Question

$\int \frac{dx}{\sin x-\cos x+\sqrt{2}}$ is equal to.

• A. $-\frac{1}{\sqrt{2}}\tan \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$
• B. $\frac{1}{\sqrt{2}}\tan \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$
• C. $\frac{1}{\sqrt{2}}\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$
• D. $-\frac{1}{\sqrt{2}}\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$

Answer 1

The correct option is D $-\frac{1}{\sqrt{2}}\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$

Let $I=\int \frac{dx}{\sin x-\cos x+\sqrt{2}}$

$=\int \frac{dx}{\sqrt{2}\left(\sin x\cdot \sin \frac{\pi }{4}-\cos x\cdot \cos \frac{\pi }{4}+1\right)}$

$=\frac{1}{\sqrt{2}}\int \frac{d}{1-\cos (x+\frac{\pi }{4})}$

$=\frac{1}{\sqrt{2}}\int \frac{dx}{1-\cos 2\left(\frac{x}{2}+\frac{\pi }{8}\right)}$

$=\frac{1}{\sqrt{2}}\int \frac{dx}{2{\sin }^2\left(\frac{x}{2}+\frac{\pi }{8}\right)}$

$=\frac{1}{2\sqrt{2}}\int cose{c}^2\left(\frac{x}{2}+\frac{\pi }{8}\right)dx$

$=\frac{1}{2\sqrt{2}}\cdot \frac{-\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)}{\frac{1}{2}}+C$

$=-\frac{1}{\sqrt{2}}\cdot \cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$.