$\int \left(\frac{\cos x - \sin x}{1 + \sin 2x}\right) dx$ is equal to

The integral is $\int \frac{\cos x - \sin x}{1 + \sin 2x} dx$

Using the identity $\sin 2x = 2\sin x \cos x$:

$1 + \sin 2x = 1 + 2\sin x \cos x$

Since $\sin^2 x + \cos^2 x = 1$:

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x$

$= (\sin x + \cos x)^2$

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The integral becomes:

$\int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} dx$

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Let $t = \sin x + \cos x$

Differentiating both sides:

$dt = (\cos x - \sin x) dx$

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Substituting into the integral:

$\int \frac{1}{t^2} dt$

$= \int t^{-2} dt$

$= \frac{t^{-1}}{-1} + C$

$= -\frac{1}{t} + C$

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Substituting back $t = \sin x + \cos x$:

$-\frac{1}{\sin x + \cos x} + C$

Therefore, the answer is $\frac{-1}{\cos x + \sin x} + C$