The problem asks to find the value of the integral $\int \frac{1}{\sin x - \cos x} dx$.

1. Problem Description

The problem asks to find the value of the integral $\int \frac{1}{\sin x - \cos x} dx$.

2. Solution Steps

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

So the integral becomes:

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

Using the trigonometric identity $\sin(x) = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})$, we have

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

Also, we know that $\csc(x) = \frac{1}{\sin(x)}$. Therefore,

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

The integral of $\csc(u)$ is known to be $-\ln|\csc(u) + \cot(u)| + C$ or $\ln|\tan(\frac{u}{2})| + C$.

Using the second form:

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

3. Final Answer

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