Find the integral of the function $\frac{\cos x-\sin x}{1+\sin 2 x}$.

(N/A) We have the integral $I = \int \frac{\cos x-\sin x}{1+\sin 2 x} dx$.
Using the identities $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$,the denominator becomes:
$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.
Thus,the integral is $I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t^2} = \int t^{-2} dt$.
Integrating with respect to $t$:
$I = \frac{t^{-2+1}}{-2+1} + C = -t^{-1} + C = -\frac{1}{t} + C$.
Substituting back $t = \sin x + \cos x$:
$I = -\frac{1}{\sin x + \cos x} + C$,where $C$ is an arbitrary constant.