Question

Solve: $\int \frac{dx}{\cos x-\sin x}$

Answer 1

we can write $\cos x-\sin x=\sqrt{2}(\frac{\cos x}{\sqrt{2}}-\frac{\sin x}{\sqrt{2}})=\sqrt{2}\sin (x+\frac{3\pi }{4})$

Hence, $\int \frac{dx}{\cos x-\sin x}=\int \frac{1}{\sqrt{2}}\frac{dx}{\sin (x+\frac{3\pi }{4})}$

let $u=x+\frac{3\pi }{4}du=dx$

$=\frac{1}{\sqrt{2}}\int \frac{du}{\sin u}=\frac{1}{\sqrt{2}}\int cosecudu=\frac{1}{\sqrt{2}}(-\ln |cosecu+\cot u|)+C=\frac{-1}{\sqrt{2}}\left(\ln |cosec\right(x+\frac{3\pi }{4})+\cot (x+\frac{3\pi }{4}\left)|\right)+C$