We are asked to solve the integration: $\int \frac{1}{\sin x - \cos x} dx$.

1. Problem Description

We are asked to solve the integration: $\int \frac{1}{\sin x - \cos x} dx$.

2. Solution Steps

We can rewrite the denominator using the identity $a \sin x - b \cos x = R \sin(x - \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan \alpha = \frac{b}{a}$. In our case, $a = 1$ and $b = 1$, so $R = \sqrt{1^2 + 1^2} = \sqrt{2}$ and $\tan \alpha = \frac{1}{1} = 1$, which means $\alpha = \frac{\pi}{4}$.

Therefore, we have:

$\sin x - \cos x = \sqrt{2} \sin(x - \frac{\pi}{4})$.

Thus, the integral becomes:

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

We know that

$\int \csc(u) du = -\ln |\csc(u) + \cot(u)| + C$.

So,

$\frac{1}{\sqrt{2}} \int \csc(x - \frac{\pi}{4}) dx = \frac{1}{\sqrt{2}} (-\ln |\csc(x - \frac{\pi}{4}) + \cot(x - \frac{\pi}{4})|) + C$.

Also, we have another form for the integral of $\csc u$:

$\int \csc(u) du = -\ln |\tan(\frac{u}{2})| + C$

So,

$\frac{1}{\sqrt{2}} \int \csc(x - \frac{\pi}{4}) dx = \frac{1}{\sqrt{2}} (-\ln |\tan(\frac{x - \frac{\pi}{4}}{2})|) + C = -\frac{1}{\sqrt{2}} \ln |\tan(\frac{x}{2} - \frac{\pi}{8})| + C$.

3. Final Answer

The solution to the integral is:

$-\frac{1}{\sqrt{2}} \ln |\tan(\frac{x}{2} - \frac{\pi}{8})| + C$.

Alternatively,

$-\frac{1}{\sqrt{2}} \ln |\csc(x - \frac{\pi}{4}) + \cot(x - \frac{\pi}{4})| + C$.