$\cfrac{{\cos x - \sin x}}{{1 + \sin 2x}}$

: Let $I = \int {\cfrac{{\cos x - \sin x}}{{1 + \sin 2x}}dx} = \int {\cfrac{{\cos x - \sin x}}{{1 + 2\sin x\cos x}}} dx$

$= \int {\cfrac{{\cos x - \sin x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx}$

Put $\cos x + \sin x = t$ $\Rightarrow$ $\left( { - \sin x + \cos x} \right)dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{{t^2}}}} = \cfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \cfrac{{ - 1}}{t} + C = - \cfrac{1}{{\cos x + \sin x}} + C$