2} - x} \right)\cos \left( {\cfrac{\pi }{2} - x} \right)}}dx}$

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

….(ii)
Adding (i) and (ii),

we get
$2I = \int\limits_0^{\pi /2} {\left( {\cfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}} + \cfrac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \right)} dx$

$= \int\limits_0^{\pi /2} {\cfrac{{\sin x - \cos x + \cos x - \sin x}}{{1 + \sin x\cos x}}} dx = \int\limits_0^{\pi /2} {0dx} = 0$

$\Rightarrow$ $I = 0$