$\int \frac{d x}{\cos x-\sin x}$ is equal to
$\int \frac{d x}{\cos x-\sin x}$ is equal to
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$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{\pi}{8}\right)\right|+C$
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$\frac{1}{\sqrt{2}} \log \left|\cot \left(\frac{x}{2}\right)\right|+C$
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$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}-\frac{3 \pi}{8}\right)\right|+C$
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$\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$
Solution
$\int \frac{d x}{\cos x-\sin x}=\frac{1}{\sqrt{2}} \int \frac{1}{\cos \left(x+\frac{\pi}{4}\right)} d x=\frac{1}{\sqrt{2}} \int \sec \left(x+\frac{\pi}{4}\right) d x=\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C$
Asked in: JEE Main 2004
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