$\int \frac{1}{\cos x-\sin x} d x$ is equal to |
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{3 \pi}{8}\right)\right|+C$ $\frac{1}{\sqrt{2}} \log \left|\cot \frac{x}{2}\right|+C$ $\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{8}\right)\right|+C$ $\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$ |
$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$ |
We have, $I=\int \frac{1}{\cos x-\sin x} d x=\frac{1}{\sqrt{2}} \int \frac{1}{\cos x \cos \frac{\pi}{4}-\sin x \sin \frac{\pi}{4}} d x$ $\Rightarrow I =\frac{1}{\sqrt{2}} \int \frac{1}{\cos \left(x+\frac{\pi}{4}\right)} d x=\frac{1}{\sqrt{2}} \int \sec \left(x+\frac{\pi}{4}\right) d x$ $\Rightarrow I=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}+\frac{\pi}{8}\right)\right|+C=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$ |