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



$\frac{\cos x-\sin x}{1+\sin 2 x}=\frac{\cos x-\sin x}{\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cos x}$    $\left[\sin ^{2} x+\cos ^{2} x=1 ; \sin 2 x=2 \sin x \cos x\right]$

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

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

$\therefore(\cos x-\sin x) d x=d t$

$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$

$=\int \frac{d t}{t^{2}}$

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



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

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