Question

$\int \frac{dx}{cosx-sinx}$ is equal to

• A. $-\frac{1}{\sqrt{2}}log\left|tan[\frac{x}{2}-\frac{\pi }{8}]\right|+C$
• B. $\frac{1}{\sqrt{2}}log\left|cot\left[\frac{x}{2}\right]\right|$ $+C$
• C. $\frac{1}{\sqrt{2}}log\left|tan[\frac{x}{2}-\frac{3\pi }{8}]\right|+C$
• D. $\frac{1}{\sqrt{2}}log\left|tan[\frac{x}{2}+\frac{\pi }{8}]\right|+C$

Answer 1

The correct option is A $-\frac{1}{\sqrt{2}}log\left|tan[\frac{x}{2}-\frac{\pi }{8}]\right|+C$

$\int \frac{dx}{\cos x-\sin x}=\frac{1}{\sqrt{2}}\int \frac{dx}{\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x}$

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

$=\frac{1}{\sqrt{2}}\int \frac{dx}{\sin (\frac{\pi }{4}-x)}$

$=-\frac{1}{\sqrt{2}}\int \frac{dx}{\sin (x-\frac{\pi }{4})}$

$=-\frac{1}{\sqrt{2}}ln\left|\tan (\frac{x}{2}-\frac{\pi }{8})\right|+c$.