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

Hint:

For $\sin x \pm \cos x$, convert into single sine or cosine using $\sqrt{2}$ identity.

Answer 1

The Correct Option is D

Concept: Use identity: \[ \sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) \]

Step 1: Simplify denominator.
\[ \sin x - \cos x + \sqrt{2} = \sqrt{2}\left[\sin\left(x - \frac{\pi}{4}\right) + 1\right] \]

Step 2: Rewrite integral.
\[ \int \frac{dx}{\sqrt{2}[1 + \sin(x - \frac{\pi}{4})]} \]

Step 3: Use standard identity.
\[ \frac{1}{1+\sin \theta} = \frac{1-\sin \theta}{\cos^2 \theta} \]

Step 4: Substitute $\theta = x - \frac{\pi}{4}$.
Integral reduces to cotangent form.

Step 5: Final integration.
\[ = -\frac{1}{\sqrt{2}}\cot\left(\frac{x}{2} + \frac{\pi}{8}\right) + C \] Conclusion:
Answer = Option (D)