$\int \frac{1}{\cos x-\sin x} d x$ is equal to

$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$

We have,

$I=\int \frac{1}{\cos x-\sin x} d x=\frac{1}{\sqrt{2}} \int \frac{1}{\cos x \cos \frac{\pi}{4}-\sin x \sin \frac{\pi}{4}} d x$

$\Rightarrow I =\frac{1}{\sqrt{2}} \int \frac{1}{\cos \left(x+\frac{\pi}{4}\right)} d x=\frac{1}{\sqrt{2}} \int \sec \left(x+\frac{\pi}{4}\right) d x$

$\Rightarrow I=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{\pi}{8}\right)\right|+C=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$