Question

$\int \frac{dx}{\sin x-\cos x+\sqrt{2}}$ equals to

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

Answer: (D)

Let $I=\int \frac{dx}{\sin x-\cos x+\sqrt{2}}$
$=\int \frac{dx}{\sqrt{2}\left(\sin x\sin \frac{\pi }{4}-\cos x\cos \frac{\pi }{4}+1\right)}$
$=\frac{1}{\sqrt{2}}\int \frac{dx}{1-\cos \left(x+\frac{\pi }{4}\right)}$
$=\frac{1}{\sqrt{2}}\int \frac{dx}{2{\sin }^2\left(\frac{x}{2}+\frac{\pi }{8}\right)}$
$=\frac{1}{2\sqrt{2}}\int cose{c}^2\left(\frac{x}{2}+\frac{\pi }{8}\right)dx$
$=\frac{1}{2\sqrt{2}}\frac{-\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)}{\frac{1}{2}}+C$
$=-\frac{1}{\sqrt{2}}\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C$