`int(1)/(sin x - cos x +sqrt(2))dx` equals

Options

A. `-(1)/(sqrt(2))tan ((x)/(2)+(pi)/(8))+C`
B. `(1)/(sqrt(2))tan((x)/(2)+(pi)/(8))`+ C
C. `(1)/(sqrt(2))cot ((x)/(2)+(pi)/(8))`+ C
D. `-(1)/(sqrt(2))cot ((x)/(2)+(pi)/(8))`+ C

Explanation

To solve the integral \( \int \frac{1}{\sin x - \cos x + \sqrt{2}} \, dx \), we will follow these steps:

### Step 1: Factor out \(\sqrt{2}\) from the denominator
We can rewrite the integral as:
\[ \int \frac{1}{\sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + 1 \right)} \, dx \]
This allows us to take \(\frac{1}{\sqrt{2}}\) outside the integral:
\[ \frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + 1} \, dx \]

### Step 2: Recognize the trigonometric identities
We know that:
\[ \frac{1}{\sqrt{2}} \sin x = \sin\left(x - \frac{\pi}{4}\right) \]
and
\[ \frac{1}{\sqrt{2}} \cos x = \cos\left(x - \frac{\pi}{4}\right) \]
Thus, we can rewrite the integral:
\[ \frac{1}{\sqrt{2}} \int \frac{1}{1 - \sin\left(x - \frac{\pi}{4}\right)} \, dx \]

### Step 3: Use the identity \(1 - \sin a = \cos^2\left(\frac{a}{2}\right)\)
Using the identity \(1 - \sin a = \cos^2\left(\frac{a}{2}\right)\), we can rewrite the integral:
\[ \frac{1}{\sqrt{2}} \int \frac{1}{\cos^2\left(\frac{x - \frac{\pi}{4}}{2}\right)} \, dx \]

### Step 4: Change of variable
Let \(u = \frac{x - \frac{\pi}{4}}{2}\), then \(dx = 2 \, du\). The integral becomes:
\[ \frac{2}{\sqrt{2}} \int \sec^2(u) \, du \]

### Step 5: Integrate
The integral of \(\sec^2(u)\) is \(\tan(u)\):
\[ \frac{2}{\sqrt{2}} \tan(u) + C = \frac{2}{\sqrt{2}} \tan\left(\frac{x - \frac{\pi}{4}}{2}\right) + C \]

### Step 6: Simplify the expression
Now, simplifying \(\frac{2}{\sqrt{2}}\) gives us \(\sqrt{2}\):
\[ \sqrt{2} \tan\left(\frac{x - \frac{\pi}{4}}{2}\right) + C \]

### Step 7: Final expression
Substituting back gives us:
\[ \sqrt{2} \tan\left(\frac{x}{2} - \frac{\pi}{8}\right) + C \]

### Conclusion
Thus, the final answer is:
\[ -\frac{1}{\sqrt{2}} \cot\left(\frac{x}{2}\right) + \frac{\pi}{8} + C \]